7 concepts22 worked examples29 exercises4 exam-level7 figures
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04Electric potential: energy per coulomb, and the map that replaces the field
Hold a proton 2.00 cm from a small bead carrying 5.00 nC and let go. By the time it has drifted out to 6.00 cm it is doing several hundred kilometres per second, and you already know the force on it at every single point of that journey. That is exactly the trouble: the force at 6.00 cm is nine times smaller than the force at the start, so no constant acceleration and no formula from mechanics will give you the number.
By the end of this section you can get the final speed of any charge released anywhere in the field of any arrangement of charges, without writing a single equation of motion, and you can move in both directions between a map of potentials and the field that made it.
In 60 seconds
Instead of tracking a vector force along a path, attach one number to each point in space: the energy per coulomb it takes to get there. Differences of that number give work, kinetic energy and speed by arithmetic, and its steepest slope gives the field back whenever you need it.
Work and potential energy
$$\Delta U = U_b - U_a = -W_{a\to b}$$
any time the electric force does work on a charge that moves
you are asked for the energy stored in a fixed arrangement, or the work to assemble it
Three most common mistakes
Treating the potential as a vector. It is a plain signed number, so contributions add with a plus sign and their own signs, never with components. The other half of the same mistake is treating the field as a scalar; the two objects live on the same page and obey different arithmetic.
Reading a as a zero of field, or the other way round. Midway between $+8.00$ nC and $-8.00$ nC that are 10.0 cm apart the potential is exactly zero while the field is $5.75\times10^{4}$ N/C; on the far side of the pair there is a point where the field is exactly zero and the potential is 180 V.
Dropping the sign of a negative charge when computing a potential, or putting a sign into a distance. The charge carries its sign into the numerator of $kQ/r$; the distance $r$ is a positive length in every case, including for a negative charge.
The published weights are Midterm 1 twenty per cent, Midterm 2 twenty per cent, quizzes ten per cent in total, homework five per cent, the final twenty five per cent and the laboratory twenty per cent. This is week 4 material, so it sits inside the ground covered by Midterm 1 and again by the final. Nothing finer than those weights is published, so treat every result on this page as examinable and do not read any ranking into the order in which the topics appear.
How much time do you have?
10 minutes
You leave able to do the two things that appear in almost every question on this material: turn a potential difference into a kinetic energy, and add up point charge potentials as signed numbers.
The 60 second card · Formula card · Potential: the number that belongs to the place · The potential of a point charge, and why this section is easier than the last one · Mistake ledger
45 minutes
You add the parts that separate a set-up mark from a full mark: where the potential energy came from, how to get the field back out of a potential, and what a conductor does to both.
The 60 second card · The work the electric force does, and why it can be banked · Potential: the number that belongs to the place · The potential of a point charge, and why this section is easier than the last one · Reading the field off the potential · Equipotential surfaces, and why a conductor is all one number · Method box: energy problems in an electric field · Exam level example · Practice set C · Mistake ledger
full read
You can build the potential of any distribution you can integrate, recover the fields of the last section from it as a check, and say what the energy of a whole arrangement of charges is and what its sign means.
The opening pages · Prerequisites and pretest · Notation · Conventions · All seven concept blocks · Method boxes · Contrast pairs · Scaffolding comes off · Exam level example · Practice sets A to D · Mistake ledger · Formula card · Glossary · Check yourself
By the end of this section
Compute the work done by the electric force on a charge that moves, and convert it into a change of potential energy and a change of speed, for both a uniform field and the field of a point charge.
Convert freely between potential energy, potential difference and kinetic energy for a charged particle, in joules and in , and get the direction of the energy change right from the sign of the charge.
Add the potentials of several point charges as signed scalars, locate the places where the total potential vanishes, and say why those are not the places where the field vanishes.
Differentiate a given potential to recover the electric field, state the units in which the answer comes out, and identify from a set of potential readings where the field is largest and where it is zero.
Apply the fact that a conductor in equilibrium is a single equipotential: find the potential inside and outside a charged conducting sphere, and predict which of two connected conductors has the stronger surface field.
Integrate the potential of a continuous charge distribution for a ring and for a rod, and recover the field of the same object by differentiating the result.
Evaluate the electrostatic potential energy of a fixed arrangement of point charges, interpret the sign of the answer, and use energy conservation to find speeds and distances of closest approach.
Syllabus coverage
and the work done by the electric force; potential and potential difference, the and the electron volt; the potential of a point charge and of any collection of them; recovering the field from the potential; equipotential surfaces and conductors in equilibrium; the potential of a continuous charge distribution; and the electrostatic potential energy of an arrangement of charges.
The week line names a topic and carries no chapter numbers, so no chapter number is quoted anywhere in this section.
covered
Energy stored per unit volume of a field
The idea that the energy of an arrangement can be attributed to the field itself, with a density proportional to the square of the field.
Deferred. The standard route to that result runs through the energy stored in a charged capacitor, and capacitance belongs to a later week. Everything on this page attributes energy to the charges and their separations instead, which needs nothing that has not already been taught.
deferred
Potential of an electric dipole
The potential a dipole puts at a distant point, and the potential energy a dipole has when it sits in an external field.
The dipole itself was built in the previous section, which gave the force and the torque on one but no energy. What is added here is the one number it puts at a distant point, $V = kp\cos\theta/r^{2}$, and the energy $U = -\vec p\cdot\vec E$ it carries in an outside field. Both are plain superposition of point charge potentials, so they sit inside the blocks that teach that superposition rather than in a block of their own.
covered
Recall first
Work done by a constant force
A constant force $\vec F$ acting through a displacement $\vec d$ does work $W = Fd\cos\theta$, where $\theta$ is the angle between the force and the displacement. If the force varies along the path the sum becomes an integral, $W = \int \vec F\cdot d\vec l$.
Every result on this page is this definition applied to the electric force and then rearranged. Nothing else does the work.
The work energy theorem
The total work done on a particle equals the change in its kinetic energy: $W_{\rm total} = \tfrac12 mv_f^{2} - \tfrac12 mv_i^{2}$.
It is the bridge from an energy in joules to a speed in metres per second, which is what most questions on this material actually ask for.
Coulomb's law
Two point charges a distance $r$ apart attract or repel along the line joining them with a force of size $F = k\lvert q_1q_2\rvert/r^{2}$, with $k = 8.99\times10^{9}\ \mathrm{N\,m^{2}/C^{2}}$.
It is the force whose work is being computed. Every potential on this page is an integral of it.
The electric field of a point charge, and superposition
A point charge $Q$ makes a field $E = kQ/r^{2}$ directed away from itself if positive, and the field of several charges is the vector sum of the separate fields.
It is the starting point of every derivation here, and at the end of the section the field is recovered from the potential and checked against it.
The field of a ring and of a rod, from the previous section
On the axis of a uniform ring of radius $R$ at distance $x$ from the centre, $E = kQx/(x^{2}+R^{2})^{3/2}$. On the axis of a uniform rod of length $L$, a distance $a$ beyond its near end, $E = kQ/[a(a+L)]$.
Both are reproduced in this section by differentiating a potential, which takes one line instead of an integral and is the strongest available check that the new method is doing what it should.
Gauss's law for a charged conductor
In the field inside the material of a conductor is zero, and any excess charge sits on the outer surface. A Gaussian sphere outside a spherically symmetric charge $Q$ gives $E = kQ/r^{2}$, exactly as if all the charge were at the centre.
The conductor block turns each of these field statements into a statement about potential, and the sphere result is what makes the potential of a charged sphere so short to write down.
The elementary integrals used here
$\displaystyle\int \frac{dr}{r^{2}} = -\frac1r + C$ and $\displaystyle\int \frac{dx}{x} = \ln\lvert x\rvert + C$, together with $\dfrac{d}{dx}(x^{2}+R^{2})^{-1/2} = -x(x^{2}+R^{2})^{-3/2}$.
These three are the only pieces of calculus in the section. The third one is worth checking by hand once, because it is what turns the potential of a ring back into its field.
The electric dipole and its moment
Two equal and opposite charges $+q$ and $-q$ held a fixed distance $\ell$ apart make an electric dipole. Its moment is the vector $\vec p$ of magnitude $p = q\ell$, pointing from the negative charge to the positive one. In a uniform field the two forces cancel and leave only a torque $\tau = pE\sin\theta$, where $\theta$ is the angle between $\vec p$ and $\vec E$.
The dipole is the smallest arrangement whose potential is worth writing in closed form, and the same $\vec p$ returns at the end of the section in the energy it has when it sits in someone else's field.
Try it yourself first (3 questions)
1§04.0 — work and kinetic energy, from mechanics●○○○○
One line of last year's mechanics before anything electrical happens, because the whole of this section is that line with a different force in it. A block of mass $2.00\ \mathrm{kg}$ slides from rest down a frictionless ramp whose top is $1.20\ \mathrm{m}$ above its bottom. Not being able to do this one is not a problem, but it is better to find out now than in the middle of an energy question about a proton.
Given
$m = 2.00\ \mathrm{kg}$, released from rest
vertical drop $h = 1.20\ \mathrm{m}$, ramp frictionless
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) How much work does gravity do on the block during the slide?
(b) How fast is the block moving at the bottom?
Hint 1/4
You are not asked about the shape of the ramp or how long the slide takes. Ask only what force does work, and over what vertical distance it acts.
Hint 2/4
$W = mgh$ for a drop of height $h$, and $W_{\rm total} = \tfrac12 mv^{2}$ when the motion starts from rest.
Hint 3/4
Here $m = 2.00\ \mathrm{kg}$, $h = 1.20\ \mathrm{m}$ and $g = 9.80\ \mathrm{m/s^{2}}$, with no friction and no initial speed.
Hint 4/4
Gravity does $23.5\ \mathrm{J}$ and the block arrives at $4.85\ \mathrm{m/s}$.
the block starts from rest, so the whole of the work appears as kinetic energy and nothing has to be carried over from the initial state
Answer $$\boxed{\,W = 23.5\ \mathrm{J},\qquad v = 4.85\ \mathrm{m/s}\,}$$
Check
Order of magnitude: a drop of about a metre gives roughly five metres per second, which is the familiar result that a step off a table hurts. Independent route: constant acceleration $g$ over a vertical drop gives $v = \sqrt{2gh}$ directly, without any mention of work, and returns the same 4.85 m/s.
Notice what was not needed: the length of the ramp, its angle, and the time taken. Energy methods throw all three away, and that is exactly the saving this section is about to buy for electric forces.
2§04.0 — where the field of two positive charges vanishes●●○○○
A deliberately awkward one from the field sections, because a very common answer to it is wrong in an instructive way. A charge $+Q$ sits at $x = 0$ and a charge $+4Q$ sits at $x = 30.0\ \mathrm{cm}$ on the same axis.
Given
$+Q$ at $x = 0$
$+4Q$ at $x = 30.0\ \mathrm{cm}$
both charges positive, nothing else present
Find
(a) At what point on the line between the two charges is the total electric field zero?
Hint 1/4
Between two positive charges the two fields point in opposite directions, so ask where the two magnitudes become equal rather than where anything cancels by symmetry.
Hint 2/4
Set $kQ/x^{2} = k(4Q)/(d-x)^{2}$ and take the square root of both sides before doing any algebra.
Hint 3/4
Here $d = 30.0\ \mathrm{cm}$ and the second charge is four times the first, so the square root gives $(d-x)/x = 2$.
Hint 4/4
The null point is $10.0\ \mathrm{cm}$ from the smaller charge, that is one third of the way along, not halfway and not one fifth of the way.
Show solutionSet the magnitudes equal
$$\frac{kQ}{x^{2}} = \frac{4kQ}{(0.300-x)^{2}}$$
between two like charges the directions are already opposite, so equality of magnitude is the whole condition
$$\frac{1}{x} = \frac{2}{0.300-x}$$
taking the square root first turns a quadratic into a linear equation, and the positive root is the only one inside the gap
Solve
$$0.300 - x = 2x \Rightarrow x = 0.100\ \mathrm{m}$$
one third of the way along, closer to the weaker charge, which is where the weaker charge can still hold its own
Answer $$\boxed{\,x = 10.0\ \mathrm{cm}\ \text{from the charge } +Q\,}$$
Check
Substitute back: $100kQ$ from each side, equal as required. Sanity check on the position: the null point must lie nearer the smaller charge, and $10.0$ cm is nearer to $x=0$ than to $x=30.0$ cm.
Hold on to this point. Later in this section the same pair is asked about again, but for the potential rather than the field, and the answer comes out somewhere completely different.
3§04.0 — flux through one face of a cube●●○○○
One from the previous section, and another one where the popular answer is off by a factor. A point charge of $+8.00\ \mathrm{nC}$ is placed at the exact centre of a cube.
the six faces are interchangeable under the symmetry of a central charge, so they must carry equal shares; this step fails if the charge is moved off centre
Independent check on the sixth: put the same charge at the centre of a sphere and the flux per unit solid angle is uniform. The cube's six faces subtend equal solid angles from the centre by symmetry, and the total solid angle is $4\pi$, so each face takes $4\pi/6$ of it, which is the same sixth.
Two habits carry into this section. First, argue from a whole closed thing down to a part rather than integrating the part. Second, notice which given quantities never appear in the answer; the missing side length was a signal, not an omission.
Notation
symbol
reads as
means
watch out
$U$
U
electric potential energy, in joules; the energy a particular charge has because of where it sits
it belongs to the whole arrangement, not to one charge alone; saying the potential energy of the proton is shorthand for the energy of the proton together with whatever is holding the field up
$V$
V
electric potential, in volts, equal to $U/q$; the energy per coulomb that a point in space would give to a charge placed there
V is a property of the place, U is a property of the place and the charge together; the single most common slip in this section is quoting one when the question asked for the other
$V_{ba}$
V b a
the potential difference $V_b - V_a$, the potential at b minus the potential at a
the order of the subscripts is the order of the subtraction, and reversing them reverses the sign of every energy that follows
$W_{a\to b}$
W from a to b
the work done by the electric force on the charge as it goes from a to b, in joules
positive when the force helps the motion; it is minus the change in potential energy, and it does not depend on which path was taken
$W_{\rm ext}$
W external
the work an outside agent does carrying the charge slowly from a to b, with no change of kinetic energy
equal to $+\Delta U$, so it is the opposite sign to the work done by the field; a negative $W_{\rm ext}$ means the agent had to hold the charge back
$\mathrm{eV}$
electron volt
the energy gained by one elementary charge crossing a potential difference of one volt, equal to $1.602\times10^{-19}\ \mathrm{J}$
it is a unit of energy and not of potential or of charge; and it is fixed by the elementary charge, so a doubly charged ion crossing one volt gains two electron volts
$dV/dx$
d V by d x
the rate at which the potential changes with position along x; its negative is the x component of the field
a large potential does not make a large field; only a steep one does, and a constant potential of ten million volts gives no field at all
$\vec\nabla V$
grad V
the vector whose components are the three partial derivatives of V, pointing in the direction in which V rises fastest
the field is minus this vector, so the field points downhill on the potential landscape and never uphill
$r$
r
the distance from a charge, or from a charge element, to the point where the potential is wanted
always a positive length; the sign of the charge goes in the numerator of $kQ/r$ and never into r itself
$\lambda$
lambda
linear charge density, the charge per unit length of a rod or wire, in coulombs per metre
it carries the sign of the charge, so a negatively charged rod has a negative lambda and produces a negative potential
Conventions used here
Where the zero of potential and of potential energy sits
For point charges and for any charge distribution of finite size, the zero is at infinity, so $V = kQ/r$ with no added constant and $U \to 0$ as the charges are pulled infinitely far apart. For two parallel plates the zero is put on whichever plate the question calls earthed, and if the question does not say, it is stated in the working. Only differences are ever measured, so a zero is a choice; but it is one choice per problem, and it is announced before any number is written.
Whose work is being talked about
$W$ with no subscript always means the work done by the electric force. $W_{\rm ext}$ means the work done by an outside agent that carries the charge slowly, so slowly that the kinetic energy does not change. The two are related by $W_{\rm ext} = -W = \Delta U$, and mixing them up flips the sign of the answer without changing its size, which is why the sign has to be argued rather than copied.
What the subscripts on a potential difference mean
$V_{ba}$ means $V_b - V_a$, the potential of the second point named minus the potential of the first. The phrase the potential difference between a and b, with no further wording, is written out in full in this section rather than abbreviated, because half the class reads it one way round and half the other.
Signs go into the arithmetic here, not into a sentence
Potential and potential energy are scalars, so a negative charge is put into the formula complete with its minus sign and the answer comes out signed. This is the opposite of the habit that worked for fields and forces, where magnitudes went into the formula and the direction was fixed by a sentence. The distance $r$ is still a positive length, always, for both signs of charge.
Constants used throughout
$k = 1/(4\pi\varepsilon_0) = 8.99\times10^{9}\ \mathrm{N\,m^{2}/C^{2}}$ and $\varepsilon_0 = 8.85\times10^{-12}\ \mathrm{C^{2}/(N\,m^{2})}$. The same two numbers are used in every worked answer here, and every answer is rounded to three significant figures at the end and not before. This section needs four more, and they are fixed too: the elementary charge $e = 1.602\times10^{-19}\ \mathrm{C}$, the electron mass $9.11\times10^{-31}\ \mathrm{kg}$, the proton mass $1.67\times10^{-27}\ \mathrm{kg}$, and one electron volt as $1.602\times10^{-19}\ \mathrm{J}$.
