Kevin Zhou Physics Olympiad Handouts

Electromagnetism II: Electricity

Chapters 3 and 4 of Purcell cover the material presented here, as does chapter 6 of Wang and
Ricardo, volume 2. Image charges are covered in more detail in section 3.2 of Griffiths. For an array
of interesting physical examples, see chapters II-6 through II-9 of the Feynman lectures. There is a
total of 77 points.

1 The Method of Images

Idea 1
The method of images can be used in some highly symmetric situations to compute the
electric field in the vicinity of a conductor. Specifically, consider any configuration of static
charges and take any equipotential surface containing some of the charges. Then the resulting
field configuration outside that surface is the field configuration we would have if that surface
bounded a conductor. This is simply because it has constant potential on the conductor
surface, so it must be the right answer by the uniqueness theorem.

[4] Problem 1. The simplest application of the method of images is the case of a charge q a distance a
from an infinite grounded conducting plane. This problem explores some of its subtleties, assuming
you’ve already read the basic treatment in section 3.4 of Purcell.

(a) Find the force on the charge.

(b) Find the work needed to move the charge out to infinity. (Answer: q 2 /16π0 a.)

(c) Find the total potential energy of the charges on the conducting plane, i.e. the potential energy
associated only with their interaction with each other. (Answer: q 2 /16π0 a.)

(d) Now suppose there is another parallel grounded conducting plane on the other end of the
charge, a distance b away. How many image charges are needed now? Draw some of them.

(e) A conducting plane forces the electric field to be perpendicular to it. Suppose we modified the
plane so that the electric field was instead always parallel to it. Find the force on the charge.

Example 1

Two grounded conducting half-planes intersect, so that in cylindrical coordinates, the equa-
tions describing the planes are θ = 0 and θ = θp = π/2. A charge q is placed somewhere
between the planes. Can the method of images be used to find the force on the charge? What
if θp = 2π/3, or for general θp ?

Solution
We can solve the first case with three image charges. Suppose the charge q is at (x, y). Then
we can reflect in the plane θ = 0, adding a charge −q at (x, −y) to satisfy its boundary condi-
tion. Then we can reflect both the real charge and this image charge in the plane θ = π/2 to

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satisfy that plane’s boundary condition, adding a charge −q at (−x, y) and a charge q at (x, y).

But when the other plane is at θ = 2π/3, there is no configuration of image charges that
works. For concreteness, let’s suppose the real charge is at point A, on the y-axis.

Reflecting in the θ = 0 plane forces us to have an image charge −q at D, reflecting in the

θ = 2π/3 plane yields an image charge q at E, and reflecting in the θ = 0 plane again yields
a −q charge at F , which is real since it’s in the same region as A. But this isn’t allowed:
the point of image charges is to provide an easy way of calculating the effects of screening
charges on conducting surfaces on a given set of real charges (i.e. the charge at A), so it’s
not legal to introduce new real charges in the process. We would get the same conclusion
if we reflected about the planes in a different order – we always need a charge at F . More
generally, the method of images works for this problem if and only if θp = π/n for integer n.

[4] Problem 2. In this problem you’ll develop the method of images for spheres.

(a) A point charge −q is located at x = a and a point charge Q is located at x = A. Show that
the locus of points with φ = 0 is a circle in the xy plane, and hence a spherical shell in space.

(b) Show that the center of the sphere is at the origin provided that

r2 Qr
a= , −q = −
A A
where r is the sphere’s radius. These results will be used throughout the problem set.

(c) Now suppose a point charge Q is a distance b from the center of a spherical grounded conducting
shell of radius r. Find the force on the charge, considering both the cases b < r and b > r.

(d) The case b < r is a bit confusing. On one hand, argue that the total charge is Q plus the image
charge, and hence nonzero. On the other hand, argue that the total charge must be zero, by
considering an appropriate Gaussian surface. One of these arguments is wrong – which one?

As an aside, the fundamental reason the method of images works for spheres is that electromagnetism
has conformal symmetry, a symmetry under any local rescaling of space which preserves angles.
(One example of a conformal transformation is inversion in Euclidean geometry.) The setup here is
related to the conducting plane by such a transformation.

