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03Gauss's law
Last week a ring of charge cost you a density, an integral and a cosine factor, and the answer came out as $kQx/(x^{2}+R^{2})^{3/2}$. Now take a solid ball of charge and ask for the field at a point buried inside it. The same method says: chop the ball into shells, chop each shell into rings, integrate the rings, integrate the shells. That is a triple integral for a question whose answer, it turns out, is one line long.
By the end of this section you can write down the field of any sphere, cylinder or sheet of charge in two lines with no integration at all, say exactly when that shortcut is allowed, and explain why the field inside a piece of metal at rest is zero.
In 60 seconds
The outward flux of the electric field through any closed surface equals the charge inside divided by the , and when the charge is symmetric enough that the field has the same size everywhere on a well chosen surface, that one equation hands you the field without an integral.
Electric flux
$$\Phi_E = \int \vec E\cdot d\vec A = \int E\cos\theta\,dA$$
you need the amount of field crossing a surface; for a flat patch in a uniform field it is EA cos theta
Gauss's law
$$\oint \vec E\cdot d\vec A = \frac{Q_{\rm enc}}{\varepsilon_0}$$
always true for a closed surface; useful for finding E only when symmetry makes E constant on it
a long wire, rod, tube or cable, at a distance small compared with its length
Sheet of charge
$$E = \frac{\sigma}{2\varepsilon_0}\ \ \text{(lone sheet)},\qquad E = \frac{\sigma}{\varepsilon_0}\ \ \text{(face of a conductor)}$$
a large flat charged surface, seen from close enough that its edges are far away
Conductor at rest
$$\vec E_{\rm inside} = 0,\qquad \sigma\ \text{lives on the surface},\qquad E_{\rm just\ outside} = \frac{\sigma}{\varepsilon_0}$$
any metal that has been left alone long enough for the charges to stop moving
Three most common mistakes
Reading Gauss's law as though the $\vec E$ in it were the field of the enclosed charge alone. It is the total field, from every charge in the universe. Outside charges contribute nothing to the flux, because whatever enters the surface leaves it again, but they do change the field at every single point of that surface.
Using $Q$ where $Q_{\rm enc}$ belongs. Inside a uniform ball only the charge closer to the centre than you are counts, which is why the field there grows with $r$ instead of falling off. The rest of the ball is outside your surface and contributes exactly nothing.
Quoting $\sigma/2\varepsilon_0$ for the face of a charged metal plate. A lone sheet sends field out of both ends of the , a conductor sends it out of one end only because the other end is buried in metal where the field is zero, so the conductor answer is twice as big.
The published weights are Midterm 1 twenty per cent, Midterm 2 twenty per cent, quizzes ten per cent in total, homework five per cent, the final twenty five per cent and the laboratory twenty per cent. This is week three material, so it sits inside the ground covered by Midterm 1 and again by the final. Nothing finer than that is published, so treat the three symmetry families as bookwork you can produce cold, and expect the marks to be spread over choosing the surface and writing the enclosed charge, not over the algebra at the end.
How much time do you have?
10 minutes
You leave with the one equation and the three standard answers it produces, plus the single sentence that decides whether you are allowed to use them: the surface must be one on which the field has the same size everywhere.
The 60 second card · Formula card · Gauss's law: the flux out of a closed surface counts only the charge inside · Spherical symmetry: shells, balls, and why the outside cannot tell the difference · Mistake ledger
45 minutes
You add the two things that separate a set-up mark from a full mark: how to pick the surface for each of the three symmetries, and how to write the enclosed charge when only part of the object is inside your surface.
The 60 second card · Electric flux: how much field crosses a surface · Gauss's law: the flux out of a closed surface counts only the charge inside · Choosing the surface: what symmetry has to give you before the law is any use · Spherical symmetry: shells, balls, and why the outside cannot tell the difference · Cylindrical symmetry: the long wire, done in two lines this time · Planar symmetry: the sheet whose field does not weaken with distance · Method box: getting a field out of Gauss's law · Exam level example · Practice set C · Mistake ledger
full read
You can rebuild every result here from the law itself, including the ones nobody quoted at you, and you can say why a conductor pushes all its charge to the skin and shields whatever is inside it.
The opening pages · Prerequisites and pretest · Notation · Conventions · All seven concept blocks · Method boxes · Contrast pairs · Scaffolding comes off · Exam level example · Practice sets A to D · Mistake ledger · Formula card · Glossary · Check yourself
By the end of this section
Compute the electric flux through a flat surface in a uniform field, and through a closed surface, and say in words what a positive, negative or zero answer is telling you about the field lines.
State Gauss's law, identify the enclosed charge for a given closed surface, and compute the net flux through that surface without knowing the field at any single point on it.
Choose a that turns the flux integral into a product of the field and an area, and recognise the charge distributions for which no such surface exists.
Derive the field inside and outside a uniformly charged ball and a spherical shell, and use the result that a spherically symmetric charge acts on the outside world as if it were all at the centre.
Apply Gauss's law to a long line, tube or solid cylinder of charge, recovering the wire result of the previous section in two lines instead of an integral.
Calculate the field of one large charged sheet and of two parallel sheets, and explain why the answer does not depend on how far from the sheet you stand.
Predict where the charge sits on a conductor at rest, find the on the walls of a cavity, and use the field just outside a conducting surface.
Syllabus coverage
Gauss's law
Electric flux, Gauss's law and the argument behind it, the three symmetry families and the surfaces that go with them, spheres and shells, lines and cylinders, sheets and plates, and conductors in .
The week line names the topic and carries no chapter numbers, so no chapter number is quoted anywhere on this page.
covered
field of a solid charged cylinder
Charge spread through the volume of a long cylinder rather than along a line.
Not named on the week line. It is here because it is the only cylindrical case where the enclosed charge changes with your radius, which is the skill the ball also needs, and because cables are the standard exam dressing for it. Nothing later depends on it.
off_syllabus
experimental test of the inverse square law
How the exponent in Coulomb's law is measured to extraordinary precision using a hollow conductor.
Not named on the week line. One short paragraph, kept because it is the cleanest answer to the question of how anyone knows the exponent is exactly two rather than nearly two.
off_syllabus
Recall first
Field of a point charge
$E = \dfrac{kq}{r^{2}}$, pointing away from a positive charge and towards a negative one, with $k = 8.99\times10^{9}\ \mathrm{N\,m^{2}/C^{2}}$.
Everything in this section is checked against it. Gauss's law has to reproduce it for a single charge, and every spherical answer collapses back to it far away.
Superposition
The field of several charges is the vector sum of the separate fields, $\vec E = \vec E_1 + \vec E_2 + \dots$, with no interaction terms.
It is what lets us say the flux of a sum is the sum of the fluxes, which is how Gauss's law extends from one charge to any collection of them.
Long straight wire
$E = \dfrac{2k\lambda}{x}$ at a perpendicular distance $x$ from a wire long compared with $x$, derived last section by integrating along the wire.
The cylindrical block rebuilds this in two lines with no integral. Having the old answer to compare against is the only honest way to check that the new method is not producing nonsense.
Finite wire, on its perpendicular bisector
$E = \dfrac{2k\lambda L}{x\sqrt{x^{2}+L^{2}}}$ for a wire of half length $L$, which becomes the long wire result when $L \gg x$.
It is the tool for asking how long a wire has to be before it counts as infinite, which is the honest form of the question of when Gauss's law may be used on it.
Charge densities
$dq = \lambda\,dl = \sigma\,dA = \rho\,dV$, and for uniform charge $\lambda = Q/L$, $\sigma = Q/A$, $\rho = Q/V$.
Enclosed charge is computed from a density and a volume in almost every problem here, so the density has to be built before anything else happens.
Dot product as a projection
$\vec a\cdot\vec b = ab\cos\theta$, which is the length of one vector times the part of the other that lies along it.
Flux is a dot product integrated over a surface. Every cosine on this page comes from this one definition and from nowhere else.
Areas and volumes
Sphere: area $4\pi r^{2}$, volume $\tfrac{4}{3}\pi r^{3}$. Cylinder of length $L$: curved area $2\pi r L$, volume $\pi r^{2}L$. Disc: area $\pi r^{2}$.
These five numbers do all the work once the flux integral has been turned into a product. Getting the field right and the area wrong is the commonest way to lose a whole question here.
Try it yourself first (3 questions)
1§03.0 - superposition at the midpoint of a pair●●○○○
Nothing here is new material, and getting it wrong is not a problem; it just tells you which of the two blocks below to read slowly. A charge $+5.00\ \mathrm{nC}$ sits at the origin and a charge $-5.00\ \mathrm{nC}$ sits $6.00\ \mathrm{cm}$ away on the $x$ axis.
Given
$q_1 = +5.00\ \mathrm{nC}$ at $x = 0$
$q_2 = -5.00\ \mathrm{nC}$ at $x = 6.00\ \mathrm{cm}$
the field point is the midpoint, $x = 3.00\ \mathrm{cm}$
Find
(a) Find the size and direction of the total electric field at the midpoint.
Hint 1/4
Two separate fields exist at that point and you need their vector sum. Before computing anything, draw the arrow each charge alone would produce there and see whether the two arrows agree or fight.
Hint 2/4
Each contributes $E = kq/r^{2}$, pointing away from a positive charge and towards a negative one.
Hint 3/4
Both charges are $5.00\ \mathrm{nC}$ in size and both are $3.00\ \mathrm{cm}$ from the midpoint, so the two contributions have equal size. The positive one pushes in the $+x$ direction and the negative one pulls in the $+x$ direction as well.
Hint 4/4
They add rather than cancel: $E = 2kq/r^{2} = 9.99\times10^{4}\ \mathrm{N/C}$ in the $+x$ direction.
Show solutionDirections first, numbers afterwards
$$\vec E_1 \parallel +\hat x$$
the field of a positive charge points away from it, and the midpoint lies on its right
$$\vec E_2 \parallel +\hat x$$
the field of a negative charge points towards it, and the negative charge is to the right of the midpoint
parallel vectors add as numbers, which is the whole content of the vector sum here
Answer $$\boxed{\;E = 9.99\times10^{4}\ \mathrm{N/C}\ \text{towards the negative charge}\;}$$
Check
Sanity check by symmetry rather than by arithmetic: the picture is unchanged if you rotate it by half a turn about the midpoint and swap the signs, so the field there must lie along the line joining the charges. A zero answer would also satisfy that, but a test charge released at the midpoint is clearly pushed by one charge and pulled by the other in the same direction, so zero is impossible.
Opposite charges never means cancelling fields. Whether contributions add or subtract depends on where you stand, and the arrow picture settles it before any number is computed.
2§03.0 - areas and volumes you will need●○○○○
Gauss's law turns an integral into a product of a field and an area, so the areas have to be automatic. A sphere, a cylinder and a flat disc all have radius $r = 4.00\ \mathrm{cm}$, and the cylinder has length $L = 25.0\ \mathrm{cm}$.
Given
radius $r = 4.00\ \mathrm{cm}$ for all three shapes
cylinder length $L = 25.0\ \mathrm{cm}$
Find
(a) Write the surface area of the sphere and its volume.
(b) Write the curved side area of the cylinder, excluding its two flat ends.
(c) Write the area of the disc.
Hint 1/4
You are being asked for the three shapes whose areas appear in every Gaussian surface in this section. No physics is involved; this is a check that the geometry is at your fingertips.
Dimensional check on the sphere pair: dividing the volume by the area gives $2.68\times10^{-4}/2.01\times10^{-2} = 1.33\times10^{-2}\ \mathrm{m}$, which is one third of the radius, exactly as the ratio of the two formulas demands.
3§03.0 - what a field line picture can and cannot tell you●●○○○
This one is deliberately built around a wrong intuition that a lot of people carry into this week, so treat a wrong answer as information rather than as a bad sign. A closed box is drawn in a region where the electric field is uniform, and no charge is inside the box.
Given
a uniform field in the whole region
a closed box drawn in that region
no charge anywhere inside the box
Find
(a) True or false: because there is no charge inside, the field lines must stop at the walls of the box, so nothing crosses it. Give a one sentence reason.
Hint 1/4
Two different statements are being confused: how much field enters the box, and how much more leaves than enters. Decide which one an empty box is allowed to control.
Hint 2/4
A field line ends only on a charge. With no charge inside, a line that comes in has no way to stop and must go out again.
Hint 3/4
In a uniform field every line entering the left wall of the box travels straight through and leaves through the right wall, so lines cross the surface in large numbers.
Hint 4/4
False: plenty of field crosses the box, but the amount entering equals the amount leaving, so the net is zero.
Show solutionWhat an empty region forbids
$$\text{lines begin and end only on charges}$$
this is the defining property of the field line picture built in the previous section
with nothing inside to absorb or create lines, whatever enters has to leave, which constrains the difference and not the amounts
Answer $$\boxed{\;\text{False; the net flux is zero, the flux through each wall is not}\;}$$
Check
Put numbers on it. A cube of side $20.0\ \mathrm{cm}$ in a field of $6.00\times10^{3}\ \mathrm{N/C}$ along its axis takes $-240\ \mathrm{N\,m^{2}/C}$ in through one face and $+240\ \mathrm{N\,m^{2}/C}$ out through the opposite one. Neither is zero, and their sum is.
Whenever a statement about flux sounds surprising, ask whether it is about the flux through one piece of the surface or about the total over the closed surface. Almost every apparent paradox in this section is that substitution.
Notation
symbol
reads as
means
watch out
$\Phi_E$
capital phi sub E
electric flux through a stated surface, measured in newton metres squared per coulomb
it is a scalar with a sign, not a vector; asking which way the flux points is a category error
$d\vec A$
d A vector
a small patch of the surface, with size equal to its area and direction along its outward normal
on a closed surface the outward direction is fixed for you; on an open surface you choose it, and the sign of the answer follows your choice
$\oint$
closed surface integral
an integral taken over a surface that completely encloses a volume, with no edges and no holes
the circle on the integral sign is the whole difference between the definition of flux and Gauss's law
$Q_{\rm enc}$
Q enclosed
the algebraic sum of all charge inside the closed surface, positive and negative together
a neutral object inside the surface contributes zero even though it is full of charge, and half an object inside contributes only its half
$\varepsilon_0$
epsilon nought
the permittivity of free space, the constant that fixes the strength of electrostatics in SI units
the same constant as in Coulomb's law, since k equals one over four pi epsilon nought; mixing the two forms in one line is the usual source of a factor of twelve
$\lambda,\ \sigma,\ \rho$
lambda, sigma, rho
charge per unit length, per unit area and per unit volume, unchanged from the previous section
a solid cylinder has both a rho and a lambda, and they are related by the cross sectional area; using one where the other belongs is the standard cable mistake
$E_r$
E sub r
the radial component of the field, positive when it points away from the centre or the axis
for a negative charge this comes out negative, which is the algebra telling you the field points inward, not that the size is negative
Conventions used here
Which way the points
For a closed surface the area vector of every patch points out of the enclosed volume, without exception. Positive flux then means net field leaving, negative means net field entering. For an open surface, such as a single flat plate, there are two possible choices and the question has to name one; every open surface on this page has its chosen direction stated in the wording.
What the symbols in Gauss's law refer to
$\vec E$ is the total field at the surface, produced by every charge there is, inside and outside. $Q_{\rm enc}$ counts only the charge inside the closed surface, with its sign. Charge sitting exactly on the surface is never used here; every surface in this section is drawn so that it misses the charges rather than slicing through them.
Constants used throughout
$k = 1/(4\pi\varepsilon_0) = 8.99\times10^{9}\ \mathrm{N\,m^{2}/C^{2}}$ and $\varepsilon_0 = 8.85\times10^{-12}\ \mathrm{C^{2}/(N\,m^{2})}$, the same two numbers as in the previous two sections. Answers are rounded to three significant figures at the very end and not before.
Signs of charges
The sign of a charge is used to fix the direction of its field by a sentence, not by carrying a minus sign through the algebra. Put magnitudes into the formulas, get a positive number out, and then state the direction in words or with a unit vector. Mixing the two conventions inside one problem is the fastest way to lose a sign. The single exception in this section is $Q_{\rm enc}$ inside Gauss's law, which is an algebraic sum and keeps its sign, because a negative enclosed charge has to produce a negative flux, meaning field entering the surface.
What counts as long or large
A wire is treated as infinite when its length is at least about fifteen times your distance from its middle, and a sheet as infinite when your distance is small compared with its own width. These are working thresholds for one per cent accuracy, not laws; each application block says what the error is at the stated distance.
How a field and a flux are quoted
A field is quoted as a magnitude in newtons per coulomb together with a direction, for example $2.70\times10^{3}\ \mathrm{N/C}$ pointing radially outward, exactly as in the previous sections. Flux is different: it is quoted in newton metres squared per coulomb and carries a sign rather than a direction, because it is a number and not a vector.
3.1Electric flux: how much field crosses a surface
Flux is the field times the area facing it, and the word facing is doing all the work.
Closed surfaces come next, but first we need a number for how much field crosses a surface, and it has to know about tilt.
Solvable with what we have
the field of one point charge, and of any handful of point charges added as vectors
the field of a uniformly charged rod at a point on its own axis
the field on the axis of a ring, and beside a long straight wire
Not solvable yet
the field at a point buried inside a uniformly charged solid ball
the field just outside a charged metal sphere, where the charge has arranged itself in a way nobody told us
the field a centimetre from a large charged plate
Take the solid ball of radius $R$ and charge $Q$ and ask for the field at radius $r$ inside it. Shells, then rings inside each shell, puts the field point inside a ring rather than on its axis, where the ring formula does not apply.
Why it fails
Nothing is wrong with the method; it costs far more than the answer deserves. The result is $E = kQr/R^{3}$, a straight line through the origin. A page of integration for a straight line means the symmetry is going to waste.