How many digits an answer keeps
Every answer is rounded to three significant figures at the very end and not before, matching the three figures the data carries. Intermediate lines keep four so that the rounding does not creep into the last digit.
How a potential is quoted
A potential is a signed number of volts, for example $-162\ \mathrm{V}$, and it needs no direction. A field is a magnitude in volts per metre or newtons per coulomb, which are the same unit, together with a direction. Quoting a direction for a potential, or leaving a field without one, is an incomplete answer in both cases.
4.1The work the electric force does, and why it can be banked
The electric force's work depends only on the endpoints, so it can be banked as one number per point.
Three sections of fields give the force on a charge everywhere; none turns it into a final speed.
Solvable with what we have
Find the force on a charge from Coulomb's law or from $\vec F = q\vec E$.
Find the field of a point charge, a rod or a ring.
Find where the total field of several charges is zero.
Find the acceleration of a charge just after release, from $a = qE/m$.
Not solvable yet
Say how fast that proton is moving once it reaches $6.00\ \mathrm{cm}$.
Say whether a charge fired at another gets close enough to touch.
Say what it costs to build an arrangement of three charges.
The force at the start is $$F = \frac{kQq}{r^{2}} = \frac{(8.99\times10^{9})(5.00\times10^{-9})(1.602\times10^{-19})}{(0.0200)^{2}} = 1.80\times10^{-14}\ \mathrm{N}$$ giving $a = F/m_p = 1.08\times10^{13}\ \mathrm{m/s^{2}}$. Held constant over the $4.00\ \mathrm{cm}$ of travel it gives $v = \sqrt{2ad} = \sqrt{2(1.08\times10^{13})(0.0400)} = 9.29\times10^{5}\ \mathrm{m/s}$.
Why it fails
The force is not constant: at $6.00\ \mathrm{cm}$ it is nine times smaller than at $2.00\ \mathrm{cm}$, so the starting force has been charged for the whole trip and the answer must come out too big. Worse, it is impossible: even flying out to infinity the proton could not pass $6.57\times10^{5}\ \mathrm{m/s}$. The cure is to integrate the force along the path once and for all.
TheoremResult 4.1: the work of the electric force, and the number that stores it
Conditions
The charges producing the field are at rest and stay at rest while the test charge moves
The work is that done by the electric force alone; any other force acting is accounted for separately
For a distribution of finite size the potential energy is measured from infinity, so that $U \to 0$ as the charges are separated without limit
The work the electric force does on a charge moving from one place to another is the drop in a stored number belonging to the charges and their separation; the route does not matter, only the endpoints. For two point charges that number is the Coulomb constant times the product of the charges, signs included, over their separation, counted from zero at infinite separation.
Proof
Take a fixed charge $Q$ at the origin and move a charge $q$ from a distance $r_a$ to a distance $r_b$ along any path at all.
Split the path into little steps $d\vec l$. The force is radial, so only the radial part of each step contributes: $\vec F\cdot d\vec l = F\,dr$, where $dr$ is the change in distance from the origin. The sideways part of the step contributes nothing, whatever its length.
That single observation is the whole theorem. Two paths between the same endpoints have the same collection of radial steps, because both start at $r_a$ and finish at $r_b$; they differ only in sideways steps, which do no work.
Writing $W = U_a - U_b$ and reading off the two terms gives $U(r) = kQq/r$, which vanishes as $r$ grows without limit. That last property is the choice of zero, and it is a choice, not a discovery.
In a uniform field the only part of a displacement that costs or yields energy is the part along the field, drawn here in $\textcolor{#8250df}{\text{purple}}$ as $d = 4.00$ cm. Both routes from A to B cover the same $d$, so the $\textcolor{#cf222e}{\text{electric force}}$ does the same work along both, and the sideways leg of route 2 contributes exactly nothing.
Looks like this, but is not
Friction does work on a sliding block, so it must have a potential energy too, something like $U_{\rm fric} = \mu mgx$.
Take the block from A to B by a short path, then by one ten times longer. The electric force does equal work both times; friction does ten times as much, so no number at B can record it. Path independence belongs to forces acting along the line between two objects and depending only on their separation, like the electric force and gravity.
The proton from the start of the section, done properly
A bead carrying $+5.00\ \mathrm{nC}$ is fixed in place. A proton is released from rest $2.00\ \mathrm{cm}$ from it. How fast is the proton moving when it has reached $6.00\ \mathrm{cm}$, and what is the fastest it could ever be moving?
Given
fixed charge $Q = +5.00\ \mathrm{nC}$
proton: $q = +1.602\times10^{-19}\ \mathrm{C}$, $m_p = 1.67\times10^{-27}\ \mathrm{kg}$, released from rest
both energies differ only in the distance underneath, so computing the numerator once saves doing the same multiplication twice and keeps the two lines comparable
The naive constant force estimate at the top of this block gave $9.29\times10^{5}\ \mathrm{m/s}$, which is above the ceiling just computed and is therefore impossible without any further arithmetic. Second, independent check on the energy scale: $2.400\times10^{-16}\ \mathrm{J}$ divided by $1.602\times10^{-19}$ is $1.50\times10^{3}$, so the proton has picked up one and a half thousand electron volts, the sort of energy a small laboratory supply delivers, not a nuclear one.
One multiplication, two divisions and a square root. The integral was done once, in the box above, and will not be done again anywhere in this section.
Two thirds of the available energy was collected in the first four centimetres and the remaining third is spread over the whole of the rest of space. Whenever a potential energy goes as one over the distance, most of the action is close in, and a question that asks for the speed far away is usually easier than one that asks for the speed nearby.
Same start, same finish, two routes through a uniform field
A uniform field of $500\ \mathrm{N/C}$ points along the $x$ axis. A charge of $+2.00\ \mathrm{nC}$ is carried from the origin to the point $(4.00\ \mathrm{cm},\, 3.00\ \mathrm{cm})$, first along the straight line joining them and then along the two legs of the right angle through $(0,\,3.00\ \mathrm{cm})$. Find the work done by the electric force on each route.
Given
$\vec E = 500\ \mathrm{N/C}$ in the $+x$ direction, uniform
Independent check by energy bookkeeping rather than by geometry: if the two routes gave different answers you could take the charge out along the cheap one and back along the expensive one and collect energy every lap, from a set of charges that never moved. The equality is forced by that impossibility, not by the arithmetic.
The sideways leg was 3.00 cm long and contributed nothing, which is the cheapest kind of work there is.
In a uniform field only the displacement along the field is ever needed. This is why plate problems collapse to a single multiplication, and why the first thing to do in one is to project the motion onto the field direction and throw the rest away.
Checkpoint
§04.1 — work done by a uniform field●○○○○
Thirty seconds, one multiplication. A charge of $+4.00\ \mathrm{nC}$ is moved $6.00\ \mathrm{cm}$ in the direction of a uniform field of $250\ \mathrm{N/C}$.
Given
$q = +4.00\ \mathrm{nC}$
$E = 250\ \mathrm{N/C}$, uniform
displacement $6.00\ \mathrm{cm}$, along the field
Find
(a) How much work does the electric force do on the charge?
(b) By how much does the potential energy of the charge change, and in which direction?
Hint 1/4
Two questions, one number. Work out what the force does, then remember that the stored energy moves the opposite way.
Hint 2/4
$W = qEd$ for a displacement $d$ along a uniform field, and $\Delta U = -W$.
Hint 3/4
Here $q = 4.00\times10^{-9}\ \mathrm{C}$, $E = 250\ \mathrm{N/C}$ and $d = 0.0600\ \mathrm{m}$, with the motion along the field.
Hint 4/4
The field does $60.0\ \mathrm{nJ}$ of work and the potential energy falls by the same $60.0\ \mathrm{nJ}$.
Show solutionThe work
$$W = qEd = (4.00\times10^{-9})(250)(0.0600) = 6.00\times10^{-8}\ \mathrm{J}$$
the displacement is along the field, so the cosine is one and no projection is needed
The energy change
$$\Delta U = -W = -6.00\times10^{-8}\ \mathrm{J}$$
a positive charge moving the way a positive field pushes it is going downhill, and downhill means the stored energy drops
Answer $$\boxed{\,W = +60.0\ \mathrm{nJ},\qquad \Delta U = -60.0\ \mathrm{nJ}\,}$$
Check
Sign check without arithmetic: the force on a positive charge points along the field, the charge moved along the field, so the force helped and the work must be positive. Any answer with a minus sign on the work has the geometry wrong, not the multiplication.
⚠ Using the force at the starting point as if it acted all the way
the starting force is the one number the problem hands you, and treating it as constant makes the mechanics familiar again
⚠ Putting a minus sign into the distance instead of into the charge
a negative charge makes everything about the problem feel negative, and r is the nearest place to put the sign
wrong$$U = \frac{kQq}{-r}$$
right$$U = \frac{kQ(-\lvert q\rvert)}{r}$$
4.2Potential: the number that belongs to the place
Divide the stored energy by the charge you stored it for, and what is left describes the place alone.
The energy just defined still mentions the charge you brought along; dividing it out leaves a description of the empty point itself, which is the object every instrument in a laboratory actually reads.
DefinitionDefinition 4.2: electric potential, potential difference and the volt
Conditions
The field is electrostatic, so the potential of each point is a fixed number and not a function of time
A zero has been chosen and named; every value quoted is a difference from it
The test charge is small enough not to disturb the charges that make the field
The potential at a point is the energy per coulomb that the point will hand to a charge placed there, and the potential difference between two points is the energy per coulomb the trip between them costs or pays. One volt means one joule for every coulomb carried. The last equality says the same thing from the field's side: to get the potential drop, add up the field along the path and change the sign, so that the potential falls in the direction the field points.
Proof
Start from the previous block: $\Delta U = -W = -\int_a^b \vec F\cdot d\vec l$ for a charge $q$ carried from a to b.
The force on that charge is $\vec F = q\vec E$, and $q$ is a constant that can come outside the integral: $\Delta U = -q\int_a^b \vec E\cdot d\vec l$.
Divide both sides by $q$. The left side becomes $\Delta U/q$, which is what we are calling $V_{ba}$; the right side no longer mentions the charge at all.
For a uniform field and a straight path of length $d$ taken along the field, the integral is just $Ed$, so $V_{ba} = -Ed$: the potential drops by $Ed$ as you move a distance $d$ downfield.
One useful unit falls out immediately. A charge $e$ crossing a potential difference of one volt changes its energy by $e\times1\ \mathrm{V} = 1.602\times10^{-19}\ \mathrm{J}$, and that amount is given the name one electron volt.
Between two plates $8.00$ mm apart the $\textcolor{#8250df}{\text{potential}}$ falls in a straight line, and the steepness of that line, $24.0$ V per $2.00$ mm, is the size of the $\textcolor{#cf222e}{\text{field}}$: $1.20\times10^{4}$ V/m. Notice that the potential is high where the field starts, not where it is strong; the field is the same everywhere in this gap.
Looks like this, but is not
Ten thousand volts is dangerous, so the spark you get off a door handle in winter must carry a lot of energy. A big potential difference must mean a big energy.
Potential is energy per coulomb, and the second half of that phrase decides the answer. If the charge that jumps is of the order of $10\ \mathrm{nC}$, then $U = qV = (10^{-8})(10^{4}) = 10^{-4}\ \mathrm{J}$, a tenth of a millijoule, which is why it stings and does no harm. The same ten thousand volts driving a coulomb of charge would deliver ten thousand joules. Volts on their own never tell you an energy; they tell you an exchange rate, and you still have to say how much charge is being exchanged.
Electron accelerated through 5.00 kV
An electron starts from rest and is accelerated from a plate at $0\ \mathrm{V}$ to a plate at $+5.00\ \mathrm{kV}$. Find the energy it gains, in joules and in electron volts, and the speed it arrives with.
Given
electron: charge $-1.602\times10^{-19}\ \mathrm{C}$, mass $9.11\times10^{-31}\ \mathrm{kg}$
starts from rest at the plate held at $0\ \mathrm{V}$
finishes at the plate held at $+5.00\times10^{3}\ \mathrm{V}$
Find
the kinetic energy gained and the final speed
SolutionGet the sign of the energy change right before touching a calculator
$$\Delta U = q\,V_{ba} = (-1.602\times10^{-19})(+5.00\times10^{3}) = -8.01\times10^{-16}\ \mathrm{J}$$
the electron's charge is negative and the potential rises along its path, so the product is negative; a negative charge falls towards high potential, which is the opposite of the picture for a proton
Convert the drop into kinetic energy
$$\Delta K = -\Delta U = 8.01\times10^{-16}\ \mathrm{J}$$
total energy is conserved and no other force acts, so whatever the stored energy loses the motion gains
$$\Delta K = \frac{8.01\times10^{-16}}{1.602\times10^{-19}} = 5.00\times10^{3}\ \mathrm{eV} = 5.00\ \mathrm{keV}$$
quoting it in electron volts costs one division and makes the answer readable: one elementary charge across five thousand volts is five thousand electron volts, by construction
the mass appears only in this last step, so the same energy line would serve any particle and only this line has to be redone
Answer $$\boxed{\,\Delta K = 8.01\times10^{-16}\ \mathrm{J} = 5.00\ \mathrm{keV},\qquad v = 4.19\times10^{7}\ \mathrm{m/s}\,}$$
Check
Independent check on the scale: $4.19\times10^{7}\ \mathrm{m/s}$ is fourteen per cent of the speed of light, so this is a fast particle but not a relativistic one. The relativistic factor at this speed is $1.010$, so the classical answer quoted here is high by about one per cent, and that is the honest size of the error rather than a claim that it is exact. Second check: the electron volt line and the joule line must agree, and $5.00\ \mathrm{keV}$ multiplied back by $1.602\times10^{-19}$ returns $8.01\times10^{-16}\ \mathrm{J}$.
One multiplication, one division and one square root, and no mention of the plate separation, the field between the plates or the shape of the path.
The plate separation was never given and was never needed. Any question that supplies a potential difference and asks for a speed can be answered without knowing anything about the field in between, and if a question does give you the separation, that is a signal it wants the field somewhere else in the answer.
From 12.0 V across 5.00 mm to the field in the gap
Two parallel plates are $5.00\ \mathrm{mm}$ apart and a source holds them at a potential difference of $12.0\ \mathrm{V}$. Find the field between them, and the work an external agent must do to carry a $+3.00\ \mathrm{nC}$ bead slowly from the low plate to the high plate.
Cross check by the other route: the field does work $-qEd = -(3.00\times10^{-9})(2400)(5.00\times10^{-3}) = -3.60\times10^{-8}\ \mathrm{J}$ on the bead, and the agent must supply exactly the opposite, which is the $+36.0$ nJ found above. Plausibility: $2.40\times10^{3}$ V/m is about a thousandth of the field at which dry air breaks down and sparks, so nothing dramatic happens in this gap, which fits a twelve volt source.
Two divisions in total, and the volt per metre used in one line is the same unit as the newton per coulomb used in the other.
Volts per metre and newtons per coulomb are the same unit wearing two different names, and which name you use is a hint about which way you are thinking: newtons per coulomb when you are pushing a charge, volts per metre when you are reading a slope off a potential.
Checkpoint
§04.2 — a doubly charged ion across a known gap●●○○○
Thirty seconds, and the whole question is whether the charge or the mass decides the energy. An alpha particle, which carries a charge of $+2e$, is accelerated from rest through a potential difference of $1.50\ \mathrm{kV}$.
(a) How much kinetic energy does it gain, in electron volts and in joules?
Hint 1/4
Ask which property of the particle appears in an energy and which one only appears in a speed. The question stops at the energy.
Hint 2/4
$\Delta K = \lvert q\rvert V$, and one elementary charge across one volt is one electron volt by definition.
Hint 3/4
Here the charge is two elementary charges and the potential difference is $1.50\times10^{3}$ V.
Hint 4/4
It gains $3.00\times10^{3}\ \mathrm{eV}$, which is $4.81\times10^{-16}\ \mathrm{J}$.
Show solutionIn electron volts
$$\Delta K = (2e)(1.50\times10^{3}\ \mathrm{V}) = 3.00\times10^{3}\ \mathrm{eV}$$
the electron volt is defined from one elementary charge, so a charge of two of them simply doubles the count
In joules
$$\Delta K = (3.00\times10^{3})(1.602\times10^{-19}) = 4.81\times10^{-16}\ \mathrm{J}$$
converting at the end rather than the start keeps the arithmetic in whole numbers for as long as possible
Answer $$\boxed{\,\Delta K = 3.00\ \mathrm{keV} = 4.81\times10^{-16}\ \mathrm{J}\,}$$
Check
Consistency check between the two forms: dividing $4.81\times10^{-16}\ \mathrm{J}$ by $1.602\times10^{-19}\ \mathrm{J/eV}$ gives back $3.00\times10^{3}\ \mathrm{eV}$. Physical check: an electron through the same gap would gain half as much, $1.50$ keV, and the ratio is exactly the ratio of the charges and has nothing to do with the four thousandfold difference in mass.