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[2] Problem 3. An infinite grounded conducting plane at z = 0 is deformed with a hemispherical

bump of radius R centered at the origin, as shown. A charge q is placed at z = a as shown.

Can the method of images be used to find the potential in the region with the charge? If so, specify
the image charges; if not, explain why not.

[2] Problem 4 (Purcell 3.50). A point charge q is located a distance b > r from the center of a
nongrounded conducting spherical shell of radius r, which also has charge q. When b is close to r,
the charge is attracted to the shell because it induces negative charge; when b is large the charge is
clearly repelled. Find the value of b so that the point charge is in equilibrium. (Hint: you should
have to solve a difficult polynomial equation. You can either use a computer or calculator, or use
the fact that it contains a factor of 1 − x − x2 .)

[2] Problem 5. Consider a neutral spherical conductor of radius R placed in a uniform external field
E0 . By using the method of images, argue that the field created by the sphere is the same as
the field of an infinitesimal dipole at the center of the sphere, and find its dipole moment. (Hint:
suppose the external field is sourced by two point charges that are very far away.)

[5] Problem 6 (Purcell 3.45). [A] Consider a point charge q located between two parallel infinite
grounded conducting planes. The planes are a distance ` apart, and the point charge is a distance
b from the left plane. The goal of this problem is to find the total charge induced on each plane.

(a) Argue that the total charge on each plane would not change if we replaced the point charge q
with two point charges q/2, both a distance b from the left plane. By iterating this process,
convert the point charge into a uniformly charged plane, and use this to get the answer.

(b) Alternatively, using image charges, show that the electric field on the inside surface of the
left plane, perpendicular to the plane, at a point a distance r from the axis containing all the
image charges, satisfies
∞
X 2q(2n` + b)
4π0 E⊥ = .
n=−∞
((2n` + b)2 + r2 )3/2

(c) Since σ = −0 E⊥ , we can integrate both sides to find the total charge on the left plane.
However, the integral of each term by itself is simply q, so the series doesn’t converge. To get
the result, do the following steps in this specific order: group the terms ±n together, then
integrate them only out to a distance R  b, then sum over the values of |n|, then take the
limit R → ∞. You should get a finite result that matches that of part (a). As you’ll probably
see in the process, if you do the steps in any other order, you’ll get a nonsensical answer.

Those concerned with mathematical rigor might be bothered by the many choices made in part (c).
You might ask, couldn’t we have gotten a different result by changing how we did the computation?
In fact, by the Riemann rearrangement theorem, we could have gotten almost any result. But the

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way we did it is the physically correct way. It roughly sums the terms “in to out”, which respects
the fact that real plates are finite. Closely related ideas are used to “cancel infinities” in quantum
field theory, in a process known as renormalization. We’ll see another example in X1.

2 Capacitors
Idea 2
There are multiple definitions of capacitance. For a single, isolated conductor with charge Q,
the self-capacitance is defined as
Q = Cφ
where φ is the potential difference between the conductor and infinity. But for a set of two
isolated conductors with charges ±Q, you can also define a “mutual” capacitance by

Q = Cφ

where φ is the potential difference between the two conductors. When someone talks about
a “capacitance” without specification, such as in idea 4, they probably mean this latter one.

Idea 3
The definitions of C above are only useful when you have only one or two conductors in the
problem, respectively. In a situation with more than two, it’s very tricky to use the above
definitions, because all the conductors will affect each other; even a neutral conductor will
have an effect since there will be induced charges on its surface.

Instead, it’s better to revert to more general principles. The underlying principle behind
capacitance is linearity: the charges are linearly related to the fields. For multiple capacitors,
the most general possible linear relation is
X
Qi = Cij φj
j

where conductor i has charge Qi and potential φi , the potential is taken to be zero at infinity,
and the Cij are called general coefficients of capacitance, or in electrical engineering, the
Maxwell capacitance matrix. Similarly, inverting this relation,
X
φi = pij Qj
j

where the pij are called coefficients of potential.