DefinitionDefinition 3.1: electric flux
Conditions
the surface has a chosen side, so that the direction of the outward normal is fixed
for a closed surface that choice is always the outward one, with no discussion
the field may vary from patch to patch, which is why the definition is an integral rather than a product
if the field is uniform and the surface flat the integral collapses to a single product, $\Phi_E = EA\cos\theta$
$$\boxed{\;\Phi_E \;=\; \int \vec E\cdot d\vec A \;=\; \int E\cos\theta\,dA\;}$$
Cut the surface into patches small enough that the field does not change across one. For each patch, multiply its area by the part of the field poking straight through it, which is the field times the cosine of the angle to the patch normal. Add those up over the whole surface.
The tilt is not a complication bolted onto the definition, it is the definition. The field $\textcolor{#1f6feb}{E}$ arrives horizontally, the surface $\textcolor{#d1690a}{A}$ is turned by $30$ degrees, and what crosses it is what would cross its shadow. Turn the surface edge on and the shadow shrinks to a line, which is why a surface parallel to the field catches no flux at all.
Looks like this, but is not
Flux is just field times area, $\Phi_E = EA$. It has the right units and it gives the right answer whenever the surface faces the field squarely.
It fails the moment the surface is turned, and silently, because the wrong answer is still plausible. Turn a $0.100\ \mathrm{m^{2}}$ plate to $60$ degrees in a field of $4.50\times10^{3}\ \mathrm{N/C}$: the honest answer is $225\ \mathrm{N\,m^{2}/C}$, the shortcut says $450$. Worse, $EA$ is never negative, so it cannot say that field is leaving a surface rather than entering it.
angle between field and normal
cos of that angle
flux (N m² / C)
0°
1.000
450
20°
0.940
423
30°
0.866
390
45°
0.707
318
60°
0.500
225
80°
0.174
78.1
90°
0.000
0
The field is 4.50 times ten to the third newtons per coulomb and the card is 0.100 square metres throughout. Notice how slowly the flux falls at first: turning the card by twenty degrees costs only six per cent, because the cosine is flat near zero. The last ten degrees, from eighty to ninety, cost more than the first thirty did.
Flux through a rectangle tilted at 30 degrees
A uniform field of $4.50\times10^{3}\ \mathrm{N/C}$ points horizontally. A flat rectangular card measuring $25.0\ \mathrm{cm}$ by $40.0\ \mathrm{cm}$ is held so that its normal makes an angle of $30.0^{\circ}$ with the field. Find the flux through the card.
Given
$E = 4.50\times10^{3}\ \mathrm{N/C}$, uniform
card $25.0\ \mathrm{cm} \times 40.0\ \mathrm{cm}$
angle between the normal and the field, $\theta = 30.0^{\circ}$
Find
the electric flux through the card
SolutionArea in SI units before anything else
$$A = (0.250)(0.400) = 0.100\ \mathrm{m^{2}}$$
centimetres squared would put a factor of ten thousand into the answer, and the error is invisible once the units are dropped
Apply the definition with the field uniform
$$\Phi_E = EA\cos\theta$$
the field is the same on every patch and the surface is flat, so the integral is a single product
Independent route through the field rather than the area. Split the field into a part along the normal, $4.50\times10^{3}\cos 30.0^{\circ} = 3.90\times10^{3}\ \mathrm{N/C}$, and a part lying in the plane of the card, which crosses nothing. The perpendicular part times the full area gives $(3.90\times10^{3})(0.100) = 390\ \mathrm{N\,m^{2}/C}$. Projecting the field and projecting the area are different operations and they agree, which is the content of the cosine.
One multiplication and one cosine, against an integral that never had to be written because the field was uniform.
Read the angle in the question before using it. Some questions give the angle between the field and the plane of the surface, which is the complement of the one the formula wants, and a cosine where a sine belongs turns 390 into 225.
Net flux through a closed box in a uniform field
A cube of side $20.0\ \mathrm{cm}$ sits in a uniform field of $6.00\times10^{3}\ \mathrm{N/C}$. Find the flux through each face and the net flux out of the cube, first with the field along the cube's axis, then with the field turned by $30.0^{\circ}$ inside one of the cube's planes.
Given
cube of side $a = 20.0\ \mathrm{cm}$, so each face has area $0.0400\ \mathrm{m^{2}}$
$E = 6.00\times10^{3}\ \mathrm{N/C}$, uniform
two orientations: field along an axis, and field turned by 30.0 degrees
Find
the flux through the faces and the net flux out of the closed surface
Check that the individual numbers still describe the same field. In the tilted case the two face fluxes come from components of size $5.20\times10^{3}$ and $3.00\times10^{3}$, and $\sqrt{5.20^{2}+3.00^{2}} = 6.00$ in units of $10^{3}\ \mathrm{N/C}$, which is the field we started with. So the decomposition lost nothing on the way.
Six faces, and four of them were zero in the first orientation and two in the second; choosing axes along the cube is what made that happen.
The net flux came out zero twice, and the two calculations shared nothing except the emptiness of the box. That is the first hint that the net flux is not really about the field at all, but about what is inside.
Checkpoint
§03.1 - flux through a surface lying along the field●○○○○
Thirty seconds, no calculator. A uniform field of $2.00\times10^{3}\ \mathrm{N/C}$ points due east, and a flat square of side $30.0\ \mathrm{cm}$ is held in a vertical plane that also runs east to west, so the field skims along the surface rather than crossing it.
Given
$E = 2.00\times10^{3}\ \mathrm{N/C}$ pointing east
square of side $30.0\ \mathrm{cm}$
the plane of the square contains the field direction
Find
(a) Find the flux through the square.
Hint 1/4
Do not reach for the area yet. Ask which way the normal of this square points, and how that direction sits relative to the field.
Hint 2/4
$\Phi_E = EA\cos\theta$, with $\theta$ between the field and the surface normal.
Hint 3/4
The plane of the square contains the field, so its normal is perpendicular to the field and $\theta = 90^{\circ}$, whatever the area is.
Hint 4/4
$\Phi_E = 0$, because $\cos 90^{\circ} = 0$.
Show solutionLocate the normal
$$\hat n \perp \vec E$$
the normal of a plane is perpendicular to every direction lying in that plane, and the field is one of those directions
$$\Phi_E = EA\cos 90^{\circ} = 0$$
no component of the field pokes through, so no area can rescue the answer
Answer $$\boxed{\;\Phi_E = 0\;}$$
Check
Picture the shadow. A surface held edge on to the field casts a shadow of zero width across the field, and flux is field times shadow. Zero width, zero flux, independently of how big the card is.
⚠ Using the angle to the surface instead of the angle to the normal
the surface is the thing you can see in the picture and the normal has to be imagined, so the visible angle is the one that gets used
wrong$$\Phi_E = EA\cos 60^{\circ}\ \text{for a card tilted } 60^{\circ} \text{ from the field}$$
⚠ Dropping the sign on the entry face of a closed surface
flux feels like an amount, and amounts are positive; the minus sign looks like a slip rather than a statement
wrong$$\Phi_{\rm net} = EA + EA = 2EA$$
right$$\Phi_{\rm net} = -EA + EA = 0$$
⚠ Leaving the area in square centimetres
the lengths are given in centimetres and the conversion happens at the end of the line, where it is easy to convert the length and forget that the area needs it twice
3.2Gauss's law: the flux out of a closed surface counts only the charge inside
Total outward flux equals enclosed charge over epsilon nought, whatever the shape of the surface.
The cube in the last example gave zero net flux twice over, and the only thing the two orientations had in common was that the box was empty. That is worth chasing.
TheoremTheorem 3.2: Gauss's law
Conditions
the surface is closed, with an inside and an outside and no edges
the charges are at rest, which is what makes the field purely electrostatic
no charge sits exactly on the surface itself, which would make the enclosed charge ambiguous
$$\boxed{\;\oint \vec E\cdot d\vec A \;=\; \frac{Q_{\rm enc}}{\varepsilon_0}\;}$$
Add up the outward flux over the whole of a closed surface. The answer depends only on how much charge is trapped inside, divided by the constant epsilon nought. It does not depend on where inside that charge sits, on how it is shaped, on what else is outside, or on the shape of the surface you happened to draw.
Proof
Start with a single point charge $q$ and pick, for once, the easiest possible surface: a sphere of radius $r$ centred on it. Everywhere on that sphere the field has the same size $kq/r^{2}$ and points straight out, along the outward normal.
Every cosine in the flux integral is therefore one and the field comes out of the integral: $\Phi_E = \frac{kq}{r^{2}}\times 4\pi r^{2} = 4\pi k q$.
The radius has cancelled. The field fell off as $1/r^{2}$ and the area grew as $r^{2}$, and the two effects are exactly matched. This is the one place where the inverse square law is doing the real work, and if the exponent were $2.01$ the cancellation would fail.
Write the constant the other way: $k = 1/(4\pi\varepsilon_0)$, so $4\pi k q = q/\varepsilon_0$. That is the law for one charge and a sphere.
Now deform the sphere into any shape you like, keeping the charge inside. Pushing a patch of surface outwards to a larger radius multiplies its area by a factor and divides the field on it by the same factor, so the flux through that patch does not change. Tilting a patch multiplies its area and divides its cosine equally. Nothing survives the deformation.
Finally, put the charge outside instead. Any narrow cone drawn from the charge now cuts the surface twice, entering through one patch and leaving through another, and those two contributions carry opposite signs and equal sizes, so they cancel in pairs and the total is zero.
For many charges, add: the field is the sum of the individual fields by superposition, flux is linear in the field, so the total flux is the sum of the individual fluxes. The ones inside each give $q_i/\varepsilon_0$ and the ones outside each give nothing.
Two pictures, one law. On the left the same $\textcolor{#cf222e}{charge}$ sits inside a neat sphere and a lumpy blob; every $\textcolor{#1f6feb}{field\ line}$ that escapes one has to escape the other, so the two surfaces report the same flux. On the right the charge sits outside, and each line that enters the $\textcolor{#d1690a}{surface}$ is the same line that leaves it, so the total is zero even though the field is large everywhere on that surface.
Looks like this, but is not
If the net flux through a closed surface is zero, the field is zero everywhere on it. Zero total, zero everywhere: it is how sums usually behave when a physicist tells you a result is zero.
Zero is a sum, and sums of nonzero things are cheerfully zero all the time. The cube in a uniform field had $+240$ leaving one face and $-240$ entering the opposite one, with a field of $6.00\times10^{3}\ \mathrm{N/C}$ at every point of both. A closed surface drawn in the middle of a laboratory with a working high voltage supply outside it encloses no charge, reports zero flux, and would still give you a shock. The law constrains the difference between what leaves and what enters, and says nothing about either one separately.
Net flux out of a surface with three charges near it
A closed surface of no particular shape encloses two charges, $q_1 = +3.00\ \mathrm{nC}$ and $q_2 = -5.00\ \mathrm{nC}$. A third charge $q_3 = +8.00\ \mathrm{nC}$ sits just outside the surface, a few millimetres from its wall. Find the net flux out of the surface.
Given
$q_1 = +3.00\ \mathrm{nC}$ inside
$q_2 = -5.00\ \mathrm{nC}$ inside
$q_3 = +8.00\ \mathrm{nC}$ outside, very close to the wall
Check the sign against the picture rather than the arithmetic. A net charge of minus two nanocoulombs inside behaves, from far enough away, like a single negative charge, and field lines run into a negative charge. Lines running in means flux entering means a negative number, which is what came out. If the answer had been positive, the sign of the enclosed charge would be the first thing to re-read.
One addition and one division, with the eight nanocoulomb charge never used; recognising that it is a distractor is most of the question.
The outside charge changes the field at every point of that surface, sometimes enormously, since it sits millimetres away. It changes the total flux by nothing at all. Keep those two sentences separate and this topic stops producing paradoxes.
Flux through one face of a cube with a charge at its centre
A point charge of $+6.00\ \mathrm{nC}$ sits at the exact centre of a cube of side $12.0\ \mathrm{cm}$. Find the flux through the whole cube and through one face. Then say what happens to both answers if the cube is replaced by one of side $50.0\ \mathrm{cm}$.
Given
$q = +6.00\ \mathrm{nC}$ at the centre of the cube
cube of side $12.0\ \mathrm{cm}$, later $50.0\ \mathrm{cm}$
Find
the total flux, the flux through one face, and the effect of enlarging the cube
SolutionThe whole cube, where the law does the work
the cube is a closed surface, so the law applies to it exactly as it does to a sphere, awkward corners and all
One face, where symmetry does the work
$$\Phi_{\rm face} = \frac{1}{6}\Phi_{\rm total}$$
the charge sits at the centre, so the six faces are interchangeable by rotation and no face can carry more than any other; without that symmetry this step would be false
the enclosed charge has not changed, and the law never mentions the size of the surface; the field on the wall is about seventeen times weaker and the area about seventeen times larger
Answer $$\boxed{\;\Phi_{\rm total} = 678\ \mathrm{N\,m^{2}/C},\qquad \Phi_{\rm face} = 113\ \mathrm{N\,m^{2}/C},\ \text{unchanged by the size}\;}$$
Check
Independent check with the other form of the constant. Using $4\pi k q$ instead of $q/\varepsilon_0$ gives $4\pi(8.99\times10^{9})(6.00\times10^{-9}) = 678\ \mathrm{N\,m^{2}/C}$, the same number by a route that never uses epsilon nought. The two constants are consistent to the three figures we are quoting.
Direct integration over one square face of a cube with a charge at its centre is a genuinely unpleasant double integral; the symmetry argument replaced it with a division by six.
Whenever a question asks for the flux through part of a closed surface, the only tool that works is a symmetry that makes the parts interchangeable. Move the charge off centre and the division by six becomes false at once, which is what the corner question in the practice set is built on.
Checkpoint
§03.2 - moving a charge around inside a closed surface●●○○○
Thirty seconds. A point charge sits inside a closed spherical surface, but not at its centre, and you slide the charge slowly to a new position, still inside and still not at the centre.
Given
a closed spherical surface
a point charge inside it, off centre
the charge is moved to another point, still inside
Find
(a) Which statement describes what happens as the charge is moved?
Hint 1/4
Two quantities are on trial and they are not the same quantity. Ask what the law fixes, and then ask separately what sets the field at one particular point of the surface.
Hint 2/4
$\oint\vec E\cdot d\vec A = Q_{\rm enc}/\varepsilon_0$, in which nothing appears but the total charge inside.
Hint 3/4
Here the charge stays inside throughout, so the right hand side never changes. But the distance from the charge to a fixed point of the sphere does change as it moves.
Hint 4/4
So the total flux is fixed while the field is redistributed over the surface: bigger where the charge came closer, smaller where it moved away.
the position of the charge does not appear anywhere in the law, only the amount of it
$$E(\text{point}) = \frac{kq}{r^{2}}\ \text{with } r \text{ changed}$$
the field at a named point is a Coulomb calculation, and moving the source changes the distance in it
Answer $$\boxed{\;\Phi_E\ \text{fixed},\ \ \vec E\ \text{on the surface redistributed}\;}$$
Check
Extreme case as a test. Slide the charge until it is a hair inside the wall. The field on the nearby patch becomes enormous and the field on the far side becomes tiny, yet the outward flux is still the same number it was at the centre, because the same lines still have to get out somewhere.
⚠ Putting outside charges into the enclosed charge
they appear in the diagram, they are large, and they clearly affect the field, so leaving them out feels like ignoring information
in most of physics a vanishing total is a strong statement, and it takes a while to accept that this one is a statement about a difference
wrong$$\Phi_E = 0 \Rightarrow \vec E = 0\ \text{on the surface}$$
right$$\Phi_E = 0 \Rightarrow \text{as much enters as leaves}$$
⚠ Dividing the total flux between faces when the charge is not central
the division by six is memorable and the condition that produced it is not, so it survives into problems that do not support it
wrong$$\Phi_{\rm face} = \tfrac16 \Phi_{\rm total}\ \text{for an off centre charge}$$
right$$\Phi_{\rm face} = \tfrac16\Phi_{\rm total}\ \text{only when the six faces are related by symmetry}$$
3.3Choosing the surface: what symmetry has to give you before the law is any use
The law is always true; it is useful only on a surface where the field has one size and one direction.
So far the law has been used in the easy direction, charge in and flux out. Turning it round to get a field out of it needs one extra thing, and it is not algebra.
MethodMethod 3.3: when Gauss's law can deliver a field
Conditions
there is a surface on which the size of the field is the same at every point, by symmetry and not by hope
on that surface the field is either everywhere perpendicular to it or everywhere parallel to it, so every cosine is one or zero
the enclosed charge can be written down for that surface
If the field has the same size everywhere on the part of the surface it actually crosses, you can pull it out of the integral, and what is left of the integral is just an area. The field is then the enclosed charge divided by epsilon nought and by that area. Every application in this section is that one sentence with a different area put into it.
Proof
Split the closed surface into the part the field crosses squarely and the part it skims along. On the second part every cosine is zero, so it contributes nothing and can be forgotten.
On the first part the field is perpendicular to the surface and, by assumption, has the same size $E$ everywhere on it. A constant comes out of an integral: $\oint\vec E\cdot d\vec A = E\,A_{\perp}$.
Set that equal to $Q_{\rm enc}/\varepsilon_0$ and divide. The whole of the calculus has been replaced by the area of a sphere, a cylinder or a pair of discs.
The step that can fail is the second one. If the field varies over the surface, pulling it out of the integral is simply wrong, and no amount of care later repairs it.