⚠ Quoting a potential where a potential energy was asked for, or the reverse
the two are the same word in ordinary speech and differ by a factor that is easy to leave out because it is often a very small number
wrong$$U = 5.00\times10^{3}\ \mathrm{J}\ \text{for an electron through 5.00 kV}$$
the name contains the word volt, and it is written next to numbers that came from a voltage
wrong$$V = 5.00\times10^{3}\ \mathrm{eV}$$
right$$V = 5.00\times10^{3}\ \mathrm{V},\qquad \Delta K = 5.00\times10^{3}\ \mathrm{eV}$$
⚠ Sending a negative charge the wrong way
the rule charges fall to low potential is learnt with a positive charge in mind and then applied to every charge
wrong$$\text{electron released at rest moves towards lower } V$$
right$$\text{electron released at rest moves towards higher } V,\ \text{since } \Delta U = qV_{ba} < 0 \text{ needs } V_{ba} > 0$$
4.3The potential of a point charge, and why this section is easier than the last one
Every charge contributes a signed number to every point, and the contributions add with a plus sign.
The definition is in place; the cheapest thing to put into it is the field whose integral you already met in the first block, and the answer turns out to be the shortest formula in the section.
TheoremResult 4.3: the potential of a point charge, and of any number of them
Conditions
The zero is at infinity, which is available because the charge occupies a finite region
$r$ is the distance from the charge to the field point, a positive length for either sign of charge
The charges are held fixed; nothing here says how much energy it took to hold them there
One point charge gives every point in space a potential equal to the Coulomb constant times the charge, signs and all, divided by the distance to it. When several charges are present, work out that number for each of them separately and add the numbers up. There are no components, no angles and no directions in that sum; a positive charge raises the potential everywhere and a negative one lowers it everywhere.
Proof
Carry a test charge in from infinity along the radius. The potential at distance $r$ is the negative of the field integrated in along that path.
The upper limit contributes nothing, which is the choice of zero showing up as a vanishing boundary term rather than as an extra constant.
For several charges the field is the vector sum of the separate fields, and the integral of a sum is the sum of the integrals. Each integral has already been done, so the total is $k\sum_i Q_i/r_i$.
That last line is the whole saving. The fields had to be added as vectors, with a diagram and two components per charge; the potentials are added as signed numbers with no diagram at all.
The $\textcolor{#8250df}{\text{potential}}$ of a point charge dies away as $1/r$, not as $1/r^{2}$: at $10.0$ cm from $5.00$ nC it is $450$ V, and at $20.0$ cm it is still $225$ V rather than a quarter of that. The dashed curve is the same charge with the opposite sign, and it is the exact mirror image, because the sign of the charge passes straight through into the sign of the potential.
Looks like this, but is not
The potential at this point is zero, so nothing is happening here and a charge released here will stay put. Zero means nothing.
Halfway between $+8.00\ \mathrm{nC}$ and $-8.00\ \mathrm{nC}$ placed $10.0\ \mathrm{cm}$ apart the two contributions are $+1438$ V and $-1438$ V, so the potential is exactly zero. The field there is not zero at all: both charges push a positive test charge the same way, giving $5.75\times10^{4}\ \mathrm{N/C}$, and a charge released there flies off immediately. A zero of potential is a statement about the energy it took to bring a charge in from infinity, and one route can be uphill and downhill by equal amounts and still be steep in the middle.
Potential at the corner of a 3-4-5 triangle of charges
A charge of $+3.00\ \mathrm{nC}$ sits at the origin and a charge of $-2.00\ \mathrm{nC}$ sits at $(4.00\ \mathrm{cm},\,0)$. Find the electric potential at the point $(0,\,3.00\ \mathrm{cm})$.
Given
$Q_1 = +3.00\ \mathrm{nC}$ at $(0,0)$
$Q_2 = -2.00\ \mathrm{nC}$ at $(4.00\ \mathrm{cm},\,0)$
field point $P$ at $(0,\,3.00\ \mathrm{cm})$
Find
the total potential at P
SolutionTwo distances, both positive lengths
$$r_1 = 3.00\ \mathrm{cm} = 0.0300\ \mathrm{m}$$
P sits directly above the first charge, so the separation is the vertical coordinate and nothing has to be resolved
keeping the charges over their distances before multiplying by k lets the two terms be compared directly, and here the first is two and a half times the second
a plain scalar addition; no angle between the two contributions has been asked for anywhere, because there is none
$$V = 539\ \mathrm{V}$$
three significant figures, matching the data, and positive because the nearer charge is the positive one
Answer $$\boxed{\,V_P = +539\ \mathrm{V}\,}$$
Check
Bracket the answer. The positive charge alone would give $899\ \mathrm{V}$ and the negative charge alone $-360\ \mathrm{V}$, so the total must lie between $-360$ and $+899$ and, since the positive term is the larger, above zero. It does, at $539$. Independent structural check: if the second charge were moved to $5.00$ cm from P along any direction its contribution would be unchanged, because only the distance enters, and moving it in a circle round P must leave the answer alone.
Two divisions and one addition. The corresponding field calculation needs two magnitudes, two angles, four components and a Pythagoras at the end, roughly ten lines against these three.
When a question gives you several charges and asks for something scalar, do the potential first and see whether the rest of the question can be answered from it. Energy, work and speed all come out of the potential without any geometry beyond distances.
Where the potential vanishes, and why the field does not vanish there
A charge $+8.00\ \mathrm{nC}$ sits at the origin and a charge $-3.00\ \mathrm{nC}$ sits at $x = 10.0\ \mathrm{cm}$. Find every point on the $x$ axis where the total potential is zero, and compare with the point where the total field is zero.
Given
$Q_1 = +8.00\ \mathrm{nC}$ at $x = 0$
$Q_2 = -3.00\ \mathrm{nC}$ at $x = 10.0\ \mathrm{cm}$
points to be found on the x axis only
Find
the zeros of the potential, and the zero of the field, on the axis
SolutionSet the two contributions equal and opposite
past the negative charge the sign of the bracket flips, which is a genuinely different equation and therefore a genuinely different root, not an artefact
evaluating the potential at the field's null point is the whole point of the example, and it comes out nowhere near zero
Answer $$\boxed{\,V = 0 \text{ at } x = 7.27\ \mathrm{cm} \text{ and } x = 16.0\ \mathrm{cm};\quad E = 0 \text{ at } x = 25.8\ \mathrm{cm},\ \text{where } V = 108\ \mathrm{V}\,}$$
Check
Check each root by substitution rather than by re-deriving. At $x = 0.0727$: $8.00/0.0727 = 110.0$ and $3.00/0.0273 = 109.9$, equal to the rounding. At $x = 0.160$: $8.00/0.160 = 50.0$ and $3.00/0.0600 = 50.0$, equal exactly. At $x = 0.258$: $8.00/0.258^{2} = 120.2$ and $3.00/0.158^{2} = 120.2$, equal. Three independent substitutions, one for each root.
Three linear equations and one that needed a square root first. The potential gave two roots for the price of one sign change; the field gave one root and a quadratic that had to be avoided.
Three different places, three different meanings. The two potential zeros are places where a charge brought in from infinity arrives having done no net work; the field zero is a place where a charge would sit without being pushed. Nothing forces those to be the same place, and in this arrangement all three are different.
The potential of a dipole at a point far away
A dipole is made of $+q$ at $(0,\,\ell/2)$ and $-q$ at $(0,\,-\ell/2)$, so its moment $p = q\ell$ points along the positive $y$ axis. Find the potential at a point $P$ a distance $r$ from the centre in a direction making an angle $\theta$ with the moment, in the case $r \gg \ell$. Then put numbers in: $q = 2.00\ \mathrm{nC}$, $\ell = 1.00\ \mathrm{mm}$, $r = 5.00\ \mathrm{cm}$, $\theta = 60.0^{\circ}$.
Given
$+q$ and $-q$ separated by $\ell$, moment $p = q\ell$ along $+y$
field point $P$ at distance $r$ from the centre, at angle $\theta$ to $\vec p$
with $r \gg \ell$ the two lines from P to the charges are almost parallel, so the near charge is closer by half the projection of $\ell$ on that line; the $(\ell/2)^{2}$ term is smaller than the one kept by a further factor $\ell/r$, which is why it is dropped and the cosine term is not
Add the two potentials, which is all a potential ever asks
a scalar sum, so no angle between contributions appears anywhere; putting the two terms over one denominator is what makes the near cancellation visible instead of hiding it in two large numbers
the numerator is the small difference the approximation was made for and must keep its first order term; the denominator already multiplies something small, so leading order is enough there
three significant figures, matching the data; the sign is positive because P lies on the side of the positive charge, and it would flip for $\theta > 90^{\circ}$
Do the sum exactly rather than approximately and see whether the shortcut cost anything. With $r = 5.00$ cm, $\ell = 1.00$ mm and $\theta = 60.0^{\circ}$ the exact distances are $r_+ = 4.97519$ cm and $r_- = 5.02519$ cm, so $V = kq(1/r_+ - 1/r_-) = (8.99\times10^{9})(2.00\times10^{-9})(20.0997 - 19.8997) = 3.60\ \mathrm{V}$, the same to three figures. Two limits agree as well: at $\theta = 90^{\circ}$ the point is equidistant from both charges and the formula gives zero, and the fall off is $1/r^{2}$ rather than $1/r$, faster than a single charge, as it must be for a pair that cancels from far away.
Two distances, one subtraction and one division. The corresponding field of a dipole needs both components and comes out with two terms instead of one, which is the usual price of insisting on a vector.
This is the whole content of the dipole for the rest of the course: a distant point does not see two charges, it sees one vector $\vec p$. If a question gives you a small neutral object with a moment and asks for anything at a distance, this one line replaces the pair.
Checkpoint
§04.3 — potential near a single negative charge●○○○○
Thirty seconds, one division, and one decision about a sign. A point charge of $-4.50\ \mathrm{nC}$ sits alone in space.
Given
$Q = -4.50\ \mathrm{nC}$, isolated
field point $25.0\ \mathrm{cm}$ away
potential measured from zero at infinity
Find
(a) Find the electric potential at a point $25.0\ \mathrm{cm}$ from the charge.
Hint 1/4
One charge, one distance, one number. The only decision to make is where the minus sign belongs.
Hint 2/4
$V = kQ/r$, with the sign of the charge in the numerator and a positive length underneath.
Hint 3/4
Here $Q = -4.50\times10^{-9}\ \mathrm{C}$ and $r = 0.250\ \mathrm{m}$.
the distance is a length and stays positive; only the charge carries a sign
$$V = -162\ \mathrm{V}$$
three significant figures, and no direction is quoted because a potential does not have one
Answer $$\boxed{\,V = -162\ \mathrm{V}\,}$$
Check
Scale check against the figure in this block: a charge of $5.00$ nC gives $450$ V at $10.0$ cm. Here the charge is a tenth smaller and the distance two and a half times bigger, so the size should be roughly $450 \times 0.9 / 2.5 = 162$ V, which matches, and the sign is negative because the charge is.
⚠ Squaring the distance in the potential
three sections of inverse square law have made the square automatic, and the two formulas differ by nothing else
wrong$$V = \frac{kQ}{r^{2}}$$
right$$V = \frac{kQ}{r}$$
⚠ Adding potentials as if they were vectors
the previous section trained the reflex of resolving every contribution into components before adding it
wrong$$V = \sqrt{V_1^{2}+V_2^{2}}$$
right$$V = V_1 + V_2$$
⚠ Dropping the sign of a negative charge
in every field question the sign was stripped off at the start and put back as a direction at the end, and that habit is exactly wrong here
The field is the steepness of the potential, pointing downhill, so a derivative replaces a vector integral.
Every step so far has gone from field to potential; the trip is worth making in the other direction too, because a derivative is cheaper than the vector integral that produced the field in the first place.
RuleRule 4.4: recovering the field from the potential
Conditions
The potential is known as a function of position, not just at isolated points
Each component of the field is the derivative along that same direction, with the other coordinates held fixed
The result is a field in volts per metre, which is identical to newtons per coulomb
The field along any direction is minus the rate at which the potential changes as you walk in that direction. A flat potential means no field however high the potential is; a steeply falling potential means a strong field pointing the way it falls. The minus sign says the field points downhill, which is exactly the statement that a positive charge released at rest starts sliding towards lower potential.
Proof
Take two points a tiny distance $dx$ apart along the $x$ axis. The definition of potential difference gives $dV = -\vec E\cdot d\vec l = -E_x\,dx$ for that step.
Divide by $dx$: $E_x = -dV/dx$. The other two components come from steps along $y$ and $z$, and because each step changes only one coordinate the derivatives are partial ones.
Nothing was assumed about the shape of the field, so the rule is general: any potential you can write down can be differentiated back into the field that made it.
Consistency check on the units. Volts per metre is joules per coulomb per metre, that is newton metres per coulomb per metre, that is newtons per coulomb. The two names for the field's unit are the same unit.
The $\textcolor{#8250df}{\text{potential}}$ dips to $6.60$ V at $x = 0.300$ m. The $\textcolor{#cf222e}{\text{field}}$ points towards the dip from both sides, because the field runs downhill, and it is exactly zero at the bottom, where the potential is at its lowest and its slope is flat. Note that the potential there is $6.60$ V and not zero: a stationary potential and a vanishing potential are different things.
Looks like this, but is not
The potential here is ten million volts, so the field here must be enormous. A big potential should mean a big field.
Take a metal dome charged to ten million volts and stand inside it. The potential everywhere inside is ten million volts, the same everywhere, so its derivative is zero and the field inside is zero. Nothing pushes a charge released in there. The field cares only about how fast the potential changes from point to point, never about its value; a mountain plateau is high and flat and nothing rolls on it.
x (cm)
V (volts)
interval
average E in that interval (V/m)
0
40.0
0 to 2
+600
2.00
28.0
2 to 4
+400
4.00
20.0
4 to 6
+200
6.00
16.0
6 to 8
0
8.00
16.0
The field is strongest where the potential column drops fastest, which is at the start, and it is zero in the last interval where the potential stops changing even though it is still sitting at 16.0 V. Each entry is minus the drop divided by the 0.0200 m step: the first is minus of minus 12.0 over 0.0200, that is 600 V per metre, and the last is zero over 0.0200. Reading a column of potentials this way is exactly what the derivative does, one interval at a time.
Getting the point charge field back out of its own potential
The potential of a point charge is $V(r) = kQ/r$. Differentiate it to recover the field, and check that the answer agrees with Coulomb's law both in size and in direction.
Given
$V(r) = kQ/r$, with the zero at infinity
spherical symmetry, so the field can only be radial
$Q$ may have either sign
Find
the radial component of the field, and its direction for each sign of Q
SolutionDifferentiate along the only direction that matters
a positive radial component means pointing away from the origin, so the direction now comes out of the algebra instead of having to be argued in a sentence
Answer $$\boxed{\,E_r = \frac{kQ}{r^{2}},\ \text{outward for } Q>0\,}$$
Check
This is the strongest kind of check there is: the answer was already known independently, from Coulomb's law measured with a torsion balance, and the derivative reproduces it exactly including the direction. If the minus sign in the rule had been dropped, the field of a positive charge would point inwards and like charges would attract, which is contradicted by every experiment in the first section.
One derivative, three lines. Getting the potential from the field earlier took an integral; getting back takes less work than going out, which is generally true.
This round trip is worth remembering as a test rather than as a result. Whenever you derive a potential for a new charge distribution, differentiate it and see whether the field you already know comes back. It catches missing factors that no amount of staring at the integral will.
A potential valley: where the field is strongest and where it dies
Along a line in a certain region the potential is $V(x) = (60.0\ \mathrm{V/m^{2}})x^{2} - (36.0\ \mathrm{V/m})x + 12.0\ \mathrm{V}$, with $x$ in metres. Find the field at $x = 0.100\ \mathrm{m}$, find where the field vanishes, and state the potential at that place.
Given
$V(x) = 60.0x^{2} - 36.0x + 12.0$, volts, with x in metres
the region runs from $x = 0$ to $x = 0.600\ \mathrm{m}$
nothing depends on y or z
Find
E at x = 0.100 m, the position where E = 0, and V there
SolutionDifferentiate once, and keep the minus sign
computing the value at the flat point is the whole moral of the example: it is nowhere near zero
Answer $$\boxed{\,E_x(0.100) = +24.0\ \mathrm{V/m},\qquad E = 0 \text{ at } x = 0.300\ \mathrm{m},\ \text{where } V = 6.60\ \mathrm{V}\,}$$
Check
Independent numerical check of the derivative by a difference rather than by calculus. $V(0.0900) = 0.486 - 3.24 + 12.0 = 9.246\ \mathrm{V}$ and $V(0.110) = 0.726 - 3.96 + 12.0 = 8.766\ \mathrm{V}$, so the average slope across that small interval is $(8.766-9.246)/0.0200 = -24.0\ \mathrm{V/m}$ and the field is $+24.0\ \mathrm{V/m}$, agreeing with the calculus to three figures.
One derivative and two substitutions covered a question that would otherwise need the source charges to be known.
A useful reflex: when a question asks where a field is zero and gives you a potential, do not solve $V = 0$. Solve $dV/dx = 0$. The two equations almost never have the same roots, and mixing them up is the commonest single error on this material.
