This is problem 1463 from the Russian magazine Quant. Unfortunately my Russian is not good enough, so I apologize if the translation of the problem contains errors.

Problem. Inside a flat charge capacitor with distance between the plates $d$ a weakly conducting slab of thickness $h<d$ and resistivity $\rho$ is inserted. The area of the plates and the slab is $S$ (as always it is assumed $d\ll\sqrt{S}$). The capacitor is charged to potential $U_0$. Then the plates of the capacitor are short circuited. Find the maximal current across the slab.

Answer. $I_{max}=U_0 S/\rho h$

I obtain a completely different answer $I_{max}=U_0 S/\rho d$ and think that their solution is flawed.

Let's show that their solution doesn't make any sense. In the official solution they draw the picture

enter image description here

where $C=\varepsilon_0S/(d-h)$ and $R=\rho h/S$, claiming that it is equivalent to the initial system. Then they say that after the short circuit the capacitor $C$ (which initially has been charged to voltage $U_0$) will discharge through resistance $R$. This means that it will take approximately time $RC$ for the capacitor to discharge.

Let's return to the initial system. We are free to assume that $\rho\to\infty$, $h\to 0$, so that $R$ is finite. Technically, this means that the slab is a very thin insulator placed inside the capacitor. So when the capacitor is short circuited, plate charges will disappear instantly due to current spike in the short circuit. This contradicts the official solution, where it takes time $RC$ to discharge the capacitor.

Q: So which of the answers is correct?

My solution is as follows:

The charges on the faces of the slab can not change abruptly, but the charges on the plates of the capacitor will change abruptly as a result of the short circuit. So the words "right after the short circuit" here means that we are considering the system when this spike of current across the wire connecting the plates has quickly adjusted the plate charges.

Let the charges on the plates and the faces of the slab (right after short circuit) be $q_1,Q,-Q,q_2$ (see the picture). Then we calculate initial charges on the surfaces of the slab $\pm Q$, where $$ Q=\frac{U_0\varepsilon_0 S}{d-h}. $$ Then calculate charges on the plates (right after short circuit) so that voltage drop between the plates is $0$: $$ \frac{q_1}{2\varepsilon_0 S}d+\frac{Q}{2\varepsilon_0 S}h-\frac{-Q}{2\varepsilon_0 S}h-\frac{q_2}{2\varepsilon_0 S}d=0, $$ and the electroneutrality is satisfied $$ q_1+q_2=0. $$ Thus $$ q_{1,2}=\mp Q\frac{h}{d}. $$ Then we calculate the electric field strength inside the slab, $$ E=\frac{q_1+Q+Q-q_2}{2\varepsilon_0 S}=\frac{U_0}{d}. $$ Thus current density is $j=E/\rho$ and the current $I=jS=\frac{U_0 S}{\rho d}$.

enter image description here

  • $\begingroup$ @Hans How do you arrive at the conclusion that $q_{2,1}=\pm Q\frac{h}{d}$? $\endgroup$
    – Crimson
    Feb 13, 2018 at 15:49
  • $\begingroup$ @Crimson first electroneutrality condition $q_1+q_2=0$ coupled with the condition that voltage drop between the plates $\frac{q_1d-q_2d+2Qh}{2\varepsilon_0 S}$ is $0$. Solving this equations one concludes $q_{2,1}=\pm Q\frac{h}{d}$. $\endgroup$
    – Hans
    Feb 13, 2018 at 15:54
  • $\begingroup$ @Hans I agree with the neutrality condition. How do you arrive at this expression for the voltage drop between the plates? $\endgroup$
    – Crimson
    Feb 13, 2018 at 16:08
  • $\begingroup$ @Crimson there are 4 plates with charges, $q_1,+Q,-Q,q_2$. I just calculate electric field by using the well known formula $\frac{\sigma}{2\varepsilon_0}$, where $\sigma=Q/S$ is surface charge density. Then I sum the corresponding voltage drops between the plates of the capacitor generated by this charges. $\endgroup$
    – Hans
    Feb 13, 2018 at 16:17
  • 1
    $\begingroup$ Hi Hans. Welcome to Phys.SE. 1. The SE software does not allow any users (not even trusted users) to see who voted what. 2. As for the protection, see this meta post. $\endgroup$
    – Qmechanic
    Feb 15, 2018 at 9:40

3 Answers 3


After giving it some thought, here is a new answer:

The answer from the book is incorrect. Your derivation is correct.

They probably made the same mistake as I did in my other answer, which I now believe is incorrect. The mistake I made is that I ignored the capacitance of the resistor.

I thought it would be easier to see the system as two capacitors with a resistor in between. However it only makes things more complicated. It is thus easier to follow your derivation using Gauss' law. Still, in this answer I will show that also when you consider the system to consist of two capacitors with a resistor in between, the answer you obtain differs from that in the book.

When a voltage is put over a resistor, there is a small surface charge at the interface between the leads and the resistor. This surface charge is given by:

$$\rho=\frac{\epsilon_0 U S}{L},$$

with $L$ the lenght of the resistor. This causes a resistor to act as a smalll capacitor. This paracitic capacitance and the corresponding charge are typically very small for common resistors. However, in this specific case, the resistor formed by the slab has a shape with $\sqrt{S}>>L$. This causes the resistor to have a large capacitance. The slab should therefore be represented as an ideal resistor with a capacitance placed parallel to it.

Our total system thus consists two capacitors formed by the gaps between the slab and the plate, one ideal resistor in between them and parallel to this resistor, the capacitance of the slab.