In Olympiad physics, you’ll almost never want to compute the Cij or pij explicitly. Instead,
the point here is that if you’re given the charges and want the potentials, or vice versa, you
can build up the answer you want using the principle of superposition, computing all the
fields you need, e.g. using Gauss’s law.

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Remark
General capacitance coefficients are discussed further in section 3.6 of Purcell. One nontrivial
fact is that Cij = Cji , which is proven by energy conservation in problem 3.64 of Purcell.
Capacitance coefficients can be clunky to work with. For example, suppose you want to
compute the familiar capacitance of a system of two conductors. By definition, we have

Q1 = C11 φ1 + C12 φ2 , Q2 = C21 φ1 + C22 φ2 .

An ordinary two-plate capacitor corresponds to the special case of opposite charges on the
plates, so we write Q = Q1 = −Q2 . There is a potential difference V across the plates, so
φ1 = φ2 + V , and plugging this in gives

Q = (C11 + C12 )φ2 + C11 V, −Q = (C22 + C21 )φ2 + C21 V.

Eliminating φ2 from the system of equations above, we find the familiar mutual capacitance

Q 2
C11 C22 − C12
C= =
V C11 + C22 + 2C12
where we used C12 = C21 . This is quite an inconvenient formula, so as a result we won’t
consider general capacitance coefficients any further, except briefly for practice in problem 8.

[2] Problem 7 (Purcell 3.21). Consider a capacitor made of four parallel plates with large area A,
evenly spaced with small separation s. The first and third are connected by a wire, as are the
second and fourth. What is the capacitance of the system?

[3] Problem 8. Consider two concentric spherical metal shells, with radii a < b.

(a) Compute their capacitance using Gauss’s law.

(b) Compute their capacitance by computing the four capacitance coefficients, verifying that
C12 = C21 along the way, and using the result for C above.

[3] Problem 9 (MPPP 152). Four identical non-touching metal spheres are positioned at the vertices
of a regular tetrahedron, as shown.

A charge 4q given to sphere A raises it to a potential V . Sphere A can also be raised to potential V
if it and one of the other spheres are each charged with 3q. What must be the size of equal charges
given to A and two other spheres for the potential of A to again be raised to V ? What if all four
spheres are used?

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[3] Problem 10. 01W USAPhO 2008, problem A1.

Idea 4
A two-plate capacitor with voltage difference V and mutual capacitance C stores energy
1 1
U = QV = CV 2 .
2 2
Many circuits have multiple two-plate capacitors. In general, these need to be handled with
the capacitance coefficients introduced in idea 3. But in practice, capacitors used in circuits
are designed to produce fields confined within themselves, so that different capacitors don’t
interact with each other. In that case, we can just use mutual capacitance throughout, and
C adds in parallel, while 1/C adds in series. (But this not work if, e.g. you put one capacitor
inside another, in which case you should think about the charges and fields directly.)

[2] Problem 11 (Purcell 3.24). Some estimates involving capacitance.

(a) Estimate the capacitance of the Earth.

(b) Make a rough estimate of the capacitance of the human body.

(c) By shuffling over a nylon rug on a dry winter day, you can easily charge yourself up to a
couple of kilovolts, as shown by the length of the spark when your hand comes too close to a
grounded conductor. How much energy would be dissipated in such a spark?

[2] Problem 12. The total energy can also be found by integrating the electric field energy,
Z
0
U= E 2 dV.
2

(a) Show that this agrees with U = CV 2 /2 for a parallel plate capacitor.

(b) Show that this agrees with U = CV 2 /2 for a capacitor made of concentric spheres.

The general proof is more advanced, but if you’re interested, one slick method is given in problem
1.33 of Purcell.

[3] Problem 13 (Purcell 3.26). A parallel-plate capacitor consists of a fixed plate and a movable plate
that is allowed to slide in the direction parallel to the plates. Let x be the distance of overlap.

The separation between the plates is fixed. Let C(x) be the capacitance.

(a) Assume the plates are electrically isolated, so that their charges ±Q are constant. By differ-
entiating the energy, find the leftward force on the movable plate in terms of Q and C(x).

(b) Now assume the plates are connected to a battery, so that their potential difference φ is held
constant. Find the leftward force on the movable plate, in terms of φ and C(x).