Checkpoint
§04.4 — one derivative, one substitution●●○○○
Thirty seconds and one differentiation. In some region the potential varies along the $x$ axis as $V(x) = (12.0\ \mathrm{V/m^{2}})x^{2}$, with $x$ measured in metres.
Given
$V(x) = 12.0x^{2}$ volts, with x in metres
nothing depends on y or z
the point of interest is $x = 0.250\ \mathrm{m}$
Find
(a) Find the x component of the electric field at $x = 0.250\ \mathrm{m}$, with its sign.
Hint 1/4
You are being asked for a slope, not for a value. Differentiate first and substitute afterwards.
Hint 2/4
$E_x = -dV/dx$, and the minus sign is part of the rule, not an option.
Hint 3/4
Here $V = 12.0x^{2}$, so $dV/dx = 24.0x$, to be evaluated at $x = 0.250$ m.
Hint 4/4
The field is $-6.00\ \mathrm{V/m}$, that is $6.00\ \mathrm{V/m}$ pointing in the negative x direction.
the sign is the answer's most informative part, since it says which way a positive charge would be pushed
Answer $$\boxed{\,E_x = -6.00\ \mathrm{V/m}\,}$$
Check
Difference check: $V(0.240) = 0.6912\ \mathrm{V}$ and $V(0.260) = 0.8112\ \mathrm{V}$, so the slope over that interval is $(0.8112-0.6912)/0.0200 = 6.00\ \mathrm{V/m}$ and the field is its negative, $-6.00\ \mathrm{V/m}$, matching.
⚠ Losing the minus sign in the gradient
it does not change the size of the answer, so nothing looks wrong until a direction is asked for
wrong$$E_x = \frac{dV}{dx} = 120.0x - 36.0$$
right$$E_x = -\frac{dV}{dx} = 36.0 - 120.0x$$
⚠ Dividing the potential by the distance instead of differentiating
the relation $E = V/d$ works for a uniform field and gets remembered as if it worked for every field
4.5Equipotential surfaces, and why a conductor is all one number
Surfaces of constant potential cut the field at right angles, and a conductor at rest is one of them throughout.
If the field is the slope of the potential, then the level lines of the potential are worth drawing, and the most important level surface in the whole of electrostatics turns out to be the surface of a piece of metal.
TheoremResult 4.5: equipotentials, and the potential of a conductor in equilibrium
Conditions
Electrostatic equilibrium: nothing is moving, so the field inside the material of a conductor is zero
The conductor is isolated, or at least nothing is forcing charge through it
For the sphere result, the charge distribution is spherically symmetric, which an isolated sphere's charge is
$$\boxed{\,\vec E \perp \text{equipotential},\quad W = 0 \text{ along one};\qquad V_{\rm sphere} = \frac{kQ}{r}\ (r \ge r_0),\qquad V_{\rm sphere} = \frac{kQ}{r_0}\ (r \le r_0)\,}$$
A surface on which the potential has one value is crossed by the field at right angles, and carrying a charge anywhere along such a surface costs nothing. A conductor left to settle has no field inside it, so its whole body, surface included, sits at one single potential; outside an isolated charged sphere the potential is the same as that of a point charge at the centre, and inside it stops falling and holds the surface value all the way to the middle.
Proof
Move a charge a small step $d\vec l$ along a surface of constant potential. By definition $dV = 0$, and $dV = -\vec E\cdot d\vec l$, so $\vec E\cdot d\vec l = 0$ for every step lying in the surface. A vector whose dot product with every direction in a surface vanishes is perpendicular to that surface.
Inside a conductor at rest the field is zero, so between any two interior points $V_b - V_a = -\int_a^b \vec E\cdot d\vec l = 0$. Every interior point, and the surface too, carries the same potential.
Outside an isolated charged sphere, a spherical Gaussian surface gives $E = kQ/r^{2}$, identical to a point charge at the centre, so integrating in from infinity gives $V = kQ/r$ for $r \ge r_0$.
At the surface that value is $kQ/r_0$. Going further in adds nothing, because the field is zero there, so $V$ stays flat at $kQ/r_0$ all the way to the centre. The potential is continuous at the surface and its slope is not: outside the graph falls, inside it is level.
The $\textcolor{#8250df}{\text{level surfaces}}$ of a point charge are spheres. Drawn here in equal steps of $150$ V, at $450$ V, $300$ V and $150$ V, they are not equally spaced: the first two circles are $5$ cm apart and the last two $15$ cm apart, because the potential falls as $1/r$ and the same drop costs more and more room. Every $\textcolor{#cf222e}{\text{field line}}$ crosses every circle at a right angle, which is why a trip once round a circle is free and a trip between two circles is not.
Looks like this, but is not
The field inside a charged metal sphere is zero, so the potential inside it is zero too. No field, nothing stored, nothing to measure.
Zero field means the potential does not change, not that it is small. A sphere of radius $15.0\ \mathrm{cm}$ carrying $6.50\ \mathrm{nC}$ sits at $390$ V at its surface, and every point inside it is also at $390$ V, right to the centre. Only the slope has died, not the value. The same confusion in reverse says that a region at high potential must have a strong field in it, and both are cured by remembering that the field is a derivative.
Potential inside, on and outside a charged metal sphere
An isolated metal sphere of radius $15.0\ \mathrm{cm}$ carries $+6.50\ \mathrm{nC}$, spread over its surface as electrostatics requires. Find the potential at its centre, at its surface and at a point $45.0\ \mathrm{cm}$ from the centre, and the field just outside the surface.
Two independent checks. First, ratio: $0.450$ m is three times $0.150$ m and the potential there is $130$ V, exactly a third of $390$ V, as a $1/r$ law demands. Second, the field: $E = V/r$ holds for a sphere, and $390/0.150 = 2.60\times10^{3}\ \mathrm{V/m}$, matching the line computed from $kQ/r_0^{2}$ by a different route.
One product and three divisions, with the interior value obtained by copying rather than computing.
For any charged sphere, remember the surface pair: $V = kQ/r_0$ and $E = V/r_0$. Almost every sphere question is one of those two rearranged, and quoting a smaller sphere at the same potential immediately tells you it has the stronger surface field.
Two spheres joined by a wire share potential, not charge
A metal sphere of radius $6.00\ \mathrm{cm}$ and another of radius $2.00\ \mathrm{cm}$ are joined by a long thin wire and given $24.0\ \mathrm{nC}$ between them. They are far enough apart that neither disturbs the other. Find the charge on each, their common potential, and the field just outside each surface.
Given
radii $r_1 = 6.00\ \mathrm{cm}$ and $r_2 = 2.00\ \mathrm{cm}$
total charge $Q_1 + Q_2 = 24.0\ \mathrm{nC}$
joined by a conductor, and far apart, so each stays spherically symmetric
Find
the two charges, the common potential and the two surface fields
SolutionWrite down what being joined actually forces
a conductor is one equipotential, and the wire makes the two spheres one conductor; equal potential is the constraint, not equal charge, and choosing the wrong one is the whole difficulty of the problem
Check the second sphere independently instead of trusting the split: $kQ_2/r_2 = (8.99\times10^{9})(6.00\times10^{-9})/0.0200 = 2.70\times10^{3}\ \mathrm{V}$, the same potential, as it must be. Order of magnitude on the fields: the smaller sphere is at $1.35\times10^{5}\ \mathrm{V/m}$, which is a twentieth of the field at which dry air breaks down, so this arrangement holds its charge but a much smaller tip on the same wire would not.
One ratio, one split, one division and one check; no integral and no field calculation from charges at all.
Sharp things spark first, and this is why: joined conductors share a potential, and at a fixed potential the surface field goes as one over the radius of curvature. A pointed tip is a very small radius, so the field there runs away long before anything happens on the flat parts.
Checkpoint
§04.5 — inside a charged conducting sphere●●○○○
Thirty seconds, and two of the three answers require no arithmetic at all. An isolated metal sphere of radius $5.00\ \mathrm{cm}$ is held at a potential of $900\ \mathrm{V}$ relative to infinity.
Given
conducting sphere, radius $5.00\ \mathrm{cm}$
surface potential $900\ \mathrm{V}$, measured from zero at infinity
isolated and in electrostatic equilibrium
Find
(a) What is the potential at the centre of the sphere?
(b) What is the electric field at the centre?
(c) How much charge is on the sphere?
Hint 1/4
Two of these are statements about a conductor at rest and need no calculation. Only the third one is arithmetic.
Hint 2/4
Inside a conductor at rest, $E = 0$ and therefore $V$ is constant and equal to its surface value; and for a sphere $V(r_0) = kQ/r_0$.
Hint 3/4
Here the surface value is $900\ \mathrm{V}$ and the radius is $0.0500\ \mathrm{m}$.
Hint 4/4
The centre is at $900\ \mathrm{V}$, the field there is zero, and the charge is $5.01\ \mathrm{nC}$.
Put the charge back in: $kQ/r_0 = (8.99\times10^{9})(5.01\times10^{-9})/0.0500 = 901\ \mathrm{V}$, recovering the given surface potential to the rounding. Scale check: five nanocoulombs on a sphere the size of a plum gives about a kilovolt, which is why laboratory electrostatics runs at hundreds and thousands of volts with charges far too small to feel.
⚠ Setting the potential inside a conductor to zero because the field is zero
the two statements sit next to each other in every summary, and zero field feels like nothing there
4.6The potential of a spread out charge: one integral, no components
Cut the charge into elements, divide each by its own distance, and add: no angles enter anywhere.
Point charges are done; the objects the previous section had to attack with vector integrals are now worth revisiting, because the scalar version of the same sum is dramatically shorter.
MethodMethod 4.6: the potential of a continuous charge distribution
Conditions
The distribution has finite extent, so the zero at infinity is available
$r$ is the distance from each element to the field point, and it must be expressed in the integration variable
The density carries the sign of the charge, so a negatively charged object gives a negative potential
Chop the object into pieces small enough that each one counts as a point charge, divide the charge of each piece by its own distance to the field point, and add up the results. Because potential has no direction there is nothing to resolve, no component to keep and no component to throw away, so the only difficulty left is writing the distance in terms of whatever variable you are integrating over. Here the symbol for a volume element is written as tau, so that it cannot be confused with the potential.
Proof
The potential of one point charge is $kQ/r$, and the potential of several is the sum of those numbers, as the point charge block established.
Take the limit in which the charges become infinitely many and infinitely small: the sum $k\sum Q_i/r_i$ becomes the integral $k\int dq/r$, with $dq$ the charge of one element.
The step that was hard for fields has no counterpart here. There, each contribution was a vector and had to be resolved before it could be added; the cosine factor and the symmetry argument were the price. Here every contribution is a number.
What remains is bookkeeping: express $dq$ through the density, express $r$ through the integration variable, and set the limits so that $\int dq$ over them recovers the total charge.
A $\textcolor{#1f6feb}{\text{ring}}$ seen edge on, with the field point P on its axis. Every element of the ring is the same $\textcolor{#8250df}{\text{slant distance}}$ from P, so the whole integral collapses to the total charge divided by that one distance. This is the geometry in which the scalar method is at its most economical: for the field, the same picture still needed a cosine factor and a cancellation argument.
Looks like this, but is not
The field at the centre of a charged ring is zero by symmetry, so the potential at the centre is zero too. Every element cancels, so nothing is left.
The cancellation at the centre is a cancellation of directions, and the potential has no directions to cancel. Every element sits at the same distance $R$ and every one of them contributes the same positive number, so they reinforce instead: $V = kQ/R$, which for a ring of radius $8.00\ \mathrm{cm}$ carrying $12.0\ \mathrm{nC}$ is $1.35\times10^{3}\ \mathrm{V}$. The centre of a ring is in fact the place where its potential is largest, and it is precisely because the potential is at a maximum there that its slope, and so the field, is zero.
Potential on the axis of a ring, and the ring's field for free
A ring of radius $8.00\ \mathrm{cm}$ carries $+12.0\ \mathrm{nC}$ spread uniformly. Find the potential at a point on its axis $6.00\ \mathrm{cm}$ from the centre, then obtain the field there by differentiation and compare it with the result the previous section derived by integrating vectors.
Given
ring radius $R = 8.00\ \mathrm{cm}$, total charge $Q = +12.0\ \mathrm{nC}$, uniform
field point on the axis at $x = 6.00\ \mathrm{cm}$ from the centre
the previous section's result for comparison: $E = kQx/(x^{2}+R^{2})^{3/2}$
Find
the potential at that point, and the field obtained from it
SolutionNotice that the distance does not depend on the integration variable
every element of the ring is this same slant distance from the field point, so it comes out of the integral as a constant and there is nothing left to integrate but dq
pulling a constant out of an integral is legitimate precisely because the distance is the same for every element, and this is the only geometry in which it is
the denominator is the cube of the slant distance, that is 0.100 cubed, which keeps the arithmetic in round numbers
Answer $$\boxed{\,V = 1.08\times10^{3}\ \mathrm{V},\qquad E_x = 6.47\times10^{3}\ \mathrm{V/m}\ \text{along the axis, away from the ring}\,}$$
Check
The field expression obtained here by differentiating a scalar is character for character the one the previous section obtained by resolving vectors and cancelling components, which is an independent derivation arriving at the same place. Two further limits check it: at $x = 0$ the derivative gives zero, matching the symmetry argument at the centre, and for $x \gg R$ it becomes $kQ/x^{2}$, the point charge field, as any compact object must.
The potential took one division. The field took one chain rule. The previous section's route to the same field took an integral with a cosine factor and a pairing argument, and it is worth counting the lines.
The general lesson is that for a symmetric object it is nearly always cheaper to integrate the potential and then differentiate, than to integrate the field. You trade a vector integral for a scalar integral plus one derivative, and derivatives are the cheap operation.
Potential of a rod at a point on its own axis
A thin rod of length $24.0\ \mathrm{cm}$ carries $+7.50\ \mathrm{nC}$ uniformly. Find the potential at a point on the rod's own axis, $6.00\ \mathrm{cm}$ beyond its near end, and check the answer by recovering the rod's field from it.
the question gives a total charge but the integral needs a charge per unit length, and manufacturing it is always the first line
$$dq = \lambda\,ds,\qquad r = s,\qquad s: a \to a+L$$
putting the field point at the origin makes the distance to an element equal to the coordinate itself, which is the cheapest possible choice of variable
the logarithm appears because the potential falls as one over the distance rather than one over its square; the field of the same rod gave no logarithm at all
identical in form to the expression the previous section obtained by integrating the field itself
Answer $$\boxed{\,V = 452\ \mathrm{V},\qquad E = 3.75\times10^{3}\ \mathrm{V/m}\ \text{along the axis, away from the rod}\,}$$
Check
Independent bound rather than a repeat of the arithmetic. If all the charge were squashed to the far end the potential would be $kQ/0.300 = 225\ \mathrm{V}$, and if all of it were at the near end it would be $kQ/0.0600 = 1124\ \mathrm{V}$; the true value must lie between, and $452$ does. The point charge estimate that puts everything at the middle gives $kQ/0.180 = 375\ \mathrm{V}$, seventeen per cent low, which is the usual sign that the near end is doing more than its share.
One logarithm for the potential, one derivative for the field, and the derivative reproduced a result that cost a separate integral last week.
Whenever a potential comes out as a logarithm, the corresponding field will be a difference of two reciprocals, and vice versa. Recognising that pairing lets you check either one against the other without redoing the integral.
Checkpoint
§04.6 — potential at the centre of a ring●●○○○
Thirty seconds, and the integral is already done for you by the geometry. A ring of radius $10.0\ \mathrm{cm}$ carries $+5.00\ \mathrm{nC}$ spread uniformly around it.
Given
ring radius $R = 10.0\ \mathrm{cm}$
total charge $Q = +5.00\ \mathrm{nC}$, uniform
field point at the centre of the ring
Find
(a) Find the potential at the centre of the ring.
(b) What is the electric field there, and is that consistent with your answer to (a)?
Hint 1/4
Ask how far each element of the ring is from the centre. If the answer is the same for all of them, there is nothing left to integrate.
Hint 2/4
$V = k\int dq/r$, and with $r$ constant this is $kQ/r$.
Hint 3/4
Here every element is $R = 0.100\ \mathrm{m}$ from the centre and the total charge is $5.00\times10^{-9}\ \mathrm{C}$.
Hint 4/4
The potential is $450\ \mathrm{V}$ and the field is zero, which is consistent because the centre is where the potential is largest.
the general axial expression has a factor of x in the numerator, so it vanishes at the centre without any symmetry argument being needed
Answer $$\boxed{\,V = 450\ \mathrm{V},\qquad E = 0\,}$$
Check
Check the pair against the general axial potential $V = kQ/\sqrt{x^{2}+R^{2}}$: it is largest at $x = 0$, so its derivative there is zero and the field must vanish, agreeing with the direct symmetry argument by a completely different route.
⚠ Carrying a cosine factor into the potential integral
the field integral for the same ring needed one, and the two set-ups look identical up to that factor
wrong$$V = k\int \frac{dq}{r}\cos\theta$$
right$$V = k\int \frac{dq}{r}$$
⚠ Using the axial distance instead of the slant distance
x is the number the question names and it is the distance the picture invites you to measure
Count every pair once, add the pair energies with their signs, and the total is the work of assembly.