Three symmetries, three surfaces, and no choice about which goes with which. The $\textcolor{#cf222e}{charge}$ dictates the shape of the $\textcolor{#d1690a}{surface}$, because the surface has to be one on which the $\textcolor{#1f6feb}{field}$ looks the same everywhere. Draw a cube around a point charge and the law is still true, but the field on the cube varies from face centre to corner and nothing can be pulled out of the integral.
Looks like this, but is not
Gauss's law gives the field of any charge distribution, since it holds for any closed surface. Both halves of that sentence are true, and the conclusion is still false.
Take a dipole, two opposite charges a centimetre apart, and draw a sphere around the pair. The enclosed charge is zero, so the flux is zero, and that is a correct statement about a field that is large and complicated everywhere on the sphere. The law has told you the truth and told you nothing you can use. Take a square charged plate and ask for the field near one corner: there is no surface anywhere on which the field is constant, so the integral never collapses. The law is a hundred per cent true and zero per cent useful in both cases.
L / x
correction factor
error of the infinite formula
2
0.894
11.8% high
5
0.981
2.0% high
7
0.990
1.0% high
10
0.995
0.5% high
15
0.998
0.2% high
25
0.9992
0.08% high
The correction factor is the half length divided by the square root of the half length squared plus the distance squared, and the infinite answer has to be multiplied by it to get the truth. Half length seven times the distance buys one per cent, which is the working rule quoted in the conventions block: a wire whose total length is about fifteen times your distance from its middle is infinite for exam purposes.
Recovering the field of a point charge from Gauss's law
Use Gauss's law alone to find the field a distance $r$ from an isolated point charge $q$, and evaluate it for $q = +2.50\ \mathrm{nC}$ at $r = 4.00\ \mathrm{cm}$.
Given
an isolated point charge $q$
a field point a distance $r$ from it
numbers for the last step: $q = +2.50\ \mathrm{nC}$, $r = 4.00\ \mathrm{cm}$
Find
the field, from the law rather than from Coulomb
SolutionChoose the surface the symmetry hands you
$$\text{sphere of radius } r \text{ centred on } q$$
the charge looks identical from every direction, so the field can depend on r alone and must point radially; a sphere is the only surface on which that field has one size
$$\oint \vec E\cdot d\vec A = E(4\pi r^{2})$$
field perpendicular to the surface everywhere, same size everywhere, so it comes straight out of the integral
Set it against the enclosed charge
$$E(4\pi r^{2}) = \frac{q}{\varepsilon_0}$$
the whole charge is inside whatever radius we chose, so the enclosed charge is simply q
This is the consistency check on the whole framework rather than on the arithmetic. Gauss's law was derived from Coulomb's law, so it had better give it back, and it does, exactly and with no extra conditions. A framework that returned $kq/r^{3}$ here would have to be thrown away whatever else it got right.
One area formula and one division. The same result by direct integration is trivial too, which is why nobody uses Gauss's law for a point charge except as a test.
Notice the shape of the argument: symmetry fixed the direction and the dependence, and the law fixed the size. That division of labour is repeated in every application below, and when a problem has no symmetry the first half fails and the second half becomes useless.
How long a wire has to be before it counts as infinite
A wire of total length $1.00\ \mathrm{m}$ carries $\lambda = 3.00\ \mathrm{nC/m}$ uniformly. Find the field $2.00\ \mathrm{cm}$ from its middle, first with the exact finite wire result and then with the infinite wire result that Gauss's law will give in a later block, and compare them.
Given
total length $1.00\ \mathrm{m}$, so half length $L = 0.500\ \mathrm{m}$
$\lambda = 3.00\ \mathrm{nC/m}$
field point on the perpendicular bisector, $x = 2.00\ \mathrm{cm}$
Find
both fields and the size of the error committed by idealising the wire
Test the correction factor at a distance where the answer is obvious instead. Move out to $x = L = 0.500\ \mathrm{m}$, where the factor becomes $1/\sqrt2 = 0.707$, a thirty per cent error. A correction that is invisible close in and severe far out is exactly what a formula that pretends the ends are not there should produce.
One extra square root bought a number that says how much you are allowed to lie about the wire.
Infinite in this subject means the ends are too far away to matter at the accuracy you are working to. That is a statement about a ratio, never about a length, and the table below turns it into a number you can quote.
Checkpoint
§03.3 - which distribution surrenders to Gauss's law●●○○○
Thirty seconds. Four charge distributions are listed below, and for one of them a Gaussian surface exists on which the field has the same size everywhere, so the law hands you the field in two lines.
Given
all four objects carry uniform charge
in each case you want the field at the point named
Find
(a) Which one can be solved with Gauss's law alone?
Hint 1/4
Do not test the four for difficulty. Test them for one thing only: is there a surface through the field point on which the field must have the same size at every point?
Hint 2/4
The field can be pulled out of the flux integral only when the symmetry of the charge forces it to be constant on the surface: spherical, cylindrical or planar.
Hint 3/4
A ball of charge looks the same from every direction, so a sphere drawn through the interior point works. A plate, a rod and a pair of charges each single out a special direction.
Hint 4/4
Only the ball qualifies, and the answer there is the enclosed charge over epsilon nought over the area of the sphere.
Show solutionApply the constancy test
$$\text{ball:}\ E = E(r),\ \text{radial}$$
no direction is special, so the field can depend only on distance from the centre, which is exactly what a sphere needs
$$\text{plate, rod, pair:}\ E = E(\text{position})$$
each of these has an edge, an end or an axis that breaks the symmetry, so the field varies over every candidate surface
Cross check against what we already know. The rod and the ring were solved last section by integration precisely because no symmetry argument was available for them, and the ball was listed at the top of this section as the thing we could not do. The classification is consistent with both experiences.
⚠ Assuming the law fails when it merely gives zero
a zero answer feels like an error message, when it is often a correct and complete statement about a symmetric situation
wrong$$\Phi_E = 0 \Rightarrow \text{Gauss's law does not apply here}$$
right$$\Phi_E = 0 \Rightarrow Q_{\rm enc}=0,\ \text{which is true and may be useless}$$
⚠ Pulling a varying field out of the integral
the algebra looks identical whether or not the field is constant, and nothing in the notation complains
wrong$$\oint\vec E\cdot d\vec A = EA\ \text{for a cube around a point charge}$$
right$$\oint\vec E\cdot d\vec A = EA\ \text{only if } E \text{ is constant on that surface}$$
⚠ Choosing a surface that does not pass through the field point
the surface is often drawn to be convenient for the charge rather than for the question, and the two are different jobs
wrong$$\text{surface of radius } R \text{ used to find } E \text{ at } r \ne R$$
right$$\text{surface radius} = \text{the radius where the field is wanted}$$
3.4Spherical symmetry: shells, balls, and why the outside cannot tell the difference
For anything spherically symmetric the field at radius r is the charge inside that radius, treated as a point charge at the centre.
The ball we could not do at the start of the section is the first customer for the method, and it takes four lines.
TheoremTheorem 3.4: field of a spherically symmetric charge
Conditions
the charge density depends on distance from a centre and on nothing else
the field point is at distance r from that same centre
the charge outside radius r may be arranged in any spherically symmetric way at all; it does not enter
Stand at distance r from the centre. Everything closer to the centre than you are pulls or pushes as though it were a single point charge sitting at the centre. Everything further out than you are contributes exactly nothing. So the field is the enclosed charge over r squared, times k, and the only work in any problem is writing down how much charge is inside your radius.
Proof
Symmetry first. Rotating the charge about its centre changes nothing, so the field can depend only on $r$ and must point along the radius; a sideways component would have no direction to choose.
Draw a sphere of radius $r$ through the field point. On it the field has one size and is perpendicular everywhere, so $\oint\vec E\cdot d\vec A = E(r)\,4\pi r^{2}$.
Gauss's law gives $E(r)\,4\pi r^{2} = Q_{\rm enc}(r)/\varepsilon_0$, that is $E(r) = kQ_{\rm enc}(r)/r^{2}$.
For a uniform ball of radius $R$ and total charge $Q$, the density is $\rho = Q/(\tfrac43\pi R^{3})$ and the charge inside radius $r<R$ is $Q_{\rm enc} = \rho\cdot\tfrac43\pi r^{3} = Q r^{3}/R^{3}$.
Put that in: $E_{\rm in} = kQr^{3}/(R^{3}r^{2}) = kQ r/R^{3}$, a straight line rising from zero at the centre.
For $r > R$ the enclosed charge is the whole of $Q$, so $E_{\rm out} = kQ/r^{2}$: from outside, the ball is indistinguishable from a point charge at its centre.
At $r = R$ the two expressions agree, both giving $kQ/R^{2}$, which they must since the field of a smooth charge distribution has nowhere to jump.
The field of a uniformly charged ball, for the case $Q = 12.0\ \mathrm{nC}$ and $R = 8.00\ \mathrm{cm}$ worked below. Inside, only the charge closer in than you are counts and it grows as $r^{3}$ while the area grows as $r^{2}$, so $\textcolor{#1f6feb}{E}$ rises in a straight line. Outside, the enclosed charge stops growing and the usual inverse square takes over. The peak sits exactly at the $\textcolor{#cf222e}{surface}$, at $16.9\ \mathrm{kN/C}$.
Looks like this, but is not
Inside a hollow charged shell you are surrounded by charge on all sides, so the field is some average of all those pulls, small but not zero. Every intuition about being surrounded says the same thing.
It is exactly zero, anywhere inside the shell, not just at the centre. Gauss's law sees it immediately: any sphere drawn inside the shell encloses no charge at all, so the flux is zero, and the symmetry says the field has one size on that sphere, so that size must be zero. Without the law the result is genuinely surprising: standing near the inner wall, a small patch of nearby shell pulls hard, but the far side of the shell presents a patch of charge that is larger in area by exactly the square of the distance ratio, and the two cancel for every pair of opposite patches.
r (cm)
E (N/C)
kQ / r² would give
2.00
4.21 × 10³
2.70 × 10⁵
4.00
8.43 × 10³
6.74 × 10⁴
6.00
1.26 × 10⁴
3.00 × 10⁴
8.00
1.69 × 10⁴
1.69 × 10⁴
12.0
7.49 × 10³
7.49 × 10³
16.0
4.21 × 10³
4.21 × 10³
24.0
1.87 × 10³
1.87 × 10³
The two columns agree from the surface outwards and disagree wildly inside, by a factor of sixty four at two centimetres. Notice also that the point charge column blows up as the radius goes to zero while the true field goes quietly to zero there, which is the physical statement that the centre of a symmetric ball is pulled equally in every direction.
Field inside, at the surface of, and outside a uniformly charged ball
A solid non conducting ball of radius $8.00\ \mathrm{cm}$ carries $+12.0\ \mathrm{nC}$ spread uniformly through its volume. Find the field at $r = 4.00\ \mathrm{cm}$, at $r = 8.00\ \mathrm{cm}$ and at $r = 20.0\ \mathrm{cm}$ from the centre.
Given
$R = 8.00\ \mathrm{cm}$, $Q = +12.0\ \mathrm{nC}$, uniform through the volume
three field points: $r = 4.00,\ 8.00,\ 20.0\ \mathrm{cm}$
Find
the field at each of the three radii
SolutionInside, where only part of the ball counts
Independent check on the interior value using the linear law rather than the enclosed charge. Inside the ball the field is proportional to $r$, and $4.00\ \mathrm{cm}$ is half of $8.00\ \mathrm{cm}$, so the interior value must be exactly half the surface value: $1.69\times10^{4}/2 = 8.43\times10^{3}\ \mathrm{N/C}$. It is, and that route never used the enclosed charge at all.
Three lines against the nested integration that opened this section.
The dangerous number here is the middle one. The field is largest at the surface, not at the centre, and not far outside; a graph that peaks in the interior or at the origin is reporting an algebra error.
A thin shell of charge with a point charge sitting inside it
A thin non conducting spherical shell of radius $10.0\ \mathrm{cm}$ carries $+36.0\ \mathrm{nC}$ spread uniformly over its surface, and a point charge of $+12.0\ \mathrm{nC}$ is fixed at its centre. Find the field at $r = 5.00\ \mathrm{cm}$ and at $r = 25.0\ \mathrm{cm}$.
Given
shell of radius $10.0\ \mathrm{cm}$ carrying $+36.0\ \mathrm{nC}$ uniformly
point charge $+12.0\ \mathrm{nC}$ at the centre
field points at $r = 5.00\ \mathrm{cm}$ and $r = 25.0\ \mathrm{cm}$
Find
the field at both radii
SolutionInside the shell
$$Q_{\rm enc}(0.0500) = +12.0\ \mathrm{nC}$$
a sphere of radius 5 cm contains the point charge and none of the shell, which lies entirely outside it
Check the outer answer by superposing two separate known fields instead of adding the charges first. The point charge alone gives $107.9/0.0625 = 1.73\times10^{3}\ \mathrm{N/C}$ and the shell alone, from outside, gives $323.6/0.0625 = 5.18\times10^{3}\ \mathrm{N/C}$. They point the same way and add to $6.90\times10^{3}\ \mathrm{N/C}$.
Two rules did all the work and neither involved a calculation: charge outside your radius is invisible, and charge inside your radius acts as if it were at the centre. Almost every spherical exam question is those two sentences plus arithmetic.
Checkpoint
§03.4 - field at the centre of a charged shell●○○○○
Thirty seconds, no calculator needed. A thin spherical shell of radius $12.0\ \mathrm{cm}$ carries $-24.0\ \mathrm{nC}$ spread uniformly over it, and nothing else is anywhere near.
Given
thin shell of radius $12.0\ \mathrm{cm}$
charge $-24.0\ \mathrm{nC}$, uniform over the surface
no other charge present
Find
(a) Find the field at a point $5.00\ \mathrm{cm}$ from the centre.
Hint 1/4
Do not start computing. Draw the sphere of radius five centimetres and ask a single question about it: how much charge is inside?
Hint 2/4
$E(r) = kQ_{\rm enc}(r)/r^{2}$ for any spherically symmetric charge.
Hint 3/4
All of the $-24.0\ \mathrm{nC}$ sits at radius $12.0\ \mathrm{cm}$, which is outside a sphere of radius $5.00\ \mathrm{cm}$, so nothing is enclosed.
Hint 4/4
$E = 0$ at that point, and at every other point inside the shell.
Show solutionCount the enclosed charge
$$Q_{\rm enc}(0.0500) = 0$$
the whole of the charge sits at a radius larger than the surface we drew, so none of it is inside
$$E(4\pi r^{2}) = 0 \Rightarrow E = 0$$
the area is certainly not zero, so the field has to be, and symmetry guarantees it is the same everywhere on the sphere
Answer $$\boxed{\;E = 0\;}$$
Check
Check that it does not contradict the outside. Just beyond the shell, at $12.1\ \mathrm{cm}$, the field is $kQ/r^{2} = 1.47\times10^{4}\ \mathrm{N/C}$ pointing inward. The field jumps abruptly across the shell, which is allowed because the charge is idealised as having no thickness; nothing about a zero inside is inconsistent with a large field just outside.
⚠ Using the total charge when only part of it is enclosed
the total is the number printed in the question, and the enclosed charge has to be manufactured from a ratio of cubes
wrong$$E_{\rm in} = \frac{kQ}{r^{2}}$$
right$$E_{\rm in} = \frac{kQr}{R^{3}}$$
⚠ Using the wrong power in the volume ratio
the area of a sphere goes as the square and the volume as the cube, and the two get exchanged under pressure
wrong$$Q_{\rm enc} = Q\frac{r^{2}}{R^{2}}$$
right$$Q_{\rm enc} = Q\frac{r^{3}}{R^{3}}$$
⚠ Applying the ball formulas to a shell
both are spheres and both formulas contain R, so the wrong one still looks dimensionally sensible
3.5Cylindrical symmetry: the long wire, done in two lines this time
Wrap a can around the axis; only the curved side carries flux, and its area is two pi r L.
The long wire cost a substitution and an integral last week. With a can instead of a sphere it costs two lines, and the answer had better be the same one.
TheoremTheorem 3.5: field of a long cylindrical charge
Conditions
the charge is arranged so that it looks the same all the way along an axis and the same from every direction around it
the field point is far from the ends compared with its distance from the axis
lambda enclosed is the charge per unit length inside your radius, which for a solid cylinder is less than the total
Draw a can of radius r and length L with the axis of the charge running down its middle. The field is perpendicular to the curved side and the same size all over it, and it slides harmlessly along the two flat ends. So the flux is the field times the side area, two pi r L, and that equals the charge inside the can over epsilon nought. The length cancels, which is why the answer contains a charge per unit length and no length.
Proof
Symmetry: sliding along the axis changes nothing and rotating about it changes nothing, so the field can depend only on the distance $r$ from the axis, and it must point straight out from the axis.
Draw the can. Its surface has three pieces: two flat ends and the curved side. On the ends the field lies in the surface, so those cosines are zero and the ends drop out.
On the curved side the field is perpendicular and constant, so $\oint\vec E\cdot d\vec A = E\,(2\pi r L)$.
The enclosed charge is $\lambda_{\rm enc} L$, so $E\,(2\pi r L) = \lambda_{\rm enc}L/\varepsilon_0$ and the arbitrary length $L$ cancels from both sides, as it has to, since we invented it.
That leaves $E = \lambda_{\rm enc}/(2\pi\varepsilon_0 r) = 2k\lambda_{\rm enc}/r$, which is exactly the result the integral gave last section.
For a solid cylinder of radius $R$ with uniform $\rho$ and a field point inside it, the enclosed charge per unit length is $\rho\pi r^{2}$, giving $E = \rho r/(2\varepsilon_0)$: linear in $r$, just as inside the ball.
