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What are you here for?
05Review and Consolidation I: electric field, Gauss's law, and potential
Four weeks in, the questions stop telling you what to use. A small bead of known charge is let go from rest a few centimetres from something charged, and the last line asks how fast it is moving later on. Nothing names a formula. Most of the marks are decided in the first thirty seconds, before any algebra happens.
By the end of this section you can look at any electrostatics question from the first four weeks, say in one sentence which of the three routes it wants and why the other two cost more, carry it through, and run a check that catches a sign or a factor before you hand the paper in.
In 60 seconds
Everything so far is three objects, the charge, the field it makes and the potential that goes with that field, plus four arrows between them; the whole skill is picking the cheap arrow.
Field of a point charge
$$E = \dfrac{kq}{r^{2}},\qquad k = 8.99\times10^{9}\ \mathrm{N\,m^{2}/C^{2}}$$
any charge small enough to treat as a point, and the building block of every superposition sum
Superposition, in its two flavours
$$\vec E = \sum_i \vec E_i,\qquad V = \sum_i \dfrac{kq_i}{r_i}$$
more than one charge is present; the first sum needs components, the second needs only signs
Gauss's law
$$\oint \vec E\cdot d\vec A = \dfrac{Q_{\rm enc}}{\varepsilon_0}$$
one shape with spherical, cylindrical or planar symmetry, and you want the field
a long charged wire or a large charged sheet is in the picture and you do not want to integrate
Three most common mistakes
Adding potentials as if they were vectors, or adding field magnitudes as if they were numbers. The potential needs signs and nothing else; the field needs components.
Putting $r^{2}$ under the potential or $r$ under the field. Every second lost mark on this material is one power of $r$.
Reading zero field as zero potential inside a conductor. The field is zero there, the potential is whatever the outside world set it to, and it is almost never zero.
The published weights for this course are Midterm 1 twenty per cent, Midterm 2 twenty per cent, quizzes ten per cent, homework five per cent, the final twenty five per cent and the laboratory twenty per cent. Everything reviewed here belongs to the first midterm block, and nothing on this page is new material.
How much time do you have?
10 minutes
You leave with the map, the decision rule and the table of standard results, which between them decide the opening line of almost every question in this block.
The 60 second card · Formula card · The three objects and the four arrows between them · Choosing the cheap route · Mistake ledger
45 minutes
You add the two places where marks actually leak: mixing the vector sum with the scalar sum, and forgetting what a metal does to a problem. Then you run the ladder to see whether you can do it without the scaffolding.
The 60 second card · The three objects and the four arrows between them · Choosing the cheap route · Vectors for the field, plain numbers for the potential · The standard shapes, field and potential side by side · A conductor in equilibrium: one fact and its consequences · Fading ladder · Mistake ledger
full read
Everything above plus the energy route, reading a field off a potential graph, the exam length worked example and the four practice tiers. The interleaved set at the end is the one that predicts your exam mark, because there nobody tells you which tool to use.
Hook and card · Pre test · All seven concepts · Method boxes · Contrast pairs · Fading ladder · Exam length example · Practice A to D · Check yourself
By the end of this section
Translate between charge, field and potential in either direction, and name which of the four connecting formulas you are using.
Choose, before writing anything, between a superposition sum, Gauss's law and an energy argument, and justify the choice by cost.
Combine several charges correctly, adding fields as vectors and potentials as signed numbers, and locate the points where either one vanishes.
Apply the standard results for a ball, a shell, a long line and a large sheet to both the field and the potential, inside and outside.
Predict where charge sits on a conductor, what the field is on each side of its surface, and what the potential of the whole piece of metal is.
Extract the field from a potential, whether the potential comes as a formula or as a graph, and read off where the field vanishes.
Compute speeds, works and distances of closest approach from energy conservation instead of from forces.
Syllabus coverage
Catch up and Review
The four weeks behind us: charge and Coulomb's law, the electric field of point charges and of continuous distributions, Gauss's law, and electric potential. No new law is introduced here; what is new is the decision procedure that picks between them and the cross checks that catch a wrong answer.
The week line carries no chapter numbers, so none are quoted anywhere on this page. Everything reviewed is drawn from the material of the first four sections of the course.
covered
Recall first
Coulomb's law and the field it makes
$F = \dfrac{k\vert q_1q_2\vert}{r^{2}}$ along the line joining the two charges, and dividing by the test charge gives $E = \dfrac{k\vert q\vert}{r^{2}}$, pointing away from a positive source and towards a negative one.
Every result on this page is checked against it. If a formula you have written does not collapse to this one when the object shrinks to a point, the formula is wrong.
Superposition
The field of several charges is the vector sum $\vec E = \sum_i \vec E_i$ of what each one would make on its own, and the potential is the algebraic sum $V = \sum_i kq_i/r_i$.
It is the only reason we are allowed to solve one charge at a time and add up, which is what every multi charge question secretly asks.
Field of a continuous distribution
Cut the object into pieces $dq$, write $dE = k\,dq/r^{2}$ for one piece, and integrate the components, using symmetry to kill whichever component has to cancel.
This is the expensive route, and half of the skill in this section is recognising the cases where you do not have to take it.
Electric flux and Gauss's law
$\Phi_E = \int \vec E\cdot d\vec A$, and for any closed surface $\oint \vec E\cdot d\vec A = Q_{\rm enc}/\varepsilon_0$.
It converts a surface integral into a division, but only when symmetry makes $E$ constant over the surface and either along it or straight through it.
The three symmetric standard fields
Outside a spherical charge, $E = kQ_{\rm enc}/r^{2}$; outside a long line, $E = \lambda/(2\pi\varepsilon_0 r)$; near a large sheet, $E = \sigma/(2\varepsilon_0)$.
These three cover most of what an exam can ask with Gauss's law, and knowing them cold saves the two or three minutes that decide a paper.
Electric potential
$V_b - V_a = -\int_a^b \vec E\cdot d\vec l$, and for a point charge measured from infinity $V = kq/r$ with the sign of $q$ carried through.
It replaces three component integrals by one integral of a number, which is why anything involving energy or speed goes through $V$ rather than through $\vec E$.
Potential energy and work
A charge $q$ at a place where the potential is $V$ has $U = qV$; moving it between two points changes that energy by $\Delta U = q\,\Delta V$, and with no other force acting $\Delta K = -\Delta U$.
Every how fast and every how close question in this block is this one line plus arithmetic.
What a conductor does in equilibrium
The field inside the metal is zero, all excess charge sits on surfaces, and the field just outside a surface is $\sigma/\varepsilon_0$ and perpendicular to it.
The word metal in a question changes the answer for the same shape and the same charge, and it is the single most reliable way an exam separates a memorised formula from an understood one.
Try it yourself first (3 questions)
1§05.0 - potential where the field cancels●●○○○
Nothing here is new, and getting it wrong is not a problem; it tells you which paragraph below to read first. Two point charges of $+5.00\ \mathrm{nC}$ each sit on a line $8.00\ \mathrm{cm}$ apart. Look at the point exactly halfway between them, where the two field arrows are equal and opposite, so the net field really is zero.
Given
two charges of $+5.00\ \mathrm{nC}$
separation $8.00\ \mathrm{cm}$
the point of interest is the midpoint of the line joining them
Find
(a) What is the electric potential at that midpoint?
Hint 1/4
You are being asked for a number, not for a direction. Decide first whether the quantity in the question is the kind of thing that can cancel by pointing opposite ways.
Hint 2/4
Potentials from several charges add algebraically: $V = \sum_i kq_i/r_i$. There are no components and no angles in that sum.
Hint 3/4
Each charge is $5.00\ \mathrm{nC}$ and each sits $4.00\ \mathrm{cm}$ from the midpoint, so each contributes $kq/r = (8.99\times10^{9})(5.00\times10^{-9})/0.0400$.
Hint 4/4
The two contributions are equal and both positive, so $V = 2\times1.12\times10^{3} = 2.25\times10^{3}\ \mathrm{V}$.
Show solution
We add potentials rather than fields because the question asks for a number; going through the field here would mean computing two vectors and then integrating them back, for no gain.
Sign check without arithmetic: bringing a positive test charge in from infinity to a spot squeezed between two positive charges clearly costs work, so $V$ there has to be positive and comfortably large. A zero would mean it costs nothing, which is plainly false.
2§05.0 - net flux with a charge just outside●●○○○
A closed box has a $+3.00\ \mathrm{nC}$ charge and a $-7.00\ \mathrm{nC}$ charge sitting inside it. A third charge of $+12.0\ \mathrm{nC}$ is held a centimetre outside one of its walls, close enough that the field it makes at the wall is large.
Given
charges inside: $+3.00\ \mathrm{nC}$ and $-7.00\ \mathrm{nC}$
Separate two questions that feel like one: what the field on the wall is, and what the net flux through the whole closed surface is. Only the second one is being asked.
Hint 2/4
$\oint\vec E\cdot d\vec A = Q_{\rm enc}/\varepsilon_0$, where $Q_{\rm enc}$ counts only charge inside, with sign.
Hint 3/4
Inside the box we have $+3.00\ \mathrm{nC}$ and $-7.00\ \mathrm{nC}$, so $Q_{\rm enc} = -4.00\ \mathrm{nC}$; the $+12.0\ \mathrm{nC}$ outside is not part of that total.
Hint 4/4
So $\Phi = -4.00\times10^{-9}/8.85\times10^{-12} = -452\ \mathrm{N\,m^{2}/C}$.
Show solution
We read Gauss's law as bookkeeping rather than as a way to find a field, because the flux itself is what is asked; hunting for $\vec E$ on a box with no symmetry would be work with no payoff.
The sign is checkable on its own: the net charge inside is negative, so field lines end inside the box, so more flux enters than leaves, so the outward flux must come out negative.
Nothing about this answer would change if the outside charge were a coulomb instead of a nanocoulomb, or if it moved. That insensitivity is the whole point of the law.
3§05.0 - work done on a negative charge●●●○○
This one is designed to catch a habit rather than a gap. A negative charge is carried slowly, with no change of speed, from a point where the potential is $-40\ \mathrm{V}$ to a point where it is $+60\ \mathrm{V}$.
Given
charge $q < 0$
start at $V_i = -40\ \mathrm{V}$, finish at $V_f = +60\ \mathrm{V}$
moved slowly, so the kinetic energy does not change
Find
(a) True or false: you must do positive work to carry it there.
Hint 1/4
Do not decide from the words higher potential. Decide from the energy of this particular charge, which depends on the sign of $q$ as well as on $V$.
Hint 2/4
The energy of a charge in a potential is $U = qV$, so $\Delta U = q\,\Delta V$, and the work you must supply when the speed does not change is exactly $\Delta U$.
Hint 3/4
Here $\Delta V = +60 - (-40) = +100\ \mathrm{V}$ and $q$ is negative, so $\Delta U = q(+100\ \mathrm{V})$ is a negative number.
Hint 4/4
$\Delta U < 0$, so the work you do is negative: the field pulls the charge there and you have to hold it back.
Show solution
We put the sign into $\Delta U = q\,\Delta V$ instead of arguing from force directions, because the field pattern is never given and the energy statement needs only two signs.
Put the sign of the charge into the energy, not into the potential
$$\Delta U = q\,\Delta V = (\text{negative})(+100\ \mathrm{V}) < 0$$
the belongs to the space, the sign of the charge belongs to the object being moved
$$W_{\rm you} = \Delta U < 0$$
at constant speed the kinetic energy does not change, so everything you supply goes into potential energy, and here that quantity decreases
Answer $$\boxed{\;W_{\rm you} < 0\ \text{, the statement is false}\;}$$
Check
Independent check with a concrete case: put an electron next to a positively charged plate. The plate is at high potential, the electron is pulled straight at it, and nobody has to push.
High potential is only downhill for positive charges. Half the sign errors in this block come from forgetting that the hill is inverted for the other sign.
Notation
symbol
reads as
means
watch out
$\vec E$
E vector
the electric field at a stated point, a vector measured in newtons per coulomb
$E$ without the arrow is its magnitude, never negative; a minus sign in front of $E$ is a statement about direction, not about the size of the field
$V$
V
the electric potential at a stated point, a signed number measured in
it carries no direction at all, so asking which way $V$ points is asking about something that does not exist
$V_{ab}$
V a b
the potential difference $V_a - V_b$ between two named points
the order of the subscripts is the whole content of the symbol; swapping them changes the sign
$U$
U
, in joules, of a charge sitting in a potential or of a pair of charges
$U = qV$ for one charge in an external potential, but $U = kq_1q_2/r$ for a pair; using the pair formula twice double counts the same energy
$Q_{\rm enc}$
Q enclosed
the net charge inside a chosen closed surface, with its sign
charge outside the surface still makes a field on it, but contributes nothing to the enclosed total
$\sigma,\ \lambda,\ \rho$
sigma, lambda, rho
charge per unit area, per unit length and per unit volume
on a conductor $\sigma$ is the local surface density, which varies from place to place unless the surface is a sphere
$\varepsilon_0$
epsilon nought
the permittivity of free space, $8.85\times10^{-12}\ \mathrm{C^{2}/(N\,m^{2})}$
$k = 1/(4\pi\varepsilon_0)$, so a stray factor of $4\pi$ is the usual sign that the two constants got mixed
$\nabla V$
grad V
the vector of the three partial derivatives of $V$, whose negative is the field
in one dimension it is just $dV/dx$; the minus sign in $\vec E = -\nabla V$ is not optional and is the single most dropped sign in this block
Conventions used here
Where the zero of potential is
For any charge distribution of finite size, $V = 0$ infinitely far away, and every potential quoted on this page is measured from there. The infinite line and the infinite sheet are the two exceptions: their potential does not settle down at infinity, so for those only differences between two named points are quoted, never a single value of $V$.
How a potential difference is written
$V_{ab}$ means $V_a - V_b$, the potential of the first point minus that of the second. When a charge moves, $\Delta V = V_{\rm final} - V_{\rm initial}$ and $\Delta K = -q\,\Delta V$. Reading $\Delta V$ backwards flips the sign of the answer and nothing else, which is why it is so easy to miss.
Constants and significant figures on this page
$k = 8.99\times10^{9}\ \mathrm{N\,m^{2}/C^{2}}$, $\varepsilon_0 = 8.85\times10^{-12}\ \mathrm{C^{2}/(N\,m^{2})}$, $e = 1.60\times10^{-19}\ \mathrm{C}$, $m_p = 1.67\times10^{-27}\ \mathrm{kg}$, $m_e = 9.11\times10^{-31}\ \mathrm{kg}$. Answers are quoted to three significant figures.
Where the sign of a charge goes
Signs go into the picture and into the potential sum. In a field magnitude the charge enters as $\vert q\vert$ and the direction is decided by looking at the diagram: away from positive, towards negative. Writing a negative magnitude is the fastest way to lose a direction mark.
Angles and directions
Angles are measured anticlockwise from the $+x$ axis unless the question says otherwise, and vector components are always listed in the order $(x, y, z)$. In flux the angle is measured from the outward normal of the surface, not from the surface itself.
What a metal does to the problem
Conducting, metal and metallic all mean the same thing here: charge is free to move and has already finished moving. Non conducting, insulating, plastic and glass all mean the charge stays where it was put. Every question on this page states which one it is, because the two give different answers for the same shape and the same charge.
Units and how an answer is quoted
A field is quoted as a magnitude in $\mathrm{N/C}$ (the same unit as $\mathrm{V/m}$) together with a direction in words. A potential is quoted as a signed number in volts, with no direction, because it does not have one. An energy is quoted in joules, and in as well whenever a single elementary charge is involved.
5.1The three objects and the four arrows between them
Charge makes a field, the field carries a potential, and every problem is a trip between two of the three.
Three quantities are on the table and nothing yet says how they sit relative to each other.
Solvable with what we have
the force between point charges, and the field of a collection of them
the field of a ring, rod or disc, by cutting it up and integrating components
the field of a sphere, long cylinder or large sheet, in one line, from a closed surface
the potential of several point charges, by adding numbers
Not solvable yet
how fast a released charge is moving later on
how close a fast charge gets to another before it stops
the work to move a charge along a curved path in an awkward field
the field of an object whose potential you know but whose charge you do not
Try the first of those with force alone. A proton released from rest $2.00\ \mathrm{cm}$ from a fixed $+45.0\ \mathrm{nC}$ charge starts with acceleration $9.69\times10^{13}\ \mathrm{m/s^{2}}$. Put that into $v^{2} = 2a\,\Delta r$ over the $6.00\ \mathrm{cm}$ to the finish line and you get $3.41\times10^{6}\ \mathrm{m/s}$.