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(c) If the movable plate is held in place, the two answers above should be equal because nothing
is moving. Verify that this is the case, being careful with signs.

(d) In terms of electric fields, why is there a force on the movable plate? Does the effect invoked
in the answer to this part change the conclusion of parts (a) through (c) at all?

Idea 5: Dielectrics
While we’re on the subject of capacitors, it’s useful to introduce dielectrics, which will be
important later. All electrical insulators are dielectrics. In the presence of an electric field, a
dielectric will polarize, with positive charges displaced slightly along the field. The resulting
electric dipoles distributed throughout the material in turn create a field that tends to
weaken the original applied field.

Each part of a dielectric polarizes based on the local electric field, but that electric field
depends on the applied field, and the polarization of every other piece of the dielectric. Thus,
solving for the electric field for a general dielectric geometry is very difficult, and usually
not possible in closed form, just like how it’s usually not possible to solve for the field of a
charged conductor. In Olympiad physics, you will almost always consider highly symmetric
situations, where a dielectric simply reduces the applied electric field everywhere by a factor
of κ, called the dielectric constant. (We’ll consider some trickier situations in E8.)

Consider a parallel plate capacitor with charge ±Q on each plate. If a dielectric is inserted
with the charge kept the same, then the field inside is reduced by a factor of κ. Thus, the
capacitance C = Q/V increases by a factor of κ. Dielectrics may increase the amount of
energy that can be stored in a capacitor, which is typically limited by the voltage V0 where
electrical breakdown occurs. So if V0 stays the same, the maximal stored energy U = CV02 /2
goes up by a factor of κ.

Plugging in the definition of C, this result implies that the energy density in the capacitor is
κ0 E 2 . But we showed in E1 that the energy density of the electric field is only 0 E 2 . The
extra energy is stored in the dielectric material itself: it takes energy to separate positive
and negative charges within the dielectric, as if we were stretching many microscopic springs.
This potential energy is released when the capacitor is discharged.

3 Tricky Problems
Example 2: PPP 151

A closed body with conducting surface F has self-capacitance C. The surface is now dented
so that the new surface F ∗ is entirely inside F . Prove that the capacitance has decreased.

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Solution

The energy stored in the capacitor is U = Q2 /2C. Therefore, if we give the capacitor a fixed
charge Q, proving that F ∗ has lower C is equivalent to showing that it takes positive work
to dent the foil from F to F ∗ . It’s cleaner to show the other direction, i.e. that starting from
F ∗ , we can get to F while only lowering the energy.

Suppose without loss of generality that F is infinitesimally larger than F ∗ . (We can break
any finite change into infinitesimal stages and repeat this argument.) We can go from
F ∗ to F by just taking each charge on the surface and moving it outward until it hits F .
This lowers the energy because the electric field is always directed outward, as we proved in E1.

At this point, the charges lie on F , but they don’t have the right distribution, i.e. F is not
an equipotential. Now we let the charges spontaneously redistribute themselves so that F is
again an equipotential. This again lowers the energy, proving the desired result.

Example 3

Are there charge distributions that aren’t spherically symmetric, but which produce an exact
r̂/r2 field outside of them?

Solution
If you know a bit about the multipole expansion, this might seem like a daunting question.
To make the field exactly r̂/r2 , you need to make sure the charge distribution has no dipole
moment, no quadrupole moment, no octupole moment, and so on to infinity, and it seems
impossible to satisfy all of these constraints without spherical symmetry. But we have
already seen an example of such a charge distribution earlier in the problem set!

Recall that when we treated the method of images for spheres, we found that in some
situations, the complicated charge densities on conducting spheres were exactly the same as
those produced by a fictitious image charge inside the sphere, and generally away from its
center. If we place the origin at that image charge, then we have an example of a charge
distribution that is perfectly r̂/r2 far away, but which isn’t spherically symmetric. (The
general solution is given here.)

