Kvant Physics Problem 17

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Verdicts: PASS + PASS
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Problem

A charged plate carrying charge $q$ is placed between the plates of a short-circuited parallel-plate capacitor. The plate is moved parallel to itself through a distance $x$ (Fig. 6). What charge passes through the external circuit of the capacitor during this process if the distance between its plates is $d$?

Fig. 6

Setup and Assumptions

A parallel-plate capacitor has plate separation $d$. Its plates are connected by an external conductor, so the capacitor is short-circuited and the potential difference between the plates is always zero.

A thin conducting plate carrying a total charge $q$ is placed between the capacitor plates. The charged plate is moved parallel to the capacitor plates from one position to another, its displacement measured along the direction perpendicular to the capacitor plates. Let the distance from the left capacitor plate to the charged plate be $x$, so the distance from the charged plate to the right capacitor plate is $d-x$.

The required quantity is the total charge that passes through the external circuit while the charged plate moves through a distance $x$.

The capacitor plates are assumed very large compared with the separation $d$, so edge effects are neglected. The charged plate is also assumed parallel to the capacitor plates. Electrostatic equilibrium is maintained throughout the motion, and the connecting wire is assumed to have negligible resistance so that the capacitor plates remain at the same potential.

Physical Principles

For conductors in electrostatic equilibrium, the electric field inside each conductor is zero.

The electric field produced by an infinite charged sheet with surface charge density $\sigma$ has magnitude

$$E=\frac{\sigma}{2\varepsilon_0}$$

on each side of the sheet.

The electric field between parallel charged sheets is obtained by superposition.

Since the capacitor plates are short-circuited, the potential difference between them must satisfy

$$V=0.$$

The charge passing through the external circuit equals the change of charge on either capacitor plate.

Derivation

Let the charges induced on the left and right capacitor plates be $Q_1$ and $Q_2$.

Because the capacitor plates are connected together and initially neutral,

$$Q_1+Q_2=-q.$$

Introduce the corresponding surface charge densities

$$\sigma_1=\frac{Q_1}{S},\qquad \sigma_2=\frac{Q_2}{S},\qquad \sigma=\frac{q}{S},$$

where $S$ is the area of the plates.

Consider the electric field in the region between the left capacitor plate and the charged plate. By superposition,

$$E_L=\frac{\sigma_1}{2\varepsilon_0} -\frac{\sigma_2}{2\varepsilon_0} -\frac{\sigma}{2\varepsilon_0}.$$

Similarly, in the region between the charged plate and the right capacitor plate,

$$E_R=\frac{\sigma_1}{2\varepsilon_0} -\frac{\sigma_2}{2\varepsilon_0} +\frac{\sigma}{2\varepsilon_0}.$$

The potential difference between the capacitor plates is

$$V=-E_L x-E_R(d-x).$$

Since the capacitor is short-circuited,

$$E_L x+E_R(d-x)=0.$$

Substituting the expressions for the fields gives

$$\left(\sigma_1-\sigma_2-\sigma\right)x + \left(\sigma_1-\sigma_2+\sigma\right)(d-x) =0.$$

After expansion,

$$(\sigma_1-\sigma_2)d+\sigma(d-2x)=0.$$

Hence

$$\sigma_1-\sigma_2 = -\sigma\frac{d-2x}{d}.$$

Together with

$$\sigma_1+\sigma_2=-\sigma,$$

this yields

$$2\sigma_1 = -\sigma-\sigma\frac{d-2x}{d} = -\sigma\left(2-\frac{2x}{d}\right),$$

so

$$\sigma_1 = -\sigma\left(1-\frac{x}{d}\right).$$

Therefore

$$Q_1 = -q\left(1-\frac{x}{d}\right).$$

Likewise,

$$Q_2 = -\frac{q,x}{d}.$$

Suppose the charged plate starts adjacent to the left capacitor plate, corresponding to $x=0$. Then

$$Q_1(0)=-q, \qquad Q_2(0)=0.$$

After the plate has been displaced by a distance $x$,

$$Q_1(x) = -q\left(1-\frac{x}{d}\right).$$

The charge that has flowed through the external circuit is the decrease in the magnitude of the charge on the left plate,

$$\Delta Q = Q_1(x)-Q_1(0) = \frac{q,x}{d}.$$

Its direction is from the left plate to the right plate for $q>0$.

Result

The amount of charge passing through the external circuit during the displacement is

$$\boxed{\Delta Q=\frac{q,x}{d}}.$$

No numerical values are given in the statement. The result remains symbolic:

$$\boxed{\Delta Q=\frac{q,x}{d}}.$$

The unit is the coulomb, since

$$[\Delta Q] = [q]\frac{[x]}{[d]} = \mathrm{C}.$$

Sanity Checks

If $x=0$, the charged plate has not moved. The formula gives

$$\Delta Q=0,$$

which is consistent with physical intuition.

If $x=d$, the charged plate moves from one capacitor plate to the other. The formula gives

$$\Delta Q=q.$$

In this case a total charge equal to the charge carried by the moving plate passes through the external connection.

The ratio $x/d$ is dimensionless, so $\Delta Q$ has the same dimension as $q$, namely charge. Dimensional consistency is satisfied.

The most common source of error is the sign of the electric field contributions from the three charged sheets when forming $E_L$ and $E_R$. A sign mistake at that stage changes the relation between $Q_1$ and $Q_2$ and leads to an incorrect induced charge transfer.