Why it fails
That acceleration held for one instant and no other: by $8.00\ \mathrm{cm}$ the force is a sixteenth of its starting value. Done properly the trip comes out twice as slow. Integrating the varying force fixes it, and that integral is what the potential already is.
RuleRule 5.1: the two bridges between field and potential
Conditions
the path from $a$ to $b$ in the first formula may be any path at all, because an electrostatic field does no net work around a closed loop
the derivative in the second formula is taken at one point, holding the other two coordinates fixed
both statements describe the same field and the same potential, so a dropped minus sign in one of them shows up as a dropped minus sign in the other
To get the potential, walk between the two points and add up the field along the way, negative walking with the field and positive against it. To come back, ask how steeply the potential falls in a direction; that slope with its sign flipped is the field component there.
Proof
In one dimension the first bridge reads $V(x) - V(x_0) = -\int_{x_0}^{x} E_x\,dx'$.
Differentiate both sides with respect to $x$. The left side gives $dV/dx$; the right side gives $-E_x$ by the fundamental theorem of calculus.
So $E_x = -dV/dx$, which is the second bridge. The two statements are one statement, read in opposite directions.
The two arrows along the bottom are the cheap ones, and most revision questions are solved by finding a way to use one of them instead of climbing down from the charge twice.
Looks like this, but is not
The potential difference between two points is the field times the gap, $\Delta V = Ed$. The units are right, and it is exact between two large parallel sheets, where most people first meet it.
It holds only where the field is the same all along the path. In the table below the potential drops $4495\ \mathrm{V}$ from $1.00$ to $2.00\ \mathrm{cm}$ while $Ed$ gives $8990\ \mathrm{V}$ with the near field and $2250\ \mathrm{V}$ with the far one. The field changed fourfold on the way and the shortcut cannot know that.
distance r (cm)
field E (N/C)
potential V (V)
1.00
8.99 × 10⁵
8990
2.00
2.25 × 10⁵
4495
5.00
3.60 × 10⁴
1798
10.0
8.99 × 10³
899
20.0
2.25 × 10³
450
Doubling the distance from 1.00 cm to 2.00 cm quarters the field and halves the potential, and every further doubling does the same again. If your two columns ever fall off at the same rate, one of them has the wrong power of r in it.
Speed of a proton released 2.00 cm from a fixed 45.0 nC charge
A $+45.0\ \mathrm{nC}$ charge is clamped in place. A proton is released from rest $2.00\ \mathrm{cm}$ away from it and flies off along the line joining them. How fast is it moving when it is $8.00\ \mathrm{cm}$ from the clamped charge?
released from rest at $r_i = 2.00\ \mathrm{cm}$, measured at $r_f = 8.00\ \mathrm{cm}$
Find
the speed at 8.00 cm
Solution
Energy rather than force, because the force changes by a factor of sixteen over the trip and the question never asks for the time; the moment a question asks how fast rather than how long, the force route is the expensive one.
Get the potential difference the proton falls through
same formula at the finish line; note that only the difference will matter, so the choice of zero cancels out
$$\Delta V = V_f - V_i = -1.52\times10^{4}\ \mathrm{V}$$
final minus initial, in that order, because that is what the energy statement is written in terms of
Turn the potential difference into kinetic energy
$$\Delta K = -q\,\Delta V = -(1.60\times10^{-19})(-1.52\times10^{4}) = 2.43\times10^{-15}\ \mathrm{J}$$
the proton is positive and it moves to lower potential, so the two minus signs leave a gain, which is what we expect for a repelled particle flying away
$$K_f = \Delta K = 2.43\times10^{-15}\ \mathrm{J}$$
it started from rest, so the change in kinetic energy is the whole of it
Two independent checks. In electron volt units the proton fell through $1.52\times10^{4}\ \mathrm{V}$, so it carries $15.2\ \mathrm{keV}$, and the standard shortcut $v = 1.384\times10^{4}\sqrt{K/\mathrm{eV}}\ \mathrm{m/s}$ reproduces $1.71\times10^{6}\ \mathrm{m/s}$. And $v/c = 0.0057$, so treating the proton non relativistically was legitimate.
Three divisions and one square root, against an integral of a varying force that nobody wants to do under exam conditions.
This is the loop closing on the failure at the top of the section. The naive constant acceleration answer was $3.41\times10^{6}\ \mathrm{m/s}$, twice too big, because it used the starting force all the way. The potential is the machine that does that integral once and for all.
Field on the axis of a charged ring, obtained from its potential
A thin ring of radius $6.00\ \mathrm{cm}$ carries $25.0\ \mathrm{nC}$ spread evenly around it. Find the potential and the field at a point on the axis $8.00\ \mathrm{cm}$ from the centre, without doing a vector integral.
Given
ring radius $R = 6.00\ \mathrm{cm}$, total charge $Q = 25.0\ \mathrm{nC}$
field point on the axis at $x = 8.00\ \mathrm{cm}$
Find
the potential and the axial field there
Solution
We build the potential first and differentiate it, because the direct field integral would need the axial component of every ring element and a factor $\cos\theta$; the potential sum has no components in it at all.
Add the potential of every piece, which needs no components
Check the formula at a place where the answer is known for another reason: at $x=0$ it gives $E_x = 0$, which symmetry demands, and $V = kQ/R = 3.75\times10^{3}\ \mathrm{V}$, which is what you get by adding up a ring of charge all at distance $R$. Far away it gives $E \to kQ/x^{2}$; at $x = 8.00\ \mathrm{cm}$ the point charge formula would give $3.51\times10^{4}\ \mathrm{N/C}$, a factor 1.95 too big, so we are correctly not yet in the far field.
One derivative instead of one integral over the ring with a cosine factor in it.
Whenever a distribution has a symmetry axis, the potential on that axis is usually a one line sum and the field is its derivative. Look for that route before setting up components.
Checkpoint
§05.1 - potential of a single negative point charge●○○○○
Thirty seconds, no calculator needed beyond one division. A point charge of $-12.0\ \mathrm{nC}$ sits alone in space.
Given
$q = -12.0\ \mathrm{nC}$
the field point is $5.00\ \mathrm{cm}$ away
$V = 0$ infinitely far from the charge
Find
(a) What is the potential at that point?
Hint 1/4
You are asked for a signed number, so the answer has one sign and no direction attached to it.
Hint 2/4
For a single point charge measured from infinity, $V = kq/r$, with the sign of $q$ carried straight through.
Hint 3/4
Here $q = -12.0\ \mathrm{nC}$ and $r = 5.00\ \mathrm{cm} = 0.0500\ \mathrm{m}$, so $V = (8.99\times10^{9})(-12.0\times10^{-9})/0.0500$.
Hint 4/4
$V = -2.16\times10^{3}\ \mathrm{V}$.
Show solution
Straight from $V = kq/r$ with the sign carried into the numerator, since one point charge needs neither superposition nor an integral.
: a nanocoulomb at a few centimetres gives kilovolts, which is the familiar scale of a charged comb or a van de Graaff terminal at arm's length.
⚠ Putting the square in the potential
the field formula is the one you meet first and the hand writes it automatically
wrong$$V = \frac{kq}{r^{2}}$$
right$$V = \frac{kq}{r}$$
⚠ Taking the magnitude of the charge in a potential
in field problems you are trained to use $\vert q\vert$ and put the direction in by hand, and the habit carries over
wrong$$V = \frac{k\vert q\vert}{r}$$
right$$V = \frac{kq}{r}\ \text{, sign of }q\text{ kept}$$
⚠ Dropping the minus sign when going from V back to E
the derivative is the visible operation and the sign in front of it looks decorative
wrong$$E_x = \frac{dV}{dx}$$
right$$E_x = -\frac{dV}{dx}$$
5.2Choosing the cheap route
Read what the last line of the question asks for, and let that pick the tool before you write anything down.
The map tells you which trips exist; it does not tell you which one this particular question wants, and that is where the time goes.
MethodRule 5.2: what the question is asking for decides the tool
Conditions
Gauss's law delivers a field only when one shape carries all the charge and the symmetry is spherical, cylindrical or planar
the energy route needs no symmetry at all, but it can only answer questions about energy, speed, work or distance
when neither of the first two applies, superposition always works and always costs more
$$\boxed{\;\text{speed or work}\ \Rightarrow\ V\ ;\qquad \text{field with symmetry}\ \Rightarrow\ \oint \vec E\cdot d\vec A = \frac{Q_{\rm enc}}{\varepsilon_0}\ ;\qquad \text{otherwise}\ \Rightarrow\ \textstyle\sum\;}$$
If the question ends in a speed, an energy or an amount of work, go through the potential. If it ends in a field and there is one symmetric object, draw a closed surface and divide. If it ends in a field and there is no symmetry, you are adding contributions, and you should expect that to take longer.
Work down the chart and stop at the first box you can honestly enter. Entering a box you are not entitled to is the most expensive mistake available in this material, because it produces a clean looking wrong answer.
Looks like this, but is not
There is a closed surface in the picture and a known enclosed charge, so Gauss's law gives the field. The law is certainly true here, and the flux is certainly $Q_{\rm enc}/\varepsilon_0$.
True and useful are different things. The law relates the field to an integral of it, and only symmetry lets you pull $E$ out of that integral. Put a point charge off centre inside a sphere and the flux is unchanged while the field on the surface varies from $3.67\times10^{3}$ to $2.00\times10^{4}\ \mathrm{N/C}$; one equation cannot deliver a function.
question asked
route
what it costs
field at r greater than R
Gauss, spherical surface
one division
potential at r greater than R
point charge result
one division
speed of a released proton
potential difference, then energy
two divisions and a square root
field at a point off the symmetry axis of a rod
superposition integral
one integral with components
field inside the metal of a shell
the conductor rule
no arithmetic at all
Four of these five are one liners, and the one that is not is the one where the shape has no symmetry to exploit. Under exam pressure the useful habit is to assume you are in a one liner until the geometry proves otherwise.
Field 3.00 cm from a long charged wire, the cheap way and the expensive way
A very long straight wire carries $3.50\ \mathrm{nC/m}$ uniformly. Find the field $3.00\ \mathrm{cm}$ from the wire, well away from either end.
Given
$\lambda = 3.50\ \mathrm{nC/m}$
distance from the wire $r = 3.00\ \mathrm{cm}$
the wire is long enough that the ends do not matter
Find
the field magnitude and direction
Solution
Gauss's law, because the question wants a field and the object is one long straight thing, which is the textbook trigger for a cylindrical surface.
Take the cheap route: a cylinder around the wire
$$E\,(2\pi r L) = \frac{\lambda L}{\varepsilon_0}$$
on a coaxial cylinder the field has the same size everywhere and points straight through the curved wall, so E comes out of the integral; the two flat ends contribute nothing because the field runs along them
Independent arithmetic on the other form of the constant. The denominator is $2\pi\varepsilon_0 r = 1.67\times10^{-12}$, and $3.50\times10^{-9}$ divided by it is $2.10\times10^{3}\ \mathrm{N/C}$, which agrees and confirms that $k$ and $\varepsilon_0$ were not mixed up.
One line against one improper integral. The saving is entirely due to the symmetry, not to any cleverness.
Why the same cylinder is useless for a 20.0 cm rod
Now shorten the wire to a rod of length $20.0\ \mathrm{cm}$ carrying the same $3.50\ \mathrm{nC/m}$, and ask for the field $3.00\ \mathrm{cm}$ from its midpoint, on the perpendicular bisector.
Given
$\lambda = 3.50\ \mathrm{nC/m}$, rod length $L = 20.0\ \mathrm{cm}$
field point $y = 3.00\ \mathrm{cm}$ from the midpoint, perpendicular to the rod
Find
the field there, and how far the long wire answer is off
Solution
We spend the first half showing why the cylinder stops working instead of integrating immediately, because the decision is the lesson here; the integral is the price of a symmetry this rod does not have.
Show why the closed surface stops helping
$$\oint \vec E\cdot d\vec A = \frac{\lambda L}{\varepsilon_0}$$
the law is still true; a cylinder around the rod still encloses the same charge
$$E \ne \text{constant on the wall}$$
near the ends the field tilts and weakens, so E cannot be taken outside the integral and one equation is left with two unknowns
Do the integral that the symmetry no longer does for you
Answer $$\boxed{\;E = 2.01\times10^{3}\ \mathrm{N/C}\ \text{, perpendicular to the rod}\;}$$
Check
The finite answer must be smaller than the infinite one, and it is: the ratio of the two is exactly $(L/2)/\sqrt{y^{2}+(L/2)^{2}} = 0.958$, so the rod is 4.2 per cent weaker than an endless wire. That number is a pure geometry factor and can be written down before any arithmetic.
One integral, unavoidable. The reward is knowing exactly how good the long wire approximation is instead of hoping.
Long means long compared with the distance to the field point, not long in centimetres. Here the half length is just over three times the distance and the error is already down to four per cent.
Checkpoint
§05.2 - picking the tool for a plate●○○○○
Thirty seconds, no arithmetic. A large flat non conducting sheet carries charge spread evenly over it, and you are asked for the field at a point $2.00\ \mathrm{mm}$ from the middle of the sheet, far from any edge.
Given
a large uniformly charged flat sheet
field point $2.00\ \mathrm{mm}$ from the sheet, near the middle
no other charge anywhere
Find
(a) Which route gets you the answer fastest?
Hint 1/4
Ask two questions in order: is a field being requested, and is there a single object with a symmetry that a closed surface can exploit.
Hint 2/4
A flat sheet has planar symmetry, so a box straddling the sheet has field only through its two faces.
Hint 3/4
That box gives $2EA = \sigma A/\varepsilon_0$ for the sheet in question, with the area cancelling, and the distance $2.00\ \mathrm{mm}$ never entering.
Hint 4/4
So the answer is Gauss's law with a pillbox, and it takes one line.
Show solution
The pillbox wins because planar symmetry makes $E$ constant and perpendicular on exactly two faces; adding the sheet up element by element would cost a double integral for the same one line answer.
the field can only point straight away from the sheet, because tilting it would break the mirror symmetry of the sheet about its own plane
$$2EA = \frac{\sigma A}{\varepsilon_0}\ \Rightarrow\ E = \frac{\sigma}{2\varepsilon_0}$$
two faces carry flux, the curved side carries none, and the area cancels
Answer $$\boxed{\;E = \frac{\sigma}{2\varepsilon_0}\ \text{, from Gauss's law}\;}$$
Check
Consistency check on the shape of the answer: no distance appears in it, which is only believable because the sheet is large compared with the 2.00 mm gap. Move a metre away from a sheet of finite size and the formula stops being usable.
⚠ Using Gauss's law on a shape with no symmetry
the law holds for every closed surface, so it feels applicable to every closed surface
wrong$$E = \frac{Q_{\rm enc}}{\varepsilon_0 A}\ \text{for any closed surface}$$
right$$E = \frac{Q_{\rm enc}}{\varepsilon_0 A}\ \text{only if }E\text{ is constant on }A$$
⚠ Going after the field when the question wanted a speed
the field is the quantity you practised most, so the hand starts drawing arrows before the question has been read to the end
wrong$$a = \frac{qE}{m}\ \text{, then kinematics with varying }E$$
right$$\Delta K = -q\,\Delta V$$
⚠ Treating a finite rod as an infinite line without checking
the two formulas look alike and the infinite one is shorter
wrong$$E = \frac{2k\lambda}{y}\ \text{for a rod of length }L$$
5.3Vectors for the field, plain numbers for the potential
Two charges give two arrows and two numbers, and the arrows and the numbers cancel in different places.
Both sums were introduced separately, and the exam habit worth building is doing them in the same breath so that neither one borrows the other's rules.
RuleRule 5.3: the two superposition sums
Conditions
the field sum needs components, so a set of axes has to be chosen and stated before the first term is written
the potential sum needs the sign of each charge and nothing else, and the answer is a single signed number
both sums assume the charges do not disturb each other, which is exactly true for fixed point charges
$$\boxed{\;\vec E = \sum_i \frac{k\vert q_i\vert}{r_i^{2}}\,\hat r_i \;,\qquad V = \sum_i \frac{k q_i}{r_i}\;}$$
For the field, work out what each charge would do on its own, draw that arrow, break it into components and add the components. For the potential, work out $kq/r$ for each charge with its own sign and add the numbers; there is nothing to resolve and no angle to look up.
The same two charges, the same point, two different answers about what cancels. A pair of equal and opposite charges is the cleanest place to see it: the numbers cancel on the whole perpendicular bisector, the arrows never do.
Looks like this, but is not
The field is zero here, so the potential is zero here too. It is tempting because both quantities come from the same charges and both are being summed.