[2] Problem 14. Consider a set of n conducting, very large parallel plates, placed in zero external
electric field. The plates are given charges Qi . If the left ends of the plates are at locations xi , and
the plates have thickness di , what is the total charge on the left end of the leftmost plate, and the
right end of the rightmost plate?
[2] Problem 15 (Purcell 3.9). A conducting spherical shell has charge Q and radius R1 . A larger
concentric conducting spherical shell has charge −Q and radius R2 .
(a) If the outer shell is grounded, explain why nothing happens to the charge on it.
(b) If instead the inner shell is grounded, e.g. by connecting it to ground by a very thin wire that
passes through a very small hole in the outer shell, find its final charge.

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(c) It’s not so clear why charge would leave the inner shell in part (b), thinking in terms of forces.
A small bit of positive charge will certainly want to hop on the wire and follow the electric
field across the gap to the larger shell. But when it gets to the larger shell, it seems like it
has no reason to keep going to infinity, because the field is zero outside. And, even worse, the
field will point inward once some positive charge has moved away from the shells. So it seems
like the field will drag back any positive charge that has left. Does charge actually leave the
inner shell? If so, what’s wrong with the above reasoning?
[2] Problem 16. The usual expression for the capacitance of a parallel-plate capacitor is A0 /d.
However, in reality the field within the capacitor is not perfectly uniform, and there are fringe fields
outside. Taking these effects into account, is the true capacitance higher or lower than A0 /d?
R
[2] Problem 17 (Purcell 4.16). In a parallel plate capacitor, the quantity E · ds should be equal to
V for any path that connects the two plates.
A charged capacitor can be discharged R by attaching a wire to the external surfaces of the plates.
No matter how one attaches the wire, E · ds along the wire should be equal to V . And as we’ve
argued in E2, this is sufficient to cause charges to move along the wire, even if the electric field
points in the “wrong” direction at some points along the wire, because the wire has negligible
capacitance: charges within it move rigidly, each pushing the next one and pulling the previous one.
But it’s puzzling how this works for a capacitor, because the electric field is supposed to be
essentially zero just outside it. Consider two possible limiting cases for the wire’s shape.

R
In each case, explain qualitatively how E · ds can be equal to V . In particular, how large are the
contributions from the distinct segments of the wire (the horizontal and vertical parts in the first
case, and the straight and curved parts in the second)?
[3] Problem 18. Consider two conducting spheres of radius r separated by a distance a  r. The
spheres can be thought of as the two plates of a parallel plate capacitor.
(a) By assuming the surface charge density on each sphere is uniform, estimate the capacitance.

(b) In reality, the surface charges on each sphere will be distorted by the other sphere. The surface
charges assumed in part (a) will induce changes in the surface charges, represented by an
image charge for each sphere. But these image charges will induce further image charges, and
so on, yielding an infinite series for the charge distribution and hence the capacitance. Find
the first two terms in the series for the capacitance.

01^‚
(c) If a/r = 10, roughly estimate the error in neglecting the other terms.

[3] Problem 19. USAPhO 2022, problem A2. A computational problem involving surface tension.

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Example 4

Find the leading interaction force between a dipole of dipole moment p and a grounded
conducting sphere of radius r, separated by a distance R  r. What if the sphere is
electrically neutral instead?

Solution
Place the origin at the center of the sphere and orient the z-axis to pass through the dipole.
We can regard the dipole p = qd as a combination of two charges

q at z = R, −q at z = R + d

where d is very small. In the grounded case, this induces two image charges in the sphere,
qr r2 qr r2
at z = , − at z =
R R R+d R+d
approximately separated by r2 /R2 . We can now use Coulomb’s law four times, but that’s a
bit tedious. Instead, decompose the image charges into a dipole moment and a net charge,
pr3 qr qr pr
p0 = , Q0 = − ≈ 2.
R3 R R+d R
We can place both of these at the origin, because this slight displacement will only affect the
answer by subleading terms in r/R. Then the corresponding fields, far along the z-axis, are

2kpr3 kpr
Ep0 (z) = , EQ0 (z) = .
R3 z 3 R2 z 2
The first term is negligible compared to the first, due to the many powers of R and z in the
denominator. Thus, keeping only the second term, the force on the original dipole is
d 2kp2 r
F =p E(z) =−
dz z=R R5
which falls off very quickly with distance. This derivation illustrates a common subtlety: it
might not always be obvious how far to approximate. We threw away terms subleading in
r/R, because we only wanted the leading contribution. But if we had applied that principle
to the image charges at the first step, we would have thrown out the tiny net charge Q0 ,
which actually provides the dominant contribution to the force, because of how tiny p0 is.