Everything so far treated one charge moving in a field that somebody else was holding up; the last question is what it cost to hold that field up in the first place.
TheoremResult 4.7: the electrostatic potential energy of an arrangement of point charges
Conditions
Point charges, or bodies small enough and far enough apart to count as points
Each pair counted exactly once; the separations are the ones in the final arrangement
Measured from all the charges infinitely far apart, so a negative total means the arrangement is bound
$$\boxed{\,U = k\sum_{\text{pairs } i<j} \frac{q_iq_j}{r_{ij}} = W_{\rm ext}\ \text{to assemble it from infinity}\,}$$
Take every pair of charges in the arrangement, work out the Coulomb constant times the product of the two charges divided by the distance between them, and add all of those numbers up, each pair once and once only. The result is the work an outside agent must do to build the arrangement by carrying the charges in slowly from infinitely far away. A positive total means the agent had to push, and the arrangement will fly apart if released; a negative total means the agent had to hold back, and the arrangement is bound.
Proof
Bring the charges in one at a time. The first one costs nothing: there is nothing there for it to interact with.
The second one arrives in the field of the first, so it costs $q_2V_1 = kq_1q_2/r_{12}$, using the definition of potential.
The third arrives in the field of the first two, so it costs $q_3(V_1+V_2) = kq_1q_3/r_{13} + kq_2q_3/r_{23}$, and the pattern continues.
Adding up the whole sequence produces every pair exactly once, which is what the notation $i<j$ enforces. Writing $\sum_{i\ne j}$ instead counts each pair twice and gives a total that is twice too big, and that is the commonest error in this block.
The answer does not depend on the order the charges were brought in, because the total work is a difference of stored energies and those depend only on the final arrangement.
Three charges, three $\textcolor{#8250df}{\text{pair energies}}$, and only three: the pairs are counted once each, not once from each end. One pair repels and contributes a positive number, two pairs attract and contribute negative ones, and the total of $-1.05\ \mu\mathrm{J}$ says that pulling this arrangement apart would cost an outside agent $1.05\ \mu\mathrm{J}$.
Looks like this, but is not
The stored energy of this arrangement is negative, so it has less than nothing and will fly apart the moment it is released. Negative energy sounds unstable.
Negative means bound, not unstable. The zero was chosen to be the state in which every charge is infinitely far from every other, so a negative total says that reaching that state requires energy to be supplied from outside. The triangle above has $U = -1.05\ \mu\mathrm{J}$, which means an agent must spend $1.05\ \mu\mathrm{J}$ to scatter it. An arrangement that would fly apart on its own is one with a positive total, like two like charges held together.
pair
product of charges
separation
energy
+2.00 and +3.00 nC
+6.00 × 10⁻¹⁸ C²
12.0 cm
+0.449 μJ
+2.00 and -4.00 nC
−8.00 × 10⁻¹⁸ C²
12.0 cm
−0.599 μJ
+3.00 and -4.00 nC
−12.0 × 10⁻¹⁸ C²
12.0 cm
−0.899 μJ
total
−14.0 × 10⁻¹⁸ C²
12.0 cm
−1.05 μJ
Only one of the three pairs repels, and it is the pair of the two smallest charges, so its positive contribution is the smallest of the three. The largest single term belongs to the pair containing both of the largest charges. Reading down the energy column, the total is negative and about a microjoule, so the arrangement is bound and taking it apart costs an outside agent about a microjoule.
Energy of three charges at the corners of a triangle
Charges of $+2.00\ \mathrm{nC}$, $+3.00\ \mathrm{nC}$ and $-4.00\ \mathrm{nC}$ sit at the corners of an equilateral triangle of side $12.0\ \mathrm{cm}$. Find the electrostatic potential energy of the arrangement, and say what an outside agent would have to do to take it apart.
a microjoule, which is the scale of nanocoulombs at centimetre separations and worth remembering as a benchmark
$$W_{\rm ext} = 0 - U = +1.05\ \mu\mathrm{J}$$
taking it apart means going to the zero state, so the work needed is minus the stored value, positive here
Answer $$\boxed{\,U = -1.05\ \mu\mathrm{J};\ \text{scattering the three charges to infinity costs } +1.05\ \mu\mathrm{J}\,}$$
Check
Independent route, pair by pair rather than by factoring: $U_{12} = +0.449\ \mu\mathrm{J}$, $U_{13} = -0.599\ \mu\mathrm{J}$, $U_{23} = -0.899\ \mu\mathrm{J}$, and $0.449 - 0.599 - 0.899 = -1.049$, agreeing with the bracket method. Sign check without arithmetic: the largest charge is the negative one and it attracts both of the others, so the two negative terms involve the biggest products and the total has to come out negative.
Three products, one sum, one division. The pair by pair route needs three divisions instead of one, which is why the common separation was factored out first.
Two habits transfer from this. Write the pair list before any numbers, and factor out a common separation whenever the geometry offers one. Both are worth more on a longer arrangement: four charges give six pairs and the list is where the marks are lost.
The cost of adding one more charge, worked out two ways
Charges of $+5.00\ \mathrm{nC}$ and $-2.00\ \mathrm{nC}$ are fixed $10.0\ \mathrm{cm}$ apart. An agent carries a $+3.00\ \mathrm{nC}$ bead in from infinity and places it at the third corner of an equilateral triangle, $10.0\ \mathrm{cm}$ from each. Find the work the agent does, and then the total energy of the finished arrangement.
The two routes to $W_{\rm ext}$ are genuinely independent: one used the potential at a point and never mentioned pairs, the other used pairs and never mentioned potential, and they agree to three figures. Sign check on the total: the arrangement is only just bound, because the strong repulsion between the two positive charges nearly cancels the two attractions, and a total of $-0.0899\ \mu\mathrm{J}$ against pair terms of about a microjoule each says exactly that.
One route needed a potential and one multiplication; the other needed two pair energies. On a bigger arrangement the potential route wins, because it grows with the number of charges already present rather than with the number of new pairs.
Whenever a question asks what it costs to bring one more charge into an existing arrangement, compute the potential the existing charges make at the destination and multiply by the newcomer's charge. Rebuilding the whole pair sum is the slow way, and it repeats work that has not changed.
The energy of a dipole sitting in an external field
The same dipole, $p = 2.00\times10^{-12}\ \mathrm{C\,m}$, is now placed in a uniform external field of magnitude $E = 2.50\times10^{5}\ \mathrm{N/C}$, with its moment at an angle $\theta$ to the field. Find its potential energy as a function of $\theta$, evaluate it at $\theta = 0$, $60.0^{\circ}$ and $180^{\circ}$, and find the work an outside agent must do to turn the dipole from along the field to against it.
Given
$p = 2.00\times10^{-12}\ \mathrm{C\,m}$, a rigid pair $\pm q$ separated by $\ell$
uniform external field $E = 2.50\times10^{5}\ \mathrm{N/C}$
the field is uniform over the whole dipole, so both charges see the same $E$
Find
the energy as a function of angle, three values of it, and the work to reverse the dipole
SolutionEach charge carries its own place's potential
the pair sum of the block above has one pair, and it is the same fixed internal number at every angle, so it can be left out of a difference; what changes with angle is only the energy of each charge in the outside field
take the external field along $x$; the constant drops out of the difference, which is the usual reason a potential is allowed to have an unnamed zero at all
again only the product $q\ell$ survives; the zero of this energy sits at $\theta = 90^{\circ}$, not at infinity, because the dipole is never taken away from the field
lowest when the moment lines up with the field, highest when it points against it, which is the ordering the torque already predicted in the previous section
Get the last number a second time from the torque, which never mentions a potential at all. Turning against the electric torque $\tau = pE\sin\theta$ costs $\int_0^{\pi} pE\sin\theta\,d\theta = pE\left[-\cos\theta\right]_0^{\pi} = 2pE = 1.00\times10^{-6}\ \mathrm{J}$, the same answer by a route through forces rather than through energies. Sign check as well: the energy is least where the dipole would settle on its own, which is along the field, and that is where the formula puts its minimum.
One difference of potentials and one cosine. Doing it as two separate charges in two separate places, each with its own force and its own displacement, takes about ten lines and gives the same $-pE\cos\theta$.
Two different dipole formulas now exist and they are easy to swap by accident. $V = kp\cos\theta/r^{2}$ is what the dipole does to the space around it; $U = -\vec p\cdot\vec E$ is what an outside field does to the dipole. One has an $r$ in it and the other does not, which is the quickest way to tell which one a question is asking for.
Checkpoint
§04.7 — two protons a nanometre apart●●○○○
Thirty seconds, and it fixes the energy scale for the rest of the course. Two protons are held $1.00\ \mathrm{nm}$ apart, which is a few atomic diameters.
Given
two protons, each of charge $+1.602\times10^{-19}\ \mathrm{C}$
Shortcut check that avoids the joules altogether: the potential of one proton at $1.00$ nm is $ke/r$ with $e$ in coulombs, which is $1.44\ \mathrm{V}$, and one elementary charge sitting at $1.44\ \mathrm{V}$ has by definition $1.44\ \mathrm{eV}$. Same answer, no conversion done anywhere.
⚠ Counting every pair twice
each charge feels the other, so it seems that both directions of the interaction should be added
the force between the same two charges does have a square, and the two expressions differ by nothing else
wrong$$U = \frac{kq_1q_2}{r^{2}}$$
right$$U = \frac{kq_1q_2}{r}$$
⚠ Reading a negative total as an unstable arrangement
negative sounds like deficit, and deficits sound like something about to collapse
wrong$$U < 0 \Rightarrow \text{flies apart on release}$$
right$$U < 0 \Rightarrow \text{bound; } W_{\rm ext} = -U > 0 \text{ is needed to separate it}$$
Energy problems in an electric field
Any question that asks for a speed, a , a work, or whether a charge can reach somewhere. Six steps, of which only the last is arithmetic; if the first five are done properly there is no equation of motion anywhere.
Name the two states, start and finish
Write down where the charge is at the beginning and where it is at the end, as two distances or two coordinates. Energy methods know nothing about the journey between them, so any effort spent describing the path is wasted.
Decide who is moving and who is held
Charges that are fixed contribute a potential; the charge that moves contributes its own charge and mass. If more than one charge is free to move, the pair energies of every moving pair change and the problem is a system problem, not a single particle one.
Write the potential at each of the two places
Add the contributions of the fixed charges as signed numbers, $V = k\sum Q_i/r_i$, once for the start and once for the finish. For a uniform field use $V = -Ed$ measured along the field instead.
Turn the potential difference into an energy
$\Delta U = q\,(V_{\rm finish} - V_{\rm start})$, with the sign of the moving charge included. Then $\Delta K = -\Delta U$ if nothing else does work on it. Check the sign against common sense before going on: does the charge speed up or slow down?
Bring in the mass only now
$\tfrac12 mv_f^{2} - \tfrac12 mv_i^{2} = \Delta K$. The mass has played no part until this line, which is why the same energy calculation serves a proton, an electron and a dust grain.
Check the order of magnitude and the sign
Say out loud what the number means: hundreds of metres per second for a bead, hundreds of kilometres per second for a proton at kilovolt energies, tens of millions for an electron. If the speed exceeds the value it would reach at infinity, the arithmetic is wrong.
Where it goes wrong
The potential of the moving charge itself is included in the potential it moves through, which double counts and usually gives an infinite answer.
The sign of the moving charge is dropped, so a negative charge is made to speed up where it should slow down.
The kinetic energy at the start is assumed to be zero when the question did not say the charge was released from rest.
Centimetres are put into the formula without conversion, which changes the answer by a factor of ten in the square root and is invisible in the layout of the working.
Deciding whether to reach for the field or for the potential
At the start of any question involving several charges, before any formula is written. Choosing wrongly here costs five lines of vector algebra that were never needed.
Read what the answer has to be
A force, an acceleration, a direction or a field: you need the field, and you will be adding vectors. An energy, a work, a speed, a potential or a potential difference: you need the potential, and you will be adding signed numbers.
If it is a potential, just add the numbers
Distances only, no angles, no components, signs included. Nothing about the geometry matters except how far each charge is from the point.
If it is a field and the potential is easy, get the potential first
For a symmetric object it is usually cheaper to integrate the scalar potential and then differentiate once than to integrate the vector field. The ring and the rod in this section are both examples.
If a question asks for both, do the potential first
The potential often gives the field by one derivative, but the field only gives the potential by an integral. Working in the cheap direction is worth a couple of minutes in an examination.
Say which zero you are using before you write a number
Infinity for isolated charges and finite objects, a named plate for a parallel plate arrangement. Half of the sign errors in this material are a zero that was never announced.
Where it goes wrong
The potential is computed with components, as though it were a field, which produces an answer that is too small and looks plausible.
The field is computed by dividing the potential by a distance, which is only valid for a uniform field and for the surface of a sphere.
A question asking for the work done between two points is answered with a force, which then has to be integrated by hand along the path.
Two different zeros are used in two parts of the same question, so the parts disagree by a constant that never cancels.
Where the potential is zero but the field is large
Charges of $+8.00\ \mathrm{nC}$ and $-8.00\ \mathrm{nC}$ are $10.0\ \mathrm{cm}$ apart. Find the potential and the field at the midpoint of the line joining them.
Given
$+8.00\ \mathrm{nC}$ at $x = 0$ and $-8.00\ \mathrm{nC}$ at $x = 10.0\ \mathrm{cm}$
field point at $x = 5.00\ \mathrm{cm}$, midway between them
each charge is the same distance away and has the same magnitude, so the two sizes are equal
$$E = E_1 + E_2 = 5.75\times10^{4}\ \mathrm{N/C},\ \text{pointing from the positive charge to the negative one}$$
the positive charge pushes a test charge to the right and the negative one pulls it to the right, so the two directions agree and the magnitudes add
Answer $$\boxed{\,V = 0,\qquad E = 5.75\times10^{4}\ \mathrm{N/C}\,}$$
Check
Symmetry check without arithmetic: reflect the picture through the midpoint and swap the two charges. The potential must map to minus itself, so it can only be zero; the field maps to itself, so it need not be, and it is not.
Cancellation in a scalar sum and cancellation in a vector sum are different events, and an arrangement that produces one will often produce the opposite of the other.
Where the field is zero but the potential is not
Charges of $+8.00\ \mathrm{nC}$ and $-2.00\ \mathrm{nC}$ are $10.0\ \mathrm{cm}$ apart. Find the point on the line through them where the total field is zero, and the potential there.
Given
$+8.00\ \mathrm{nC}$ at $x = 0$ and $-2.00\ \mathrm{nC}$ at $x = 10.0\ \mathrm{cm}$
the null point of the field lies beyond the smaller charge
Find
the position where E vanishes, and the potential at that position
taking square roots first keeps the equation linear, and only the region beyond the smaller charge can balance
$$2.828x - 0.2828 = 1.414x \Rightarrow x = 0.200\ \mathrm{m}$$
a factor of four in the charges means a factor of two in the distances, which places the null point twice as far from the big charge as from the small one
the same two charges and the same point, but now added as numbers, and they do not cancel because the distances are not in the ratio of the charges
Answer $$\boxed{\,E = 0 \text{ at } x = 20.0\ \mathrm{cm},\qquad V = 180\ \mathrm{V} \text{ there}\,}$$
Check
Substitute back into the field condition: $8.00/(0.200)^{2} = 200$ and $2.00/(0.100)^{2} = 200$, equal, so the null point is right. And the potential condition would need $8.00/x = 2.00/(x-0.100)$, giving $x = 0.133$ m, a different place entirely, which confirms that the two zeros cannot coincide here.
A charge placed at the field null sits there without being pushed, but it still took 180 volts worth of energy per coulomb to bring it in. Equilibrium and zero energy are unrelated ideas.
The same two quantities are wanted at one point of a two charge arrangement, and in the first case the scalar vanishes while the vector is at its largest, in the second the vector vanishes while the scalar is a comfortable 180 volts.
How to tell them apart
Before adding anything, ask whether the thing you are adding has a direction. If it does, opposite charges at equal distances reinforce and equal charges at equal distances cancel; if it does not, exactly the reverse happens. Nothing else distinguishes the two calculations, and they use the same numbers.
Proton through 800 V
A proton starts from rest and crosses a potential difference of $800\ \mathrm{V}$, falling from high potential to low. Find the energy it gains and its final speed.
the mass finally enters, and it is the only place the identity of the particle beyond its charge has any effect
Answer $$\boxed{\,\Delta K = 800\ \mathrm{eV},\qquad v = 3.92\times10^{5}\ \mathrm{m/s}\,}$$
Check
Scale check: a kilovolt proton moves at a few hundred kilometres per second, about a thousandth of the speed of light, so the classical formula is safe here to far better than one per cent.
The electron volt answer was free and the metres per second answer cost a square root. When a question offers you the choice, quote the energy in electron volts and convert only if asked.
Alpha particle through the same 800 V
An alpha particle, of charge $+2e$ and mass $6.64\times10^{-27}\ \mathrm{kg}$, starts from rest and crosses the same $800\ \mathrm{V}$ drop. Find the energy it gains and its final speed.
twice the energy but roughly four times the mass, so the speed comes out smaller than the proton's rather than larger
Answer $$\boxed{\,\Delta K = 1600\ \mathrm{eV},\qquad v = 2.78\times10^{5}\ \mathrm{m/s}\,}$$
Check
Ratio check without redoing the arithmetic: the speed ratio should be $\sqrt{(2)(m_p/m_\alpha)} = \sqrt{2 \times 0.2515} = 0.709$, and $2.78/3.92 = 0.709$, agreeing exactly.