Halfway between two equal positive charges the two arrows point opposite ways and cancel, while the two numbers are both positive and add. For two charges of $+5.00\ \mathrm{nC}$ a distance $8.00\ \mathrm{cm}$ apart, the midpoint has $E = 0$ and $V = 2.25\times10^{3}\ \mathrm{V}$. Swap one sign and it goes the other way round: at the midpoint of a positive and a negative charge, $V = 0$ while the field is a healthy $4.32\times10^{4}\ \mathrm{N/C}$.
position x (cm)
net field Ex (N/C)
potential V (V)
-12.0
0
-300
-4.00
-1.69 × 10⁴
0
+2.40
+7.80 × 10⁴
0
+6.00
+4.99 × 10⁴
-1798
+18.0
-3.89 × 10⁴
-2198
The field vanishes once, at 12.0 cm to the left of the small charge, and the potential there is -300 V. The potential vanishes twice, at 4.00 cm to the left of the small charge and at 2.40 cm to its right, and at neither of those places is the field anywhere near zero. Three separate points answering three separate questions, and no two of them coincide.
Field and potential at the fourth corner of a square
A square has sides of $5.00\ \mathrm{cm}$. Charges of $+8.00\ \mathrm{nC}$, $-5.00\ \mathrm{nC}$ and $+8.00\ \mathrm{nC}$ are fixed at three of its corners, taken in order around the square, so that the negative charge is diagonally opposite the empty corner. Find the field and the potential at the empty corner.
Given
square of side $a = 5.00\ \mathrm{cm}$
$+8.00\ \mathrm{nC}$ at $(0,0)$ and at $(a,a)$, $-5.00\ \mathrm{nC}$ at $(a,0)$
the empty corner is at $(0,a)$
Find
the field vector and the potential at the empty corner
Solution
We do the scalar sum first even though the field is the harder half, because three divisions give a number we can lean on while checking the vector work.
Do the easy sum first, to have a number to lean on
Independent route to the same magnitude. The two positive charges are mirror images in the diagonal, so their fields add to $\sqrt2\times2.88\times10^{4} = 4.07\times10^{4}\ \mathrm{N/C}$ straight along that diagonal, and the negative charge subtracts $8.99\times10^{3}$ along the same line. $4.07\times10^{4} - 0.899\times10^{4} = 3.17\times10^{4}$, with no components used anywhere.
Three field magnitudes, one factor of root two, and two component sums, against one line of arithmetic for the potential.
When a picture has a mirror symmetry, use it to combine the symmetric pair first. It replaces four component calculations with one, and it gives you the independent check for free.
Where the field cancels between +4.00 nC and -16.0 nC
Two point charges are fixed on the $x$ axis: $+4.00\ \mathrm{nC}$ at the origin and $-16.0\ \mathrm{nC}$ at $x = 12.0\ \mathrm{cm}$. Find every point on that axis where the total field is zero.
Given
$q_1 = +4.00\ \mathrm{nC}$ at $x = 0$
$q_2 = -16.0\ \mathrm{nC}$ at $x = 12.0\ \mathrm{cm}$
Find
all points on the axis where the field vanishes
Solution
We settle which region can work by direction before writing any algebra, because squaring blind produces a second root sitting in a region where cancellation is impossible and it then has to be thrown away by hand.
Decide which region can possibly work, before any algebra
$$\text{between them: both arrows point}\ +x$$
to the right of a positive charge the field points right, and to the left of a negative charge it also points right, so nothing can cancel in between
$$\text{right of }q_2\text{: the nearer charge is the bigger one}$$
the larger charge is also the closer one there, so it wins at every point and cancellation is impossible
$$\text{left of }q_1\text{: possible}$$
here the small charge is the near one and the big charge is far, so the two effects can trade off
the positive root is the only one with physical meaning, since both distances are positive
$$s = 0.120\ \mathrm{m}\ \Rightarrow\ x = -12.0\ \mathrm{cm}$$
twelve centimetres to the left of the small charge
Answer $$\boxed{\;x = -12.0\ \mathrm{cm}\ \text{, and nowhere else on the axis}\;}$$
Check
Substitute back rather than trusting the algebra. At $x = -12.0\ \mathrm{cm}$ the distances are $12.0\ \mathrm{cm}$ and $24.0\ \mathrm{cm}$, giving $k(4.00\ \mathrm{nC})/0.0144 = 2.50\times10^{3}$ and $k(16.0\ \mathrm{nC})/0.0576 = 2.50\times10^{3}\ \mathrm{N/C}$; equal sizes, opposite directions.
Two lines, once the region argument has removed two thirds of the axis.
Always find the region before solving. Squaring an equation happily hands you roots in regions where the directions never allowed a cancellation, and those roots look completely respectable.
Checkpoint
§05.3 - zero potential and zero field●●○○○
Thirty seconds. Somebody hands you a point in space and tells you the electric potential there is exactly zero.
Given
a single point in space
the potential there is zero, with infinity as the reference
nothing else is stated about the charges producing it
Find
(a) True or false: the electric field at that point must also be zero.
Hint 1/4
Ask what each of the two quantities is a sum of, and what has to happen for each sum to come out zero.
Hint 2/4
$V$ is a sum of signed numbers, $\vec E$ is a sum of arrows. A sum of numbers can vanish while a sum of arrows does not, and the other way round.
Hint 3/4
Take $+6.00\ \mathrm{nC}$ and $-6.00\ \mathrm{nC}$ a distance $10.0\ \mathrm{cm}$ apart and stand at the midpoint: the two potentials are equal and opposite, and both field arrows point the same way.
Hint 4/4
So $V = 0$ there while $E = 4.32\times10^{4}\ \mathrm{N/C}$, and the statement is false.
Show solution
One counterexample rather than a general argument: a single arrangement with $V = 0$ and a large $E$ is enough to kill a claim of the form must.
both arrows point from the positive charge towards the negative one, so they add rather than cancel
Answer $$\boxed{\;V = 0\ \text{ while }\ E = 4.32\times10^{4}\ \mathrm{N/C}\ \text{: false}\;}$$
Check
Check by the other bridge. The potential is zero all along the perpendicular bisector, but the field is minus the slope of $V$ measured across that surface, and the potential certainly changes as you step off it. A flat value along one line says nothing about the slope perpendicular to it.
⚠ Adding field magnitudes without components
the potential sum has just been done and it was a straight addition, so the hand repeats it
Outside any ball of charge the field falls off like one over distance squared, exactly as if the whole charge sat at the centre. Beside a long charged line it falls off like one over distance. Beside a large charged sheet it does not fall off at all, as long as you stay close enough that the sheet still looks infinite.
A uniformly charged insulating ball, plotted twice. The field has a corner at the surface and the potential does not, which is a general fact: the potential is the integral of the field, and integrating smooths a corner into a bend.
Looks like this, but is not
Inside a uniformly charged ball the potential is constant, because inside a conductor it is. Both objects are spheres, both are charged, and the sentence about conductors is one you have certainly learned.
Constant potential inside means zero field inside, and inside a uniformly charged insulator the field is $kQr/R^{3}$, which is not zero anywhere except at the exact centre. The potential there runs from $1.5\,kQ/R$ at the centre down to $kQ/R$ at the surface. The word that changes the answer is conducting, not spherical.
shape
field E
potential V
point charge q
kq/r²
kq/r
ball of charge Q, radius R, outside
kQ/r²
kQ/r
same ball inside, insulating
kQr/R³
kQ(3R² - r²)/2R³
same ball inside, conducting
0
kQ/R, the surface value
thin shell radius R, inside
0
kQ/R, the surface value
long line, charge per length λ
λ/2πε₀r
differences only, logarithmic
large sheet, charge per area σ
σ/2ε₀
differences only, linear in distance
Two rows carry a zero field and a non zero potential, and those two rows are where most of the marks are lost. A zero in the middle column never forces a zero in the right hand one, because the potential at a point records the whole journey in from infinity, not the local field.
Field and potential inside and outside a 4.00 cm ball of 12.0 nC
A solid non conducting ball of radius $4.00\ \mathrm{cm}$ carries $+12.0\ \mathrm{nC}$ spread evenly through its volume. Find the field and the potential at $r = 2.00\ \mathrm{cm}$ and at $r = 10.0\ \mathrm{cm}$ from the centre.
Given
radius $R = 4.00\ \mathrm{cm}$, charge $Q = +12.0\ \mathrm{nC}$, spread evenly through the volume
the two field points are at $2.00\ \mathrm{cm}$ and $10.0\ \mathrm{cm}$ from the centre
Find
the field and the potential at both radii
Solution
Gauss's law for the fields because the symmetry is spherical, then the standard integrated results for the potentials rather than redoing the integral.
Outside, where the ball is disguised as a point charge
integrating the inside field from r out to the surface and adding the surface value; the potential keeps rising inwards even though the field falls
Answer $$\boxed{\;r = 2.00\ \mathrm{cm}:\ E = 3.37\times10^{4}\ \mathrm{N/C},\ V = 3.71\times10^{3}\ \mathrm{V}\;}$$
Check
The two pieces must agree at the surface, and they do: both formulas give $E = kQ/R^{2} = 6.74\times10^{4}\ \mathrm{N/C}$ and $V = kQ/R = 2.70\times10^{3}\ \mathrm{V}$ at $r = R$. The inside potential must also lie between the surface value and the central value $1.5\,kQ/R = 4.05\times10^{3}\ \mathrm{V}$, and $3.71\times10^{3}$ does.
Four substitutions. The expensive part is remembering which of the four formulas belongs to which region, which is why the graph above is worth thirty seconds.
Notice that the field at 2.00 cm is three times the field at 10.0 cm, while the potential at 2.00 cm is only three and a half times the potential there. Different powers of $r$ mean the two quantities never scale together.
Potential difference across a 2.00 mm gap between charged sheets
Two large parallel sheets face each other across $2.00\ \mathrm{mm}$. One carries $+35.0\ \mathrm{nC/m^{2}}$ and the other $-35.0\ \mathrm{nC/m^{2}}$, spread evenly. Find the field between them and the potential difference across the gap.
Given
$\sigma = \pm35.0\ \mathrm{nC/m^{2}}$ on the two sheets
separation $d = 2.00\ \mathrm{mm}$
the sheets are large compared with the gap
Find
the field between the sheets and the potential difference
Solution
We add the two sheet fields and then multiply by the gap, because the field is uniform there; writing $\int \vec E\cdot d\vec l$ would be the same product dressed up.
uniform everywhere in the gap, which is what makes the next step a multiplication rather than an integral
Turn a uniform field into a potential difference
$$\Delta V = E\,d$$
the field does not change along the path, so the integral collapses to a product; this is the one case where the V equals E times d shortcut is legitimate
$$\Delta V = (3.95\times10^{3})(2.00\times10^{-3}) = 7.91\ \mathrm{V}$$
the positive sheet is the higher potential one
Answer $$\boxed{\;E = 3.95\times10^{3}\ \mathrm{N/C},\qquad \Delta V = 7.91\ \mathrm{V}\;}$$
Check
Unit check that is also a physics check: $\mathrm{(N/C)\times m = N\,m/C = J/C = V}$, so a field times a distance really is a voltage, and the newton per coulomb and the volt per metre are the same unit written two ways.
One addition and one multiplication, with no integral anywhere, entirely because the field is uniform.
Eight volts across two millimetres is a mild laboratory number, and it is worth carrying as a scale: a few tens of nanocoulombs per square metre give you single figure volts across a millimetre gap.
Checkpoint
§05.4 - field inside a solid metal ball●○○○○
Thirty seconds, and the arithmetic is a trap rather than the point. A solid ball of metal with radius $5.00\ \mathrm{cm}$ carries $+20.0\ \mathrm{nC}$ of excess charge and has been left alone long enough for everything to settle.
Given
solid conducting ball, radius $5.00\ \mathrm{cm}$
excess charge $+20.0\ \mathrm{nC}$, in equilibrium
the field point is $3.00\ \mathrm{cm}$ from the centre
Find
(a) What is the electric field there?
Hint 1/4
Before reaching for a formula, decide which of the two spheres you are dealing with: the insulating one where charge is frozen in place, or the metal one where it is free to move.
Hint 2/4
In a conductor at equilibrium the field inside the metal is zero, because any field would push the free charges and they have already stopped moving.
Hint 3/4
A spherical surface at $3.00\ \mathrm{cm}$ lies inside the metal, and it encloses no charge at all, because all $20.0\ \mathrm{nC}$ has gone to the outer surface.
Hint 4/4
So $E = 0$, exactly, and no arithmetic is needed.
Show solution
We answer from the equilibrium property instead of drawing a Gaussian sphere and computing, because zero field inside is already the result Gauss's law would hand back.
Use the property of the metal, not the geometry of the ball
$$\vec E = 0\ \text{everywhere inside the metal}$$
equilibrium means the free charges have stopped, and they only stop when there is no field left to push them
$$Q_{\rm enc} = 0\ \text{for any surface inside}$$
consistent with the first line through Gauss's law, and it tells you where the charge went: to the outer surface
Answer $$\boxed{\;E = 0\;}$$
Check
Cross check from the other side. Just outside the ball the field is $kQ/R^{2} = 7.19\times10^{4}\ \mathrm{N/C}$, so the field jumps discontinuously at the surface. That jump is exactly $\sigma/\varepsilon_0$, which is what the surface charge is there to produce.
⚠ Using the insulating ball formula for a metal one
both are spheres with a radius and a total charge, and the formula with $r/R^{3}$ in it feels more advanced
5.5A conductor in equilibrium: one fact and its consequences
Charge in a metal stops moving only when the field inside vanishes, and everything else follows from that.
Metals turned up in the Gauss's law week as an application; they turn up in exams as the thing that changes the answer, so they get a block of their own here.
TheoremTheorem 5.5: what equilibrium forces on a conductor
Conditions
the conductor has been left alone long enough for the charges to stop moving, which is the meaning of electrostatic equilibrium
the statements are about the metal itself, not about cavities inside it or the space around it
$\sigma$ in the last statement is the local surface density at the point you are looking at, not the total charge over the total area, unless the surface is a sphere
Free charges keep moving until there is nothing left to push them, so the field in the metal is zero. Since the field is the slope of the potential, a zero field means the potential does not change anywhere in the metal, so the whole lump is at one value. Since the field is zero, Gauss's law applied to any small surface buried in the metal says there is no charge in there, so all the excess has gone to the surfaces. And a flat box straddling the surface, with no flux on the inside face, gives the field just outside as the local surface density over epsilon nought.
Proof
Suppose the field inside the metal were not zero somewhere. Free electrons there would feel a force and move, which contradicts the assumption that nothing is moving. So $\vec E = 0$ throughout the metal.
Take any two points $a$ and $b$ in the metal. Then $V_b - V_a = -\int_a^b \vec E\cdot d\vec l = 0$, so the whole conductor is one value of $V$.
Take any closed surface entirely inside the metal. The flux through it is zero because $\vec E$ is zero on it, so $Q_{\rm enc} = 0$: no net charge hides in the bulk.
Take a flat box with one face just inside and one just outside a piece of the surface, of area $A$. The inside face gets no flux, the sides get none because the outside field is perpendicular to the surface, and the outside face gets $EA$. So $EA = \sigma A/\varepsilon_0$.
A charge sealed in a inside a metal shell. The inner wall grabs exactly enough charge to hide the contents from the metal, and whatever is left over is pushed to the outer wall, where the outside world sees it.
Looks like this, but is not
The field inside the metal is zero, so the potential inside the metal is zero. It reads like a direct consequence, and the first half of it is certainly true.
Zero field means the potential does not change, not that it is zero. A metal ball of radius $4.00\ \mathrm{cm}$ carrying $+12.0\ \mathrm{nC}$ has $E = 0$ everywhere inside and $V = kQ/R = 2.70\times10^{3}\ \mathrm{V}$ everywhere inside, including the exact centre. A constant is only zero if something makes it zero, and here nothing does.
r (cm)
plastic ball, E (N/C)
metal ball, E (N/C)
plastic V (V)
metal V (V)
0
0
0
4045
2697
2.00
3.37 × 10⁴
0
3708
2697
4.00
6.74 × 10⁴
6.74 × 10⁴
2697
2697
10.0
1.08 × 10⁴
1.08 × 10⁴
1079
1079
Both balls have radius 4.00 cm and carry 12.0 nC. From 4.00 cm outwards the two objects are indistinguishable, which is the shell theorem doing its work. Inside, the metal is dead flat at 2697 V while the plastic climbs to 4045 V at the centre, and the metal reaches a potential that is a full third lower than the plastic ball's centre for exactly the same charge.
