Question

The plates of a parallel-plate capacitor are maintained with a constant voltage by a battery as they are pushed together, without touching. How is the amount of charge on the plates affected during this process?
The plates of a parallel-plate capacitor are maintained with a constant voltage by a battery as they are pushed together, without touching. How is the amount of charge on the plates affected during this process?
The amount of charge remains constant.
The amount of charge on the plates decreases during this process.
The amount of charge on the plates becomes zero.
The amount of charge on the plates increases during this process.

Answer 1

The amount of charge on the plates increases during this process.

Explanation:

The relationship between charge and potential difference through a capacitor is

$C=\frac{Q}{\Delta V}$

where

C is the capacitance

Q is the charge stored

$\Delta V$ is the potential difference

The capacitance of a parallel-plate capacitor is given by

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0$ is the vacuum permittivity

A is the area of the plates

d is the distance between the plates

Combining the two equations, we get:

$Q=\frac{\epsilon_0 A \Delta V}{d}$

In this problem:

- The potential difference between the plates, $\Delta V$, is kept constant

- The area of the plates, A, remains  constant

- The distance between the plates, d, is decreased

Since Q is inversely proportional to d, this means that as the plates are pushed together, the amount of charge on the plates increases during the process.

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