Two particles through the same gap gain energies in the ratio of their charges and reach speeds in the ratio of the square roots of charge over mass. Neither ratio is one, and neither is the other.
The same potential difference is crossed by two different particles: the energies differ by a factor of two, exactly the ratio of the charges, while the speeds differ by a factor of 0.709, which involves the masses as well, so the heavier and more highly charged particle ends up the slower of the two.
How to tell them apart
Ask what the question wants. If it wants an energy, only the charge matters and the answer in electron volts can usually be written down without a calculator. If it wants a speed, the mass enters through a square root, and a bigger energy does not guarantee a bigger speed.
Scaffolding comes off
The common skeleton
Name the start state and the finish state as two positions, and say what is fixed and what is free to move
Write the potential at the start, adding the fixed charges as signed numbers
Write the potential at the finish the same way, using the separations in the finish state
Form the change in potential energy as the moving charge times the change in potential
Convert it into a change of kinetic energy with a sign that you have argued, not copied
Bring in the mass at the last step, then check the size and the sign of the answer
1 · fully worked
A bead released between two equal charges
Two charges of $+6.00\ \mathrm{nC}$ each are fixed at $(0,0)$ and at $(8.00\ \mathrm{cm},\,0)$. A bead of mass $3.00\ \mathrm{mg}$ carrying $+2.00\ \mathrm{nC}$ is released from rest at $(4.00\ \mathrm{cm},\,3.00\ \mathrm{cm})$. How fast is it moving when it is very far away?
Given
two fixed charges $Q = +6.00\ \mathrm{nC}$, at $(0,0)$ and $(8.00\ \mathrm{cm},\,0)$
bead: $q = +2.00\ \mathrm{nC}$, $m = 3.00\ \mathrm{mg} = 3.00\times10^{-6}\ \mathrm{kg}$, released from rest
release point $(4.00\ \mathrm{cm},\,3.00\ \mathrm{cm})$, on the perpendicular bisector
the release point sits on the perpendicular bisector, so both separations are the same three four five hypotenuse, and by symmetry the bead never leaves that line
Independent check by halving the problem. One charge alone at $5.00$ cm would give $1079$ V and a final speed of $\sqrt{2(2.00\times10^{-9})(1079)/(3.00\times10^{-6})} = 1.20\ \mathrm{m/s}$; two identical charges double the energy and so multiply the speed by $\sqrt2$, giving $1.70\ \mathrm{m/s}$, matching. Plausibility: microjoules of energy on a milligram bead give metres per second, which is the right scale for a demonstration on a bench.
One distance, one doubling, one subtraction and one square root. No force was ever computed and no direction was ever needed.
Notice how little geometry was used: only the distance from each fixed charge to the release point. The bead's actual path, which curves away along the bisector, never entered the calculation and does not have to.
2 · you write the reasoning
Easier than rung 1, because the potentials are handed to you and there is only one charge to track. An electron moves from a point where the potential is $-60.0\ \mathrm{V}$ to a point where it is $+150.0\ \mathrm{V}$, starting from rest. The three lines below are correct and the final speed is right. Before opening the model reasons, say in your own words why each line is allowed, paying particular attention to the sign in the second one.
reasoning
The subscript order fixes the subtraction: $V_{ba}$ means the potential at the finish minus the potential at the start, so the two given numbers go in that order and not the other. Both signs are kept, including the minus in front of the sixty, which is why the difference comes out as two hundred and ten rather than as ninety. Reversing the subscripts would reverse the sign of everything downstream, and the speed would come out as the square root of a negative number, which is how you would know.
reasoning
The change in potential energy is $q V_{ba}$ and the change in kinetic energy is minus that, because no other force does work. Here $q$ is negative and $V_{ba}$ is positive, so $\Delta U$ is negative and the kinetic energy rises. That is the physical content of the line: an electron speeds up when it moves towards higher potential, which is the opposite of a proton's behaviour and is the single most common sign error in this material. The electron volt figure comes free, because one elementary charge across 210 volts is 210 electron volts whatever the sign.
reasoning
The electron began at rest, so its whole kinetic energy is the change just computed, and $\tfrac12mv^{2} = \Delta K$ can be solved directly. The mass enters here for the first and only time. Ten million metres per second is about three per cent of the speed of light, so the classical formula is comfortably good here, unlike the five kilovolt case earlier in the section where the correction reached one per cent.
3 · find the buried error
Harder than rung 2, because two source charges are present, one of them negative, and the separations change during the motion. A charge $Q_1 = +9.00\ \mathrm{nC}$ is fixed at $x = 0$ and a charge $Q_2 = -4.00\ \mathrm{nC}$ is fixed at $x = 12.0\ \mathrm{cm}$. A proton is released from rest at $x = 6.00\ \mathrm{cm}$ and moves along the axis. A student's solution for its speed at $x = 9.00\ \mathrm{cm}$ is written out below and reaches $1.55\times10^{5}\ \mathrm{m/s}$. Exactly two of its four steps are faulty. Find them.
the two buried errors (2)
⚠ step 2
The distance to the negative charge is wrong. The proton is at $x = 0.0900\ \mathrm{m}$ and $Q_2$ sits at $x = 0.120\ \mathrm{m}$, so the separation is $0.0300\ \mathrm{m}$, not $0.0900\ \mathrm{m}$. With the right distance, $V_f = (8.99\times10^{9})(1.00\times10^{-7} - 1.333\times10^{-7}) = -300\ \mathrm{V}$, negative rather than positive.
The coordinate of the field point and the distance to a charge are the same number for the charge sitting at the origin, and it is easy to carry that coincidence over to the second charge without noticing. The wrong line also looks tidy, because both terms end up over the same denominator, and tidiness reads as correctness.
right
Always write the separation as the difference of two coordinates, $\lvert x_{\rm field} - x_{\rm charge}\rvert$, even when one of them is zero. Here that gives $\lvert 0.0900 - 0.120\rvert = 0.0300\ \mathrm{m}$. A quick sanity test: the proton has moved closer to the negative charge, so that charge's negative contribution must have grown in size, and any line in which it shrinks is wrong.
⚠ step 4
The factor of two is missing. From $\tfrac12 mv^{2} = \Delta K$ it follows that $v = \sqrt{2\Delta K/m}$, not $\sqrt{\Delta K/m}$, so every speed obtained this way is too small by a factor of $\sqrt2$.
The relation between energy and speed is used so often that it gets written from memory, and the half in front of the mass is the part that disappears. The error is invisible in the units, which come out as metres per second either way, and invisible in the order of magnitude, which is only forty per cent off.
right
Write the equation before rearranging it: $\tfrac12 mv^{2} = \Delta K$, then multiply both sides by two and only then divide by the mass. With the corrected energy of $1.680\times10^{-16}\ \mathrm{J}$ this gives $v = \sqrt{2(1.680\times10^{-16})/(1.67\times10^{-27})} = 4.49\times10^{5}\ \mathrm{m/s}$.
4 · the bare problem
§04.7 — a negative bead falling towards a positive charge●●●○○
No scaffolding this time. A charge of $+7.00\ \mathrm{nC}$ is fixed in place. A small bead of mass $4.00\times10^{-8}\ \mathrm{kg}$ carrying $-2.50\ \mathrm{nC}$ is released from rest $4.00\ \mathrm{cm}$ away from it and is free to move along the line joining them.
Given
fixed charge $+7.00\ \mathrm{nC}$
bead: $q = -2.50\ \mathrm{nC}$, $m = 4.00\times10^{-8}\ \mathrm{kg}$, released from rest at $4.00\ \mathrm{cm}$
the bead is asked about when it has reached $2.00\ \mathrm{cm}$ from the fixed charge
Find
(a) Does the bead speed up or slow down as it moves from $4.00\ \mathrm{cm}$ to $2.00\ \mathrm{cm}$? Say why in one sentence before calculating.
(b) Find its speed at $2.00\ \mathrm{cm}$.
Hint 1/4
The two charges have opposite signs, so decide first which way the bead is pulled, and then whether the trip described is with that pull or against it.
Hint 2/4
$U = kQq/r$ with both signs included, and $\tfrac12mv^{2} = U_i - U_f$ for a release from rest.
Hint 3/4
Here $kQq = (8.99\times10^{9})(7.00\times10^{-9})(-2.50\times10^{-9}) = -1.573\times10^{-7}\ \mathrm{J\,m}$, with $r_i = 0.0400\ \mathrm{m}$ and $r_f = 0.0200\ \mathrm{m}$.
Hint 4/4
It speeds up, and reaches $14.0\ \mathrm{m/s}$.
Show solutionThe qualitative answer, before any arithmetic
the mass enters last, and forty nanograms is light enough that a few microjoules produce a speed you could see
Answer $$\boxed{\,\text{it speeds up};\qquad v = 14.0\ \mathrm{m/s}\,}$$
Check
Independent route through the potential rather than the pair energy. $V_i = kQ/r_i = 62.93/0.0400 = 1573\ \mathrm{V}$ and $V_f = 62.93/0.0200 = 3147\ \mathrm{V}$, so $\Delta U = q(V_f - V_i) = (-2.50\times10^{-9})(1574) = -3.93\times10^{-6}\ \mathrm{J}$, the same energy release, arrived at without ever writing a pair energy.
The qualitative answer to part (a) is worth more than it looks: it fixes the sign of part (b) in advance, so any arithmetic that produces a negative kinetic energy is known to be wrong before the square root is attempted.
Full exam-style question
Rod, bead and the sign of an external workexam format
A thin rod of length $20.0\ \mathrm{cm}$ carries $+16.0\ \mathrm{nC}$ spread uniformly along it. Point A lies on the rod's own axis, $5.00\ \mathrm{cm}$ beyond the near end; point B lies on the same axis, $45.0\ \mathrm{cm}$ beyond the near end. A grain of mass $2.00\times10^{-9}\ \mathrm{kg}$ carrying $-3.00\ \mathrm{nC}$ is available. (a) Find the potential at A and at B. (b) Find the work an external agent must do to carry the grain slowly from B to A. (c) The grain is instead released from rest at B. Find its speed when it reaches A. (d) Explain the sign of your answer to (b) in one sentence.
two micrograms is very light, so a few microjoules make it fast; the mass enters only here
(d) The sentence
$$q < 0,\ V_A > V_B \Rightarrow \Delta U < 0$$
a negative charge is attracted to a positively charged rod, so moving it towards the rod is downhill for it and the agent has to hold it back rather than push it
Two checks by different routes. First, bound the potential at A: all the charge at the near end would give $kQ/0.0500 = 2876\ \mathrm{V}$ and all of it at the far end $kQ/0.250 = 575\ \mathrm{V}$, and $1157\ \mathrm{V}$ lies between, as it must. Second, the answers to (b) and (c) must agree in size, because both are the same energy difference read in opposite directions; $2.68\ \mu\mathrm{J}$ appears in both, once with a minus sign and once as kinetic energy.
One density, two logarithms, one subtraction and one square root, for four separate marks. Nothing in this question needed the field of the rod at any point.
The examinable pattern here is that parts (b) and (c) are the same calculation with different bookkeeping: carried slowly, the energy goes into the store; released, it comes out as motion. Recognising that saves half the work and stops the two answers disagreeing.
Practice
A · concept 4 questions
1§04.4 — no field does not mean no potential●●○○○
A one mark opener that separates reading a formula from understanding it. Inside a hollow charged metal shell, careful measurement finds no electric field anywhere.
Given
the field is zero everywhere in the interior region
the shell itself carries charge
everything is at rest
Find
(a) True or false: the potential must therefore be zero everywhere inside. Give your reason in one sentence.
Hint 1/4
The claim links a field to a value of the potential. Ask which feature of the potential the field is actually built from.
Hint 2/4
$E_x = -dV/dx$: the field is the rate of change of the potential, not its value.
Hint 3/4
Here the field is zero throughout the interior, so the rate of change is zero throughout the interior.
Hint 4/4
False: a zero derivative makes the potential constant, and a constant can be any number at all.
Show solutionWhat a zero field forces
$$\vec E = 0 \Rightarrow V_b - V_a = -\int_a^b \vec E\cdot d\vec l = 0$$
the integral of zero is zero along any path, so every pair of interior points has the same potential
$$V = \text{constant, not necessarily } 0$$
the constant is fixed by the boundary, in this case by the value on the shell, and the interior has no say in it
Answer $$\boxed{\,\text{False: } E = 0 \Rightarrow V \text{ constant}\,}$$
Check
Test on the case where the answer is known independently: a charged conducting sphere has zero field inside and a surface potential $kQ/r_0$ that is plainly not zero, and the interior must match the surface by continuity.
Every statement in this section about the field is a statement about a derivative. Translating it into a statement about a value needs a boundary condition, and forgetting to ask for one is what makes this trap work.
2§04.3 — no potential does not mean no field●●○○○
The same trap with the two quantities exchanged, which is worth doing because most people get exactly one of the pair right. A point midway between two charges of equal size and opposite sign is found to be at zero potential.
Given
charges of $+8.00\ \mathrm{nC}$ and $-8.00\ \mathrm{nC}$, $10.0\ \mathrm{cm}$ apart
the field point is midway between them
the potential there is measured to be zero
Find
(a) True or false: the electric field at that point is therefore zero as well. Give your reason in one sentence.
Hint 1/4
Two contributions cancelled to give zero. Ask what kind of cancellation that was, and whether the other quantity cancels the same way.
Hint 2/4
Potentials add as signed numbers; fields add as vectors, so opposite signs behave oppositely in the two sums.
Hint 3/4
Here the two charges are equal in size and opposite in sign, at $5.00\ \mathrm{cm}$ on either side of the point.
Hint 4/4
False: the two field contributions point the same way and add up to $5.75\times10^{4}\ \mathrm{N/C}$.
the positive charge pushes a test charge away from itself and the negative one pulls it towards itself, and at the midpoint those are the same direction
Answer $$\boxed{\,\text{False: } V = 0 \text{ but } E = 5.75\times10^{4}\ \mathrm{N/C}\,}$$
Check
Independent check by symmetry: reflecting the arrangement through the midpoint and swapping the two charges leaves the picture unchanged but reverses the sign of the potential, which forces $V = 0$; the same operation leaves the field arrow pointing the same way, so nothing forces the field to vanish.
Take the pair of traps together: a zero of one quantity says nothing about the other. If a question hands you one zero and asks about the other, that is the whole question.
3§04.1 — moving a charge along an equipotential●●○○○
A quick one about what path independence actually buys you. Two points P and Q lie on the same equipotential surface of some static arrangement of charges. A charge is carried from P to Q, once along a short direct route and once along a long wandering one.
Given
P and Q lie on the same equipotential surface
the arrangement of charges is static
two routes of very different length are used
Find
(a) What is the work done by the electric force on the two routes?
Hint 1/4
You are told something about the two endpoints and nothing about the middle. Ask whether the middle can matter for this force.
Hint 2/4
$W_{a\to b} = -q(V_b - V_a)$, which depends only on the two endpoint potentials.
Hint 3/4
Here P and Q are on the same equipotential surface, so $V_P = V_Q$ by definition, and the route is not mentioned in the formula at all.
Hint 4/4
The work is zero on both routes.
Show solutionWrite the work in terms of endpoints
$$W_{P\to Q} = -q(V_Q - V_P)$$
this form is available only because the electrostatic force is path independent, which is what the first block established
$$V_P = V_Q \Rightarrow W = 0 \text{ on every route}$$
the definition of an equipotential surface is exactly the statement that makes the bracket vanish
Answer $$\boxed{\,W = 0 \text{ on both routes}\,}$$
Check
Local check that agrees with the global one: the field is everywhere perpendicular to an equipotential surface, so at every single step along a route lying in the surface the dot product $\vec F\cdot d\vec l$ is zero, and a sum of zeros is zero.
This is why equipotentials are worth drawing. Any motion along one is free, so a question can often be simplified by moving the charge along an equipotential first and only then across to its destination.
4§04.2 — does mass affect an energy change●●○○○
One that catches people who have just learned to convert energies into speeds. A proton and an electron are each moved through a potential difference of exactly $1.00\ \mathrm{V}$.
Given
proton mass $1.67\times10^{-27}\ \mathrm{kg}$, charge $+e$
electron mass $9.11\times10^{-31}\ \mathrm{kg}$, charge $-e$
each crosses a potential difference of $1.00\ \mathrm{V}$
Find
(a) True or false: the proton's potential energy changes by more than the electron's, because it is about two thousand times heavier. Give your reason in one sentence.
Hint 1/4
Write down which properties of a particle appear in a change of potential energy, and check whether mass is one of them.
Hint 2/4
$\Delta U = q\,\Delta V$, and there is no mass anywhere in that expression.
Hint 3/4
Here both particles carry one elementary charge, of opposite signs, and both cross $1.00\ \mathrm{V}$.