Charge in a cavity of a shell that carries -2.00 nC of its own
A point charge of $+6.00\ \mathrm{nC}$ sits at the centre of a spherical cavity inside a metal shell. The cavity has radius $a = 3.00\ \mathrm{cm}$, the shell's outer radius is $b = 5.00\ \mathrm{cm}$, and the shell itself carries a net charge of $-2.00\ \mathrm{nC}$. Find the charge on each surface, the field at $8.00\ \mathrm{cm}$ from the centre, and the potential of the shell.
Given
point charge $+6.00\ \mathrm{nC}$ at the centre of the cavity
shell from $a = 3.00\ \mathrm{cm}$ to $b = 5.00\ \mathrm{cm}$, net charge $-2.00\ \mathrm{nC}$
field point at $r = 8.00\ \mathrm{cm}$ from the centre
Find
the two surface charges, the field at 8.00 cm and the potential of the metal
Solution
The conductor rule rather than Gauss's law used blindly, because the useful fact is not the flux through some surface but that the field inside the metal is already known to be zero.
Let the zero field inside the metal do the bookkeeping
$$Q_{\rm enc} = 0\ \text{for a sphere drawn inside the metal}$$
the field is zero on that sphere, so the flux through it is zero, so whatever it encloses must add to nothing
$$q_{\rm inner} = -6.00\ \mathrm{nC}$$
it encloses the point charge and the inner wall, and those two have to cancel
Independent check through the surface density. The outer surface carries $4.00\ \mathrm{nC}$ over $4\pi b^{2}$, giving $\sigma = 1.27\times10^{-7}\ \mathrm{C/m^{2}}$, and $\sigma/\varepsilon_0 = 1.44\times10^{4}\ \mathrm{N/C}$. The point charge formula at $r = b$ gives $kQ/b^{2} = 1.44\times10^{4}\ \mathrm{N/C}$ as well. That density result also checks the potential without reusing our division: for a sphere $V(b) = E(b)\,b = (1.44\times10^{4})(0.0500) = 7.2\times10^{2}\ \mathrm{V}$, which is the $719\ \mathrm{V}$ above. Two different laws, one number, and the volts fall out of the same density.
No integrals and no calculus of any kind; three applications of one idea and two divisions.
The three step pattern here is worth memorising as a pattern: inner wall cancels the cavity contents, outer wall takes the remainder, outside world sees only the total.
Two metal spheres joined by a wire share 40.0 nC unequally
A metal sphere of radius $8.00\ \mathrm{cm}$ and another of radius $2.00\ \mathrm{cm}$ stand far apart and are joined by a long thin wire. A total charge of $+40.0\ \mathrm{nC}$ is put on the pair. How does the charge divide, and where is the surface field larger?
Given
spheres of radius $R_1 = 8.00\ \mathrm{cm}$ and $R_2 = 2.00\ \mathrm{cm}$
joined by a wire, far enough apart that neither disturbs the other
total charge $+40.0\ \mathrm{nC}$
Find
the charge on each sphere and the field at each surface
Solution
We start from equal potentials rather than from an equal split of charge, because the wire is what fixes $V$ and the charges are free to move; assuming a half and half split would be answering a different problem.
The wire makes them one conductor, so it fixes the potential
Check through the surface densities instead of the fields. $\sigma_1 = Q_1/4\pi R_1^{2} = 3.98\times10^{-7}$ and $\sigma_2 = 1.59\times10^{-6}\ \mathrm{C/m^{2}}$, a ratio of four again, and $\sigma/\varepsilon_0$ reproduces both surface fields to three figures.
Two equations and two divisions, once the wire has been read as the statement that the potentials are equal.
Sharp points on a conductor are small spheres in disguise, which is why the field is largest there and why lightning conductors are pointed rather than rounded.
Checkpoint
§05.5 - charge induced on a cavity wall●●○○○
Thirty seconds. A solid block of metal has a hollow cavity drilled in it. A charge of $+5.00\ \mathrm{nC}$ is placed inside the cavity, not touching the walls, and the block as a whole is electrically neutral.
Given
neutral metal block with a cavity
charge $+5.00\ \mathrm{nC}$ inside the cavity, touching nothing
everything has settled to equilibrium
Find
(a) What is the total charge on the wall of the cavity?
Hint 1/4
Draw a closed surface that lies entirely within the metal and wraps around the cavity, and ask what the field is on it.
Hint 2/4
Inside a conductor in equilibrium the field is zero, so the flux through any surface buried in the metal is zero, so by Gauss's law the charge it encloses is zero.
Hint 3/4
That surface encloses the $+5.00\ \mathrm{nC}$ in the cavity together with whatever has collected on the cavity wall.
Hint 4/4
So the wall must carry $-5.00\ \mathrm{nC}$, and the outer surface of the block carries $+5.00\ \mathrm{nC}$ to keep the block neutral.
Show solution
A Gaussian surface buried in the metal, because the zero field there turns the whole question into one line of bookkeeping instead of a study of how the induced charge arranges itself.
Consistency with charge conservation: the block started neutral and no charge entered or left it, so the $-5.00$ on the wall must be balanced by $+5.00$ somewhere, and the only other surface available is the outside.
⚠ Reading zero field inside a conductor as zero potential
zero is the most available number and the field really is zero, so the second zero arrives by association
wrong$$V_{\rm inside\ metal} = 0$$
right$$V_{\rm inside\ metal} = V_{\rm surface}\ \text{, a constant fixed from outside}$$
⚠ Splitting charge between joined spheres by area
surface charge lives on the surface, so area feels like the natural thing to share it out by
Step a small distance in the $x$ direction and see how much the potential changes; divide by the step and flip the sign, and that is the $x$ component of the field. Repeat for the other two directions. Where the potential is flat the field is zero, and where it is steepest the field is largest.
A potential built from three straight pieces, and the field that goes with it. Reading the second graph off the first is a slope calculation and nothing else, and it is a standard exam question because it needs no formula at all.
Looks like this, but is not
The potential is large here, so the field is large here. It sounds like the sort of thing that ought to be true, and on a graph the tall part does look like where the action is.
The field is the slope, not the height. On the graph above, the potential is at its largest value of $60\ \mathrm{V}$ across the whole flat stretch from $2.0$ to $5.0\ \mathrm{cm}$, and the field there is exactly zero. Meanwhile at $x = 1.0\ \mathrm{cm}$ the potential is only $30\ \mathrm{V}$ and the field is $3.00\times10^{3}\ \mathrm{V/m}$, the largest anywhere in the picture.
Field at a point from V = 5.00xy + 2.00z squared
In a certain region the potential is $V = (5.00\ \mathrm{V/m^{2}})xy + (2.00\ \mathrm{V/m^{2}})z^{2}$, with $x$, $y$ and $z$ in metres. Find the electric field at the point $(1.00, 2.00, 1.50)\ \mathrm{m}$.
Given
$V = 5.00xy + 2.00z^{2}$ volts, coordinates in metres
field point $(1.00, 2.00, 1.50)\ \mathrm{m}$
Find
the field vector and its magnitude there
Solution
Differentiation rather than any charge based method, because we are handed V directly and no charge distribution is ever mentioned.
the magnitude, quoted to three figures like the data
Answer $$\boxed{\;\vec E = (-10.0,\ -5.00,\ -6.00)\ \mathrm{V/m},\qquad E = 12.7\ \mathrm{V/m}\;}$$
Check
Check one component numerically instead of symbolically. Step $1.00\ \mathrm{mm}$ each way in $x$: $V$ changes by $5.00\times2.00\times0.00200 = 0.0200\ \mathrm{V}$ over $0.00200\ \mathrm{m}$, a slope of $10.0\ \mathrm{V/m}$, so $E_x = -10.0\ \mathrm{V/m}$, matching the derivative.
Three derivatives and one square root, against a superposition integral that would need to know where the charges are, which the question never says.
A potential given as a formula is a gift: the field falls out without ever asking what produced it.
Reading the field and a turning point off a potential graph
The graph in this block shows $V$ along the $x$ axis: it rises linearly from $0$ at $x = 0$ to $60\ \mathrm{V}$ at $x = 2.0\ \mathrm{cm}$, stays at $60\ \mathrm{V}$ until $x = 5.0\ \mathrm{cm}$, then falls linearly to $-20\ \mathrm{V}$ at $x = 9.0\ \mathrm{cm}$. Find the field in each region, and find where an electron released from rest at $x = 1.0\ \mathrm{cm}$ turns around.
Given
$V = 0$ at $x = 0$, rising to $60\ \mathrm{V}$ at $x = 2.0\ \mathrm{cm}$
$V = 60\ \mathrm{V}$ from $2.0$ to $5.0\ \mathrm{cm}$, then falling to $-20\ \mathrm{V}$ at $x = 9.0\ \mathrm{cm}$
electron: $q = -1.60\times10^{-19}\ \mathrm{C}$, $m = 9.11\times10^{-31}\ \mathrm{kg}$, released from rest at $x = 1.0\ \mathrm{cm}$
Find
the field in the three regions and the
Solution
We read slopes straight off the graph rather than fit a formula for $V(x)$, because the field only needs the slope; the turning point then comes from energy conservation with no force calculation anywhere.
Independent check on the turning point using speed instead of position. At the top of the flat stretch the electron has fallen through $30\ \mathrm{V}$, so it carries $30\ \mathrm{eV} = 4.80\times10^{-18}\ \mathrm{J}$ and moves at $3.25\times10^{6}\ \mathrm{m/s}$. It then loses energy at $2.00\times10^{3}\ \mathrm{eV}$ per metre, so it travels $30/2000 = 0.015\ \mathrm{m}$ past $x = 5.0\ \mathrm{cm}$ before stopping, which is $6.5\ \mathrm{cm}$.
Three slopes and one linear equation, with no charges and no distances between charges anywhere.
An electron released on a hill runs uphill in potential, because its charge is negative. Whenever the particle is an electron, expect every direction argument to come out backwards from the positive case.
Checkpoint
§05.6 - field from two potential readings●○○○○
Thirty seconds with one division. Along a straight line the potential is measured at two points and found to change uniformly between them.
Given
$V = 200\ \mathrm{V}$ at $x = 2.00\ \mathrm{cm}$
$V = 140\ \mathrm{V}$ at $x = 5.00\ \mathrm{cm}$
the potential varies linearly between the two points
Find
(a) What is $E_x$ in that stretch?
Hint 1/4
You are being asked for a component with a sign, so decide which way the potential is going before dividing anything.
Hint 2/4
$E_x = -dV/dx$, and for a straight line the derivative is just the change in $V$ over the change in $x$.
Hint 3/4
Here $\Delta V = 140 - 200 = -60\ \mathrm{V}$ across $\Delta x = 0.0300\ \mathrm{m}$.
One difference quotient, since the potential is linear across the interval and the derivative is exactly the slope; nothing is gained by writing an integral.
Direction check without arithmetic: a positive charge released here should move towards lower potential, which is towards larger $x$, so $E_x$ must be positive. It is.
⚠ Dropping the minus sign in the gradient
the derivative is the visible work and the sign in front looks like decoration
wrong$$E_x = \frac{dV}{dx}$$
right$$E_x = -\frac{dV}{dx}$$
⚠ Using the value of V instead of its slope
on a graph the height is what the eye reads first
wrong$$E \propto V$$
right$$E = -\frac{dV}{dx}\ \text{, so }E=0\text{ wherever }V\text{ is flat}$$
⚠ Dividing by a distance in centimetres
the graph is labelled in centimetres and the number is sitting right there
5.7Energy: the shortest route to a speed or a distance
Two lines of energy bookkeeping replace every varying force problem in this block.
The map's cheapest arrow gets used properly here, on the family of questions that a force calculation cannot finish without an integral.
RuleRule 5.7: energy of a charge, and of a pair
Conditions
$U = qV$ measures the energy of one charge sitting in a potential produced by everything else, so it must not be used for the charge that produced the potential
$U = kq_1q_2/r$ is the energy of a pair, counted once; for three charges there are three pairs and each is counted once
with no other force acting, $\Delta K + \Delta U = 0$, and the path taken between the two points never enters
Multiply the charge by the potential where it sits and you have its energy. Let it move, and whatever potential energy it loses turns into kinetic energy. For a set of charges, add up one term for every pair, with the signs of both charges in each term.
The energy of a pair against their separation, drawn for both sign combinations. A particle fired at a like charge climbs the solid curve until it runs out of kinetic energy; where the horizontal line meets the curve is as close as it gets.
Looks like this, but is not
For three charges the energy is $U = q_1V_1 + q_2V_2 + q_3V_3$, adding each charge's energy in the potential of the others. Every term in it is correct on its own.
Each pair gets counted twice that way. The pair $q_1q_2$ appears once in $q_1V_1$ and once again in $q_2V_2$, so the sum comes out exactly twice too big. Either halve it, or do what is safer under exam conditions and write one term per pair: three charges give three terms, four charges give six.
r (cm)
V (V)
K (eV)
speed (m/s)
fraction of final speed
1.50
23973
0
0
0.00
2.00
17980
5993
1.07 × 10⁶
0.50
3.00
11987
11987
1.52 × 10⁶
0.71
6.00
5993
17980
1.86 × 10⁶
0.87
infinity
0
23973
2.14 × 10⁶
1.00
Half the final speed is reached after moving only 0.50 cm, and three quarters of it within a couple of centimetres, because the energy is delivered where the field is strong. The last three quarters of the journey is almost coasting, which is why answering this with a constant acceleration would be so badly wrong.
How close a 5.00 MeV alpha particle gets to a gold nucleus
An alpha particle carrying charge $+2e$ and kinetic energy $5.00\ \mathrm{MeV}$ is fired straight at a gold nucleus, charge $+79e$, heavy enough to treat as fixed. How close does it get before it stops and comes back?
head on collision, so the whole kinetic energy is available
Find
the
Solution
Energy conservation rather than force, because the question asks how far and not how long, and the force changes by many orders of magnitude during the approach.
Say what stopping means in energy terms
$$K_i + U_i = K_f + U_f$$
no other force acts, so the total is fixed throughout the approach
$$K_i + 0 = 0 + \frac{k(2e)(79e)}{r_{\min}}$$
the alpha starts effectively infinitely far away where U is zero, and stops at closest approach where K is zero
Independent route in electron volts, which avoids joules entirely. The potential of the nucleus at $r$ is $k(79e)/r$, so an alpha of charge $2e$ has energy $k\cdot158\,e/r$ electron volts, that is $2.27\times10^{-7}/r$. Setting that to $5.00\times10^{6}\ \mathrm{eV}$ gives $r = 4.55\times10^{-14}\ \mathrm{m}$ again.
One division. The force route would need the whole trajectory and would still not answer the question asked.
A nucleus is a few times $10^{-15}\ \mathrm{m}$ across, so at $4.55\times10^{-14}\ \mathrm{m}$ the alpha never reaches it, and treating both particles as point charges the whole way in was legitimate. That check is what makes the answer trustworthy rather than merely arithmetically correct.
Energy of three charges at the corners of a 10.0 cm triangle
Charges of $+5.00\ \mathrm{nC}$, $+5.00\ \mathrm{nC}$ and $-3.00\ \mathrm{nC}$ are held at the corners of an equilateral triangle of side $10.0\ \mathrm{cm}$. How much work would have been needed to assemble this arrangement from charges that started infinitely far apart?
equilateral triangle, side $a = 10.0\ \mathrm{cm}$
all three start infinitely far apart and at rest
Find
the total potential energy of the arrangement
Solution
We sum one term per pair rather than assemble the charges one at a time through potentials, because with three equal separations the common factor $k/a$ comes straight out and only three products are left.
Build it one charge at a time and see if the same number appears. Bringing in the first charge is free. The second costs $kq_1q_2/a = +2.25\times10^{-6}\ \mathrm{J}$. The third arrives at a place where the potential is $k(q_1+q_2)/a = 899\ \mathrm{V}$, so it costs $q_3\times899 = -2.70\times10^{-6}\ \mathrm{J}$. The total is $-4.50\times10^{-7}\ \mathrm{J}$, matching.
Three products and one division, and the order in which the charges are imagined to arrive makes no difference to the total.
A negative total means the arrangement is bound: you would have to put in $4.50\times10^{-7}\ \mathrm{J}$ to pull the three charges back to infinity.
Checkpoint
§05.7 - energy gained falling through 500 volts●○○○○
Thirty seconds, and the second half is a unit conversion you should be able to do in your head. An alpha particle, charge $+2e$, starts from rest and moves through a potential drop of $500\ \mathrm{V}$.
Given
charge $q = +2e = 3.20\times10^{-19}\ \mathrm{C}$
potential drop $\Delta V = -500\ \mathrm{V}$ along its path
starts from rest, no other force acts
Find
(a) How much kinetic energy has it gained, in electron volts and in joules?