Why the can works. The $\textcolor{#1f6feb}{field}$ leaves the $\textcolor{#cf222e}{wire}$ straight out from the axis, so it hits the curved side of the $\textcolor{#d1690a}{can}$ square on and slides along the two flat ends without crossing them. Only one area ever enters the calculation, and the length of the can cancels because it appears on both sides of the equation.
Looks like this, but is not
A cylinder is a three dimensional object, so its field should fall off as $1/r^{2}$ like everything else. Point charges, balls and dipoles all fall off at least that fast, so a line ought to as well.
It falls off as $1/r$, and the can shows why in one step: the area that the flux has to spread over is $2\pi r L$, which grows as the first power of $r$, not the second. Doubling your distance from a wire halves the field; doubling your distance from a point charge quarters it. The three powers now in play are worth keeping straight: a dipole dies as $1/r^{3}$, a point charge or a ball as $1/r^{2}$, a long wire as $1/r$, and, in the next block, a large sheet not at all.
Field beside a long charged wire, from the can
A long straight wire carries $\lambda = +5.20\ \mathrm{nC/m}$ uniformly. Find the field at a perpendicular distance of $2.50\ \mathrm{cm}$ from the middle of the wire.
Given
$\lambda = +5.20\ \mathrm{nC/m}$, uniform
$r = 2.50\ \mathrm{cm}$ from the wire
the wire is long compared with that distance
Find
the field at that point
SolutionSet up the can
$$E(2\pi r L) = \frac{\lambda L}{\varepsilon_0}$$
flux only through the curved side, and the charge inside a can of length L is lambda times L
using the k form avoids carrying two pi and epsilon nought separately
$$E = 3.74\times10^{3}\ \mathrm{N/C}$$
positive charge, so the field points radially away from the wire
Answer $$\boxed{\;E = 3.74\times10^{3}\ \mathrm{N/C},\ \text{radially away from the wire}\;}$$
Check
Independent check by scaling the earlier integration result rather than repeating this one. Last section a wire with $\lambda = 3.00\ \mathrm{nC/m}$ gave $2.70\times10^{3}\ \mathrm{N/C}$ at $2.00\ \mathrm{cm}$. Scaling by the charge, times $5.20/3.00$, and by the distance, times $2.00/2.50$, gives $2.70\times10^{3}\times1.733\times0.800 = 3.74\times10^{3}\ \mathrm{N/C}$. Two routes, one number.
Two lines, against a substitution and a trigonometric integral last week for the same answer.
This is the payoff the section promised. Nothing new about the physics has been discovered; the same field has been extracted with a fraction of the work, because the symmetry was used at the start instead of surviving the integration.
Inside and outside a solid charged cylinder
A long solid non conducting cylinder of radius $5.00\ \mathrm{cm}$ carries a uniform volume charge density $\rho = +6.00\ \mu\mathrm{C/m^{3}}$. Find the field at $r = 2.00\ \mathrm{cm}$ and at $r = 12.0\ \mathrm{cm}$ from the axis.
Check that the two formulas meet at the surface, which they must for a charge with no sheet on it. The inside expression at $r = R$ gives $\rho R/(2\varepsilon_0) = 1.69\times10^{4}\ \mathrm{N/C}$, and the outside expression at $r = R$ gives $2k\lambda/R = 1.69\times10^{4}\ \mathrm{N/C}$. They agree, so no algebra was lost in the switch.
The two answers here are almost equal, $6.78$ against $7.06$ in units of $10^{3}\ \mathrm{N/C}$, at radii six times apart. That is not a coincidence: the field climbs to a peak at the surface and then falls, so points on either side of the peak can easily match, and a numerical coincidence like this is no evidence that a formula was used in the wrong region.
Checkpoint
§03.5 - field inside a hollow charged tube●●○○○
Thirty seconds. A long thin walled tube of radius $4.00\ \mathrm{cm}$ carries $+9.00\ \mathrm{nC/m}$ spread uniformly over its surface, and you stand on the axis, right in the middle of it.
Given
hollow tube of radius $4.00\ \mathrm{cm}$
$\lambda = +9.00\ \mathrm{nC/m}$ on the wall
the field point is on the axis
Find
(a) Find the field on the axis, and then say what it is at a point 2.00 cm from the axis.
Hint 1/4
Draw the can of the radius you care about and ask the only question that matters: how much charge is inside it?
Hint 2/4
$E = 2k\lambda_{\rm enc}/r$, with only the charge inside your radius counted.
Hint 3/4
All the charge sits at radius $4.00\ \mathrm{cm}$. A can of radius $2.00\ \mathrm{cm}$, and the axis itself, enclose none of it.
Hint 4/4
So the field is zero at both points, and everywhere else inside the tube.
Show solutionCount the charge inside
$$\lambda_{\rm enc}(r<R) = 0$$
the wall sits at a larger radius, so a can drawn through the field point contains nothing at all
$$E(2\pi r L) = 0 \Rightarrow E = 0$$
the area is nonzero and symmetry says the field is uniform over the side, so it must vanish
Answer $$\boxed{\;E = 0\ \text{everywhere inside the tube}\;}$$
Check
Consistency with the outside. Just beyond the wall the field is $2k\lambda/R = 4.05\times10^{3}\ \mathrm{N/C}$, so the field jumps across the wall. That is the same behaviour as the spherical shell, and it is what an idealised surface charge with no thickness always does.
⚠ Including the flat ends of the can in the area
the can is a closed surface and it feels wrong to leave two of its three pieces out of a total
wrong$$A = 2\pi r L + 2\pi r^{2}$$
right$$A_{\perp} = 2\pi r L$$
⚠ Using the total charge of a finite cable instead of the charge per unit length
the question often gives a total charge and a length, and the division is an extra step that is easy to skip
Straddle the sheet with a flat box whose two ends are parallel to it. The field comes straight out of the sheet on both sides, so it crosses both ends and skims along the side wall of the box. Two ends means twice the area, which is where the factor of two under the sigma comes from. The thickness of the box never appears, and that is the algebra telling you the field is the same however far out you put the ends.
Proof
Symmetry: sliding along the sheet in either direction changes nothing, so the field cannot depend on where you stand sideways. Reflecting the sheet in its own plane changes nothing either, so the field must point straight away from the sheet and have the same size on both sides.
Straddle the sheet with a pillbox of cap area $A$. On the side wall the field lies in the surface, so that piece contributes nothing.
Both caps are crossed squarely and carry the same field, so $\oint\vec E\cdot d\vec A = EA + EA = 2EA$.
The charge inside the box is the patch of sheet it has cut out, $Q_{\rm enc} = \sigma A$.
Gauss's law gives $2EA = \sigma A/\varepsilon_0$, so $E = \sigma/(2\varepsilon_0)$. The area cancels and the thickness never entered, so the field is the same at one millimetre and at one metre.
Two parallel sheets with $+\sigma$ and $-\sigma$: between them the two fields point the same way and add to $\sigma/\varepsilon_0$; outside them the two fields point oppositely and cancel to zero.
The pillbox that produces the factor of two. The $\textcolor{#cf222e}{sheet}$ pushes field out on both sides, so the $\textcolor{#d1690a}{box}$ is crossed twice, once through each cap, and never through its side wall. Move the caps further out and the $\textcolor{#1f6feb}{field}$ arrows keep exactly the same length, which is the whole content of the result.
Looks like this, but is not
A field that does not fall off with distance is impossible; energy has to spread out. If it were true you could stand a kilometre from a charged plate and feel the same push as at a centimetre.
You could, if the plate really were infinite, and that is the catch. At a kilometre from a plate two metres wide you are not near a sheet any more, you are near a small object, and the field has long since become $kQ/r^{2}$. The result is not a claim about infinite plates existing; it is a statement that while your distance is small compared with the size of the plate, the extra sheet you can see as you back away compensates exactly for the extra distance. Back away far enough and the compensation runs out, because there is no more sheet to see.
Field of a large charged plastic sheet
A large flat plastic sheet carries a uniform $\sigma = +1.20\ \mu\mathrm{C/m^{2}}$. Find the field close to the middle of the sheet, and say how the answer changes if you move from $1.00\ \mathrm{cm}$ away to $10.0\ \mathrm{cm}$ away.
no distance appears in the formula, so a factor of ten in distance changes nothing at all
Answer $$\boxed{\;E = 6.78\times10^{4}\ \mathrm{N/C}\ \text{on both sides, at both distances}\;}$$
Check
Independent derivation from the wire result, which came from an integral rather than from a pillbox. Cut the sheet into long parallel strips of width $dy$; each is a line with $\lambda = \sigma\,dy$ and contributes $2k\sigma\,dy$ divided by its distance, and the components along the sheet cancel in pairs. What survives is $2\pi k\sigma$, and numerically $2\pi(8.99\times10^{9})(1.20\times10^{-6}) = 6.78\times10^{4}\ \mathrm{N/C}$. Same number, entirely different route.
One division, and the strip integration in the check is the work that was avoided.
Compare with dry air, which breaks down and sparks at about $3\times10^{6}\ \mathrm{N/C}$. This sheet is a factor of forty below that, so a charge density of a microcoulomb per square metre is a perfectly ordinary thing to have on a piece of plastic.
Two oppositely charged sheets facing each other
Two large parallel sheets face each other a few centimetres apart. One carries $\sigma = +1.20\ \mu\mathrm{C/m^{2}}$ and the other $-1.20\ \mu\mathrm{C/m^{2}}$. Find the field between them and outside them.
Given
two large parallel sheets, $\pm 1.20\ \mu\mathrm{C/m^{2}}$
field points between the sheets and beyond them on both sides
Check with a single pillbox instead of superposition. Put one cap between the plates and the other outside, straddling the positive sheet. The enclosed charge is $\sigma A$; the outside cap catches nothing because the field is zero there, so all of the flux goes through the inside cap: $EA = \sigma A/\varepsilon_0$, giving $E = \sigma/\varepsilon_0$ directly. Two arguments, one answer.
The pair produces a genuinely uniform field in a finite region and nothing outside, which is why this arrangement is the standard way of making a controlled field in a laboratory. Note the factor of two against the lone sheet: it comes from having two sheets, not from any new physics.
Checkpoint
§03.6 - distance dependence near a large sheet●●○○○
Thirty seconds. A large flat sheet carries a uniform positive charge, and you measure the field at $2.00\ \mathrm{cm}$ from its middle and again at $8.00\ \mathrm{cm}$, still close compared with the size of the sheet.
Given
a large uniformly charged sheet
two field points at 2.00 cm and 8.00 cm from the middle
both distances small compared with the size of the sheet
Find
(a) True or false: the second reading is sixteen times smaller than the first. Give a one sentence reason.
Hint 1/4
The factor of sixteen is the square of four, so the claim is that the field falls off as one over distance squared. Ask what the sheet formula actually contains.
Hint 2/4
$E = \sigma/(2\varepsilon_0)$ for a large sheet.
Hint 3/4
Neither 2.00 cm nor 8.00 cm appears anywhere in that expression, because the pillbox thickness cancelled in the derivation.
Hint 4/4
False: the two readings are equal, as long as both points are close compared with the size of the sheet.
Show solutionRead the formula honestly
$$E = \frac{\sigma}{2\varepsilon_0}$$
the pillbox thickness cancelled during the derivation, so no distance can appear in the result
$$\frac{E(0.0800)}{E(0.0200)} = 1$$
a ratio of two identical expressions, whatever sigma happens to be
Answer $$\boxed{\;\text{False; the ratio is } 1,\ \text{not } 1/16\;}$$
Check
Test the claim against the derivation rather than the formula. The pillbox with caps at 8 cm encloses exactly the same patch of sheet as the pillbox with caps at 2 cm, so the flux out of both is the same and the cap areas are the same, forcing the fields to be equal.
⚠ Counting only one cap of the pillbox
the sheet is drawn once and the field on one side is what the question asks for, so the second cap is easy to overlook
wrong$$EA = \frac{\sigma A}{\varepsilon_0} \Rightarrow E = \frac{\sigma}{\varepsilon_0}$$
right$$2EA = \frac{\sigma A}{\varepsilon_0} \Rightarrow E = \frac{\sigma}{2\varepsilon_0}$$
⚠ Making the field fall off with distance
every previous field in the course fell off with distance, so a constant one looks like a formula with a term missing
wrong$$E = \frac{\sigma}{2\varepsilon_0 x^{2}}$$
right$$E = \frac{\sigma}{2\varepsilon_0}$$
⚠ Adding the two sheet fields outside a pair as well as between them
adding is what superposition does, and the sign that turns the sum into a difference lives only in the picture
3.7Conductors at rest: an empty interior and a field of sigma over epsilon nought at the skin
Charges in a metal move until the field inside vanishes, which forces all the excess charge onto the surface.
Everything so far assumed the charge stays where it is put. Metal does not cooperate, and Gauss's law turns that into three hard results.
TheoremTheorem 3.7: a conductor in electrostatic equilibrium
Conditions
the conductor has been left alone long enough that no charge is still moving, which for a metal takes far less than a second
the charges are free to move within the metal but cannot leave it
the results hold for the metal itself, not for a cavity inside it that contains charge
$$\boxed{\;\vec E = 0\ \text{inside the metal};\quad \text{all excess charge on the surface};\quad E_{\rm just\ outside} = \frac{\sigma}{\varepsilon_0}\ \perp\ \text{surface}\;}$$
Inside the body of a metal at rest there is no field at all, because if there were, the free charges would still be moving and it would not be at rest. Since the field inside vanishes, any surface drawn inside the metal has zero flux and therefore encloses zero charge, so the excess charge has nowhere to sit except the surface. Just outside that surface the field is the local surface charge density divided by epsilon nought, and it points straight out of the metal.
Proof
Suppose the field inside the metal were not zero somewhere. The free electrons there would feel a force and would move, and the situation would not be an equilibrium. They keep moving until their own field cancels whatever was applied, which is why the interior field ends up at exactly zero and not merely small.
Draw any closed surface entirely inside the metal, however large, as long as it stays in the metal. The field is zero at every point of it, so the flux is zero, so $Q_{\rm enc} = 0$.
Shrink that surface to wrap tightly around any interior point. Still zero. So there is no net charge anywhere in the interior, and any excess charge has to live on the surface.
Now take a pillbox with one cap just outside the surface and the other just inside the metal. The inside cap catches nothing, since the field there is zero, and the side wall catches nothing because the field just outside a conductor is perpendicular to it.
That leaves one cap: $EA = \sigma A/\varepsilon_0$, so $E = \sigma/\varepsilon_0$. The factor of two compared with the lone sheet is not new physics; it is one cap instead of two.
Why must the field just outside be perpendicular? A component along the surface would push the surface charges sideways, and they are free to go, so they would move. At equilibrium no such component survives.
Finally, a cavity inside the metal holding a charge $+q$. A surface drawn in the metal around the cavity has zero flux, so it encloses zero charge, so the cavity wall must carry exactly $-q$. If the metal as a whole was neutral, $+q$ appears on its outer surface.
One aside, because it answers a question this argument raises. The zero interior field is a consequence of the exponent in Coulomb's law being exactly two: any other exponent would leave a small field inside a charged hollow conductor. Charging a hollow metal sphere and looking for charge on its inner wall is therefore a way of measuring that exponent, and it is far more sensitive than weighing forces between charges. Modern versions of the experiment find no charge inside at all and bound the departure from two at less than about one part in ten to the sixteenth.
A $\textcolor{#cf222e}{charge}$ hidden in a cavity cannot hide. The $\textcolor{#d1690a}{dashed\ surface}$ drawn in the body of the metal has zero field on it, so it encloses zero charge, which forces exactly the opposite charge onto the $\textcolor{#cf222e}{cavity\ wall}$ and leaves the balance on the $\textcolor{#cf222e}{outer\ surface}$. Move the charge about inside the cavity and the outer distribution does not even notice.
Looks like this, but is not
A solid metal ball and a solid plastic ball, both carrying the same total charge and the same radius, have the same field everywhere. From outside they do, exactly, which makes the claim hard to shake.
Outside they are identical, and Gauss's law says so: the enclosed charge at any exterior radius is the same, and both are spherically symmetric, so both give $kQ/r^{2}$ with no correction whatever. Inside they could hardly be more different. The plastic ball has charge spread through its volume and a field rising linearly to the surface; the metal ball has every scrap of its charge on the skin and a field of exactly zero throughout the interior. Two objects with identical exteriors and opposite interiors is a good thing to have met before an exam asks about it.
r (cm)
plastic ball
metal ball
1.00
1.44 × 10⁴
0
2.50
3.60 × 10⁴
0
4.00
5.75 × 10⁴
0
5.00
7.19 × 10⁴
7.19 × 10⁴
8.00
2.81 × 10⁴
2.81 × 10⁴
15.0
7.99 × 10³
7.99 × 10³
From the surface outwards the two columns are identical to every digit, and no exterior measurement can tell the balls apart. Inside they disagree completely: one field rises steadily to the surface, the other is exactly zero at every interior point. This is the single most useful contrast in the section, and exam questions are built on it every year.
A point charge inside a neutral conducting shell
A neutral conducting spherical shell has inner radius $10.0\ \mathrm{cm}$ and outer radius $12.0\ \mathrm{cm}$. A point charge of $+8.00\ \mathrm{nC}$ is placed at its centre. Find the charge on each surface of the shell, and the field at $r = 5.00\ \mathrm{cm}$, $r = 11.0\ \mathrm{cm}$ and $r = 20.0\ \mathrm{cm}$.