When the total system is charged there will be a charge

$$\pm q_{gap}=\pm \frac{\epsilon_0 U_0 S}{d-h}$$

on either side of the gap capacitors. The paracitic capacitor will remain uncharged. After shorting the leads, there initially will be an voltage drop over some of the wires. Since our wires are ideal, this will be compensated by charging the paracitic capacitance and slightly discharging the the gap capacitances. This happens instantaneously and no current flows through the resistor. This proces stops when there is no longer any voltage drop over the wires, i.e., when the voltage over the paracitic capacitance matches the voltage total voltage drop over the gap capacitances.

At that point the gap capacitances will have a charge of

$$\pm q'_{gap} = \frac{\epsilon_0 U_0 S}{d-h}\frac{h}{d}$$

while the paracitic capacitor has charges.

$$\pm q'_{para} = \pm \frac{\epsilon_0 U_0 S}{d-h}\left(1-\frac{h}{d}\right)=\pm \frac{\epsilon_0 U_0 S}{d} $$

Note that in reality the surface of the slab acts as a capacitor plate for both the gap capacitance and the paracitic capacitance. The total charge on this surface does not change during the initial charge transfer.

The voltage over the paracitic capacitance is equal to the voltage over the resistor and given by:

$$U_R = U_0\frac{h}{d}$$

this leads to

$$I_{max}=\frac{U_R}{R} = U_0 \frac{S}{\rho d}$$

This is the same answer you found using Gauss' law, which in this case is much more simple. The answer in the book is thus incorrect.


The distance of current flow must be within the material containing the electrons you wish to move. You don't get a current without electrons*

I can see your line of reasoning but lets break it down a little and show you where you're going wrong.

The crux of your problem comes from thinking about a free charge between the capacitors (I imagine) a common problem set. In this case you would have electrons to move all along the distance $d$. However in this case your electrons are confined to your solid slab, $S$. Your current can only come from those electrons.

Your equation: $$I_{max} =U_{0}\frac{S}{\rho h}$$ Is describing the current from a volume of electrons with surface area S and depth h.

In case you're unaware of the background your equation is a rearrangement of the standard $V=IR$, with resistivity defined as $\rho = R \frac{A}{l}$ where $A$ is the area perpendicular to current flow and $l$ is the distance it has to flow. Rearranging we get your $I=\frac{V}{R}=V\frac{A}{\rho l}$. Since $l$ is a property of the material within which your electrons are flowing it is fixed to being a length within your slab, ie the distance $h$.

Feel like I may have laboured the point there but hopefully it got across.

*Or other charged particles.

  • 1
    $\begingroup$ My solution goes as follows. After the plates have been short circuited, I calculated the charges of the plates and charges on the faces of the slab $\frac{U_0\varepsilon_0 S}{d-h}$, so that voltage drop between the plates is $0$, and obtained that the maximal field strength inside the slab is $E=\frac{U_0}{d}$ irrespective of $h$. My answer doesn't come from assuming that electrons travel distance $d$. $\endgroup$
    – Hans
    Feb 13, 2018 at 14:15
  • 1
    $\begingroup$ However for some unknown reason they and you assume that after short circuiting all the voltage drop $U_0$ occurs across the faces of the slab. Would you mind explaining how you got this? $\endgroup$
    – Hans
    Feb 13, 2018 at 14:22
  • $\begingroup$ Also what is $V$ in your answer? Is it voltage drop between the faces of the slab and in this case how do you calculate it? $\endgroup$
    – Hans
    Feb 13, 2018 at 14:25
  • 1
    $\begingroup$ Please read the edit and the whole question carefully. There is not any assumption as the one marked in bold in your answer. $\endgroup$
    – Hans
    Feb 13, 2018 at 15:08
  • $\begingroup$ @Hans V would be the voltage drop, assuming the system starts with no potential difference. So $U_{0}$ between the two faces of the slab. I've read your solution but am unsure how you've reached all of your solutions to equations. Perhaps if you could provide a step by step difference. $\endgroup$ Feb 13, 2018 at 17:04

The answer from the book is correct.

You can consider the system to consist of two capacitors (the two gaps between the slab and the plates), with a resistor in between.

When a voltage source is connected both capacitor will be charged, until the sum of the voltages over the capacitors equals the voltage of the source.

When the source is removed and the leads are shorted, we obtain a situation of two charged capacitors in series, placed in series with a resistor. The capacitors will discharge through the resistor.

The initial voltage over the resistor will be the sum of the voltages of the capacitors. This sum will be $U_0$. The initial current will thus be.

$$I(0)=\frac{U_0}{R}=\frac{U_0 S}{\rho h}$$

Note that the current through both capacitors is the same. Therefore the buildup of charge is identical for both. Therefore,if we start with a neutral system with uncharged capacitors, we will allways have $|q_1| = |q_2| = |Q_1| = |Q_2|$.


I now believe this answer to be incorrect.

  • $\begingroup$ I really don't understand your solution. Very often such solutions contain errors, because you assume that something must be true. But a "microscopic" solution shows that the assumption wan not correct. Is this the case here? $\endgroup$
    – Hans
    Feb 13, 2018 at 16:26
  • $\begingroup$ Doesn't this solution ignore the fact that there are charges on the faces of the slab? $\endgroup$
    – Hans
    Feb 13, 2018 at 17:00
  • $\begingroup$ I have shown that this solution is wrong. But you still haven't shown that my solution is wrong. Also for unknown reasons my question is getting downvoted without any explanations. Probably there are some people out there so smart that everything is obvious to them, but then why still no one has given a sensible explanation what is going in this problem? $\endgroup$
    – Hans
    Feb 15, 2018 at 7:36
  • $\begingroup$ I now believe this answer is wrong. I leave it here, because I think most people will follow this line of reasoning. The problem with this answer is that it ignores the capacitance of the slab. For a detailed explanation see my other answer. $\endgroup$
    – Crimson
    Feb 15, 2018 at 7:55

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