Now, the situation for a neutral sphere is completely different. By the logic of problem 4,
there’s a third image at the center of the sphere to enforce neutrality,
pr
− at z = 0.
R2
The image charges can now be decomposed into a combination of two dipole moments. We
already saw the first one p0 above, while the second is, to leading order
pr r2 pr3
p00 ≈ =
R2 R R3

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with the same magnitude and direction as p0 . Thus, this system of image charges has
approximate dipole moment 2p0 . The corresponding force is

d 4 kpr3 12 kp2 r3
F =p =−
dz R3 z 3 z=R R7

which falls off even more quickly with distance. In this derivation, we didn’t have to worry
too much about getting p00 exactly right, because there was no net charge (“monopole”)
term that could’ve overwhelmed the dipole field, so all other field contributions are
safely suppressed by more powers of r/R. (Of course, if p00 had come out pointing the
opposite direction to p0 , so that the two almost cancelled, we would’ve had to be more careful.)

The lesson of this example is not to just use exact expressions and Taylor expand at the end.
Here, that brute force approach would have required Taylor expanding six Coulomb’s law
forces out to order 1/R7 , which is extraordinarily tedious. Instead, to approximate properly,
we have to think carefully in every case. Incidentally, when applied to a polar and neutral
nonpolar molecule, the 1/R7 force above is called the Debye force; it is one of the “van der
Waals forces” which are often vaguely described in chemistry classes.

Example 5

Estimate the interaction force between a point charge q and a thin conducting rod of length `,
which is a distance L  ` from the charge and oriented along the separation between them.

Solution
The interaction occurs because the point charge induces negative charges on the near end of
the rod, and positive charges on the far end. These charges are then acted on by the electric
field of the point charge, causing a force.

To get a very crude estimate, let’s just suppose that charge Q appears on the far end and
charge −Q appears on the near end. The resulting field produced in the middle is
kQ
E∼ .
`2
On the other hand, this needs to cancel a field from the point charge of
kq
E∼
L2
which tells us that Q ∼ (`/L)2 q. The force on the induced charges is

kqQ`2 kq 2 `4
 
1 1
F ∼ kqQ − ∼ − ∼ − .
L2 + `2 L2 L4 L6

Again, the force is attractive, and falls off quickly with distance.

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[3] Problem 20 (Physics Cup 2017). Estimate the interaction force between a point charge q and an
infinitely thin circular neutral conducting disc of radius r if the charge is at the axis of the disc,
and the distance between the disc and the charge is L  r.

Example 6

Find the charge distribution on a conducting disc of radius R and total charge Q.

Solution
In general, there are very few situations where the charge distribution on a conductor can
be found explicitly. As you’ve seen, some of the simplest examples can be solved with image
charges. Some more complex, two-dimensional examples can be solved with a mathematical
technique called conformal mapping. And this special example can be solved with a neat trick.

Consider a uniformly charged spherical shell centered on the origin, and consider a point P
inside the shell, on the xy plane. The electric field at point P is zero, by the shell theorem.
Recall that in the usual proof of the shell theorem, one draws two cones opening out of
P in opposite directions. The charges contained in each cone produce canceling electric fields.

Now imagine shrinking the spherical shell towards the xy plane, so it becomes elliptical. The
crucial insight is that the shell theorem argument above still works, for points on the xy
plane. When we squash the shell all the way down to the xy plane, it becomes a disc, with
zero electric field on it. This is thus a valid charge distribution for a disc-shaped conductor,
and by the uniqueness theorem, it’s the only one.

By√keeping track of how much charge gets squashed to radius [r, r + dr], we find σ(r) ∝
R/ R2 − r2 , and fixing the proportionality constant gives

Q
σ(r) = √ .
4πR R2 − r2

4 Electrical Conduction
We now leave the world of electrostatics and consider magnetostatics, the study of steady currents.
Idea 6
In a conductor with conductivity σ, the current density is

J = σE.

