Kvant Physics Problem 17
A parallel-plate capacitor with large identical plates of area $A$ is short-circuited, so both plates are connected by an external conducting wire and always remain at the same electric potential.
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Problem
Between the plates of a short-circuited parallel-plate capacitor, a plate carrying charge $q$ is placed. The plate is moved parallel to itself by 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 with large identical plates of area $A$ is short-circuited, so both plates are connected by an external conducting wire and always remain at the same electric potential. A thin conducting plate carrying total charge $q$ is placed between the capacitor plates, parallel to them and equidistant from both plates at the initial moment. The charged plate is then displaced parallel to itself by a distance $x$ in the plane of the capacitor plates.
Edge effects of the capacitor plates are neglected, so the electric field is assumed uniform over any region where the plates overlap. The charged plate is assumed thin compared with the separation $d$ and does not touch the capacitor plates during the motion. The motion is quasistatic, so at every instant electrostatic equilibrium is established and charge can freely flow through the external short-circuit.
The quantity to determine is the total charge that passes through the external circuit during the displacement.
Physical Principles
Electrostatic equilibrium of conductors requires that the electric field inside each conducting plate is zero and that all points of each connected conductor share the same electric potential.
Gauss’s law in integral form,
$\oint \mathbf{E}\cdot d\mathbf{S} = \frac{Q_{\text{enc}}}{\varepsilon_0},$
relates the electric flux through a closed surface to the enclosed charge and determines how induced surface charge is distributed on conductors.
Charge conservation implies that any change in induced charge on the capacitor plates must be compensated by charge flow through the external circuit because the plates are connected in a short circuit.
Superposition allows the induced charge distribution to be treated as the sum of contributions from different portions of the charged plate, with each portion acting independently in the planar approximation.
Derivation
Let the overlap region between the charged plate and the capacitor plates have area $S(x)$ when the plate has been displaced by a distance $x$ parallel to itself. The key point is that only the portion of the charged plate that lies between the capacitor plates contributes to inducing charge on the external electrodes.
For a uniformly charged plate with surface charge density $\sigma = \frac{q}{A_0}$, where $A_0$ is the total area of the charged plate, the charge located within the overlap region is
$q_{\text{in}}(x) = q \frac{S(x)}{A_0}.$
In the short-circuited configuration, the two capacitor plates together form a single conductor. The total induced charge on this conductor must cancel the charge enclosed between the plates in order to maintain zero electric field inside the conductor region accessible from the short. Therefore the charge on the external circuit is equal in magnitude to the change in induced charge on the plates, which equals the change in $q_{\text{in}}(x)$.
Thus the transferred charge is
$Q = \Delta q_{\text{in}} = q \frac{\Delta S}{A_0}.$
To determine $\Delta S$, consider that the plate of characteristic length $L$ in the direction of motion is partially shifted. As it moves by $x$, the overlap region decreases linearly,
$S(x) = A_0 \left(1 - \frac{x}{L}\right),$
valid for $0 \le x \le L$. Substituting this into the expression for transferred charge gives
$Q = q \left(1 - \frac{x}{L} - 1\right) = -q \frac{x}{L}.$
The sign indicates the direction of charge flow, while the magnitude of charge passing through the external circuit is
$|Q| = q \frac{x}{L}.$
Result
The charge that passes through the external circuit has magnitude
$Q = q \frac{x}{L}.$
Substitution of quantities leaves the result already in symbolic form, since no numerical values for $q$, $x$, or $L$ are specified. The unit is coulomb, so the expression is dimensionally consistent because $\frac{x}{L}$ is dimensionless.
Final result,
$\boxed{Q = q \frac{x}{L}\ \text{C}}.$
Sanity Checks
Dimensional consistency follows because $q$ has units of coulombs and the ratio $\frac{x}{L}$ is dimensionless, so the transferred charge has units of coulombs.
In the limit $x \to 0$, no displacement occurs and $Q \to 0$, matching the absence of change in overlap region. In the limit $x \to L$, the charged plate completely leaves the initial overlap region, giving $Q \to q$, which corresponds to the full redistribution of induced charge associated with the entire charge on the plate.
The derivation is sensitive to the assumption of uniform surface charge and negligible edge effects; violating these would break the proportionality between induced charge and overlap area. A sign error typically arises when relating induced charge change on the conductors to charge flow in the external circuit, since the external circuit current is defined opposite to the change in conductor charge.