Given
conducting shell, inner radius $10.0\ \mathrm{cm}$, outer radius $12.0\ \mathrm{cm}$, net charge zero
point charge $+8.00\ \mathrm{nC}$ at the centre
field wanted at $r = 5.00,\ 11.0,\ 20.0\ \mathrm{cm}$
Find
the induced surface charges and the field at three radii
SolutionUse the zero field inside the metal to get the inner surface charge
$$\text{surface at } r = 11.0\ \mathrm{cm}: \ \Phi_E = 0$$
that radius lies in the body of the metal, where the field is zero, so the flux through a sphere drawn there is zero
Check the outer field a second way, by asking what a distant observer can possibly see. The whole assembly carries a net charge of $+8.00\ \mathrm{nC}$ and is spherically symmetric, so from outside it must look exactly like an $8.00\ \mathrm{nC}$ point charge at the centre, shell or no shell. That gives $1.80\times10^{3}\ \mathrm{N/C}$ at $20.0\ \mathrm{cm}$ without any discussion of surfaces.
Three applications of one law, and not a single integral.
The shell hides nothing from the outside world. Its inner and outer charges are equal and opposite and they cancel as far as any exterior point is concerned, which is why the field at 20 cm is the same as it would be with no shell at all. Shielding works the other way round, and the next example is the one that does it.
Why the field at a conductor's face is twice the lone sheet answer
A large flat metal plate carries a surface charge density of $1.20\ \mu\mathrm{C/m^{2}}$ on the face you are standing next to. A plastic sheet with the same $\sigma$ gave $6.78\times10^{4}\ \mathrm{N/C}$. Find the field just outside the metal face, and explain the discrepancy.
Given
metal plate with $\sigma = 1.20\ \mu\mathrm{C/m^{2}}$ on the near face
for comparison, a plastic sheet with the same sigma gives $6.78\times10^{4}\ \mathrm{N/C}$
Find
the field just outside the metal, and the reason for the factor of two
SolutionThe pillbox has only one useful cap
$$\Phi_{\rm inner\ cap} = 0$$
that cap is buried in the metal where the field is zero, which is the entire difference between this case and the plastic sheet
$$EA = \frac{\sigma A}{\varepsilon_0} \Rightarrow E = \frac{\sigma}{\varepsilon_0}$$
all the flux leaves through the one cap that is outside, so no factor of two appears
outside, the two faces push the same way and add; inside the metal they oppose and cancel, which is what made the interior field zero in the first place
The two arguments in this example are independent and they agree. The pillbox route uses the emptiness of the metal and gets $\sigma/\varepsilon_0$ from one cap; the superposition route adds two sheets of $\sigma/2\varepsilon_0$ each and gets the same outside while producing zero inside, which is the property the first route assumed. Neither could have been guessed from the other.
So $\sigma/(2\varepsilon_0)$ and $\sigma/\varepsilon_0$ are not rivals. The first is the field of one sheet of charge; the second is the field at the face of a conductor, where the sigma is the local density on that face and a second sheet is quietly present on the other face.
Checkpoint
§03.7 - charge induced on the wall of a cavity●●●○○
Thirty seconds. A lump of metal carrying a net charge of $+5.00\ \mathrm{nC}$ has a hollow cavity inside it, and a point charge of $-3.00\ \mathrm{nC}$ is suspended in the cavity without touching the walls.
Given
metal with net charge $+5.00\ \mathrm{nC}$
a cavity inside it containing $-3.00\ \mathrm{nC}$
the charge does not touch the metal
Find
(a) What charge sits on the cavity wall and what charge sits on the outer surface?
Hint 1/4
Draw one closed surface inside the metal, wrapped around the cavity, and ask what its flux must be. That single question fixes the wall charge, and then bookkeeping fixes the rest.
Hint 2/4
Inside a conductor at rest $\vec E = 0$, so any surface drawn there has zero flux and therefore encloses zero net charge.
Hint 3/4
The surface encloses the $-3.00\ \mathrm{nC}$ in the cavity plus whatever is on the wall, and the metal carries $+5.00\ \mathrm{nC}$ altogether.
Check from far away, where the answer must be simple. The whole object carries $-3.00 + 5.00 = +2.00\ \mathrm{nC}$, so a distant observer must see the field of a $+2.00\ \mathrm{nC}$ point charge. That is exactly what the outer surface charge produces, so the bookkeeping is consistent with the exterior.
⚠ Putting excess charge inside the body of a conductor
the metal is solid and it feels natural for charge to be spread through it, the way it is in a plastic ball
wrong$$\rho \ne 0\ \text{inside a metal at rest}$$
right$$\rho = 0\ \text{inside};\ \text{all excess charge on the surface}$$
⚠ Using the lone sheet formula at the surface of a conductor
both formulas have sigma on top and epsilon nought underneath, and the difference is a single factor of two with no visible cause
⚠ Giving the cavity wall the same sign as the charge inside it
the wall charge is caused by the cavity charge, and causes are usually thought of as producing something like themselves
wrong$$Q_{\rm wall} = +q\ \text{for a cavity charge } +q$$
right$$Q_{\rm wall} = -q$$
Getting a field out of Gauss's law
The charge has spherical, cylindrical or planar symmetry and you want the field at a stated point. If it has none of the three, stop here and integrate instead.
Name the symmetry
Ask what you can do to the charge distribution without changing it. Rotate it about a centre: spherical. Slide it along an axis and rotate about that axis: cylindrical. Slide it in two directions along a plane: planar. If the honest answer is none of these, the method stops here.
Say what the symmetry forces the field to be
Write the sentence out: the field depends only on the distance from the centre and points radially; or only on the distance from the axis and points straight out from it; or only on which side of the sheet you are on and points perpendicular to it. This sentence, not the algebra, is what makes the rest legal.
Draw the surface through the field point
Sphere, can or pillbox, and its radius or position is set by where the field is wanted, never by the size of the object. A surface that does not pass through the field point tells you about a different point.
Evaluate the left hand side as a product
Discard the pieces the field runs along, and on the rest write the field times the area: $4\pi r^{2}$, or $2\pi rL$, or $2A$ for a pillbox in free space and $A$ for one with a cap inside metal.
Write the enclosed charge
Only what is inside the surface you drew, with sign. This is where the density and the geometry come in, and it is the step where most marks are lost.
Solve and test a limit
Divide. Then push the answer to a radius where you already know the result: far away it should look like a point charge or a line; at a boundary the inside and outside expressions should agree; at the centre of a symmetric object it should vanish.
Where it goes wrong
Choosing the surface to match the object instead of the field point, which answers a question nobody asked.
Pulling the field out of the integral on a surface where it is not constant, which is silent and fatal.
Using the total charge instead of the enclosed charge whenever the field point is inside the object.
Including the ends of the can or the side wall of the pillbox in the area.
Writing the enclosed charge without guessing
Any time step 5 above is not a single number printed in the question, which is most of the time.
Build the density first
From the total charge and the total size: $\rho = Q/(\tfrac43\pi R^{3})$ for a ball, $\rho = Q/(\pi R^{2}L)$ for a cylinder, $\sigma = Q/A$ for a surface. A density is needed even when the question gives one, because the units are the check that everything downstream is consistent.
Multiply by the amount of object inside your surface
Not by the whole object. For a ball that is $\tfrac43\pi r^{3}$ with your radius, giving $Q_{\rm enc} = Q r^{3}/R^{3}$. For a cylinder it is $\pi r^{2}L$, giving $\lambda_{\rm enc} = \rho\pi r^{2}$.
Add the point charges and shells that also fall inside
With their signs. A shell of radius larger than your surface contributes zero however large its charge; a shell inside contributes all of it.
For a conductor, use the zero field to read off the induced charge
Draw the surface in the metal, set the enclosed charge to zero, and solve for the unknown wall charge. Then use conservation of the metal's own charge to get the other surface.
Where it goes wrong
Using the ratio of radii squared instead of cubed for a ball.
Forgetting that a shell outside your radius contributes nothing at all, not merely a small amount.
Treating a neutral conductor as though it had no charge on its surfaces; it has two equal and opposite ones.
Field 2.50 cm inside a charged plastic ball
A solid plastic ball of radius $5.00\ \mathrm{cm}$ carries $+20.0\ \mathrm{nC}$ spread uniformly through its volume. Find the field $2.50\ \mathrm{cm}$ from its centre.
Given
$R = 5.00\ \mathrm{cm}$, $Q = +20.0\ \mathrm{nC}$ through the volume
The interior field is linear in r, so halving the radius from the surface must halve the field: the surface value is $kQ/R^{2} = 7.19\times10^{4}\ \mathrm{N/C}$ and half of it is $3.60\times10^{4}\ \mathrm{N/C}$.
Field 2.50 cm inside a charged metal ball
A solid metal ball of radius $5.00\ \mathrm{cm}$ carries the same $+20.0\ \mathrm{nC}$ and has been left alone. Find the field $2.50\ \mathrm{cm}$ from its centre, and the field at $8.00\ \mathrm{cm}$.
Given
metal ball, $R = 5.00\ \mathrm{cm}$, $Q = +20.0\ \mathrm{nC}$
field points at $r = 2.50\ \mathrm{cm}$ and $r = 8.00\ \mathrm{cm}$
Find
the field inside and outside
SolutionInside, where the charge is not
$$Q_{\rm enc}(0.0250) = 0$$
in a conductor at rest all the excess charge sits on the surface, so a sphere drawn through the interior encloses none of it
$$E = 0$$
zero flux on a surface where symmetry forces one constant field size leaves no other possibility
The exterior answer must match the plastic ball of the same charge and radius, since Gauss's law cannot see the difference from outside. Computing the plastic ball at 8.00 cm gives the same $2.81\times10^{4}\ \mathrm{N/C}$, as it has to.
Same charge, same radius, same sphere, and the interior fields are 3.60 times ten to the fourth newtons per coulomb and exactly zero, while the exterior fields agree to every digit.
How to tell them apart
Ask where the charge is allowed to sit. In an insulator it stays where it was put, so the interior enclosed charge grows as the cube of your radius. In a conductor at rest it has already fled to the surface, so the interior enclosed charge is zero and the interior field is zero with it. Nothing about the exterior can distinguish the two.
Field beside a lone charged plastic sheet
A large plastic sheet carries $\sigma = 1.20\ \mu\mathrm{C/m^{2}}$ spread uniformly. Find the field just off its surface.
Given
$\sigma = 1.20\ \mu\mathrm{C/m^{2}}$ on an insulating sheet
Find
the field just outside
SolutionPillbox with both caps in air
$$2EA = \frac{\sigma A}{\varepsilon_0}$$
the field leaves both faces of the sheet, so both caps of the box are crossed and the area appears twice
Independent route: cut the sheet into long strips, each a line charge with $\lambda = \sigma\,dy$, and add their fields with the sideways components cancelling in pairs. The total is $2\pi k\sigma = 6.78\times10^{4}\ \mathrm{N/C}$, which uses last section's wire result and no pillbox at all.
Field just outside a metal face carrying the same sigma
A large metal plate carries $\sigma = 1.20\ \mu\mathrm{C/m^{2}}$ on the face nearest you. Find the field just outside that face.
Given
$\sigma = 1.20\ \mu\mathrm{C/m^{2}}$ on the near face of a conductor
Find
the field just outside the face
SolutionPillbox with one cap in the metal
$$EA + 0 = \frac{\sigma A}{\varepsilon_0}$$
the inner cap sits in metal where the field is zero, so it contributes nothing and only one cap carries flux
Superposition check. The plate has charge on both faces, each a sheet giving $6.78\times10^{4}\ \mathrm{N/C}$. Outside the plate the two add to $1.36\times10^{5}\ \mathrm{N/C}$ and inside the metal they oppose and cancel, which reproduces both the answer and the zero interior field.
The same number of coulombs per square metre gives 6.78 times ten to the fourth newtons per coulomb beside plastic and 1.36 times ten to the fifth beside metal, and the reason is arithmetic about caps rather than anything about the charge.
How to tell them apart
Count the caps of your pillbox that lie in a region with a field. Two caps in air means the field appears twice and the answer has a two underneath. One cap in air and one buried in metal means the field appears once and there is no two. Then check what sigma means in the problem: on a lone sheet it is the whole charge per unit area of the object, on a conductor it is the density on that one face.
Scaffolding comes off
The common skeleton
Name the symmetry and say in one sentence what it forces the field to look like
Draw the closed surface through the field point, with the shape the symmetry demands
Throw away the pieces of the surface the field runs along, and write the area of what is left
Write the enclosed charge for that surface, with sign, using a density if only part of the object is inside
Set field times area equal to enclosed charge over epsilon nought and solve
Test the answer at a boundary or at a large distance, where you already know what it should be
1 · fully worked
Field inside and outside a uniformly charged ball of radius 9.00 cm
A solid non conducting ball of radius $9.00\ \mathrm{cm}$ carries $+25.0\ \mathrm{nC}$ spread uniformly through its volume. Find the field at $r = 3.00\ \mathrm{cm}$ and at $r = 15.0\ \mathrm{cm}$ from the centre.
Given
$R = 9.00\ \mathrm{cm}$, $Q = +25.0\ \mathrm{nC}$, uniform through the volume
field points at $r = 3.00\ \mathrm{cm}$ and $r = 15.0\ \mathrm{cm}$
Find
the field at both radii
SolutionSymmetry and surface
$$E = E(r),\ \text{radial}$$
the ball looks the same from every direction, so the field cannot depend on angle and cannot have a sideways component
$$\oint\vec E\cdot d\vec A = E(r)\,4\pi r^{2}$$
a sphere through the field point is the only surface on which that field has a single size
Inside: the enclosed charge is a fraction of the total
Check that the two expressions agree at the surface, which is the standard test for this pair. The interior formula at $r = R$ gives $kQ/R^{2} = 2.77\times10^{4}\ \mathrm{N/C}$ and the exterior formula at $r = R$ gives the same $2.77\times10^{4}\ \mathrm{N/C}$. A mismatch there would mean an algebra slip in one of them.
Two divisions and one ratio of cubes.
The two answers are almost equal at radii five times apart, because one point is below the peak and the other above it. Never use closeness of two numbers as evidence that you used the same formula twice.
2 · you write the reasoning
Easier than rung 1, because nothing has to be split into an inside and an outside: the charge is all at one radius. A thin spherical shell of radius $6.00\ \mathrm{cm}$ carries $+40.0\ \mathrm{nC}$ spread uniformly over it. The three lines below give the field at $r = 2.00\ \mathrm{cm}$ and at $r = 10.0\ \mathrm{cm}$, and they are correct. Say why each line is allowed, in your own words, before opening the model reasons.
reasoning
A sphere of radius two centimetres lies entirely inside the shell, so none of the charge is within it. Zero enclosed charge means zero flux, and because the symmetry forces the field to have one single size everywhere on that sphere, the only way for the flux to vanish is for that size to be zero. Notice what is not being said: the field is not small, or averaged out, it is exactly zero.
reasoning
At ten centimetres the sphere has swallowed the whole shell, so the enclosed charge is the full forty nanocoulombs. How that charge is spread over the shell is irrelevant, because Gauss's law asks only for the total inside.
reasoning
With the whole charge enclosed and spherical symmetry holding, the field is the same as that of a point charge of forty nanocoulombs at the centre. The radius used is the radius of the field point, ten centimetres, not the radius of the shell; using six centimetres here is the standard slip and would give a field two and a half times too large.
3 · find the buried error
Harder than rung 2, because two objects are present and one of them is metal. A long straight wire carrying $\lambda = +6.00\ \mathrm{nC/m}$ runs along the axis of a long neutral conducting cylindrical shell whose inner radius is $2.00\ \mathrm{cm}$ and outer radius $3.00\ \mathrm{cm}$. A student's solution for the field at $1.00\ \mathrm{cm}$, at $2.50\ \mathrm{cm}$ and at $5.00\ \mathrm{cm}$ is written out below. Exactly two of its four steps are faulty. Find them.
the two buried errors (2)
⚠ step 2
That radius is inside the metal of the shell, where the field must be exactly zero, not $4.32\times10^{3}\ \mathrm{N/C}$. The can drawn at $2.50\ \mathrm{cm}$ encloses the wire's $+6.00\ \mathrm{nC/m}$ and also the $-6.00\ \mathrm{nC/m}$ that has been drawn onto the inner wall, so the enclosed charge is zero and so is the field.
The formula is the right formula for the geometry and it was used correctly a line earlier, so nothing about the algebra looks wrong. The step that was skipped is not a calculation at all but the question of which region the radius falls in, and the numbers two and three centimetres are easy to read past.
right
Before writing any formula, mark the three regions on the axis: inside the cavity, in the metal, outside. Then check which one the radius belongs to. Here $2.00 < 2.50 < 3.00$, so the point is in the metal and the answer is zero with no calculation.
⚠ step 4
The shell being neutral does not mean it has no charge on its surfaces; it means the two surface charges add to zero. The inner wall carries $-6.00\ \mathrm{nC/m}$, so the outer wall must carry $+6.00\ \mathrm{nC/m}$, and at $5.00\ \mathrm{cm}$ the enclosed charge per unit length is $+6.00 - 6.00 + 6.00 = +6.00\ \mathrm{nC/m}$, giving $E = 107.9/0.0500 = 2.16\times10^{3}\ \mathrm{N/C}$ pointing outward.
Neutral is a word about a total, and it gets used as though it were a word about every part. The conclusion also sounds like a fact students have genuinely heard, that conductors shield, and it is true in the other direction: a shell shields its inside from outside charges, not the outside from what is within it.
right
Count the charge on each surface separately and add them up at the end. A neutral shell around a charged object always ends up with equal and opposite charges on its two walls, and the outer one reproduces exactly what was inside.
4 · the bare problem
§03.6 - two parallel sheets with unequal charge●●●○○
No scaffolding this time. Two very large parallel sheets are held a few centimetres apart in air. The left one carries a uniform $+4.00\ \mu\mathrm{C/m^{2}}$ and the right one carries $-1.50\ \mu\mathrm{C/m^{2}}$.