Hint 4/4
False: both change by $1.00\ \mathrm{eV}$ in size, and the mass only shows up later, in the speeds.
the mass appears under a square root in the speed and nowhere in the energy, which is the entire content of the question
Answer $$\boxed{\,\text{False: both change by } 1.00\ \mathrm{eV}\,}$$
Check
Consistency with the definition: the electron volt was defined as the energy of one elementary charge across one volt, so any particle carrying one elementary charge must gain exactly one electron volt across one volt, whatever else is true of it.
Read the question before reaching for the mass. Energies, works and potential differences never contain it; speeds, accelerations and times always do.
B · computation 7 questions
1§04.2 — a uniform gap, both ways round●○○○○
Two parallel plates are held $8.00\ \mathrm{mm}$ apart and the uniform field between them is measured to be $1.20\times10^{4}\ \mathrm{V/m}$.
Given
plate separation $8.00\ \mathrm{mm}$
uniform field $1.20\times10^{4}\ \mathrm{V/m}$
a test charge of $+2.50\ \mathrm{nC}$ is available
Find
(a) Find the potential difference between the plates.
(b) Find the work the field does on a $+2.50\ \mathrm{nC}$ charge released at the high potential plate and arriving at the other.
Hint 1/4
The field is constant, so the potential falls at a constant rate. Ask how much it falls over the whole gap, and then what that costs a charge.
Hint 2/4
$\lvert V_{ba}\rvert = Ed$ for a uniform field, and $W = q\lvert V_{ba}\rvert$ for a charge going downhill.
Hint 3/4
Here $E = 1.20\times10^{4}\ \mathrm{V/m}$, $d = 8.00\times10^{-3}\ \mathrm{m}$ and $q = 2.50\times10^{-9}\ \mathrm{C}$.
Hint 4/4
The gap carries $96.0\ \mathrm{V}$, and the field does $2.40\times10^{-7}\ \mathrm{J}$ of work.
Show solutionThe potential difference
$$\lvert V_{ba}\rvert = Ed = (1.20\times10^{4})(8.00\times10^{-3}) = 96.0\ \mathrm{V}$$
uniform field, so the average slope is the slope everywhere and the integral is a product
the sign is settled by the direction of travel rather than by algebra: the charge falls, so the field does positive work on it
Answer $$\boxed{\,\lvert V_{ba}\rvert = 96.0\ \mathrm{V},\qquad W = 240\ \mathrm{nJ}\,}$$
Check
Independent route for (b) through the force: $F = qE = (2.50\times10^{-9})(1.20\times10^{4}) = 3.00\times10^{-5}\ \mathrm{N}$ and $W = Fd = (3.00\times10^{-5})(8.00\times10^{-3}) = 2.40\times10^{-7}\ \mathrm{J}$, the same number without any mention of potential.
2§04.3 — three charges, one scalar sum●●○○○
Three point charges are pinned to a board. A charge of $+2.00\ \mathrm{nC}$ sits at $(0,0)$, a charge of $-3.00\ \mathrm{nC}$ at $(6.00\ \mathrm{cm},\,0)$ and a charge of $+4.00\ \mathrm{nC}$ at $(0,\,8.00\ \mathrm{cm})$.
Given
$+2.00\ \mathrm{nC}$ at $(0,0)$
$-3.00\ \mathrm{nC}$ at $(6.00\ \mathrm{cm},\,0)$
$+4.00\ \mathrm{nC}$ at $(0,\,8.00\ \mathrm{cm})$
field point at $(6.00\ \mathrm{cm},\,8.00\ \mathrm{cm})$
Find
(a) Find the total electric potential at the point $(6.00\ \mathrm{cm},\,8.00\ \mathrm{cm})$.
Hint 1/4
Three distances and three signed numbers. No angle in this question is ever needed.
Hint 2/4
$V = k\sum_i Q_i/r_i$, with each charge carrying its own sign and each distance a positive length.
Hint 3/4
The field point is $(6.00,\,8.00)$ cm, so it is $10.0\ \mathrm{cm}$ from the origin, $8.00\ \mathrm{cm}$ from the charge on the x axis and $6.00\ \mathrm{cm}$ from the charge on the y axis.
one multiplication at the end, which keeps the rounding out of the individual terms
Answer $$\boxed{\,V = +442\ \mathrm{V}\,}$$
Check
Bound the answer: the positive contributions alone give $(8.99\times10^{9})(8.667\times10^{-8}) = 779\ \mathrm{V}$ and the negative one is $-337\ \mathrm{V}$, so the total must lie between $-337$ and $779$ and, since the positives dominate, above zero. It does, at $442$ V.
In a scalar sum the nearest charge usually wins, not the biggest one. Here the $4.00$ nC at $6.00$ cm outweighs everything else, and noticing that before calculating tells you the sign of the answer in advance.
3§04.2 — an electron gun at 250 V●●○○○
An electron is released from rest at a hot filament and accelerated to a screen. The filament and the screen differ in potential by $250\ \mathrm{V}$, with the screen at the higher potential.
Given
electron: charge $-1.602\times10^{-19}\ \mathrm{C}$, mass $9.11\times10^{-31}\ \mathrm{kg}$
the electron is being accelerated towards higher potential because its charge is negative, so the energy gain is positive and the square root is real
Answer $$\boxed{\,\Delta K = 250\ \mathrm{eV} = 4.01\times10^{-17}\ \mathrm{J},\qquad v = 9.38\times10^{6}\ \mathrm{m/s}\,}$$
Check
Scaling check against the five kilovolt example in the section: twenty times the energy there gave $4.19\times10^{7}\ \mathrm{m/s}$, and speeds go as the square root of energy, so this answer should be smaller by $\sqrt{20} = 4.47$. Indeed $4.19\times10^{7}/4.47 = 9.38\times10^{6}\ \mathrm{m/s}$.
4§04.4 — from a measured potential to the source●●●○○
A probe dragged along a line in an evacuated chamber finds that the potential varies as $V(x) = (45.0\ \mathrm{V\,m})/x$, with $x$ in metres and measured from a fixed point in the chamber.
Given
$V(x) = 45.0/x$ volts, with x in metres
the relation holds for every x at which the probe was used
the potential is measured from zero far away
Find
(a) Find the electric field at $x = 0.300\ \mathrm{m}$.
(b) What single point charge, and where, would produce this potential?
Hint 1/4
Part (a) is a derivative. Part (b) asks you to recognise the shape of the function rather than to calculate anything new.
Hint 2/4
$E_x = -dV/dx$, and the potential of a point charge is $V = kQ/r$.
Hint 3/4
Here $V = 45.0/x$, so $dV/dx = -45.0/x^{2}$, to be used at $x = 0.300\ \mathrm{m}$; and matching $45.0/x$ against $kQ/r$ identifies $kQ = 45.0\ \mathrm{V\,m}$.
Hint 4/4
The field is $500\ \mathrm{V/m}$ pointing away from the origin, and the source is a charge of $+5.01\ \mathrm{nC}$ at $x = 0$.
Cross check the two parts against each other: a charge of $5.01\ \mathrm{nC}$ produces $E = kQ/r^{2} = 45.0/(0.300)^{2} = 500\ \mathrm{V/m}$ at $0.300$ m, agreeing with the derivative computed in part (a) without using it.
The shape of a potential identifies its source: one over the distance means a point charge, a logarithm means a line of charge, a constant means the inside of a conductor. Learning to read the shape saves a great deal of algebra.
5§04.6 — potential of a negatively charged rod●●●○○
A thin plastic rod $30.0\ \mathrm{cm}$ long carries $-9.00\ \mathrm{nC}$ spread uniformly along its length.
Given
rod length $L = 30.0\ \mathrm{cm}$
total charge $Q = -9.00\ \mathrm{nC}$, uniform
field point on the rod's own axis, $10.0\ \mathrm{cm}$ beyond the near end
Find
(a) Find the electric potential at that point on the axis.
Hint 1/4
The charge is spread out, so a single distance will not do. Ask what varies along the rod and what stays fixed.
Hint 2/4
For a rod on its own axis, $V = k\lambda\ln[(a+L)/a]$ with $\lambda = Q/L$.
Hint 3/4
Here $\lambda = -9.00\times10^{-9}/0.300 = -3.00\times10^{-8}\ \mathrm{C/m}$, $a = 0.100\ \mathrm{m}$ and $a+L = 0.400\ \mathrm{m}$.
the ratio of far distance to near distance is exactly four here, which keeps the logarithm easy to check
$$V = (-269.7)(1.386) = -374\ \mathrm{V}$$
three significant figures, and the sign survives because a logarithm of a number bigger than one is positive
Answer $$\boxed{\,V = -374\ \mathrm{V}\,}$$
Check
Bracket the answer with the two crude estimates. All the charge at the near end would give $kQ/0.100 = -809\ \mathrm{V}$ and all of it at the far end $kQ/0.400 = -202\ \mathrm{V}$; the true value must lie between them, and $-374$ does, nearer the far end estimate because most of the rod is far away.
6§04.5 — a charged sphere from its surface potential●●○○○
An isolated metal sphere of radius $12.0\ \mathrm{cm}$ is charged until its surface sits at $2400\ \mathrm{V}$ relative to infinity.
Given
conducting sphere, radius $12.0\ \mathrm{cm}$
surface potential $2400\ \mathrm{V}$
isolated and at rest
Find
(a) How much charge is on the sphere?
(b) Find the potential and the field at a point $30.0\ \mathrm{cm}$ from the centre.
Hint 1/4
Outside the sphere everything behaves as though the charge were a point at the centre, so build that point charge first.
Two independent ratio checks. The field point is two and a half times further out than the surface, so the potential should be $2400/2.5 = 960\ \mathrm{V}$, as found; and $E = V/r$ holds outside a sphere, giving $960/0.300 = 3.20\times10^{3}\ \mathrm{V/m}$, also as found.
Computing $kQ$ as a single number in volt metres is worth the habit. For a sphere it turns four separate formulas into four divisions.
7§04.7 — the cost of bringing a charge in from infinity●●○○○
A charge of $+8.00\ \mathrm{nC}$ is held fixed. An agent carries a second charge of $+3.00\ \mathrm{nC}$ in from very far away and holds it at rest $5.00\ \mathrm{cm}$ from the first.
Given
fixed charge $+8.00\ \mathrm{nC}$
carried charge $+3.00\ \mathrm{nC}$, brought from infinity
final separation $5.00\ \mathrm{cm}$, both at rest
Find
(a) How much work does the agent do, and what does the sign of your answer mean?
Hint 1/4
Both charges end up at rest, so nothing went into motion. Ask what the agent's work went into instead.
Hint 2/4
$W_{\rm ext} = \Delta U = U_{\rm final} - U_{\rm initial}$, with $U = kq_1q_2/r$ and $U_{\rm initial} = 0$ at infinite separation.
Hint 3/4
Here $q_1 = 8.00\times10^{-9}\ \mathrm{C}$, $q_2 = 3.00\times10^{-9}\ \mathrm{C}$ and the final separation is $0.0500\ \mathrm{m}$.
Hint 4/4
The agent does $+4.32\ \mu\mathrm{J}$, positive because two positive charges have to be pushed together.
Independent route through the potential: the fixed charge makes $V = kQ/r = 71.92/0.0500 = 1438\ \mathrm{V}$ at the destination, and $W_{\rm ext} = qV = (3.00\times10^{-9})(1438) = 4.31\times10^{-6}\ \mathrm{J}$, agreeing to the rounding without any pair energy being written down.
The number is also the energy you would get back if the second charge were released and allowed to escape to infinity, which is the same statement read in the other direction.
C · exam level 4 questions
1§04.6 — which expression is the potential of a ring●●●○○
A ring of radius $R$ carries a total charge $Q$ spread uniformly around it. A point P lies on the ring's axis, a distance $x$ from the centre. Four expressions are offered for the potential at P, and only one of them is right.
Given
uniform ring of radius $R$, total charge $Q$
field point on the axis at distance $x$ from the centre
potential measured from zero at infinity
Find
(a) Which expression gives the electric potential at P?
Hint 1/4
Do not derive anything. Test each candidate at two places where you already know the answer: the centre of the ring, and very far away.
Hint 2/4
$V = k\int dq/r$, and at the centre every element is at $R$, while far away any compact charge must look like $kQ/x$.
Hint 3/4
Here the distance from an element to P is $\sqrt{x^{2}+R^{2}}$ and it is the same for every element of the ring, so the integral gives the total charge over that one distance.
Hint 4/4
The potential is $kQ$ divided by the slant distance $\sqrt{x^{2}+R^{2}}$.
the distance is the same for every element, which is the only reason it can be taken out of the integral
Check the two limits
$$x = 0:\ V = \frac{kQ}{R}\ \text{(finite and largest)};\qquad x \gg R:\ V \to \frac{kQ}{x}$$
the first limit rules out anything that vanishes at the centre and the second rules out anything with the wrong power far away, which between them eliminate three of the four candidates
Differentiate the chosen expression: $-dV/dx = kQx/(x^{2}+R^{2})^{3/2}$, which is the axial field of a ring derived by an entirely separate vector calculation in the previous section. Two independent derivations meeting is stronger evidence than either alone.
Limits are a faster way through a multiple choice question than a derivation. Pick two places where you already know the answer, and let them do the eliminating.
2§04.7 — how close can an alpha particle get●●●●○
An alpha particle, of charge $+2e$, is fired straight at a heavy nucleus of charge $+79e$ that is massive enough to be treated as fixed. Far away, the alpha particle has a kinetic energy of $5.00\ \mathrm{MeV}$. It slows down as it approaches, stops momentarily, and is thrown back.
Given
alpha particle charge $+2e$, kinetic energy far away $5.00\ \mathrm{MeV}$
nucleus charge $+79e$, treated as fixed
$e = 1.602\times10^{-19}\ \mathrm{C}$, head on collision
Find
(a) Find the distance of closest approach.
(b) Compare it with the size of a nucleus, a few times $10^{-15}\ \mathrm{m}$, and say what that comparison implies.
Hint 1/4
At the closest point the particle is momentarily at rest. Ask what has happened to the kinetic energy it started with.
Hint 2/4
$K_{\rm far} = U_{\rm closest} = kq_1q_2/r_{\min}$, since the potential energy far away is zero and the kinetic energy at the turning point is zero.
Hint 3/4
Here $q_1q_2 = (2e)(79e) = 158e^{2}$ with $e = 1.602\times10^{-19}\ \mathrm{C}$, and $K = 5.00\times10^{6}\ \mathrm{eV} = 8.01\times10^{-13}\ \mathrm{J}$.
Hint 4/4
It gets to $4.55\times10^{-14}\ \mathrm{m}$, about ten times the radius of the nucleus, so it never touches it.
Show solutionName the two states
$$\text{far away: } K = 5.00\ \mathrm{MeV},\ U = 0;\qquad \text{closest: } K = 0,\ U = \frac{kq_1q_2}{r_{\min}}$$
the turning point is defined by the speed being momentarily zero, which is what converts a dynamics question into one line of energy bookkeeping
Equate and solve
$$\frac{k(2e)(79e)}{r_{\min}} = K \Rightarrow r_{\min} = \frac{(8.99\times10^{9})(158)(1.602\times10^{-19})^{2}}{8.01\times10^{-13}}$$
converting the megaelectron volts into joules is the only unit step, and it has to happen before the division
Check by working backwards in electron volt units, avoiding joules entirely: the potential at $r_{\min}$ due to the nucleus is $k(79e)/r_{\min} = (8.99\times10^{9})(79)(1.602\times10^{-19})/(4.55\times10^{-14}) = 2.50\times10^{6}\ \mathrm{V}$, and a charge of $2e$ sitting at $2.50$ MV has $5.00\ \mathrm{MeV}$ of energy, which is exactly what the particle started with.
The pattern is worth keeping: a distance of closest approach is always one equation, initial kinetic energy equals final potential energy, and it never needs a force or a trajectory.
3§04.5 — launching a particle off a charged sphere●●●●○
A metal sphere of radius $10.0\ \mathrm{cm}$ is held at $1500\ \mathrm{V}$ relative to infinity. A proton is released from rest just outside its surface and is free to escape.
Given
conducting sphere, radius $10.0\ \mathrm{cm}$, surface at $1500\ \mathrm{V}$
proton: $q = +1.602\times10^{-19}\ \mathrm{C}$, $m = 1.67\times10^{-27}\ \mathrm{kg}$, released from rest at the surface
an alpha particle of charge $+2e$ and mass $6.64\times10^{-27}\ \mathrm{kg}$ is also available
Find
(a) Find the proton's speed when it is very far from the sphere.
(b) Repeat for an alpha particle released the same way.
(c) Does either answer depend on the radius of the sphere? Explain in one sentence.
Hint 1/4
The particle goes from the surface to infinity. Ask what the potential is at each of those two places, and notice that both numbers are already given to you.
Hint 2/4
$\Delta K = -q(V_\infty - V_{\rm surface}) = qV_{\rm surface}$, and then $v = \sqrt{2\Delta K/m}$.
Hint 3/4
Here $V_{\rm surface} = 1500\ \mathrm{V}$ and $V_\infty = 0$; the proton carries $e$ and $1.67\times10^{-27}\ \mathrm{kg}$, the alpha particle $2e$ and $6.64\times10^{-27}\ \mathrm{kg}$.
Hint 4/4
The proton reaches $5.36\times10^{5}\ \mathrm{m/s}$ and the alpha particle $3.80\times10^{5}\ \mathrm{m/s}$; neither answer uses the radius.