Hint 1/4
Do not reach for the mass. Kinetic energy gained depends on the charge and the potential drop only; the mass would only be needed for a speed.
Hint 2/4
$\Delta K = -q\,\Delta V$, and one electron volt is the energy one elementary charge gains falling through one volt.
Hint 3/4
Here the charge is two elementary charges and the drop is $500\ \mathrm{V}$.
Hint 4/4
So $\Delta K = 2\times500 = 1000\ \mathrm{eV}$, which is $1000\times1.60\times10^{-19} = 1.60\times10^{-16}\ \mathrm{J}$.
Show solution
We count in electron volts and convert once at the end, because the charge is a whole number of $e$; that keeps the powers of ten out of the middle of the arithmetic, where they are easiest to lose.
Count in electron volts first, then convert once
$$\Delta K = -q\,\Delta V = (2e)(500\ \mathrm{V}) = 1000\ \mathrm{eV}$$
working in electron volts keeps the arithmetic to one multiplication and postpones the awkward power of ten
$$\Delta K = (1000)(1.60\times10^{-19}) = 1.60\times10^{-16}\ \mathrm{J}$$
one conversion at the end, which is also the only place a power of ten error can enter
Answer $$\boxed{\;\Delta K = 1.00\times10^{3}\ \mathrm{eV} = 1.60\times10^{-16}\ \mathrm{J}\;}$$
Check
Cross check on the size of the unit itself: an electron falling through one volt gains $1.60\times10^{-19}\ \mathrm{J}$ by definition, so a thousand electron volts must be a thousand times that.
⚠ Double counting pairs in the energy of a group
adding $qV$ for each charge feels symmetric and complete, and every individual term is correct
the force between the same two charges does have $r^{2}$ in it, and the two formulas sit next to each other in memory
wrong$$U = \frac{kq_1q_2}{r^{2}}$$
right$$U = \frac{kq_1q_2}{r}$$
⚠ Getting the sign of the energy change backwards
$\Delta V$ and $\Delta U$ have the same sign only when the charge is positive, and the exam question is usually about an electron
wrong$$\Delta K = +q\,\Delta V$$
right$$\Delta K = -q\,\Delta V$$
Deciding what to compute, in the first thirty seconds
Every question in this block, before the pen touches the paper. It is worth doing out loud in your head even when the answer feels obvious, because the questions that feel obvious are the ones that punish the wrong route hardest.
Read the last line first
The final sentence names the quantity you are being marked on. A speed, an energy or an amount of work sends you to the potential. A field or a force sends you to the field. A flux sends you straight to the enclosed charge and nowhere else.
Count the objects and name their symmetry
One sphere, one long cylinder or one large sheet means a closed surface will work. Two or more objects, or one object with no symmetry, means a sum or an integral.
Look for the word conducting
If the object is a metal, half the work is already done: no field inside it, all the charge on its surfaces, one value of the potential throughout. If it is an insulator, none of that applies.
Choose the zero and the axes, and write them down
For a bounded charge distribution take $V=0$ at infinity. Pick which direction is positive and say so in one line. Half of all sign errors are decided here, before any physics happens.
Write the standard result for each object present
Do not rederive anything that is in the summary table. Write the formula down with symbols first and put the numbers in afterwards; it makes the check in step six possible.
Combine, then check before moving on
Vectors add by components, potentials add as signed numbers, energies add one term per pair. Then run the checklist in the next box.
Where it goes wrong
Starting to compute a field when the question asked for a speed, which costs an integral you did not need.
Applying Gauss's law to a surface where the field is not constant, which produces a clean looking number that is simply wrong.
Using an insulating sphere formula on a metal one, or the other way round.
Never writing down where the zero of potential is, and then losing the sign of the answer with no way to find it again.
Five checks that catch a wrong electrostatics answer
On every numerical answer, before you write the box around it. Between them these five catch most of what goes wrong in this material, and each one takes a few seconds.
Units
A field must come out in $\mathrm{N/C}$, which is the same as $\mathrm{V/m}$. A potential must come out in volts. If a length is left over or missing, one power of $r$ is in the wrong place.
Push a parameter to an extreme where you already know the answer. Far from any bounded object the field must become $kQ/r^{2}$; at the centre of a symmetric arrangement the field must vanish; at the surface the inside and outside formulas must agree.
Sign and direction
Does the field point away from positive charge? Does a positive charge released from rest move towards lower potential? Does the potential of a negative charge come out negative?
Order of magnitude
A nanocoulomb at a few centimetres gives kilovolts and tens of kilonewtons per coulomb. A proton falling through a kilovolt reaches a few times $10^{5}\ \mathrm{m/s}$. If your answer is a thousand times off one of those, something is wrong before the algebra.
A second route
Get the same number a different way, even roughly. Field from $\sigma/\varepsilon_0$ and from $kQ/r^{2}$; energy in joules and in electron volts; potential by summing charges and by integrating the field. Agreement between two routes is the only check that is genuinely independent.
Where it goes wrong
Repeating the same calculation and calling it a check, which confirms arithmetic and nothing else.
Checking units on the final line only, when the power of $r$ went wrong three lines earlier and the units happened to survive.
Skipping the order of magnitude check because the number came out of a calculator and therefore looks authoritative.
Midpoint of two equal positive charges
Two charges of $+6.00\ \mathrm{nC}$ are fixed $10.0\ \mathrm{cm}$ apart. Find the field and the potential at the midpoint of the line joining them.
Given
two charges of $+6.00\ \mathrm{nC}$
separation $10.0\ \mathrm{cm}$, midpoint is $5.00\ \mathrm{cm}$ from each
Find
the field and the potential at the midpoint
Solution
We work out the field and the potential in the same breath instead of in two separate calculations, because the contrast is the point: identical geometry, zero for one quantity and not for the other.
no directions exist to cancel, so two positive charges give two positive contributions
Answer $$\boxed{\;E = 0,\qquad V = 2.16\times10^{3}\ \mathrm{V}\;}$$
Check
Sanity check by imagining the work: sliding a positive test charge in from infinity to a spot squeezed between two positive charges plainly costs energy, so $V$ has to be positive and large, while at the exact midpoint there is no net push, so $E$ is zero. Both conclusions come from the picture without arithmetic.
Midpoint of two equal and opposite charges
Now make one of them negative: $+6.00\ \mathrm{nC}$ and $-6.00\ \mathrm{nC}$, still $10.0\ \mathrm{cm}$ apart. Find the field and the potential at the same midpoint.
Given
charges $+6.00\ \mathrm{nC}$ and $-6.00\ \mathrm{nC}$
separation $10.0\ \mathrm{cm}$, midpoint is $5.00\ \mathrm{cm}$ from each
Find
the field and the potential at the midpoint
Solution
We reuse the magnitudes from the previous case and change only signs and directions, because no distance moved; recomputing from scratch would bury the one thing that actually changed.
Field: two arrows, now pointing the same way
$$E_1 = E_2 = 2.16\times10^{4}\ \mathrm{N/C}$$
the magnitudes are unchanged, because only a sign has moved
Answer $$\boxed{\;E = 4.32\times10^{4}\ \mathrm{N/C},\qquad V = 0\;}$$
Check
The zero is not an accident of the midpoint: the potential vanishes at every point of the plane that perpendicularly bisects the pair, because every point on it is equidistant from the two charges. The field is nowhere near zero on that plane.
Flipping one sign moves the zero from the field to the potential and leaves everything else in the picture untouched.
How to tell them apart
Ask what is being summed. Arrows cancel when they point opposite ways, which happens between like charges. Numbers cancel when they have opposite signs, which happens between unlike charges. The two conditions are not just different, they are opposites, which is why the zeros never coincide.
Sphere of radius 10.0 cm with the charge at its centre
A point charge of $+8.00\ \mathrm{nC}$ sits at the centre of an imaginary sphere of radius $10.0\ \mathrm{cm}$. Find the flux out of the sphere and the field on it.
Given
$q = +8.00\ \mathrm{nC}$ at the centre
imaginary sphere of radius $10.0\ \mathrm{cm}$
Find
the flux and the field on the surface
Solution
We take the flux from the law and only then the field from the symmetry, because the law on its own never delivers $E$; the symmetry is what licenses the second step, and the next example removes it.
the field has the same size at every point of this sphere, because nothing in the arrangement distinguishes one point of it from another, so E comes out of the integral
Answer $$\boxed{\;\Phi = 904\ \mathrm{N\,m^{2}/C},\qquad E = 7.19\times10^{3}\ \mathrm{N/C}\;}$$
Check
The field agrees with the direct point charge result $kq/r^{2} = 71.92/0.0100 = 7.19\times10^{3}\ \mathrm{N/C}$, which is the check that the flux was shared out correctly.
The same sphere with the charge 4.00 cm off centre
Keep the same sphere and the same charge, but slide the charge $4.00\ \mathrm{cm}$ away from the centre, still inside. What happens to the flux, and what happens to the field on the surface?
Given
$q = +8.00\ \mathrm{nC}$, now $4.00\ \mathrm{cm}$ from the centre
same imaginary sphere of radius $10.0\ \mathrm{cm}$
Find
the flux, and whether the field can still be extracted
Solution
We evaluate the field at the nearest and the farthest point rather than attempt the surface integral, because two numbers differing by a factor of five already prove $E$ cannot be pulled outside it.
at the far point, 14.0 cm away; the field varies by a factor of more than five over the surface, so it cannot be pulled out of the integral
Answer $$\boxed{\;\Phi = 904\ \mathrm{N\,m^{2}/C}\ \text{unchanged, but }E\text{ is not obtainable from it}\;}$$
Check
Consistency check on the two extreme values: the average of the field over the surface, weighted by area, still has to reproduce the same total flux. The near and far values straddle the uniform $7.19\times10^{3}\ \mathrm{N/C}$ of the centred case, as they must.
Moving the charge changes the field at every point of the surface and changes the flux through it not at all.
How to tell them apart
Gauss's law is always true and only sometimes useful. It becomes useful when you can name a symmetry that forces $E$ to have the same value everywhere on your surface. If you cannot say that sentence out loud about your surface, the law will give you a flux and nothing more.
Scaffolding comes off
The common skeleton
Name the quantity the question ends with, and let it pick the route.
State where $V = 0$ and which direction counts as positive.
Write the standard result for each object in the picture, with symbols before numbers.
Combine: vectors by components, potentials as signed numbers, energies one term per pair.
Put the numbers in, in SI units, and keep three significant figures.
Check units, a limiting case, the sign, the order of magnitude and one independent route.
1 · fully worked
Speed of a proton pulled in from 25.0 cm to a sphere of radius 5.00 cm
A sphere of radius $5.00\ \mathrm{cm}$ carries $-20.0\ \mathrm{nC}$ spread evenly over its surface and is held fixed. A proton is released from rest $25.0\ \mathrm{cm}$ from the centre. How fast is it moving when it reaches the surface?
Given
sphere: radius $5.00\ \mathrm{cm}$, charge $-20.0\ \mathrm{nC}$, held fixed
released from rest at $r_i = 25.0\ \mathrm{cm}$, arrives at $r_f = 5.00\ \mathrm{cm}$
Find
its speed on arrival
Solution
We go through the potential rather than through the force, because the force grows by a factor of twenty five along the path and would need an integral, while the potential difference is two divisions.
Route and conventions
$$\text{question ends in a speed} \Rightarrow \text{use } V$$
the force grows by a factor of twenty five on the way in, so a force calculation would need an integral we can avoid
$$V = 0 \text{ at infinity, } V = \frac{kQ}{r} \text{ for } r \ge R$$
outside a spherical charge the potential is that of a point charge at the centre, which is the standard result being quoted rather than rederived
In electron volts the proton fell through $2.88\times10^{3}\ \mathrm{V}$, so it carries $2.88\ \mathrm{keV}$, and the shortcut $v = 1.384\times10^{4}\sqrt{K/\mathrm{eV}}$ gives $7.42\times10^{5}\ \mathrm{m/s}$. The speed is a quarter of a per cent of $c$, so no relativity is needed.
Two divisions, one subtraction and one square root.
Every step here is reusable. The next three rungs use the same six lines with less and less of the scaffolding printed for you.
2 · you write the reasoning
Easier than the rung above, because there is only one point to evaluate the potential at and no subtraction of two potentials. How much work must you do to bring a proton from far away to a point $3.00\ \mathrm{cm}$ from a fixed $+15.0\ \mathrm{nC}$ charge, arriving at rest? The steps are given; supply the reason for each one before opening the model answers.
reasoning
Work and energy are what the potential exists to deliver; a force calculation would need an integral from infinity to 3.00 cm and would give the same number after much more effort.
reasoning
The source is a single point charge and the zero of potential is at infinity, so the standard result applies directly with no integration and no sign to worry about, since the charge is positive.
reasoning
The proton arrives at rest, so its kinetic energy does not change and all the work you do goes into potential energy: $W = \Delta U = q\,\Delta V$, with $\Delta V$ measured from infinity where $V$ is zero. Quoting the answer in keV as well makes the size of it recognisable.
3 · find the buried error
Harder than the rung above, because the sign of the source and the identity of the particle both matter, and two different formulas are in play. A solid non conducting ball of radius $6.00\ \mathrm{cm}$ carries $-30.0\ \mathrm{nC}$ spread evenly. An electron is released from rest at the surface. How fast is it moving when it is $18.0\ \mathrm{cm}$ from the centre? Below is a student's solution. Exactly two of its four steps are wrong. Find them.
the two buried errors (2)
⚠ step 2
The potential of a point charge is $kQ/r$, not $kQ/r^{2}$; the field formula has been used where the potential formula belongs.
The two formulas differ by one power of $r$ and the field version is the one practised most, so the hand writes it automatically. It also survives a units check if the units are only looked at on the last line.
The mass used is the proton mass. The particle in this question is an electron, which is about eighteen hundred times lighter.
The previous problem in almost every problem set involves a proton, and the number $1.67\times10^{-27}$ is already written on the page. Nothing in the algebra flags it.
right
Use $m_e = 9.11\times10^{-31}\ \mathrm{kg}$, which raises the speed by a factor of about forty three for the same energy.
4 · the bare problem
§05.7 - proton released at the surface of a charged sphere●●●○○
No scaffolding this time, and two parts rather than one. A small sphere of radius $1.50\ \mathrm{cm}$ carries $+40.0\ \mathrm{nC}$ spread evenly over its surface. A proton is released from rest right at the surface and flies away.
Given
sphere: radius $1.50\ \mathrm{cm}$, charge $+40.0\ \mathrm{nC}$, held fixed
(a) How fast is the proton moving when it is very far away?
(b) At what distance from the centre is it moving at half that final speed?
Hint 1/4
Part (a) is a fall through a potential difference with the finish line at infinity. Part (b) asks where a given fraction of that energy has been delivered, so write the kinetic energy as a function of $r$ first.
Hint 2/4
$V = kQ/r$ outside the sphere, $\Delta K = -q\,\Delta V$, and $K \propto v^{2}$, so half the final speed means a quarter of the final kinetic energy.
Hint 3/4
With $Q = 40.0\ \mathrm{nC}$ and $R = 1.50\ \mathrm{cm}$, the surface potential is $kQ/R = 2.40\times10^{4}\ \mathrm{V}$, and at radius $r$ the proton has fallen through $kQ(1/R - 1/r)$.
Hint 4/4
So $v_\infty = 2.14\times10^{6}\ \mathrm{m/s}$, and setting $1/R - 1/r = (1/R)/4$ gives $r = 2.00\ \mathrm{cm}$.
Show solution
Same energy route as the earlier rungs, but we keep $K(r)$ as a function instead of plugging numbers in early, so part (b) becomes one equation rather than a second full calculation.
the energy released between the surface and radius r, written as a function so that part (b) is one equation
$$\frac{1}{R} - \frac{1}{r} = \frac{1}{4R}$$
half the speed means a quarter of the kinetic energy, since K goes as v squared
$$\frac{1}{r} = \frac{3}{4R} \Rightarrow r = \frac{4R}{3} = 2.00\ \mathrm{cm}$$
clean because the fraction was chosen to be a quarter
Answer $$\boxed{\;v_\infty = 2.14\times10^{6}\ \mathrm{m/s},\qquad r = 2.00\ \mathrm{cm}\;}$$
Check
Two checks. In electron volts the proton falls through $24.0\ \mathrm{kV}$, and $v = 1.384\times10^{4}\sqrt{23973} = 2.14\times10^{6}\ \mathrm{m/s}$. And part (b) can be checked directly: at $r = 2.00\ \mathrm{cm}$ the potential is $1.80\times10^{4}\ \mathrm{V}$, so the proton has fallen through $5.99\times10^{3}\ \mathrm{V}$, exactly a quarter of $2.40\times10^{4}\ \mathrm{V}$.