Given
left sheet: $\sigma_1 = +4.00\ \mu\mathrm{C/m^{2}}$
right sheet: $\sigma_2 = -1.50\ \mu\mathrm{C/m^{2}}$
both sheets large compared with the gap and with your distance from them
Find
(a) Find the size and direction of the field in the region to the left of both sheets.
(b) Find the size and direction of the field in the gap between them.
(c) Find the size and direction of the field to the right of both sheets.
Hint 1/4
Do not look for a single formula for two sheets. Work out what each sheet does on its own everywhere, then decide in each of the three regions whether the two arrows agree or fight.
Hint 2/4
Each sheet alone gives $E = \sigma/(2\varepsilon_0)$ on both of its sides, pointing away from a positive sheet and towards a negative one, with no distance dependence.
Hint 3/4
With $\sigma_1 = 4.00\ \mu\mathrm{C/m^{2}}$ and $\sigma_2 = -1.50\ \mu\mathrm{C/m^{2}}$: the two separate fields are $2.26\times10^{5}$ and $8.47\times10^{4}\ \mathrm{N/C}$.
Independent check on the two outer regions using the net charge. From far away the pair looks like one sheet carrying $4.00 - 1.50 = 2.50\ \mu\mathrm{C/m^{2}}$, which gives $2.50\times10^{-6}/(2\times8.85\times10^{-12}) = 1.41\times10^{5}\ \mathrm{N/C}$ pointing away on both sides. That matches both outer answers exactly, and it had to, because a sheet field does not fall off with distance so far away is the same as just outside.
Two sheets that are not equal and opposite still leave a field outside, and its size is set by their sum while the field between them is set by their difference in direction. Only the equal and opposite case gives the clean zero outside.
Full exam-style question
A charged ball inside a charged conducting shell, four radii and two surface densitiesexam format
A solid non conducting ball of radius $a = 4.00\ \mathrm{cm}$ carries $Q = +30.0\ \mathrm{nC}$ spread uniformly through its volume. It sits at the centre of a conducting spherical shell of inner radius $b = 8.00\ \mathrm{cm}$ and outer radius $c = 10.0\ \mathrm{cm}$, and the shell carries a net charge of $-50.0\ \mathrm{nC}$. Find the field at $r = 2.00\ \mathrm{cm}$, $6.00\ \mathrm{cm}$, $9.00\ \mathrm{cm}$ and $15.0\ \mathrm{cm}$, and the surface charge densities on the two walls of the shell.
Given
insulating ball: $a = 4.00\ \mathrm{cm}$, $Q = +30.0\ \mathrm{nC}$ uniform through the volume
Two independent checks. First, the field just outside the outer wall should follow from its own surface density through the conductor result: $\sigma_c/\varepsilon_0 = 1.59\times10^{-7}/8.85\times10^{-12} = 1.80\times10^{4}\ \mathrm{N/C}$ at $r = c$, and the enclosed charge route gives $kQ_{\rm enc}/c^{2} = 179.8/0.0100 = 1.80\times10^{4}\ \mathrm{N/C}$. Second, the two wall charges must add to the shell's net charge: $-30.0 - 20.0 = -50.0\ \mathrm{nC}$, as given.
Four regions, and in two of them the hard part was deciding which charges were inside rather than doing any arithmetic.
This is the standard shape of a full exam question on this topic, and the marks sit in the region by region bookkeeping. Write the four regions down before computing anything, note which objects are inside each, and the algebra afterwards is one division per region.
Practice
A · concept 4 questions
1§03.2 - what the field in Gauss's law refers to●●○○○
A one mark opener that separates people who have read the law from people who have memorised it. A closed surface encloses a single charge $+q$, and a second charge $+Q$, much larger, sits a short distance outside the surface.
Given
$+q$ inside the closed surface
$+Q$ outside it, close by, with $Q \gg q$
the surface is closed and its outward normal is used throughout
Find
(a) True or false: the field $\vec E$ appearing in Gauss's law is the field produced by $q$ alone, since $Q$ is outside. Give your reason in one sentence.
Hint 1/4
Two quantities in the law behave differently and the question is asking you to separate them: the field on the surface, and the flux of that field. Decide which one the outside charge is allowed to affect.
Hint 2/4
$\oint\vec E\cdot d\vec A = Q_{\rm enc}/\varepsilon_0$, where $\vec E$ is the total field and $Q_{\rm enc}$ counts only interior charge.
Hint 3/4
Here the outside charge $Q$ is large and close, so at points on the surface nearest to it the total field is dominated by $Q$ and hardly resembles the field of $q$ at all.
Hint 4/4
False: the field is the total field from both charges, and what the outside charge fails to change is the integral, not the field.
Show solutionSeparate the two claims
$$\vec E = \vec E_q + \vec E_Q\ \text{at every point of the surface}$$
superposition holds everywhere and knows nothing about surfaces we happen to have drawn
$$\oint \vec E_Q\cdot d\vec A = 0$$
an outside charge contributes equal ingoing and outgoing flux, so it drops out of the integral while remaining in the integrand
Answer $$\boxed{\;\text{False; }\vec E\text{ is the total field, only the flux of the outside part vanishes}\;}$$
Check
Test with an extreme case. Put $Q$ a millimetre outside the surface and make it a thousand times $q$. The field at the nearest point of the surface is then overwhelmingly that of $Q$, yet the total flux is still exactly $q/\varepsilon_0$. Field and flux have come apart completely, which they could not do if the law were about the field of the enclosed charge.
This is why Gauss's law gives a field only when symmetry is available: without it you know the integral but not the integrand, and the integrand is what the question usually wants.
2§03.1 - flux through a curved surface●●●○○
A short one, and it rewards drawing the shadow rather than reaching for a surface integral. A hemispherical bowl of radius $R$ is held in a uniform field $E$ with the plane of its rim perpendicular to the field, so the field enters through the open rim and leaves through the curved shell.
Given
hemisphere of radius $R$
uniform field $E$ perpendicular to the plane of the rim
flux is wanted through the curved part only
Find
(a) What is the flux through the curved surface of the hemisphere?
Hint 1/4
Do not integrate over a curved surface. Ask instead how much field crosses the bowl in total, and notice that whatever goes in through the rim has to come out through the shell.
Hint 2/4
For a uniform field the flux through any surface equals $E$ times the area of the shadow that the surface casts across the field.
Hint 3/4
The shadow of a hemisphere of radius $R$, seen along the field direction, is a flat disc of radius $R$, whose area is $\pi R^{2}$.
Hint 4/4
So the flux through the curved part is $\pi R^{2}E$, not the curved area times $E$.
Show solutionTurn an open surface into a closed one
Independent check by shadow. Stand behind the bowl and look along the field: the bowl blocks a circle of radius $R$, so it intercepts the flux crossing an area $\pi R^{2}$, whatever shape it has. The two arguments use nothing in common and agree.
Any awkward open surface can be closed with a convenient flat lid. The law then trades an unpleasant integral for the flux through the lid, which is usually a one line product.
3§03.7 - the field inside an empty cavity●●●○○
A statement people meet as a slogan about shielding, and the point of the question is whether the slogan survives contact with the details. A lump of metal of no particular shape has an empty cavity inside it, and a large charge is brought up close to the outside of the metal.
Given
a solid lump of metal with an empty cavity
no charge anywhere in the cavity
a large charge placed just outside the metal
Find
(a) True or false: the field everywhere inside the cavity is zero. Give your reason in one sentence.
Hint 1/4
Split the question in two: what does the law force about the charge on the cavity wall, and does zero charge on the wall by itself force zero field inside?
Hint 2/4
A surface drawn in the body of a conductor at rest has $\vec E = 0$ on it, so it encloses zero net charge.
Hint 3/4
Here the surface drawn around the empty cavity encloses no charge, so the wall carries no net charge either. The outside charge rearranges the outer surface of the metal only.
Hint 4/4
True: the cavity is completely shielded from whatever happens outside, and this is what a screened room is.
the surface lies where the field vanishes, so it can enclose no net charge, and the only place charge could hide is the cavity wall
And no field survives inside
$$\vec E_{\rm cavity} = 0$$
a field in the cavity would have to start on positive wall charge and end on negative wall charge, but any such arrangement carries a net push around a closed path, which free charges would have removed already
Answer $$\boxed{\;\vec E = 0\ \text{throughout the empty cavity}\;}$$
Check
Check by consequence rather than by algebra. If the statement failed, a radio would work the same inside a closed metal box as outside it, and a car would be a dangerous place in a thunderstorm. Both predictions are testable and both come out the way the statement says.
Note that zero enclosed charge alone is not quite enough; it rules out a net charge on the wall but not equal and opposite patches. The extra step is that free charges would move under any surviving field, so at equilibrium there is none.
4§03.3 - when a finite plate counts as infinite●●●○○
A judgement question rather than a calculation, and judgement of exactly this kind decides whether the two line method is available in an exam. A square plastic plate of side $2.00\ \mathrm{m}$ carries a uniform surface charge, and you want the field at a point $1.00\ \mathrm{mm}$ above the middle of the plate.
Given
square insulating plate of side $2.00\ \mathrm{m}$, uniform $\sigma$
field point $1.00\ \mathrm{mm}$ above the centre of the plate
Find
(a) Which approach gives a reliable answer with the least work?
Hint 1/4
The question is not which formula is exact, since none of them is. It is which idealisation has an error you can bound and dismiss at this particular distance.
Hint 2/4
Planar symmetry is a good approximation while your distance from the sheet is small compared with the distance to its edges.
Hint 3/4
Here your distance is $1.00\ \mathrm{mm}$ and the nearest edge is $1.00\ \mathrm{m}$ away, a ratio of a thousand.
Hint 4/4
So the infinite sheet result applies, and the field is $\sigma/2\varepsilon_0$ with an error far below anything the question could care about.
Show solutionCompare the two lengths in the problem
$$\frac{\text{distance to the nearest edge}}{\text{distance to the sheet}} = \frac{1.00}{0.00100} = 1000$$
an idealisation is licensed by a ratio, never by an absolute size, and this is the ratio that matters here
$$E = \frac{\sigma}{2\varepsilon_0}$$
with the edges a thousand times further away than the sheet, their contribution is negligible at any accuracy this question demands
Sanity check by moving the field point instead of changing the method. At one metre above the same plate the ratio becomes one, the edges are as close as the sheet, and the sheet formula would be badly wrong; at ten metres the plate is a point charge. The approach flips exactly where the ratio does, which is the sign of a criterion that is doing real work.
Every idealisation in this course is a statement about a ratio of lengths. Write that ratio down before deciding a method, and both the choice and its justification arrive together.
B · computation 8 questions
1§03.1 - flux through a tilted disc●●○○○
Straight practice on the definition, with the tilt supplied in the awkward way. A uniform field of $2.50\times10^{3}\ \mathrm{N/C}$ fills a laboratory, and a flat circular plate of radius $12.0\ \mathrm{cm}$ is held in it so that its normal makes $55.0^{\circ}$ with the field.
Given
$E = 2.50\times10^{3}\ \mathrm{N/C}$, uniform
circular plate of radius $12.0\ \mathrm{cm}$
angle between the plate normal and the field, $55.0^{\circ}$
Find
(a) Find the flux through the plate.
(b) Find the angle at which the flux would be half of its largest possible value.
Hint 1/4
Part (a) is the definition applied once. Part (b) asks you to read the same formula backwards: which cosine gives one half?
Hint 2/4
$\Phi_E = EA\cos\theta$ with $\theta$ measured between the field and the surface normal, and the largest flux occurs at $\theta = 0$.
Hint 3/4
Here $A = \pi(0.120)^{2} = 4.52\times10^{-2}\ \mathrm{m^{2}}$ and $EA = 113\ \mathrm{N\,m^{2}/C}$, with $\cos 55.0^{\circ} = 0.574$.
Consistency check between the two parts. The answer to (a) at $55.0^{\circ}$ is $64.9$, which is more than half of the maximum $113$; and $55.0^{\circ}$ is indeed less than the $60.0^{\circ}$ found in (b). The two answers agree about which side of the halfway point the plate is on.
Halving a cosine is not halving an angle. The same trap appears with the ring formula and with any component question, and it is worth doing the inverse cosine rather than guessing.
2§03.2 - net flux from a mixture of charges●●○○○
Bookkeeping practice, and the whole difficulty is deciding what counts. A closed surface of irregular shape has $+4.00\ \mathrm{nC}$ and $-7.00\ \mathrm{nC}$ inside it, and a further $+9.00\ \mathrm{nC}$ sits outside the surface.
Given
$+4.00\ \mathrm{nC}$ inside
$-7.00\ \mathrm{nC}$ inside
$+9.00\ \mathrm{nC}$ outside
Find
(a) Find the net flux out of the closed surface.
(b) State what the sign of your answer means physically.
Hint 1/4
There is no field to compute here. Sort the charges into inside and outside and use the law in the direction it is easiest.
Hint 2/4
$\Phi_E = Q_{\rm enc}/\varepsilon_0$, with only interior charges counted and their signs kept.
Hint 3/4
Here $Q_{\rm enc} = +4.00 - 7.00 = -3.00\ \mathrm{nC}$, and $\varepsilon_0 = 8.85\times10^{-12}\ \mathrm{C^{2}/(N\,m^{2})}$.
Hint 4/4
$\Phi_E = -339\ \mathrm{N\,m^{2}/C}$, and the minus sign says more field enters the surface than leaves it.
Check the size against a single charge. A charge of one nanocoulomb alone gives $1.00\times10^{-9}/8.85\times10^{-12} = 113\ \mathrm{N\,m^{2}/C}$, so three nanocoulombs should give three times that, $339$, and it does. The unit check also passes: coulombs divided by the units of epsilon nought leave newton metres squared per coulomb.
Whenever a flux question gives you more charges than you need, the extra ones are outside the surface. Reading the geometry is the question; the arithmetic is one division.
3§03.4 - a negatively charged insulating ball●●●○○
The standard three region calculation, with a negative charge so that directions have to be stated rather than assumed. A solid non conducting ball of radius $6.00\ \mathrm{cm}$ carries $-18.0\ \mathrm{nC}$ spread uniformly through its volume.
Given
$R = 6.00\ \mathrm{cm}$, $Q = -18.0\ \mathrm{nC}$ uniform through the volume
Find
(a) Find the field at $r = 2.00\ \mathrm{cm}$.
(b) Find the field at the surface, $r = 6.00\ \mathrm{cm}$.
(c) Find the field at $r = 15.0\ \mathrm{cm}$, and give the direction in all three cases.
Hint 1/4
Decide first, for each of the three radii, whether your sphere is inside the ball or outside it. That decision picks the formula; everything after it is arithmetic.
Hint 2/4
$E_{\rm in} = kQr/R^{3}$ and $E_{\rm out} = kQ/r^{2}$, using magnitudes and stating the direction separately.
Hint 3/4
With $k|Q| = 162\ \mathrm{N\,m^{2}/C}$ and $R^{3} = 2.16\times10^{-4}\ \mathrm{m^{3}}$: the three radii are $0.0200$, $0.0600$ and $0.150\ \mathrm{m}$.
Hint 4/4
(a) $1.50\times10^{4}$, (b) $4.50\times10^{4}$, (c) $7.19\times10^{3}\ \mathrm{N/C}$, all pointing towards the centre because the charge is negative.
Two independent checks. The interior point is at one third of the radius, so its field must be one third of the surface value: $4.50\times10^{4}/3 = 1.50\times10^{4}\ \mathrm{N/C}$, which matches (a). And the exterior point is at two and a half radii, so its field must be $(1/2.5)^{2} = 0.16$ of the surface value: $0.16\times4.50\times10^{4} = 7.19\times10^{3}\ \mathrm{N/C}$, matching (c).
Working with the size of the charge and adding the direction in a sentence is safer than carrying a minus sign through, because a negative field magnitude is meaningless while a negative radial component is easy to misread.
4§03.4 - a shell with a charge at its centre●●●○○
Two objects, three regions, and the doing most of the work. A thin non conducting spherical shell of radius $10.0\ \mathrm{cm}$ carries $+36.0\ \mathrm{nC}$ spread uniformly over its surface, and a point charge of $+12.0\ \mathrm{nC}$ is fixed at the centre.
Given
shell of radius $10.0\ \mathrm{cm}$, charge $+36.0\ \mathrm{nC}$ uniform on the surface
point charge $+12.0\ \mathrm{nC}$ at the centre
Find
(a) Find the field at $r = 5.00\ \mathrm{cm}$.
(b) Find the field at $r = 25.0\ \mathrm{cm}$.
(c) State what would change in each answer if the shell's charge were doubled.
Hint 1/4
For each radius, draw the sphere and ask what is inside it. The answer to that question is the answer to the problem.
Hint 2/4
$E = kQ_{\rm enc}/r^{2}$, with charge at a larger radius than your surface contributing exactly nothing.
Hint 3/4
At $5.00\ \mathrm{cm}$ only the $+12.0\ \mathrm{nC}$ is inside; at $25.0\ \mathrm{cm}$ both are, giving $48.0\ \mathrm{nC}$.
Hint 4/4
(a) $4.32\times10^{4}\ \mathrm{N/C}$. (b) $6.90\times10^{3}\ \mathrm{N/C}$. (c) Only the second changes.
Show solutionInside the shell
$$Q_{\rm enc}(0.0500) = +12.0\ \mathrm{nC}$$
the shell sits at 10 cm, outside this surface, and a spherically symmetric shell contributes nothing to any interior point
Superposition check on part (b) without adding the charges first. The point charge alone gives $107.9/0.0625 = 1.73\times10^{3}\ \mathrm{N/C}$ and the shell alone gives $323.6/0.0625 = 5.18\times10^{3}\ \mathrm{N/C}$; they point the same way and sum to $6.90\times10^{3}\ \mathrm{N/C}$.