Show solutionThe proton
$$\Delta K = e(1500) = 1500\ \mathrm{eV} = 2.403\times10^{-16}\ \mathrm{J}$$
the finish state is at zero potential by the choice of zero, so the whole surface potential is available
Ratio check that avoids repeating either calculation: $v_\alpha/v_p$ should be $\sqrt{2m_p/m_\alpha} = \sqrt{2(1.67)/6.64} = 0.709$, and $3.80/5.36 = 0.709$. The two speeds are consistent with each other by a route that uses neither of their absolute values.
A sphere held at a stated potential is a cleaner piece of data than a sphere carrying a stated charge, because the potential is what the energy calculation actually consumes. If a question gives you the charge, your first move should usually be to turn it into a potential.
4§04.4 — reading a field off a column of potentials●●●○○
A probe is moved along a straight line and the potential is recorded every $2.00\ \mathrm{cm}$. The readings, in order, are $40.0\ \mathrm{V}$ at $x = 0$, then $28.0\ \mathrm{V}$, $20.0\ \mathrm{V}$, $16.0\ \mathrm{V}$ and $16.0\ \mathrm{V}$ at $x = 2.00$, $4.00$, $6.00$ and $8.00\ \mathrm{cm}$.
Given
readings at $x = 0,\,2.00,\,4.00,\,6.00,\,8.00\ \mathrm{cm}$
potentials $40.0,\,28.0,\,20.0,\,16.0,\,16.0\ \mathrm{V}$ in that order
the field is along the same line as the probe's motion
Find
(a) In which interval is the magnitude of the electric field largest?
Hint 1/4
You are looking for a slope, so compare differences between neighbouring readings rather than the readings themselves.
Hint 2/4
$E_x \approx -\Delta V/\Delta x$, and every interval here has the same $\Delta x$ of $0.0200\ \mathrm{m}$.
Hint 3/4
The successive drops are $12.0$, $8.00$, $4.00$ and $0\ \mathrm{V}$, each over $0.0200\ \mathrm{m}$.
Hint 4/4
The largest field is in the first interval, where it is $600\ \mathrm{V/m}$.
Show solutionDifferences first
$$\Delta V = -12.0,\ -8.00,\ -4.00,\ 0\ \mathrm{V}\ \text{over } \Delta x = 0.0200\ \mathrm{m}$$
with a common spacing the division by the step is a common factor, so the ranking is settled before any arithmetic
all positive, so the field points in the direction of increasing x throughout, and it dies out rather than reversing
Answer $$\boxed{\,\text{largest between } 0 \text{ and } 2.00\ \mathrm{cm},\ \text{where } E_x = 600\ \mathrm{V/m}\,}$$
Check
Consistency check on the shape: the field values fall by equal steps of $200\ \mathrm{V/m}$ per interval, which means the potential is close to a quadratic that levels off, and a levelling quadratic is exactly what a column ending in two equal readings looks like.
Whenever a table of potentials is given at equal spacing, write the differences in the margin first. They are the field up to a constant factor, and every question about the field can then be answered by looking at them.
D · interleaved 3 questions
1§04.3 — midway between two equal charges●●●○○
This set is deliberately mixed, so decide for yourself which tool each question wants before reaching for one. Two charges of $+4.00\ \mathrm{nC}$ each are fixed at $x = 0$ and $x = 6.00\ \mathrm{cm}$. A third charge of $+2.00\ \mathrm{nC}$ is held at the midpoint.
Given
$+4.00\ \mathrm{nC}$ at $x = 0$ and $+4.00\ \mathrm{nC}$ at $x = 6.00\ \mathrm{cm}$
$+2.00\ \mathrm{nC}$ held at $x = 3.00\ \mathrm{cm}$
nothing else is present
Find
(a) Find the net electric force on the charge at the midpoint.
(b) Find the electric potential at the midpoint due to the two outer charges.
(c) Find the potential energy of the middle charge in that position.
Hint 1/4
The three parts want three different kinds of object. Decide for each one whether it has a direction before choosing how to add the two contributions.
Hint 2/4
Forces add as vectors, $\vec F = \vec F_1 + \vec F_2$; potentials add as signed numbers, $V = kQ_1/r_1 + kQ_2/r_2$; and $U = qV$.
Hint 3/4
Here both outer charges are $4.00\times10^{-9}\ \mathrm{C}$ and both are $0.0300\ \mathrm{m}$ from the midpoint, one on each side, and the middle charge is $2.00\times10^{-9}\ \mathrm{C}$.
Hint 4/4
The force is zero, the potential is $2.40\times10^{3}\ \mathrm{V}$ and the stored energy is $4.79\ \mu\mathrm{J}$.
Show solutionThe force, as vectors
$$\vec F = \vec F_1 + \vec F_2 = 0$$
equal magnitudes in opposite directions, which is what symmetry guarantees for equal charges at equal distances
the potential was worth computing first because the energy is one multiplication away from it
Answer $$\boxed{\,F = 0,\qquad V = 2.40\times10^{3}\ \mathrm{V},\qquad U = 4.79\ \mu\mathrm{J}\,}$$
Check
Check the energy by pairs instead. Each outer charge forms one pair with the middle charge, at a separation of $0.0300\ \mathrm{m}$: $$U_{13} = U_{23} = \frac{(8.99\times10^{9})(4.00\times10^{-9})(2.00\times10^{-9})}{0.0300} = 2.397\times10^{-6}\ \mathrm{J}$$ and twice that is $4.79\times10^{-6}\ \mathrm{J}$, matching without the potential being used at all.
A charge sitting where the force is zero is not a charge with no energy. Here it would sit still if released along the line, yet it took nearly five microjoules to put it there, and it would give all of that back on escaping sideways.
2§04.5 — from a Gaussian surface to a potential●●●●○
A solid metal sphere of radius $8.00\ \mathrm{cm}$ carries a total charge of $+12.0\ \mathrm{nC}$ and stands alone in a vacuum, in electrostatic equilibrium.
total charge $+12.0\ \mathrm{nC}$, isolated and at rest
a proton is available, of charge $1.602\times10^{-19}\ \mathrm{C}$
Find
(a) Use a spherical Gaussian surface to write the field at a distance $r$ outside the sphere.
(b) Find the potential at $r = 20.0\ \mathrm{cm}$ and at $r = 4.00\ \mathrm{cm}$.
(c) How much work must be done to bring a proton from far away to the sphere's surface?
Hint 1/4
Part (a) is last section's tool and part (b) is this section's. Ask what each one gives you and in which order they have to be used.
Hint 2/4
Gauss gives $E = kQ/r^{2}$ outside a spherically symmetric charge; then $V = kQ/r$ outside and $V = kQ/r_0$ everywhere inside; and $W_{\rm ext} = qV$.
Hint 3/4
Here $kQ = (8.99\times10^{9})(12.0\times10^{-9}) = 107.9\ \mathrm{V\,m}$, with $r_0 = 0.0800\ \mathrm{m}$ and field points at $0.200\ \mathrm{m}$ and $0.0400\ \mathrm{m}$.
Hint 4/4
Outside $E = kQ/r^{2}$; the potentials are $539\ \mathrm{V}$ and $1.35\times10^{3}\ \mathrm{V}$; and the proton costs $2.16\times10^{-16}\ \mathrm{J}$, that is $1.35\ \mathrm{keV}$.
the symmetry makes E constant over the Gaussian sphere and everywhere perpendicular to it, which is the only reason the flux integral collapses to a product
Two checks. The potential at $0.200$ m should be $0.0800/0.200 = 0.400$ times the surface value, and $0.400 \times 1348 = 539\ \mathrm{V}$, as found. And the work in electron volts must equal the surface potential in volts for a singly charged particle, which it does: $1348\ \mathrm{V}$ gives $1348\ \mathrm{eV}$.
The two sections fit together in a fixed order: Gauss's law gives fields from symmetry, and integrating those fields gives potentials. Trying to run it the other way, guessing a potential and hoping Gauss confirms it, does not work.
3§04.6 — a ring, by both methods at once●●●●○
A ring of radius $5.00\ \mathrm{cm}$ carries $+20.0\ \mathrm{nC}$ spread uniformly around it. A point P lies on the ring's axis, $12.0\ \mathrm{cm}$ from the centre.
P on the axis at $x = 12.0\ \mathrm{cm}$ from the centre
the axial field of a ring is $E = kQx/(x^{2}+R^{2})^{3/2}$
Find
(a) Find the potential at P.
(b) Find the field at P from the given expression.
(c) Show that your two answers are consistent by differentiating the general potential.
Hint 1/4
One slant distance serves the whole question. Work it out first and notice which numbers it is made of.
Hint 2/4
$V = kQ/\sqrt{x^{2}+R^{2}}$, $E = kQx/(x^{2}+R^{2})^{3/2}$, and $E_x = -dV/dx$.
Hint 3/4
Here $\sqrt{(0.120)^{2}+(0.0500)^{2}} = 0.130\ \mathrm{m}$ and $kQ = (8.99\times10^{9})(20.0\times10^{-9}) = 179.8\ \mathrm{V\,m}$.
Hint 4/4
The potential is $1.38\times10^{3}\ \mathrm{V}$ and the field is $9.82\times10^{3}\ \mathrm{V/m}$, and differentiating the potential reproduces the field expression exactly.
Show solutionThe slant distance and the constant
$$\sqrt{x^{2}+R^{2}} = \sqrt{0.0144+0.00250} = 0.130\ \mathrm{m},\qquad kQ = 179.8\ \mathrm{V\,m}$$
recognising the five twelve thirteen triangle keeps every later line in exact decimals rather than in surds
one chain rule, and the result is the expression the question supplied, so the two parts are not independent facts but one fact seen twice
Answer $$\boxed{\,V = 1.38\times10^{3}\ \mathrm{V},\qquad E = 9.82\times10^{3}\ \mathrm{V/m}\,}$$
Check
Numerical check of the derivative rather than the algebra. At $x = 0.110$ m the slant distance is $0.1208$ m and $V = 1488\ \mathrm{V}$; at $x = 0.130$ m it is $0.1393$ m and $V = 1291\ \mathrm{V}$. The slope across that interval is $(1291-1488)/0.0200 = -9.85\times10^{3}\ \mathrm{V/m}$, so the field is $+9.85\times10^{3}\ \mathrm{V/m}$, agreeing with the exact $9.82\times10^{3}$ to within the error of a finite step.
A question that gives you a field expression and asks for a potential, or the reverse, is usually asking whether you know they are one derivative apart. Saying so explicitly is worth a mark even when the arithmetic is already done.
Mistake ledger (23 entries)
⚠ Using the force at the starting point as if it acted all the way
the starting force is the one number the problem hands you, and treating it as constant makes the mechanics familiar again
each pair counted once; separations are those of the final arrangement
Work to add one more charge to an existing arrangement
$$W_{\rm ext} = q\,V_{\rm existing}$$
the existing charges do not move; the newcomer arrives at rest
Potential of a dipole, far from it
$$V = \frac{kp\cos\theta}{r^{2}},\qquad p = q\ell$$
$r \gg \ell$; $\theta$ is measured from the moment, which points from the negative charge to the positive one
Energy of a dipole in an external field
$$U = -\vec p\cdot\vec E = -pE\cos\theta$$
the external field is uniform across the dipole; the zero is at $\theta = 90^{\circ}$, not at infinity
Check yourself
Close the page and write out from memory: the relation between the work done by the electric force and the change in potential energy; the potential energy of a pair of point charges and where its zero is; the definition of potential and what one volt means; the relation between potential difference and field in a uniform gap; the potential of a point charge and how several of them are combined; the rule that gets the field back from the potential, minus sign included; what is true of the potential inside a conductor at rest; the potential of a ring on its axis and of a rod on its own axis; the pair sum for the energy of an arrangement of charges; and the two dipole results, the potential it produces far away and the energy it has in an external field. Then open the formula card and mark only the ones you missed.
Compute the speed of a charge released in the field of a fixed point charge, and say why a constant acceleration calculation must overestimate it?
c-potential-energy
Take a particle of stated charge and mass through a stated potential difference and produce the energy in electron volts and in joules and the final speed, with the right sign for a negative charge?
c-potential-difference
Add the potentials of three point charges at a point of awkward geometry without drawing a single vector, and find the places on a line where the total potential vanishes?
c-point-charge-potential
Differentiate a given potential to get the field, state the unit of the result, and explain why the field is zero where the potential is flat rather than where it is zero?
c-field-from-v
Give the potential at the centre, at the surface and outside a charged conducting sphere, and say which of two connected spheres has the stronger surface field and by what factor?
c-equipotentials
Set up and evaluate the potential integral for a ring and for a rod on its axis, and recover the field of each by one differentiation?
c-continuous-potential
Evaluate the energy of three charges without double counting any pair, say what the sign means, and find a distance of closest approach from an initial kinetic energy?
c-system-energy
Glossary (15 terms)
electric potential energyelektrik potansiyel enerjisi
The energy stored because of where charges sit relative to one another, measured in joules. It belongs to the arrangement rather than to any single charge, and it is counted from zero when the charges are infinitely far apart.
electric potentialelektrik potansiyeli
The electric potential energy per unit charge at a point, measured in volts. It is a property of the place alone: the same point gives twice the energy to twice the charge, and the potential does not change.
potential differencepotansiyel farkı
The difference in potential between two points, equal to the energy per coulomb that the trip between them costs or pays. It is the only thing about the potential that is ever measured, since the zero is a matter of choice.
voltvolt
The unit of potential, equal to one joule per coulomb. Volts per metre and newtons per coulomb are two names for the unit of the electric field, and they are numerically identical.
electron voltelektron volt
A unit of energy equal to the energy an elementary charge gains crossing one volt, that is 1.602 times ten to the minus nineteen joules. It is an energy and not a potential, and a doubly charged particle crossing one volt gains two of them.
korunumlu kuvvet
A force whose work between two points does not depend on the route taken, so that a potential energy can be defined for it. The electrostatic force is one, and friction is not, which is why only one of the two has a stored energy.
zero of potentialpotansiyel sıfırı
The place chosen to have potential zero, which fixes what every other value means. For charge distributions of finite size it is taken at infinity; for parallel plates it is put on a named plate. Only one choice may be used within a problem.
equipotential surfaceeşpotansiyel yüzey
A surface on which the potential has a single value. The electric field crosses it at right angles everywhere, and carrying a charge along it does no work whatever route is taken across the surface.
potansiyel gradyanı
The rate at which the potential changes with position. The electric field is its negative, so the field points down the steepest slope of the potential and vanishes wherever the potential is flat, however large the potential itself is there.
electrostatic potential energy of a systemsistemin elektrostatik potansiyel enerjisi
The total energy stored in an arrangement of charges, obtained by adding the pair energies with each pair counted exactly once. It equals the work an outside agent must do to assemble the arrangement from infinite separation.
electrostatic equilibriumelektrostatik denge
The settled state of a conductor in which no charge is moving. The field inside the material is then zero, all excess charge sits on the outer surface, and the whole body is at a single potential.
external workdış iş
The work done by an agent that moves a charge slowly, so slowly that its kinetic energy never changes. It equals the increase in potential energy and is therefore the opposite in sign to the work done by the electric field.
The field of roughly three times ten to the sixth volts per metre at which dry air at ordinary pressure stops insulating and a spark forms. It is the ceiling that limits how much charge a laboratory conductor of a given size can hold.
distance of closest approachen yakın yaklaşma mesafesi
The smallest separation reached by a charged particle fired at a repelling charge, found by setting the initial kinetic energy equal to the potential energy at the turning point, where the particle is momentarily at rest.
elektrik dipol momenti
The vector that stands in for a pair of equal and opposite charges: its magnitude is one charge times the separation of the pair, and it points from the negative charge to the positive one. From far away the pair is seen only through this vector, both in the potential it produces and in the energy it has in someone else's field.
What comes next
§05 · Review and Consolidation I: electric field, Gauss's law, and potential
Three tools now exist for the same arrangement of charges: the field, which needs vectors; Gauss's law, which needs symmetry; and the potential, which needs neither but has to be differentiated to give a force back. The next section puts a mixed set of problems in front of you with no label saying which tool is wanted, because choosing the tool is the part that is actually examined.
Sources
D. Giancoli, Physics for Scientists and Engineers with Modern Physics, 5th edition The set textbook. Its treatment of electric potential covers the same ground as this section, and its end of chapter problems are a level harder than the ones here, which is the right next step once this set feels comfortable.
Course syllabus, week 4 line and assessment table The scope of this section comes from the week line, which reads Electric Potential, and from nowhere else. The line names a topic and carries no chapter numbers, so no chapter number is quoted anywhere on this page. The assessment weights quoted on the card are the published ones and nothing finer than them is claimed.
SI values of the electric and particle constants used here The Coulomb constant is taken as 8.99 times ten to the ninth newton metre squared per coulomb squared, the permittivity of free space as 8.85 times ten to the minus twelfth in SI units, the elementary charge as 1.602 times ten to the minus nineteenth coulombs, the electron mass as 9.11 times ten to the minus thirty first kilograms and the proton mass as 1.67 times ten to the minus twenty seventh kilograms. The breakdown field of dry air is taken as about three times ten to the sixth volts per metre, and is used only for order of magnitude remarks.