Alternatively, E = ρJ where ρ is the resistivity. The current and charge density satisfy
∂ρ
∇·J=− .
∂t

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The current passing through a surface S at a given time is

Z
I= J · dS.
S

Since J ∝ E, we have Ohm’s law V = IR, where V is the voltage drop across the resistor.
The power dissipated in a resistor is P = IV . The resistance R adds in series, while 1/R
adds in parallel.

[2] Problem 21 (HRK). A battery causes a current to run through a loop of wire.

(a) Suppose the wire makes a sharp corner. How do the charges know to turn around there?

(b) A copper wire with conductivity σ is joined to an iron wire with conductivity σ 0 < σ. For
the current in both sections to be the same, the electric field in the iron wire must be higher.
How does that happen?

In general, the surface charge distribution in a DC circuit can be quite complex; the aspects shown
in these two short questions are just the beginning. For some more about this, see this paper.

[2] Problem 22 (PPP 22). Two students, living in neighboring rooms, decided to economize by
connecting their ceiling lights in series. They agreed they would each install a 100 W bulb in their
own rooms and that they would pay equal shares of the electricity bill. However, both tries to get
better lighting at the other’s expense. The first student installed a 200 W bulb, while the second
student installed a 50 W bulb. Which student subsequently failed the end-of-term examinations?

To warm up for DC circuits, we’ll consider some resistor network problems.

Idea 7
If any two points in a resistor network are at the same potential, nothing will change if the
two points are connected together and treated as one. More generally, the resistance of any
resistor directly connecting the two points may be changed freely.

Example 7

Consider the 3 × 3 grid below, where every edge is a resistor R.

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Find the equivalent resistance between nodes 1 and 16.

Solution
By the above idea, we can short together nodes 2/3, and 14/15, by the diagonal symmetry
of the network. Next, we can break nodes 8 and 9 into two pieces.

This is valid because the separated nodes 8a/8b and 9a/9b still have the same potential in
the new network, by the diagonal symmetry. (This is using the above idea in reverse.) Now,
the circuit has been reduced to combinations of series and parallel resistors. The resistance
between 1 and 2/3 is R/2. The resistance between 2/3 and 14/15 is the combination of three
networks in parallel, and finally the resistance of 14/15 and 16 is R/2. Thus,
!
1 1 1 −1 1
 
1 13
Req = + + + + R = R.
2 3 3 2 2 7

You won’t see any resistor problems as complicated as this one for the rest of the training,
because they’re kind of contrived; the point of this example was just to show multiple uses
of symmetry techniques.

Example 8: PPP 23

A black box contains a resistor network and has two output terminals.

If a battery of voltage V is connected across the first terminal, the voltage across the second
terminal is V /2. If a battery of voltage V is connected across the second terminal, the voltage
across the first terminal is V . Find one possible configuration of the resistors inside the box.

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Solution
A simple configuration with two equal resistors works.

When a battery is connected across II, the horizontal resistor doesn’t do anything. When a
battery is connected across I, the two resistors comprise a voltage divider.

[2] Problem 23. 01W USAPhO 2007, problem A1.

[2] Problem 24 (IPhO 1996). Consider the following resistor network.

Find the equivalent resistance between A and B.

[3] Problem 25. Consider a cube of side length L whose edges are resistors of resistance R.
√
(a) Compute the resistance between two vertices a distance 3L apart.
√
(b) Compute the resistance between two vertices a distance 2L apart.

(c) Compute the resistance between two vertices a distance L apart.

√
(d) Generalize to vertices nL apart on an n-dimensional cube. (Give your answer in the form
of a summation.)

[2] Problem 26 (PPP 158). Consider the circuit below, where every resistor is 1 Ω.

(a) Find the equivalence resistance between the input terminals.

(b) Do the same in the case where the chain is infinitely long.

[3] Problem 27 (PPP 159-161). Superposition can be a useful trick to analyze circuit networks.

(a) Consider an infinite two-dimensional grid of identical resistors R.