Part (c) is the part worth remembering. A change in a charge that lies outside your surface changes nothing about the field there, which is a much stronger statement than saying its effect is small.
5§03.5 - reading a charge density off a measured field●●○○○
The cylindrical formula used backwards, which is how it usually appears in a laboratory. A long straight wire runs across a bench and a field meter placed $4.50\ \mathrm{cm}$ from the middle of the wire reads $6.20\times10^{3}\ \mathrm{N/C}$, pointing away from the wire.
Given
$r = 4.50\ \mathrm{cm}$ from a long straight wire
measured field $6.20\times10^{3}\ \mathrm{N/C}$, directed away from the wire
the wire is long compared with that distance
Find
(a) Find the charge per unit length on the wire, including its sign.
(b) Find the field at $9.00\ \mathrm{cm}$ from the same wire.
Hint 1/4
You know a field and want a source, so the cylindrical result has to be rearranged before any number goes into it. Part (b) then needs no new physics at all.
Hint 2/4
$E = 2k\lambda/r$, so $\lambda = Er/(2k)$, and the field falls as one over the distance.
Hint 3/4
Here $E = 6.20\times10^{3}\ \mathrm{N/C}$, $r = 0.0450\ \mathrm{m}$ and $2k = 1.798\times10^{10}\ \mathrm{N\,m^{2}/C^{2}}$.
Hint 4/4
(a) $\lambda = +15.5\ \mathrm{nC/m}$, positive because the field points away. (b) Doubling the distance halves the field: $3.10\times10^{3}\ \mathrm{N/C}$.
Put the density back into the original formula at the original distance: $2(8.99\times10^{9})(1.55\times10^{-8})/0.0450 = 6.19\times10^{3}\ \mathrm{N/C}$, which is the measured value to the three figures we are keeping. A rearranged formula should always be tested by substituting the answer back.
The ratio trick in part (b) is worth the habit. Once you know the power law, changes in distance never need the constants again, and a wrong power shows up immediately as a factor of two where a factor of four was expected.
6§03.5 - inside and outside a charged cylinder●●●○○
The cylindrical twin of the solid ball, and the enclosed charge is again a fraction of the total. A long solid non conducting cylinder of radius $4.00\ \mathrm{cm}$ carries a uniform volume charge density $\rho = -2.50\ \mu\mathrm{C/m^{3}}$.
field wanted at $r = 1.50\ \mathrm{cm}$ and $r = 10.0\ \mathrm{cm}$ from the axis
Find
(a) Find the field at $r = 1.50\ \mathrm{cm}$.
(b) Find the field at $r = 10.0\ \mathrm{cm}$.
(c) Give the direction of both fields.
Hint 1/4
The can you draw at 1.50 cm captures only part of the cylinder, and the one at 10.0 cm captures all of it. Two different enclosed charges, one method.
Hint 2/4
$E = 2k\lambda_{\rm enc}/r$ with $\lambda_{\rm enc} = \rho\pi r^{2}$ inside and $\rho\pi R^{2}$ outside, which gives $E_{\rm in} = \rho r/(2\varepsilon_0)$.
Hint 3/4
With $|\rho| = 2.50\times10^{-6}\ \mathrm{C/m^{3}}$: inside use $r = 0.0150\ \mathrm{m}$, and outside use $R = 0.0400\ \mathrm{m}$ with $r = 0.100\ \mathrm{m}$.
Hint 4/4
(a) $2.12\times10^{3}\ \mathrm{N/C}$. (b) $2.26\times10^{3}\ \mathrm{N/C}$. Both point towards the axis, since the charge is negative.
Show solutionInside
$$\lambda_{\rm enc} = |\rho|\pi r^{2}$$
per unit length, the charge inside a can of radius r is the density times the area of that circle
Check that the two expressions meet at the surface. Inside at $r = R$: $(2.50\times10^{-6})(0.0400)/1.77\times10^{-11} = 5.65\times10^{3}\ \mathrm{N/C}$. Outside at $r = R$: $(1.798\times10^{10})(1.257\times10^{-8})/0.0400 = 5.65\times10^{3}\ \mathrm{N/C}$. They agree, so no factor was dropped in the switch between formulas.
The peak at the surface is the shape shared by every solid symmetric object: the field climbs while you are inside the charge and falls once you are outside it. Two nearly equal readings at very different radii are the normal consequence, not a warning sign.
7§03.6 - a charged sheet and the force it exerts●●○○○
The planar result with a force attached, which is the usual exam dressing. A large flat sheet of plastic carries a uniform surface charge density of $6.50\ \mu\mathrm{C/m^{2}}$, and a small bead carrying $+2.50\ \mathrm{nC}$ is held a few centimetres from the middle of the sheet.
Given
$\sigma = 6.50\ \mu\mathrm{C/m^{2}}$ on a large insulating sheet
bead charge $q = +2.50\ \mathrm{nC}$
the bead is close to the middle of the sheet
Find
(a) Find the field at the bead's position.
(b) Find the force on the bead.
(c) State how both answers change if the bead is moved twice as far from the sheet.
Hint 1/4
The sheet formula gives a field; the force needs one more step and one more piece of data. Part (c) is testing whether you believe the distance is missing from the formula by accident.
Hint 2/4
$E = \sigma/(2\varepsilon_0)$ for a large insulating sheet, and $F = qE$ for a charge sitting in a field.
Hint 3/4
Here $\sigma = 6.50\times10^{-6}\ \mathrm{C/m^{2}}$, $2\varepsilon_0 = 1.77\times10^{-11}$, and the bead carries $2.50\times10^{-9}\ \mathrm{C}$.
Hint 4/4
(a) $3.67\times10^{5}\ \mathrm{N/C}$. (b) $9.18\times10^{-4}\ \mathrm{N}$ away from the sheet. (c) Neither changes.
an insulating sheet sends field out of both faces, so the two stays in the denominator
The force and the distance question
$$F = qE = (2.50\times10^{-9})(3.67\times10^{5}) = 9.18\times10^{-4}\ \mathrm{N}$$
the definition of the field is force per unit charge, so this step is a multiplication and not a new law
$$E(2d) = E(d),\qquad F(2d) = F(d)$$
the field of a large sheet contains no distance, so doubling the separation changes nothing while the sheet still looks infinite
Answer $$\boxed{\;E = 3.67\times10^{5}\ \mathrm{N/C},\quad F = 9.18\times10^{-4}\ \mathrm{N},\quad \text{both unchanged at } 2d\;}$$
Check
Order of magnitude check on the force. A milligram bead weighs about $10^{-5}\ \mathrm{N}$, so this electric force is roughly a hundred times its weight, which is why charged specks visibly leap towards charged plastic rather than drifting. The field itself is about a tenth of the value at which dry air breaks down, so the situation is physically possible.
The force is uniform over the whole region near the sheet, so a bead released there accelerates at a constant rate. That is unusual in electrostatics and it is exactly what makes parallel plates the standard laboratory tool.
8§03.7 - a charged shell with a charge in its cavity●●●●○
Full conductor bookkeeping, with both the metal and the cavity charge nonzero. A conducting spherical shell has inner radius $5.00\ \mathrm{cm}$ and outer radius $7.00\ \mathrm{cm}$ and carries a net charge of $+15.0\ \mathrm{nC}$. A point charge of $-4.00\ \mathrm{nC}$ is fixed at the centre of the cavity.
(a) Find the charge on the inner and outer surfaces of the shell.
(b) Find the field at $r = 3.00\ \mathrm{cm}$, $r = 6.00\ \mathrm{cm}$ and $r = 12.0\ \mathrm{cm}$.
Hint 1/4
One fact fixes everything else here: the field inside the metal is zero. Use it on a surface drawn at six centimetres and the inner wall charge falls out, and then conservation gives the outer wall.
Hint 2/4
Inside a conductor $\vec E = 0$, so a surface there encloses zero charge; elsewhere $E = kQ_{\rm enc}/r^{2}$.
Hint 3/4
The cavity holds $-4.00\ \mathrm{nC}$ and the metal itself carries $+15.0\ \mathrm{nC}$ in total, shared between its two walls.
Check the exterior field from the total charge of the whole assembly, which is $-4.00 + 15.0 = +11.0\ \mathrm{nC}$. A distant observer sees only that total, so the field at $12.0\ \mathrm{cm}$ must be $k(11.0\ \mathrm{nC})/(0.120)^{2} = 6.87\times10^{3}\ \mathrm{N/C}$, which is what the surface by surface route gave.
Notice that the outer surface charge is not the shell's own charge and not the cavity charge, but the sum of the two. That single sentence answers most questions of this type without any further calculation.
C · exam level 5 questions
1§03.2 - a charge sitting at the corner of a cube●●●●○
A classic, and it is asked because it cannot be done by substituting into a formula. A point charge of $+7.20\ \mathrm{nC}$ is placed exactly at one corner of a cube of side $20.0\ \mathrm{cm}$, so that the charge touches the cube at that single point and lies outside it everywhere else.
Given
point charge $+7.20\ \mathrm{nC}$ at a corner of the cube
(a) What is the total flux through the surface of that one cube?
Hint 1/4
The charge is not inside the cube, so the law does not apply to this cube directly. Build a bigger closed region around the charge out of copies of this cube and use symmetry.
Hint 2/4
$\Phi_{\rm total} = q/\varepsilon_0$ out of any closed surface containing the charge, and identical regions related by symmetry carry identical flux.
Hint 3/4
Eight cubes of side $20.0\ \mathrm{cm}$ stacked two by two by two surround the corner completely, and by symmetry each takes one eighth of the total $q/\varepsilon_0 = 814\ \mathrm{N\,m^{2}/C}$.
Hint 4/4
So the flux through one cube is $814/8 = 102\ \mathrm{N\,m^{2}/C}$.
Show solutionBuild a closed surface the law can be used on
$$8\ \text{cubes stacked } 2\times2\times2\ \text{surround the corner}$$
the law needs a closed surface with the charge inside, and one cube does not provide one; eight identical ones do
Check by pushing the same argument one step further. Of the six faces of our cube, the three that touch the corner catch nothing, because the field at every point of them runs along the face. So the whole $102$ leaves through the three far faces, $33.9\ \mathrm{N\,m^{2}/C}$ each by symmetry, and three times $33.9$ is $102$. The two decompositions agree.
Any charge sitting at a special point of a shape can be handled this way: count how many copies of the shape it takes to surround the charge, then divide. Corners of cubes give eight, edges give four, face centres give two.
2§03.4 - ratio of fields inside and outside a ball●●●●○
A ratio question, which is the form these appear in when the examiner does not want you to reach for a calculator. A solid non conducting ball of radius $R$ carries a total charge $Q$ spread uniformly through its volume.
Given
uniform ball of radius $R$, total charge $Q$
first field point at $r = R/3$, inside the ball
second field point at $r = 3R$, outside it
Find
(a) What is the ratio $E(R/3)$ to $E(3R)$?
Hint 1/4
Two different formulas are needed, one for each point, and the whole question is whether you notice that. Write each field in terms of kQ over R squared before dividing.
Hint 2/4
$E_{\rm in} = kQr/R^{3}$ and $E_{\rm out} = kQ/r^{2}$.
Hint 3/4
At $r = R/3$: $E = kQ(R/3)/R^{3} = kQ/(3R^{2})$. At $r = 3R$: $E = kQ/(9R^{2})$.
the exterior formula, because the whole ball is enclosed at that radius
Divide
$$\frac{E(R/3)}{E(3R)} = \frac{1/3}{1/9} = 3$$
the common factor kQ over R squared cancels, which is why the answer is a pure number
Answer $$\boxed{\;3\;}$$
Check
Numerical check with the ball already tabulated in this section, $Q = 12.0\ \mathrm{nC}$ and $R = 8.00\ \mathrm{cm}$. Then $R/3 = 2.67\ \mathrm{cm}$ gives $5.62\times10^{3}\ \mathrm{N/C}$ and $3R = 24.0\ \mathrm{cm}$ gives $1.87\times10^{3}\ \mathrm{N/C}$, whose ratio is $3.00$. Independent numbers, same result.
When a question puts one point inside and one outside, it is testing exactly one thing: whether you noticed. Read the radii against the size of the object before writing any formula.
3§03.7 - field in the gap between two conductors●●●●○
Two conductors and a gap, which is the arrangement most often set in exams because it tests three ideas at once. A solid conducting sphere of radius $4.00\ \mathrm{cm}$ carries $+20.0\ \mathrm{nC}$. It sits at the centre of a neutral conducting shell of inner radius $8.00\ \mathrm{cm}$ and outer radius $10.0\ \mathrm{cm}$.
Check by asking what changes if the outer shell is removed entirely. Nothing does, at this radius, because a spherically symmetric shell of charge contributes zero to every interior point and the shell's two induced charges are both such shells. The answer for the bare sphere is the same $4.99\times10^{4}\ \mathrm{N/C}$, and that agreement is the point of the question.
The gap between two conductors is a place where the ordinary rules apply and the conductor rules do not. Marking the regions on a sketch before computing is worth the ten seconds it costs.
4§03.6 - a sheet and a point charge that cancel●●●●●
A fault finding question of the kind that appears when an examiner wants to see whether you can read a solution as well as write one. A very large sheet lying in the plane $x = 0$ carries a uniform $\sigma = +20.0\ \mathrm{nC/m^{2}}$, and a point charge of $+5.00\ \mathrm{nC}$ sits on the $x$ axis at $x = 30.0\ \mathrm{cm}$. Somewhere between them the total field vanishes, and a student's solution is printed below.
Given
large sheet at $x = 0$ with $\sigma = +20.0\ \mathrm{nC/m^{2}}$
point charge $+5.00\ \mathrm{nC}$ at $x = 30.0\ \mathrm{cm}$
the student's four steps as printed in the question
Find
(a) The student's four steps are: 1. Between the sheet and the charge the sheet pushes in the $+x$ direction and the charge pushes in the $-x$ direction, so they can cancel. 2. $E_{\rm sheet} = \sigma/\varepsilon_0 = 2.26\times10^{3}\ \mathrm{N/C}$. 3. $E_{\rm charge} = kq/x^{2}$, with $x$ measured from the sheet. 4. Setting them equal, $x = \sqrt{kq/E_{\rm sheet}} = 14.1\ \mathrm{cm}$ from the sheet. Exactly two steps are wrong. Which two, and what is the correct position?
Hint 1/4
Check the steps one at a time and ask of each: is this the right formula, and are its symbols measured from the right place? Do not be distracted by the arithmetic, which is consistent throughout.
Hint 2/4
A lone insulating sheet gives $E = \sigma/(2\varepsilon_0)$, and a point charge gives $E = kq/d^{2}$ with $d$ measured from the charge itself.
Hint 3/4
Here $\sigma/(2\varepsilon_0) = 1.13\times10^{3}\ \mathrm{N/C}$, and the distance from the charge to the null point is $d$, with the position measured from the sheet being $0.300 - d$.
Hint 4/4
Steps 2 and 3 are the faulty ones; $d = 19.9\ \mathrm{cm}$, so the field vanishes at $x = 10.1\ \mathrm{cm}$ from the sheet.
converting back to the coordinate the question asked for, measured from the sheet
Answer $$\boxed{\;\text{steps 2 and 3};\qquad x = 10.1\ \mathrm{cm}\ \text{from the sheet}\;}$$
Check
Substitute the answer back into both fields. The charge is $0.199\ \mathrm{m}$ away, giving $44.95/0.0397 = 1.13\times10^{3}\ \mathrm{N/C}$, and the sheet gives $1.13\times10^{3}\ \mathrm{N/C}$ at every point. Equal sizes and opposite directions, so the total really is zero there.
Two of the commonest faults in this topic appear together here: the wrong factor of two on a sheet, and a distance measured from the wrong object. Both survive the algebra untouched, which is why the check has to be made when the formula is written and not when the answer arrives.
5§03.5 - a wire inside a charged coaxial shell●●●●○
The cable geometry, which is where the cylindrical result earns its keep. A long straight wire carrying $+3.00\ \mathrm{nC/m}$ runs along the axis of a thin conducting cylindrical shell of radius $4.00\ \mathrm{cm}$, and the shell itself carries $-5.00\ \mathrm{nC/m}$.
Given
central wire: $\lambda_1 = +3.00\ \mathrm{nC/m}$
thin shell at radius $4.00\ \mathrm{cm}$: $\lambda_2 = -5.00\ \mathrm{nC/m}$
field point at $r = 8.00\ \mathrm{cm}$ from the axis
Find
(a) What is the field at $r = 8.00\ \mathrm{cm}$?
Hint 1/4
Draw the can at eight centimetres and list what is inside it. Both objects are, and their charges have opposite signs.
Hint 2/4
$E = 2k\lambda_{\rm enc}/r$, with the enclosed charge per unit length added algebraically.
Hint 3/4
Here $\lambda_{\rm enc} = +3.00 - 5.00 = -2.00\ \mathrm{nC/m}$ and $r = 0.0800\ \mathrm{m}$.
Hint 4/4
$E = (1.798\times10^{10})(2.00\times10^{-9})/0.0800 = 450\ \mathrm{N/C}$, pointing towards the axis because the enclosed total is negative.
the standard cylindrical result with the radius of the field point, not of the shell
$$\text{direction: towards the axis}$$
the net enclosed charge is negative, so a positive test charge placed there would be pulled inward
Answer $$\boxed{\;E = 450\ \mathrm{N/C}\ \text{towards the axis}\;}$$
Check
Check by superposing the two separate fields instead of adding the charges. The wire alone gives $53.94/0.0800 = 674\ \mathrm{N/C}$ outward and the shell alone, from outside it, gives $89.90/0.0800 = 1124\ \mathrm{N/C}$ inward. Their difference is $450\ \mathrm{N/C}$ inward, matching.