Most of the speed is picked up in the first centimetre. That is the general shape of every inverse square escape problem, and it is why a constant acceleration estimate always overshoots.
Full exam-style question
A charged ball inside a charged metal shell: fields, potentials and a released electronexam format
A solid non conducting ball of radius $a = 2.00\ \mathrm{cm}$ carries $Q_1 = +9.00\ \mathrm{nC}$ spread evenly through its volume. It sits at the centre of a metal shell whose inner radius is $b = 4.00\ \mathrm{cm}$ and whose outer radius is $c = 6.00\ \mathrm{cm}$. The shell carries a net charge of $Q_2 = -5.00\ \mathrm{nC}$. Find the field at $1.00$, $3.00$, $5.00$ and $8.00\ \mathrm{cm}$ from the centre; the charge on each surface of the shell; the potential at $8.00\ \mathrm{cm}$, the potential of the shell and the potential at the centre; and the speed of an electron released from rest at $8.00\ \mathrm{cm}$ when it reaches the shell.
Given
insulating ball: $a = 2.00\ \mathrm{cm}$, $Q_1 = +9.00\ \mathrm{nC}$, uniform through the volume
metal shell: $b = 4.00\ \mathrm{cm}$, $c = 6.00\ \mathrm{cm}$, net charge $Q_2 = -5.00\ \mathrm{nC}$
electron: $q = -1.60\times10^{-19}\ \mathrm{C}$, $m = 9.11\times10^{-31}\ \mathrm{kg}$, released from rest at $8.00\ \mathrm{cm}$
Find
four fields, two surface charges, three potentials and one speed
Solution
Gauss's law for the four fields because every region is spherically symmetric, then the conductor rule for the surfaces, then a walk in from infinity for the potentials, and only then energy for the electron. Doing the potentials before the fields would mean integrating a field you have not worked out yet.
walking in from b to a through the gap, where the field is that of Q1 alone; no contribution comes from crossing the metal because there is no field in it
the electron travels from 8.00 cm to the outer surface of the shell, and only the endpoints matter
$$\Delta K = -q\,\Delta V = (1.60\times10^{-19})(150) = 2.40\times10^{-17}\ \mathrm{J}$$
a negative charge moving towards higher potential loses potential energy, so it speeds up, which is also obvious from the outward field pulling it inwards
Three independent checks. First, the potential at the centre can be assembled a completely different way, as $\tfrac32 kQ_1/a - kQ_1/b + k(Q_1+Q_2)/c = 6068 - 2023 + 599 = 4.64\times10^{3}\ \mathrm{V}$, matching. Second, the field just outside the shell from the surface density: $\sigma = 4.00\ \mathrm{nC}/(4\pi c^{2}) = 8.84\times10^{-8}\ \mathrm{C/m^{2}}$ and $\sigma/\varepsilon_0 = 9.99\times10^{3}\ \mathrm{N/C}$, which equals $k(4.00\ \mathrm{nC})/c^{2}$. Third, the electron fell through $150\ \mathrm{V}$, and the standard shortcut $v = 5.93\times10^{5}\sqrt{V}$ gives $7.26\times10^{6}\ \mathrm{m/s}$.
Four regions, two surfaces, three potentials and one energy line. The only genuinely new work is the walk inwards for the potential; everything else is a substitution.
This is the standard shape of a full exam question on this block, and it is deliberately built so that a single wrong decision early on propagates. If the ball is treated as a conductor, the first field and the last potential both go wrong. If the shell's own charge is left out of the walk in from infinity, every potential comes out too large. Getting the four fields right first, then the surfaces, then the potentials, keeps the damage contained.
Practice
A · concept 4 questions
1§05.3 - same potential, same field?●●○○○
A one mark opener that separates the people who have read the definitions from the people who have collected the formulas. Two points in some arrangement of charges are found to be at exactly the same potential.
Given
two points, call them P and Q
$V_P = V_Q$, measured with the same zero
nothing else is known about the arrangement
Find
(a) True or false: the electric field must have the same value at P and at Q.
Hint 1/4
Ask what relation ties the two quantities together, and whether it involves the value of the potential or something else about it.
Hint 2/4
The field is minus the gradient of the potential. It is a statement about how $V$ changes from point to point, not about what $V$ is at a point.
Hint 3/4
Take a solid metal ball with charge on it. Every point inside and on it has the same potential, and the field is zero inside but $\sigma/\varepsilon_0$ just outside the surface.
Hint 4/4
So equal potentials say nothing at all about the fields, and the statement is false.
Show solution
A counterexample rather than a proof, because the claim says must; one object where two points share a potential and not a field is enough, and the conductor is the cheapest such object.
Find a counterexample in the most familiar object there is
$$V_{\rm inside} = V_{\rm surface}\ \text{everywhere in a charged conductor}$$
no field inside means no change in potential inside, so every interior point qualifies as a same potential pair
two points a hair apart, at the same potential to any accuracy you like, with completely different fields
Answer $$\boxed{\;\text{False}\;}$$
Check
The general reason, not just the example: the field is a derivative of the potential, and knowing the value of a function at two points says nothing about its slope at either of them.
Whenever a claim links $V$ and $\vec E$ by their values rather than by a derivative, suspect it.
2§05.2 - what zero net flux is allowed to tell you●●○○○
A closed surface is drawn somewhere in a region full of charges, and the net electric flux through it works out to be exactly zero.
Given
a closed surface of any shape
net flux through it is zero
charges may sit inside it, outside it, or both
Find
(a) Which conclusion follows?
Hint 1/4
Write down what the law actually equates, and then ask which of the four statements is a restatement of that equation and which are extra claims.
Hint 2/4
$\oint\vec E\cdot d\vec A = Q_{\rm enc}/\varepsilon_0$. The left side is a total over the whole surface and the right side is a net charge.
Hint 3/4
Zero on the left forces zero on the right, and that is the whole content: the enclosed charges add to nothing. Nothing is said about individual charges, nor about the field at any one point.
Hint 4/4
So the safe conclusion is that the net enclosed charge is zero.
Show solution
We read the law in one direction only and stop there, because the tempting second step, from zero enclosed charge to zero field, is exactly what a dipole inside the surface refutes.
the law is an equality between two totals, so a zero on one side puts a zero on the other and stops there
$$Q_{\rm enc} = 0 \nRightarrow \vec E = 0$$
a dipole inside the surface has zero net charge and a perfectly healthy field everywhere on it
Answer $$\boxed{\;Q_{\rm enc} = 0\;}$$
Check
Test the other candidates against the dipole example: it has zero net flux, a non zero field at every point of the surface, and plenty of charge inside. Only one of the four statements survives it.
Flux is an integral. Integrals vanish for two different reasons, and the interesting one is cancellation.
3§05.5 - where the charge sits on a metal object●●●○○
An irregularly shaped lump of metal, isolated from everything else, is given an excess charge of $+30\ \mathrm{nC}$ and left until nothing is moving any more.
Given
irregular solid metal object, isolated
excess charge $+30\ \mathrm{nC}$
electrostatic equilibrium has been reached
Find
(a) Which description of the final state is correct?
Hint 1/4
Start from the one fact that equilibrium gives you and derive the rest, rather than trying to remember four separate statements.
Hint 2/4
In equilibrium the field inside the metal is zero. Gauss's law on any surface buried in the metal then forces the enclosed charge to be zero as well.
Hint 3/4
That leaves only the outer surface for the charge to sit on. The shape is irregular, so nothing forces it to spread evenly, and the field just outside is $\sigma/\varepsilon_0$ with $\sigma$ measured locally.
Hint 4/4
So it sits on the surface, unevenly, and the whole lump is at one potential.
Show solution
We lean on the zero interior field twice, once for the charge and once for the potential, rather than reach for the spherical formulas, because the lump is irregular and there is no formula for $\sigma$ on it.
nothing in the argument forces the density to be uniform, and on an irregular shape it is not
Answer $$\boxed{\;\text{surface, uneven, one potential throughout}\;}$$
Check
Check the uneven part against a case you can compute: two spheres of different radii joined by a wire end up at one potential with surface densities in the ratio of the inverse radii. A single irregular lump is the same physics with the two ends joined by metal instead of by wire.
Equal potential and equal surface density are different statements. The first is forced, the second is not.
4§05.6 - potential inside a uniformly charged insulating ball●●○○○
A ball of glass of radius $R$ has charge spread evenly through its whole volume, and it is not a conductor, so nothing moves.
Given
solid insulating ball, radius $R$, charge $Q$ uniform through the volume
the interior points of interest are at radii $0 < r < R$
Find
(a) True or false: the potential is constant everywhere inside the ball.
Hint 1/4
Constant potential and zero field are the same statement. Decide whether the field inside this object is zero.
Hint 2/4
$E_x = -dV/dx$, so a constant potential over a region means no field anywhere in that region.
Hint 3/4
Inside a uniformly charged insulator, Gauss's law with a sphere of radius $r$ gives $E = kQr/R^{3}$, which is zero only at the exact centre.
Hint 4/4
A non zero field means a changing potential, so the statement is false: $V$ runs from $1.5\,kQ/R$ at the centre down to $kQ/R$ at the surface.
Show solution
We get the field first and integrate it, because unlike a conductor this ball has a field inside; there is no shortcut that makes a non constant field give a constant potential.
Get the field, then read the potential off it
$$E(r) = \frac{kQr}{R^{3}}\quad (r<R)$$
a Gaussian sphere at radius r encloses the volume fraction of the charge, which is why the field grows linearly rather than vanishing
$$V(r) = \frac{kQ(3R^{2}-r^{2})}{2R^{3}}$$
integrating that field in from the surface; a non constant field cannot give a constant potential
$$V(0) = \frac{3kQ}{2R} = 1.5\,V(R)$$
the centre sits fifty per cent above the surface value
Answer $$\boxed{\;\text{False: }V\text{ falls from }1.5\,kQ/R\text{ to }kQ/R\;}$$
Check
Check the two ends against each other: the formula at $r = R$ gives $kQ(3R^{2}-R^{2})/2R^{3} = kQ/R$, which is the outside result at the surface, so the two pieces join up correctly.
Constant potential inside is a property of metals. For insulators, expect the potential to keep climbing all the way to the centre.
B · computation 7 questions
1§05.3 - where E vanishes and where V vanishes on the same axis●●●○○
Two point charges are fixed on the $x$ axis: $+4.00\ \mathrm{nC}$ at the origin and $-16.0\ \mathrm{nC}$ at $x = 12.0\ \mathrm{cm}$. The two questions below have three different answers between them, which is the point of asking them together.
Given
$q_1 = +4.00\ \mathrm{nC}$ at $x = 0$
$q_2 = -16.0\ \mathrm{nC}$ at $x = 12.0\ \mathrm{cm}$
both questions are about points on the $x$ axis only
Find
(a) At which point on the axis is the total field zero?
(b) At which points on the axis is the total potential zero?
Hint 1/4
For part (a) decide first which stretch of the axis could possibly work, by asking where the two arrows point in each stretch. For part (b) there is no direction to argue about, so go straight to the algebra and expect more than one root.
Hint 2/4
Field: $k\vert q_1\vert/r_1^{2} = k\vert q_2\vert/r_2^{2}$ with the arrows opposed. Potential: $kq_1/r_1 + kq_2/r_2 = 0$, signs included.
Hint 3/4
With $\vert q_2\vert = 4\vert q_1\vert$ and a separation of $12.0\ \mathrm{cm}$: the field condition needs $r_2 = 2r_1$ and the potential condition needs $r_2 = 4r_1$.
Hint 4/4
Field zero at $x = -12.0\ \mathrm{cm}$; potential zero at $x = +2.40\ \mathrm{cm}$ and at $x = -4.00\ \mathrm{cm}$.
Show solution
The field zero is found with a region argument and the potential zeros with absolute values, because the potential has no directions to rule regions out with, so both branches survive and both have to be reported.
Field: rule out two thirds of the axis first
$$\text{between them and to the right of }q_2\text{: no cancellation}$$
between the charges both arrows point towards the negative charge, and to its right the larger charge is also the nearer one
on the left of the small charge, with distances r1 and r1 + 12.0 cm; the positive root is the only physical one
$$r_1 = 0.120\ \mathrm{m} \Rightarrow x = -12.0\ \mathrm{cm}$$
twelve centimetres to the left of the origin
Potential: solve without any direction argument
$$\frac{4.00}{\vert x\vert} = \frac{16.0}{\vert x - 0.120\vert} \Rightarrow \vert x - 0.120\vert = 4\vert x\vert$$
the two contributions have opposite signs, so their magnitudes must be equal; no components and no regions to worry about
$$0.120 - x = 4x \Rightarrow x = 0.0240\ \mathrm{m}$$
taking the branch between the charges, where both absolute values open positively
$$0.120 - x = -4x \Rightarrow x = -0.0400\ \mathrm{m}$$
taking the branch to the left of the origin, which is the second root
Answer $$\boxed{\;(a)\ x = -12.0\ \mathrm{cm};\quad (b)\ x = +2.40\ \mathrm{cm}\ \text{and}\ -4.00\ \mathrm{cm}\;}$$
Check
Substitute both potential roots back: at $x = -4.00\ \mathrm{cm}$ the distances are $4.00$ and $16.0\ \mathrm{cm}$, giving $899$ and $-899\ \mathrm{V}$; at $x = +2.40\ \mathrm{cm}$ they are $2.40$ and $9.60\ \mathrm{cm}$, giving $1498$ and $-1498\ \mathrm{V}$. Both cancel exactly.
Three special points, three different answers, one pair of charges. Any question that asks for the point where nothing happens has to say which quantity it means.
2§05.4 - inside and outside a uniformly charged plastic ball●●○○○
A solid plastic ball of radius $5.00\ \mathrm{cm}$ carries $+6.00\ \mathrm{nC}$ spread evenly through its volume. It is an insulator, so the charge stays where it was put.
Given
radius $R = 5.00\ \mathrm{cm}$, charge $Q = +6.00\ \mathrm{nC}$, uniform through the volume
the two field points are at $r = 12.0\ \mathrm{cm}$ and $r = 2.50\ \mathrm{cm}$
$k = 8.99\times10^{9}\ \mathrm{N\,m^{2}/C^{2}}$
Find
(a) Find $E$ and $V$ at $r = 12.0\ \mathrm{cm}$.
(b) Find $E$ and $V$ at $r = 2.50\ \mathrm{cm}$.
Hint 1/4
Sort the two radii into outside and inside before writing anything, and remember that each region has its own pair of formulas.
Hint 2/4
Outside: $E = kQ/r^{2}$ and $V = kQ/r$. Inside a uniformly charged insulator: $E = kQr/R^{3}$ and $V = kQ(3R^{2}-r^{2})/2R^{3}$.
Hint 3/4
Here $kQ = 53.9\ \mathrm{N\,m^{2}/C}$, $R = 0.0500\ \mathrm{m}$, and the two radii are $0.120\ \mathrm{m}$ outside and $0.0250\ \mathrm{m}$ inside.
Hint 4/4
Outside: $3.75\times10^{3}\ \mathrm{N/C}$ and $450\ \mathrm{V}$. Inside: $1.08\times10^{4}\ \mathrm{N/C}$ and $1.48\times10^{3}\ \mathrm{V}$.
Show solution
We split at the surface and use the two standard results, because the enclosed charge changes character there; a single formula covering both radii does not exist.
The inside potential has to lie between the surface value $kQ/R = 1.08\times10^{3}\ \mathrm{V}$ and the central value $1.5\,kQ/R = 1.62\times10^{3}\ \mathrm{V}$, and $1.48\times10^{3}$ does. The surface field $kQ/R^{2} = 2.16\times10^{4}\ \mathrm{N/C}$ is larger than both computed fields, as it must be, since the field peaks there.
Do the region sorting first and the arithmetic second. Every wrong answer to this type comes from applying an outside formula to an inside point.
3§05.4 - potential difference near a long charged wire●●●●○
A very long straight wire carries $-2.40\ \mathrm{nC/m}$ evenly along its length. Because the wire is idealised as infinite, only differences of potential between two points are meaningful, and that is what is asked for.
Given
$\lambda = -2.40\ \mathrm{nC/m}$, wire treated as infinite
point A is $2.00\ \mathrm{cm}$ from the wire
point B is $8.00\ \mathrm{cm}$ from the wire
Find
(a) Find $V_A - V_B$.