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Find the equivalent resistance between two neighboring points by considering the superposition
of a current I flowing into one point, and an equal current I flowing out the other.

(b) What would the equivalent resistance be if the resistor directly connecting the two neighboring
points was removed?

(c) Now consider an icosahedron of identical resistors R. By superposing appropriate current

distributions, find the equivalent resistance between two neighboring vertices.

Idea 8
In a circuit of resistors and batteries, Kirchoff’s loop rule states that the sum of the voltage
drops around a loop is zero. Kirchoff’s junction rule states that the net current flowing into a
vertex is zero. (This is technically nonzero, because of the effect of problem 21, but negligible
because wires have tiny capacitance.)

Remark
If the sum of the voltage drops around a loop is zero, then why would current ever want to
flow? After all, if you had a circular tube of water, the water would never flow, because the
net drop in height along the circle is zero. The reason current flows in circuits with batteries
is that within the battery, charges are moved from lower to higher electric potential energy,
just like how a pump could be used to move water upward to start a liquid circuit, by an
“electromotive force”.

But this immediately raises the question: what is this specific force? It can’t be the electric
force, because we just established that it’s pointing the wrong way. It’s not a magnetic
effect. For some setups, it is literally a mechanical force like a pump: in the Van der
Graaff generator, a motor drives the charges on a statically charged conveyor belt to higher
potential. But that’s not how batteries work.

In a battery, there is no specific force pushing charges from low to high electric potential.
Instead, the charges just jiggle around randomly, and the result emerges from the effects of
their many collisions. To understand this, consider a gravitational analogy.

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Consider an ideal gas at temperature T released in the trough shown above. The gas
molecules will randomly collide, sometimes being propelled upward by chance. Sometimes, a
gas molecule will climb the hill and fall into the deep hole, at which point it is unlikely to
come out again. Thus, if the hole begins empty, it is energetically favorable for gas molecules
to fill it. But there is no attractive force pulling molecules up along the slope! Gravity
always points down; molecules go up the slope when they are randomly bounced that way.

This is essentially how the potential in an initially neutral battery is set up. The hole
corresponds to the lower energy state an electron can reach inside the anode, but there is no
long-range force pushing it there, just the average effect of random collisions.

[2] Problem 28 (Purcell 4.10). The basic ingredient in older voltmeters and ammeters is the gal-
vanometer, a device to measure very small currents. (It works via magnetic effects, but the exact
mechanism isn’t important here.) Inherent in any galvanometer is some resistance Rg , so a physical
galvanometer can be represented by the system shown below.

Consider a circuit such as the one shown, with all quantities unknown. We want to measure
the current flowing across point A and the voltage difference between points B and C. Given a
galvanometer with known Rg , and also a supply of known resistors (ranging from much smaller to
much larger than Rg ), how can you accomplish these two tasks? Explain how to construct your
two devices (called an ammeter and voltmeter), and also how you should insert them in the given
circuit. You will need to make sure that you (a) affect the given circuit as little as possible, and (b)
don’t destroy your galvanometer by passing more current through it than it can handle.

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Remark
Occasionally, you might see Olympiad problems where a voltmeter is connected in series.
The most common voltmeters are handheld digital multimeters, where the voltmeter setting
presents a resistance of about 10 MΩ. Thus, for such problems, you should just treat the
voltmeter like a high-resistance resistor.

Is this realistic? Well, it certainly happens every day, in almost every introductory physics
lab in the world. But no professional would ever do this on purpose, because voltmeters aren’t
designed to be used this way. There is no guarantee that the resistance of the voltmeter is a
constant. Instead, for most digital multimeters, there is a complex circuit inside that adjusts
the internal resistance depending on the input and the configuration settings. You probably
won’t break the voltmeter when you put it in series, but you won’t get reliable results either.

[2] Problem 29. 01^‚ USAPhO Quarterfinal 2009, problems 3 and 4.

[3] Problem 30. INPhO 2021, problem 1. A nice problem on practical circuit measurements. Note
that the question statement is a bit vague. You are supposed to keep track of quantities of order
RA /R and R/RV , but you are allowed to neglect quantities as small as RA /RV .

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