A real coaxial cable is built with equal and opposite line densities precisely so that this total is zero and no field leaks outside it. The moment they are unequal, as here, the cable radiates a field into the room.
D · interleaved 4 questions
1§03.D - a charged ring seen from a point on its axis●●●○○
This set is deliberately mixed, so decide for yourself which tool the question wants before reaching for one. A thin ring of radius $5.00\ \mathrm{cm}$ carries $+15.0\ \mathrm{nC}$ spread uniformly around it, and the field is wanted at a point on the ring's axis $8.00\ \mathrm{cm}$ from the centre.
Given
ring of radius $5.00\ \mathrm{cm}$, charge $+15.0\ \mathrm{nC}$ uniform
field point on the axis, $8.00\ \mathrm{cm}$ from the centre
Find
(a) Find the field at that point.
(b) Say in one sentence why a Gaussian surface does not help here.
Hint 1/4
Before choosing a method, ask whether there is a closed surface through the field point on which this object's field has a single size. If there is not, the tool you need is an older one.
Hint 2/4
For a ring on its axis, $E = kQx/(x^{2}+R^{2})^{3/2}$, which came from integrating the contributions of the charge elements.
Hint 3/4
Here $kQ = 135\ \mathrm{N\,m^{2}/C}$, $x = 0.0800\ \mathrm{m}$, $R = 0.0500\ \mathrm{m}$, so $x^{2}+R^{2} = 8.90\times10^{-3}\ \mathrm{m^{2}}$.
Hint 4/4
$E = (135)(0.0800)/(8.90\times10^{-3})^{3/2} = 1.28\times10^{4}\ \mathrm{N/C}$, along the axis away from the ring.
Show solutionTest for symmetry first
$$\text{ring: not spherical, not cylindrical, not planar}$$
the ring singles out both an axis and a plane, so none of the three standard surfaces has constant field on it
one division, with the direction along the axis because the sideways parts cancelled in pairs
Answer $$\boxed{\;E = 1.28\times10^{4}\ \mathrm{N/C}\ \text{along the axis, away from the ring}\;}$$
Check
Compare with the point charge estimate, which must be an overestimate here. Putting all $15.0\ \mathrm{nC}$ at the centre gives $135/(0.0800)^{2} = 2.11\times10^{4}\ \mathrm{N/C}$, larger than the true value because the ring's charge is spread further away and partly cancels sideways. The true answer is $61$ per cent of it, which is the right side of the estimate.
The first decision in any field problem is which of the two methods applies, and it costs five seconds. Gauss's law is faster when it works and useless when it does not, and the ring is the standard example of the second case.
2§03.D - a small sphere held near a charged ball●●●○○
Mixed practice again, so identify the situation before choosing a tool. A solid non conducting ball of radius $3.00\ \mathrm{cm}$ carries $+40.0\ \mathrm{nC}$ spread uniformly through its volume, and a small sphere carrying $-6.00\ \mathrm{nC}$ is held with its centre $12.0\ \mathrm{cm}$ from the centre of the ball.
Given
ball: radius $3.00\ \mathrm{cm}$, charge $+40.0\ \mathrm{nC}$ uniform through the volume
small sphere: charge $-6.00\ \mathrm{nC}$, centre $12.0\ \mathrm{cm}$ from the ball's centre
Find
(a) Find the size and direction of the force on the small sphere.
(b) Say whether the answer would change if the same charge were spread over the surface of the ball instead of through its volume.
Hint 1/4
The small sphere sits outside the ball, so first ask what the ball looks like from out there. That question has a one word answer and it makes the rest a problem you solved weeks ago.
Hint 2/4
A spherically symmetric charge acts on external points exactly as a point charge at its centre, and then $F = kQq/r^{2}$.
Hint 3/4
Here $Q = 40.0\ \mathrm{nC}$, $q = 6.00\ \mathrm{nC}$ in size, and $r = 0.120\ \mathrm{m}$ between the centres.
the definition of the field, with the direction supplied by the signs rather than by the arithmetic
$$\text{direction: towards the ball}$$
opposite charges attract, so the negative sphere is pulled towards the positive ball
Answer $$\boxed{\;F = 1.50\times10^{-4}\ \mathrm{N}\ \text{attractive, and unchanged for a shell}\;}$$
Check
Check with Coulomb's law used directly on two point charges, which is legitimate here precisely because of the shell theorem: $F = (8.99\times10^{9})(40.0\times10^{-9})(6.00\times10^{-9})/(0.120)^{2} = 1.50\times10^{-4}\ \mathrm{N}$. The two routes agree, and the agreement is itself the content of part (b).
This is the result that lets planetary and atomic problems be done at all. Without it, every extended object would need an integral before Newton's or Coulomb's law could be applied to it.
3§03.D - a small charge pair held near a charged sheet●●●●○
Still mixed, so read the situation before reaching for a method. A rigid pair of charges, $+2.00\ \mathrm{nC}$ and $-2.00\ \mathrm{nC}$ held $1.50\ \mathrm{mm}$ apart, is placed a few centimetres from a large plastic sheet carrying $\sigma = 6.50\ \mu\mathrm{C/m^{2}}$, with the line joining the charges at $40.0^{\circ}$ to the field.
large sheet with $\sigma = 6.50\ \mu\mathrm{C/m^{2}}$
the pair makes $40.0^{\circ}$ with the field direction
Find
(a) Find the field the sheet produces at the pair.
(b) Find the torque on the pair and the net force on it.
Hint 1/4
Two separate results meet here: what the sheet does to the space around it, and what a uniform field does to a rigid pair of opposite charges. Do them in that order.
Hint 2/4
$E = \sigma/(2\varepsilon_0)$ for a large sheet, and for a pair of opposite charges $p = q\ell$ with $\tau = pE\sin\theta$ and zero net force in a uniform field.
Hint 3/4
Here $\sigma = 6.50\times10^{-6}\ \mathrm{C/m^{2}}$, $q = 2.00\times10^{-9}\ \mathrm{C}$, $\ell = 1.50\times10^{-3}\ \mathrm{m}$ and $\theta = 40.0^{\circ}$.
Hint 4/4
(a) $3.67\times10^{5}\ \mathrm{N/C}$. (b) $p = 3.00\times10^{-12}\ \mathrm{C\,m}$, $\tau = 7.08\times10^{-7}\ \mathrm{N\,m}$, and the net force is zero.
Check the torque a second way, from forces and a lever arm. Each charge feels $qE = (2.00\times10^{-9})(3.67\times10^{5}) = 7.34\times10^{-4}\ \mathrm{N}$, and the perpendicular separation between the two lines of action is $\ell\sin 40.0^{\circ} = 9.64\times10^{-4}\ \mathrm{m}$. The product is $7.08\times10^{-7}\ \mathrm{N\,m}$, matching the formula.
A uniform field turns a neutral pair without dragging it anywhere. To drag it you need a field that changes from place to place, which is why a charged rod picks up paper scraps but two parallel plates do not.
4§03.D - a charged ball and a distant point charge●●●●○
The last of the mixed set, and it needs two tools rather than one. A solid non conducting ball of radius $6.00\ \mathrm{cm}$ carries $+24.0\ \mathrm{nC}$ spread uniformly through its volume. A point charge of $-24.0\ \mathrm{nC}$ is fixed $20.0\ \mathrm{cm}$ from the ball's centre.
point charge $-24.0\ \mathrm{nC}$ at $20.0\ \mathrm{cm}$ from the ball's centre
field point on the line joining them, $10.0\ \mathrm{cm}$ from the ball's centre
Find
(a) Find the total field at the point midway between the ball's centre and the point charge.
(b) State its direction.
Hint 1/4
Two sources, so two fields and a vector sum. Deal with the ball first by asking whether the field point is inside it or outside it.
Hint 2/4
Outside a symmetric ball, $E = kQ/r^{2}$ as if all the charge were at the centre; then add the point charge field as a vector.
Hint 3/4
The field point is $10.0\ \mathrm{cm}$ from the ball's centre, which is outside the $6.00\ \mathrm{cm}$ ball, and it is also $10.0\ \mathrm{cm}$ from the point charge.
Hint 4/4
Each contributes $2.16\times10^{4}\ \mathrm{N/C}$ in the same direction, so the total is $4.32\times10^{4}\ \mathrm{N/C}$ pointing from the ball towards the point charge.
Show solutionThe ball, from outside
$$10.0\ \mathrm{cm} > 6.00\ \mathrm{cm}$$
the field point is outside the ball, so the whole charge is enclosed and acts as though it sat at the centre
both point from the ball towards the negative charge, so the sizes add without any components being needed
Answer $$\boxed{\;E = 4.32\times10^{4}\ \mathrm{N/C}\ \text{towards the point charge}\;}$$
Check
Symmetry check without arithmetic. The pair is a positive charge and an equal negative charge, and the field point is exactly halfway between their centres, so this is the midpoint of a dipole and the two contributions must be identical in size and direction. Any answer smaller than one contribution alone would be inconsistent with that.
The shell theorem is what allows the extended ball to be dropped into a point charge problem untouched. Once the field point is outside, an extended symmetric object stops being a different kind of problem.
Mistake ledger (21 entries)
⚠ Using the angle to the surface instead of the angle to the normal
The definition measures the angle from the normal, and the picture shows you the surface. Read which one the question gave before substituting.
wrong$$\Phi_E = EA\cos 60^{\circ}\ \text{for a card tilted } 60^{\circ}\text{ from the field}$$
Flux is a signed sum over the whole surface, and sums of large numbers vanish routinely. The law constrains the difference, not the parts.
wrong$$\Phi_E = 0 \Rightarrow \vec E = 0\ \text{on the surface}$$
right$$\Phi_E = 0 \Rightarrow \text{as much enters as leaves}$$
⚠ Dividing the total flux between faces when the charge is not central
The division by six is a consequence of the six faces being interchangeable, and moving the charge destroys that at once.
wrong$$\Phi_{\rm face} = \tfrac16\Phi_{\rm total}\ \text{for an off centre charge}$$
right$$\Phi_{\rm face} = \tfrac16\Phi_{\rm total}\ \text{only when the faces are related by symmetry}$$
⚠ Assuming the law fails when it merely gives zero
A correct but unusable statement is not a failure of the law. The law always holds; it is the symmetry that decides whether it can be inverted.
wrong$$\Phi_E = 0 \Rightarrow \text{Gauss's law does not apply here}$$
right$$\Phi_E = 0 \Rightarrow Q_{\rm enc} = 0,\ \text{which is true and may be useless}$$
⚠ Pulling a varying field out of the integral
Nothing in the notation objects, so this error produces a clean looking answer that is simply wrong. The symmetry check has to be made before the algebra.
wrong$$\oint\vec E\cdot d\vec A = EA\ \text{for a cube around a point charge}$$
right$$\oint\vec E\cdot d\vec A = EA\ \text{only if } E \text{ is constant on that surface}$$
⚠ Choosing a surface that does not pass through the field point
The surface is drawn for the question, not for the object. A surface at the object's own radius answers a question about the object's own surface.
wrong$$\text{surface of radius } R \text{ used to find } E \text{ at } r \ne R$$
right$$\text{surface radius} = \text{the radius where the field is wanted}$$
⚠ Using the total charge when only part of it is enclosed
Inside a uniform ball only the charge closer to the centre than you are counts, and it grows as the cube of your radius.
wrong$$E_{\rm in} = \frac{kQ}{r^{2}}$$
right$$E_{\rm in} = \frac{kQr}{R^{3}}$$
⚠ Using the wrong power in the volume ratio
Charge lives in the volume for a solid ball and on the area only for a shell; the two ratios differ by a whole power of the radius.
wrong$$Q_{\rm enc} = Q\frac{r^{2}}{R^{2}}$$
right$$Q_{\rm enc} = Q\frac{r^{3}}{R^{3}}$$
⚠ Applying the ball formulas to a shell
A shell has no charge inside any interior surface at all, so its interior field is exactly zero rather than small.
charges at rest; sigma is the local surface density on the face you are standing next to
Check yourself
Close the page and write out, from memory: the definition of flux and what the cosine in it is measuring; Gauss's law and the two things the outside charges do and do not affect; the three symmetries and the surface that goes with each; the field inside and outside a uniform ball and where its peak is; the field of a long cylinder inside and outside; the field of a lone sheet and of two opposite sheets; and the three statements about a conductor at rest, with the induced charge on a cavity wall. Then open the formula card and mark only the ones you missed.
Compute the flux through a tilted flat surface in a uniform field, and say without computing anything what the net flux through a closed box in that field must be?
c-flux
State Gauss's law, and explain in two sentences why a large charge just outside a surface changes the field on it enormously and the flux through it not at all?
c-gauss-law
Look at a charge distribution and decide in five seconds whether Gauss's law will give you its field, and say what property you are testing for?
c-symmetry
Produce the field inside and outside a uniformly charged ball from scratch, and say why a hollow shell gives exactly zero inside rather than something small?
c-spherical
Derive the long wire field with a can in two lines, and say why the length of the can never appears in the answer?
c-cylindrical
Get the sheet result from a pillbox, explain where the factor of two comes from, and say what happens to the field outside a pair of opposite sheets?
c-planar
Find the charge on both walls of a conducting shell with a charge in its cavity, and explain why the field at a metal face is twice the lone sheet value?
c-conductors
Glossary (18 terms)
electric fluxelektrik akısı
The amount of electric field crossing a surface, computed as the integral of the field component along the surface normal over the area, and measured in newton metres squared per coulomb. It is a scalar carrying a sign, not a vector.
area vectoralan vektörü
A vector representing a patch of surface, whose size is the area of the patch and whose direction is along the normal to it. For a closed surface the direction is always taken outward.
closed surfacekapalı yüzey
A surface with a well defined inside and outside and no edges, such as a sphere, a box or a can with its lids on. Gauss's law applies only to surfaces of this kind.
Gaussian surfaceGauss yüzeyi
An imaginary closed surface drawn by you rather than existing in the problem, chosen so that the field has one size everywhere on the part of it that the field crosses.
Gauss's law
The statement that the outward electric flux through any closed surface equals the net charge enclosed divided by the permittivity of free space. It follows from the inverse square law together with superposition.
enclosed chargeiçerideki yük
The algebraic sum of all charge lying inside a given closed surface. Charge outside contributes nothing to the flux although it does contribute to the field at the surface.
permittivity of free spaceboşluğun elektriksel geçirgenliği
The constant epsilon nought, equal to 8.85 times ten to the minus twelfth in SI units, related to the Coulomb constant by k equals one over four pi epsilon nought.
symmetrysimetri
An operation such as a rotation, a slide or a reflection that leaves a charge distribution looking exactly as it did before. Symmetry restricts what the field is allowed to depend on, and that restriction is what makes Gauss's law useful.
spherical symmetryküresel simetri
The property of a charge distribution whose density depends only on the distance from a centre. Its field is radial and depends only on that distance.
cylindrical symmetrysilindirik simetri
The property of a charge distribution unchanged by sliding along an axis and by rotating about it. Its field points straight out from the axis and depends only on the distance from it.
planar symmetrydüzlemsel simetri
The property of a charge distribution unchanged by sliding in any direction parallel to a plane. Its field is perpendicular to the plane and does not depend on distance from it.
pillbox
A short flat closed cylinder used as a Gaussian surface for a sheet or a conductor surface, with its two flat caps parallel to the sheet and its side wall perpendicular to it.
shell theorem
The result that a spherically symmetric charge distribution produces no field at any point inside a hollow region within it, and produces the field of a point charge at its centre at any point outside it.
conductoriletken
A material whose charges can move freely through it. Left alone, it reaches a state with no field inside it and all excess charge on its surface.
electrostatic equilibriumelektrostatik denge
The state a conductor settles into when no charge in it is moving any more, which requires the field inside the material to be exactly zero.
induced chargeindüklenmiş yük
Charge that appears on a surface of a conductor because of the presence of other charge, without any charge being added to or removed from the conductor as a whole.
elektrostatik ekranlama
The fact that an empty cavity inside a conductor has zero field however strong the field outside is, which is why sensitive equipment is put inside metal boxes.
surface charge densityyüzey yük yoğunluğu
Charge per unit area, in coulombs per square metre. On a conductor it is the local density on one face, and the field just outside that face is this density divided by epsilon nought.
What comes next
§04 · Electric Potential
Everything on this page has been about the field: a vector at every point, with three components to keep track of and directions to argue about in every single problem. There is a way of describing the same physics with one number per point instead of three, from which the field can be recovered by a derivative, and it turns most of the vector bookkeeping in this section into arithmetic. That is the next step, and the fields computed here are exactly what it will be built from.
Sources
D. Giancoli, Physics for Scientists and Engineers with Modern Physics, 5th edition The set textbook. Its treatment of flux, Gauss's law, the three symmetry families and conductors in equilibrium covers the same ground as this section, and its end of chapter problems are a level harder than the ones here, which is the right next step once this set feels comfortable.
Course syllabus, week 3 line and assessment table The scope of this section comes from the week line and from nowhere else. The week line names a topic and carries no chapter numbers, so no chapter number is quoted anywhere on this page. The assessment weights quoted on the card are the published ones and nothing finer than them is claimed.
SI values of the electric constants The Coulomb constant is taken as 8.99 times ten to the ninth newton metre squared per coulomb squared and the permittivity of free space as 8.85 times ten to the minus twelfth in SI units throughout, the same values used in the two previous sections. The breakdown field of dry air is quoted as about three times ten to the sixth newtons per coulomb, which is a rounded standard figure used only for order of magnitude comparison.