Hint 1/4
There is no zero at infinity to lean on here, so the only route is to integrate the field between the two radii.
Here $2k\lambda = 2(8.99\times10^{9})(-2.40\times10^{-9}) = -43.2\ \mathrm{V}$ and $r_B/r_A = 8.00/2.00 = 4.00$, so $\ln 4 = 1.386$.
Hint 4/4
$V_A - V_B = (-43.2)(1.386) = -59.8\ \mathrm{V}$.
Show solution
There is no $kQ/r$ available here, since an infinite line has no usable zero at infinity, so we integrate the field between the two radii and pay for it with a logarithm.
Integrate the field, since infinity is not available
Sign check from the physics rather than the algebra: the wire is negative, so it attracts a positive test charge, so moving from $8.00\ \mathrm{cm}$ inwards to $2.00\ \mathrm{cm}$ the field does positive work, so the potential must drop. A negative answer is the only acceptable one.
The logarithm is the reason a line charge has no potential at infinity: $\ln r$ grows without limit, so there is nowhere sensible to put the zero. Quote differences only.
4§05.6 - field from a quadratic potential along a line●●●○○
Along the $x$ axis in some region the potential is measured to follow $V(x) = (150\ \mathrm{V/m^{2}})x^{2} - (80.0\ \mathrm{V/m})x$, with $x$ in metres.
Given
$V(x) = 150x^{2} - 80.0x$ volts, $x$ in metres
the region of interest is $0 \le x \le 0.500\ \mathrm{m}$
nothing is said about the charges producing this, and nothing needs to be
Find
(a) Find $E_x(x)$.
(b) Find the position where the field vanishes.
(c) Find the potential at that position.
Hint 1/4
You have $V$ as a formula, so the field is one differentiation away; do not go looking for charges.
Hint 2/4
$E_x = -dV/dx$, and a field of zero means a flat potential, which is a maximum or a minimum of $V$.
Hint 3/4
With $V = 150x^{2} - 80.0x$, the derivative is $dV/dx = 300x - 80.0$, and the data to put back in afterwards are the same two coefficients.
Hint 4/4
$E_x = 80.0 - 300x\ \mathrm{V/m}$, zero at $x = 0.267\ \mathrm{m}$, where $V = -10.7\ \mathrm{V}$.
Show solution
We differentiate rather than hunt for where $V$ vanishes, because the field is zero where the potential is flat, and those are two different questions that a hurried reader merges.
the minus sign in front is the whole physical content of the step, and the units come out as volts per metre because the coefficients were quoted per metre squared and per metre
Set it to zero and evaluate V there
$$80.0 - 300x = 0 \Rightarrow x = 0.267\ \mathrm{m}$$
the field vanishes where the potential is flat, not where the potential is zero
negative, and it is the lowest value of V in the region, since the coefficient of x squared is positive
Answer $$\boxed{\;E_x = 80.0 - 300x\ \mathrm{V/m},\quad x = 0.267\ \mathrm{m},\quad V = -10.7\ \mathrm{V}\;}$$
Check
Check the zero of the field by a different route: $V$ is a parabola with roots at $x = 0$ and $x = 80.0/150 = 0.533\ \mathrm{m}$, and a parabola turns exactly halfway between its roots, at $0.267\ \mathrm{m}$. No calculus needed for that one.
A place where $E = 0$ is a place where $V$ is stationary, not a place where $V$ is zero. Here the two are $0.267\ \mathrm{m}$ apart.
5§05.7 - energy to assemble four charges on a square●●●○○
Four charges of $+3.00\ \mathrm{nC}$ each are brought in from infinity, one at a time, and placed at the corners of a square of side $8.00\ \mathrm{cm}$.
Given
four charges, each $+3.00\ \mathrm{nC}$
square of side $a = 8.00\ \mathrm{cm}$
all four start infinitely far apart and at rest
Find
(a) How much work must be done in total?
Hint 1/4
Count the pairs before computing anything, and check the count against the picture: four corners, and every corner is joined to every other.
Hint 2/4
The energy of an arrangement is one term $kq_iq_j/r_{ij}$ per pair, each counted once. Four charges make six pairs.
Hint 3/4
Four of those six pairs are sides of length $8.00\ \mathrm{cm}$ and two are diagonals of length $8.00\sqrt2 = 11.3\ \mathrm{cm}$, and every product of charges is $(3.00\times10^{-9})^{2}$.
We count the six pairs rather than walk the charges in one by one through potentials, because the pair count is what keeps every interaction in the sum exactly once instead of twice.
Count pairs, not charges
$$\text{pairs} = \binom{4}{2} = 6:\ \text{four sides and two diagonals}$$
each pair is counted once, which is what stops the energy from coming out twice too big
Independent route, bringing them in one at a time: the first is free, the second costs $1.01\ \mu\mathrm{J}$, the third arrives next to two charges and costs $1.01(1 + 1/\sqrt2) = 1.73\ \mu\mathrm{J}$, and the fourth arrives next to three and costs $1.01(2 + 1/\sqrt2) = 2.74\ \mu\mathrm{J}$. The total is $5.48\ \mu\mathrm{J}$.
Six pairs for four charges, ten for five. The number grows faster than the number of charges, which is why these questions are usually kept to three or four.
6§05.7 - proton and alpha particle through the same voltage●●●○○
A proton and an alpha particle each start from rest and are accelerated through a potential difference of $800\ \mathrm{V}$. The alpha particle carries twice the charge and about four times the mass.
both start from rest and fall through $800\ \mathrm{V}$
Find
(a) Find the final speed of each.
(b) Find the ratio of the proton's speed to the alpha particle's.
Hint 1/4
The two particles gain different energies, because the energy depends on the charge. Work out each energy first and only then convert to a speed.
Hint 2/4
$\Delta K = -q\,\Delta V$ and $v = \sqrt{2K/m}$, so the speed goes as $\sqrt{q/m}$ for a fixed voltage.
Hint 3/4
For the proton $K = 800\ \mathrm{eV}$ and $m = 1.67\times10^{-27}\ \mathrm{kg}$; for the alpha $K = 1600\ \mathrm{eV}$ and $m = 6.64\times10^{-27}\ \mathrm{kg}$.
Hint 4/4
$v_p = 3.92\times10^{5}\ \mathrm{m/s}$, $v_\alpha = 2.78\times10^{5}\ \mathrm{m/s}$, and the ratio is $1.41$.
Show solution
We compute the two energies before either speed, because the same voltage does not mean the same energy: the alpha particle carries twice the charge, and skipping that step is the standard way this problem is lost.
Check the ratio symbolically before trusting the arithmetic: $v \propto \sqrt{q/m}$, so the ratio is $\sqrt{(e/m_p)/(2e/m_\alpha)} = \sqrt{m_\alpha/2m_p} = \sqrt{6.64/3.34} = 1.41$. It is only not exactly $\sqrt2$ because an alpha particle weighs slightly less than four protons.
Same voltage does not mean same energy, and same energy does not mean same speed. Two separate conversions, two separate places to lose a factor.
7§05.1 - a small dipole seen from two directions●●●●○
Two charges of $+3.00\ \mathrm{nC}$ and $-3.00\ \mathrm{nC}$ are held $4.00\ \mathrm{mm}$ apart. Both points asked about are $6.00\ \mathrm{cm}$ from the midpoint of the pair, which is fifteen times the separation, so the far field approximations are expected to be good.
point P is on the perpendicular bisector, $6.00\ \mathrm{cm}$ from the midpoint
point Q is on the line through both charges, $6.00\ \mathrm{cm}$ from the midpoint on the positive side
Find
(a) Find $V$ and $E$ at P.
(b) Find $V$ and $E$ at Q.
Hint 1/4
Handle the potentials first, because at one of the two points the answer needs no arithmetic at all. Then use the far field dipole expressions for the fields and check them against the exact sums.
Hint 2/4
$V = kq(1/r_+ - 1/r_-)$ exactly, and for $r \gg d$ the dipole results are $V = kp\cos\theta/r^{2}$ and, on the two special lines, $E = kp/r^{3}$ on the bisector and $E = 2kp/r^{3}$ on the axis, with $p = qd$.
Hint 3/4
Here $p = (3.00\times10^{-9})(4.00\times10^{-3}) = 1.20\times10^{-11}\ \mathrm{C\,m}$ and $r = 6.00\times10^{-2}\ \mathrm{m}$, so $kp = 0.1079\ \mathrm{V\,m^{2}}$.
Hint 4/4
At P: $V = 0$ and $E = 499\ \mathrm{N/C}$. At Q: $V = 30.0\ \mathrm{V}$ and $E = 1.00\times10^{3}\ \mathrm{N/C}$.
Show solution
We take $V$ exactly at both points and read $E$ off the dipole results, because an exact vector sum for the field would be two additions of nearly cancelling arrows, which is where the precision goes.
On the bisector, where one answer is free
$$V_P = \frac{kq}{r_+} - \frac{kq}{r_-} = 0$$
every point of the bisector is equidistant from the two charges, so the two numbers are equal and opposite; this is exact, not an approximation
exact sum, using 5.80 cm and 6.20 cm as the two distances
$$E_Q = \frac{2kp}{r^{3}} = 999\ \mathrm{N/C}$$
on the axis the two contributions point the same way and the far field result carries a factor two
Answer $$\boxed{\;P:\ V = 0,\ E = 499\ \mathrm{N/C};\qquad Q:\ V = 30.0\ \mathrm{V},\ E = 1.00\times10^{3}\ \mathrm{N/C}\;}$$
Check
Check both far field values against the exact sums. On the axis the exact field is $kq(1/0.0580^{2} - 1/0.0620^{2}) = 1.00\times10^{3}\ \mathrm{N/C}$, and the dipole formula gave $999\ \mathrm{N/C}$; on the bisector the exact sum gives $499\ \mathrm{N/C}$ against the formula's $499\ \mathrm{N/C}$. Agreement to a fraction of a per cent, which is what fifteen separations away should buy.
The dipole is the first place where $V$ and $E$ fall off at different rates from each other and from a point charge: $1/r^{2}$ and $1/r^{3}$ instead of $1/r$ and $1/r^{2}$.
C · exam level 4 questions
1§05.5 - potential inside a solid metal ball●●●●○
Exam length, one mark for the number and the rest for knowing why. A solid ball of metal with radius $4.00\ \mathrm{cm}$ carries $+12.0\ \mathrm{nC}$ and has been isolated long enough to settle.
the point of interest is $2.00\ \mathrm{cm}$ from the centre
Find
(a) What is the potential at that point?
Hint 1/4
Decide what the field is along the path from the surface to the point, and let that decide how the potential changes along it.
Hint 2/4
Inside a conductor $\vec E = 0$, so $V_b - V_a = -\int \vec E\cdot d\vec l = 0$ between any two interior points: the potential does not change, whatever its value.
Hint 3/4
That value is fixed at the surface by the outside world: $V(R) = kQ/R = (8.99\times10^{9})(12.0\times10^{-9})/0.0400$.
Hint 4/4
So the potential at $2.00\ \mathrm{cm}$ equals the surface value, $2.70\times10^{3}\ \mathrm{V}$.
Show solution
We walk in from infinity and stop where the field stops, because the potential inside the metal is fixed at the surface; there is nothing inside to integrate over.
Walk in from infinity and stop where the field stops
Check against the shape of the graph. For a conductor the potential is flat inside and falls as $1/r$ outside, so the maximum is at the surface and equals $kQ/R$. Any answer larger than $2.70\times10^{3}\ \mathrm{V}$ would put the interior above the maximum.
Two different spheres, two different answers: a plastic ball with the same charge would give $3.71\times10^{3}\ \mathrm{V}$ at this radius. The word metal is worth a thousand volts here.
2§05.5 - a student's argument about a metal ball●●●●○
A solid metal ball of radius $10.0\ \mathrm{cm}$ carries $-40.0\ \mathrm{nC}$. A student is asked for the potential $5.00\ \mathrm{cm}$ from the centre and writes three lines. Step 1: inside a conductor in equilibrium the electric field is zero. Step 2: so the potential inside is zero as well. Step 3: therefore $V = 0$ at $5.00\ \mathrm{cm}$.
the point of interest is $5.00\ \mathrm{cm}$ from the centre
the student's three steps are quoted in full above
Find
(a) Which step is the first one that is wrong, and what is the correct answer?
Hint 1/4
Check each step separately against a definition, rather than checking whether the final answer looks plausible.
Hint 2/4
$V_b - V_a = -\int_a^b \vec E\cdot d\vec l$. A zero field on the path makes the difference zero, which is a statement about two points, not about one.
Hint 3/4
With $Q = -40.0\ \mathrm{nC}$ and $R = 10.0\ \mathrm{cm}$ the surface potential is $kQ/R = (8.99\times10^{9})(-40.0\times10^{-9})/0.100$.
Hint 4/4
Step 2 is the first wrong one, and the correct answer is $V = -3.60\times10^{3}\ \mathrm{V}$.
Show solution
We test the given argument step by step instead of computing the right answer first, because the question asks where it breaks, and holding the correct number does not by itself locate the break.
Test step 1
$$\vec E = 0\ \text{inside a conductor in equilibrium}$$
correct, and it is the defining property of equilibrium; nothing to fix here
the value is fixed at the surface, where the field is not zero and the outside world has its say
Answer $$\boxed{\;\text{Step 2 is wrong};\quad V = -3.60\times10^{3}\ \mathrm{V}\;}$$
Check
Sign check, independent of the whole argument: the ball is negatively charged, so bringing a positive test charge in from infinity releases energy, so the potential everywhere in and on the ball must be negative. Zero was never a possible answer.
A zero in the middle column of the summary table almost never means a zero in the right hand one. That single sentence is worth several marks per exam.
3§05.2 - field just outside a charged metal plate●●●●○
A large isolated flat metal plate carries excess charge, and the charge density measured on its outer face is $45.0\ \mathrm{nC/m^{2}}$. You are asked for the field at a point a millimetre outside that face, well away from the edges.
Given
large flat metal plate, isolated
surface charge density on the outer face $\sigma = 45.0\ \mathrm{nC/m^{2}}$
There are two very similar looking results in your notes, one with a factor of two and one without. Decide which object you are standing next to before picking either.
Hint 2/4
A pillbox with one face inside the metal gets no flux on that face, because the field there is zero. So $EA = \sigma A/\varepsilon_0$ and $E = \sigma/\varepsilon_0$, with no two in it. The two only appears for a thin sheet with field on both sides.
Hint 3/4
With $\sigma = 45.0\ \mathrm{nC/m^{2}}$, $E = 45.0\times10^{-9}/8.85\times10^{-12}$.
Hint 4/4
$E = 5.08\times10^{3}\ \mathrm{N/C}$, perpendicular to the plate.
Show solution
We straddle the surface with a pillbox rather than quote $\sigma/2\varepsilon_0$, because one face of the box sits in the metal where the field is zero; that dead face is the entire factor of two.
Straddle the surface with a small box
$$\Phi = E A + 0 = \frac{\sigma A}{\varepsilon_0}$$
the outer face carries all the flux, the inner face carries none because the field inside the metal is zero, and the sides carry none because the field is perpendicular
Check the factor of two by a route that does not use the pillbox. The plate has charge on both faces, so a point just outside sees two sheets each of density $\sigma$: the near one gives $\sigma/2\varepsilon_0$ pushing out, and the far one gives another $\sigma/2\varepsilon_0$ in the same direction. They add to $\sigma/\varepsilon_0$, which is the same answer from a completely different picture.
Conductor face: no factor of two. Thin insulating sheet: factor of two. The difference is whether there is field on both sides of the surface you are straddling.
4§05.5 - potential inside the metal of a shell●●●●●
A point charge of $+9.00\ \mathrm{nC}$ sits at the centre of a neutral metal shell whose inner radius is $5.00\ \mathrm{cm}$ and whose outer radius is $8.00\ \mathrm{cm}$. You are asked for the potential at a point buried in the metal, $6.50\ \mathrm{cm}$ from the centre.
Given
point charge $+9.00\ \mathrm{nC}$ at the centre
neutral metal shell from $5.00\ \mathrm{cm}$ to $8.00\ \mathrm{cm}$
the point of interest is at $6.50\ \mathrm{cm}$, inside the metal
Find
(a) What is the potential there?
Hint 1/4
The metal is one equipotential, so find the potential anywhere on it and you have it everywhere on it. Pick the place where that is easiest.
Hint 2/4
Outside the whole assembly the potential is that of the total enclosed charge, so $V(r) = k(9.00\ \mathrm{nC})/r$ for $r \ge 8.00\ \mathrm{cm}$, and the outer surface is the easiest point on the conductor to evaluate.
Hint 3/4
The shell is neutral, so the total charge seen from outside is just the $9.00\ \mathrm{nC}$ at the centre, and the outer radius is $8.00\ \mathrm{cm}$.
We evaluate on the conductor, where the point charge formula still applies, and then carry the value inwards, because integrating directly to the buried point would mean building the field in the shell first.
Check that the answer is bracketed correctly. The potential falls monotonically outwards from the central charge, so the metal must sit below the value at its inner radius as computed from the centre charge alone, and above the value at $8.00\ \mathrm{cm}$ measured further out. It equals the second of those exactly, which is the flat step in the graph.
The shape of $V(r)$ here is: falling like $1/r$ from the centre out to $5.00\ \mathrm{cm}$, flat across the metal, then falling like $1/r$ again from $8.00\ \mathrm{cm}$ outwards. The flat step is what the metal contributes.
D · interleaved 4 questions
1§05.D - a charged rod seen end on●●●●○
This set is deliberately mixed, so decide for yourself which of the week's tools each one wants. A thin rod $12.0\ \mathrm{cm}$ long carries $8.00\ \mathrm{nC}$ spread evenly along it. A point lies on the line of the rod, $4.00\ \mathrm{cm}$ beyond its near end.
Given
rod length $L = 12.0\ \mathrm{cm}$, total charge $Q = 8.00\ \mathrm{nC}$, uniform
the field point is on the axis of the rod, $4.00\ \mathrm{cm}$ from the near end
$k = 8.99\times10^{9}\ \mathrm{N\,m^{2}/C^{2}}$
Find
(a) Find the potential at that point.
(b) Compare it with the value you would get by treating the rod as a point charge at its centre, and say which is larger and why.
Hint 1/4
A potential rather than a field is being asked for, so the integral has no components in it and no angles; it is a sum of numbers over the length of the rod.
Hint 2/4
$V = \int k\,dq/r$ with $dq = \lambda\,dx$ and $r$ running from $a$ to $a+L$, which gives $V = k\lambda\ln\frac{a+L}{a}$.
Hint 3/4
Here $\lambda = 8.00\ \mathrm{nC}/0.120\ \mathrm{m} = 6.67\times10^{-8}\ \mathrm{C/m}$, $a = 4.00\ \mathrm{cm}$ and $a + L = 16.0\ \mathrm{cm}$, so the ratio is $4.00$.
We integrate potentials rather than fields, because the elements sit at different distances but contribute plain numbers; the field version would need components and would still not answer what was asked.
treating the rod as a point charge at its midpoint, 10.0 cm away
Answer $$\boxed{\;V = 831\ \mathrm{V},\ \text{about 16 per cent above the point charge estimate}\;}$$
Check
Check the direction of the discrepancy without arithmetic. Half the charge sits closer than $10.0\ \mathrm{cm}$ and half sits further away, but $1/r$ rises faster as $r$ shrinks than it falls as $r$ grows, so the near half wins and the true potential must exceed the point charge value. It does.
A far away distribution can be collapsed to a point; a nearby one cannot. Here the point is only a third of the rod's length from its end, and the error is already sixteen per cent.
2§05.D - a hollow shell of charge, inside and outside●●●○○
A thin non conducting spherical shell of radius $7.00\ \mathrm{cm}$ carries $15.0\ \mathrm{nC}$ spread evenly over its surface. Nothing is inside it.
Given
thin shell, radius $R = 7.00\ \mathrm{cm}$, charge $Q = 15.0\ \mathrm{nC}$ uniform on the surface
one point is at $r = 3.00\ \mathrm{cm}$, the other at $r = 12.0\ \mathrm{cm}$
$k = 8.99\times10^{9}\ \mathrm{N\,m^{2}/C^{2}}$
Find
(a) Find $E$ and $V$ at $r = 3.00\ \mathrm{cm}$.
(b) Find $E$ and $V$ at $r = 12.0\ \mathrm{cm}$.
Hint 1/4
Two of the four numbers need no calculation at all once you have decided which region each point is in.
Hint 2/4
Inside a spherical shell there is no enclosed charge, so $E = 0$ and the potential is constant at its surface value $kQ/R$. Outside, both are the point charge results.
Hint 3/4
Here $kQ = (8.99\times10^{9})(15.0\times10^{-9}) = 135\ \mathrm{N\,m^{2}/C}$, with $R = 0.0700\ \mathrm{m}$ and the outer point at $0.120\ \mathrm{m}$.
We split at the shell radius and quote the two standard results, because the enclosed charge jumps there; deciding which side of $7.00\ \mathrm{cm}$ each radius is on is the whole question.
Inside, where the shell theorem does the work
$$Q_{\rm enc} = 0 \Rightarrow E = 0$$
a sphere of radius 3.00 cm encloses nothing, since all the charge is out at 7.00 cm
Check that the two regions join at the surface: the outside potential at $r = R$ gives $134.85/0.0700 = 1.93\times10^{3}\ \mathrm{V}$, which is exactly the constant value carried through the inside. The field, by contrast, jumps from $0$ to $kQ/R^{2} = 2.75\times10^{4}\ \mathrm{N/C}$ across the shell, as a surface charge requires.
A shell hides its charge from the inside completely as far as the field goes, and not at all as far as the potential goes. Both facts come from the same integral.
3§05.D - net force on the middle of three charges●●●○○
Three point charges sit on a line: $+2.00\ \mathrm{nC}$ at $x = 0$, $-3.00\ \mathrm{nC}$ at $x = 4.00\ \mathrm{cm}$, and $+5.00\ \mathrm{nC}$ at $x = 10.0\ \mathrm{cm}$.
Given
$q_1 = +2.00\ \mathrm{nC}$ at $x = 0$
$q_2 = -3.00\ \mathrm{nC}$ at $x = 4.00\ \mathrm{cm}$
$q_3 = +5.00\ \mathrm{nC}$ at $x = 10.0\ \mathrm{cm}$
Find
(a) Find the net electrostatic force on the middle charge, with its direction.
Hint 1/4
A force is asked for, so this is a vector sum and not an energy question. Along a line the vectors reduce to signs, but the signs still have to be argued from the picture.
Hint 2/4
$F = k\vert q_iq_j\vert/r^{2}$ for each pair, with the direction taken from the diagram: opposite signs attract, like signs repel.
Hint 3/4
The middle charge is negative, so it is pulled towards each of the positive charges: towards $-x$ by $q_1$ at $4.00\ \mathrm{cm}$, and towards $+x$ by $q_3$ at $6.00\ \mathrm{cm}$.
Hint 4/4
$F_1 = 3.37\times10^{-5}\ \mathrm{N}$ towards $-x$ and $F_3 = 3.75\times10^{-5}\ \mathrm{N}$ towards $+x$, so the net force is $3.75\times10^{-6}\ \mathrm{N}$ towards $+x$.
Show solution
This one is Coulomb's law, not potential: a force is asked for, so we add two vectors directly instead of building an energy and differentiating it back into a force.
One pair at a time, with the direction read off the picture
Check the near cancellation as a ratio before trusting the subtraction: $F_3/F_1 = (5.00/6.00^{2})/(2.00/4.00^{2}) = 0.1389/0.125 = 1.111$, so $F_3$ exceeds $F_1$ by exactly one ninth, and one ninth of $3.37\times10^{-5}$ is $3.75\times10^{-6}\ \mathrm{N}$.
When two contributions nearly cancel, compute the ratio rather than subtracting two rounded numbers. Here the difference is a tenth of each term, so rounding either one early would wreck the answer.
4§05.D - where a sheet and a point charge cancel●●●●●
A large thin non conducting sheet carries $80.0\ \mathrm{nC/m^{2}}$ spread evenly over it. A point charge of $+5.00\ \mathrm{nC}$ is held $20.0\ \mathrm{cm}$ from the sheet. Look along the line through the charge that runs perpendicular to the sheet.
Given
sheet: $\sigma = 80.0\ \mathrm{nC/m^{2}}$, large and thin, non conducting
point charge $+5.00\ \mathrm{nC}$, $20.0\ \mathrm{cm}$ from the sheet
the line of interest is perpendicular to the sheet and passes through the charge
Find
(a) Find the point on that line where the total field is zero.
(b) Say why there is no such point on the far side of the charge.
Hint 1/4
Sketch the two fields in each of the three regions of the line first, and find the one region where they can oppose each other at all.
Hint 2/4
The sheet gives $\sigma/2\varepsilon_0$ pointing away from it on both sides, and the point charge gives $kq/s^{2}$ pointing away from itself. Cancellation needs them antiparallel.
Hint 3/4
Between the sheet and the charge the sheet pushes towards the charge and the charge pushes back, so that is the only region. There $\sigma/2\varepsilon_0 = 80.0\times10^{-9}/(2\times8.85\times10^{-12}) = 4.52\times10^{3}\ \mathrm{N/C}$ and $kq = 45.0\ \mathrm{N\,m^{2}/C}$.
Hint 4/4
$s = \sqrt{45.0/4.52\times10^{3}} = 9.97\ \mathrm{cm}$ from the charge, which is $10.0\ \mathrm{cm}$ from the sheet.
Show solution
We rule two of the three regions out by direction first, because the sheet's field has the same magnitude everywhere and only its direction separates the regions; solving blind returns a distance in a region where nothing can cancel.
Rule out two of the three regions by direction alone
$$\text{beyond the charge: both fields point away from the sheet}$$
the sheet pushes outward and the charge pushes outward, so they reinforce and no cancellation is possible
$$\text{behind the sheet: both fields point away from the sheet again}$$
on that side the sheet field reverses and so does the direction from the charge, so once again they agree
measured from the charge towards the sheet; it lands inside the 20.0 cm gap, so the solution is legitimate
Answer $$\boxed{\;s = 9.97\ \mathrm{cm}\ \text{from the charge, i.e. }10.0\ \mathrm{cm}\ \text{from the sheet}\;}$$
Check
Substitute back: at that point the charge's field is $45.0/(0.0997)^{2} = 4.53\times10^{3}\ \mathrm{N/C}$ and the sheet's is $4.52\times10^{3}\ \mathrm{N/C}$, equal to within rounding and pointing opposite ways. The root also has to lie inside the gap, and $9.97 < 20.0$, so it does.
Always check that a root lands in the region you solved for. Drop the sheet to $15.0\ \mathrm{nC/m^{2}}$ and the algebra still returns a number, $23.0\ \mathrm{cm}$ from the charge, but that lands behind the sheet where the direction argument forbids it, so there would be no null point at all.
Mistake ledger (21 entries)
⚠ Putting the square in the potential
the field formula is the one you meet first and the hand writes it automatically
wrong$$V = \frac{kq}{r^{2}}$$
right$$V = \frac{kq}{r}$$
⚠ Taking the magnitude of the charge in a potential
in field problems you are trained to use $\vert q\vert$ and put the direction in by hand, and the habit carries over
wrong$$V = \frac{k\vert q\vert}{r}$$
right$$V = \frac{kq}{r}\ \text{, sign of }q\text{ kept}$$
⚠ Dropping the minus sign when going from V back to E
the derivative is the visible operation and the sign in front of it looks decorative
wrong$$E_x = \frac{dV}{dx}$$
right$$E_x = -\frac{dV}{dx}$$
⚠ Using Gauss's law on a shape with no symmetry
the law holds for every closed surface, so it feels applicable to every closed surface
wrong$$E = \frac{Q_{\rm enc}}{\varepsilon_0 A}\ \text{for any closed surface}$$
right$$E = \frac{Q_{\rm enc}}{\varepsilon_0 A}\ \text{only if }E\text{ is constant on }A$$
⚠ Going after the field when the question wanted a speed
the field is the quantity you practised most, so the hand starts drawing arrows before the question has been read to the end
wrong$$a = \frac{qE}{m}\ \text{, then kinematics with varying }E$$
right$$\Delta K = -q\,\Delta V$$
⚠ Treating a finite rod as an infinite line without checking
the two formulas look alike and the infinite one is shorter
wrong$$E = \frac{2k\lambda}{y}\ \text{for a rod of length }L$$
the speed shortcut is non relativistic and applies to a proton; for an electron the constant is $5.93\times10^{5}$ per root volt
Check yourself
Close the page and write out, from memory: the three objects and the four formulas that connect them; the three questions that decide which tool to use; the field and the potential for a ball inside and outside, for a long line and for a large sheet; the four consequences of equilibrium in a conductor; and the two energy equations. Then compare with the formula card and mark only what you missed.
Go from a charge distribution to $V$ and from $V$ back to $\vec E$, and say which formula you used for each leg?
c-map
Look at a question and say in one sentence which of the three routes it wants and why the other two cost more?
c-choose
Find the points where the field vanishes and the points where the potential vanishes for two unequal charges, and explain why they are different points?
c-superposition
Write down $E$ and $V$ inside and outside a charged ball, and beside a long line and a large sheet, without looking anything up?
c-standard
Work out the charge on each surface of a shell around a charge in its cavity, and give the potential of the metal?
c-conductors
Read the field off a potential graph, including where it is zero, and get the field from a potential given as a formula?
c-gradient
Compute a speed, a work and a distance of closest approach from energy conservation, and check the answer in electron volts?
c-energy
Glossary (16 terms)
potential differencepotansiyel farkı
The difference in potential between two points, written $V_a - V_b$: the energy per coulomb that the trip between them costs or pays. Only differences are ever measured, since the zero of potential is a matter of choice.
voltvolt
The SI unit of electric potential, equal to one joule per coulomb; a volt per metre is the same unit as a newton per coulomb.
eş potansiyel yüzey
A surface on which the potential has the same value everywhere, so that moving a charge along it costs no work; the field is always perpendicular to it.
potansiyel gradyanı
The rate at which the potential changes with position; its negative is the electric field, which is why a steep potential means a strong field.
yoldan bağımsızlık
The property that the work done by an electrostatic field between two points does not depend on the route taken, which is what allows a potential to be defined at all.
korunumlu kuvvet
A force whose work around any closed path is zero, so that a potential energy can be assigned to position alone; the electrostatic force is one.
electric potential energyelektriksel potansiyel enerji
The energy stored in the positions of charges, equal to $qV$ for a single charge in an external potential and to $kq_1q_2/r$ for an isolated pair.
electron voltelektron volt
The energy an elementary charge gains falling through one volt, equal to $1.60\times10^{-19}\ \mathrm{J}$, and the natural unit for particle energies in this course.
referans noktası
The place where the potential is declared to be zero. For any charge distribution of finite size it is taken at infinity; for an infinite line or sheet no such choice exists.
turning pointgeri dönüş noktası
The position at which a moving charge momentarily stops, reached when all of its kinetic energy has been converted into potential energy.
distance of closest approachen yakın yaklaşma mesafesi
The smallest separation a particle fired at a fixed charge reaches before it is turned back, obtained by setting its initial kinetic energy equal to the potential energy at that separation.
limiting caselimit durumu
A version of the problem in which a parameter is pushed to an extreme where the answer is already known, used to test a formula that has just been derived.
order of magnitudebüyüklük mertebesi
The size of a quantity to the nearest power of ten, used as a fast check that an answer belongs to the right physical world.
cavityboşluk
A hollow region inside a conductor; the charge induced on its wall always cancels whatever is sealed inside it, so the outside cannot tell where in the cavity the charge sits.
yük paylaşımı
The redistribution of charge between conductors joined by a wire, which settles when both are at the same potential, giving charges in proportion to their radii for spheres.
karışık pratik
Practice in which problem types are deliberately mixed rather than grouped, so that deciding which method applies becomes part of the exercise.
What comes next
§06 · Capacitance, Dielectrics, Electric Energy Storage
Everything so far has treated charge as something already placed and asked what field and potential follow. Next comes the reverse question: given two conductors and a potential difference between them, how much charge ends up where, and what happens when the space between them is filled with something other than air.
Sources
D. Giancoli, Physics for Scientists and Engineers with Modern Physics, 5th edition The set textbook for the course. Everything reviewed here comes from its treatment of charge, the electric field, Gauss's law and electric potential.
SI values of the physical constants $k = 8.99\times10^{9}\ \mathrm{N\,m^{2}/C^{2}}$, $\varepsilon_0 = 8.85\times10^{-12}\ \mathrm{C^{2}/(N\,m^{2})}$, $e = 1.60\times10^{-19}\ \mathrm{C}$, $m_p = 1.67\times10^{-27}\ \mathrm{kg}$, $m_e = 9.11\times10^{-31}\ \mathrm{kg}$, $m_\alpha = 6.64\times10^{-27}\ \mathrm{kg}$.
The course syllabus week line for this week The line reads catch up and review and carries no chapter numbers, so no chapter numbers are quoted anywhere on this page.