A parallel plate capacitor is connected to a battery that maintains a constant potential difference between the plates. If the plates are pulled away from each other, increasing their separation, what happens to the amount of charge on the plates?

Question

The amount of the charge increases, because the capacitance decreases.
The amount of the charge decreases, because the capacitance increases.
The amount of the charge increases, because the capacitance increases.
Nothing happens; the amount of charge stays the same.
The amount of the charge decreases, because the capacitance decreases.

Answer 1

In a parallel plate capacitor connected to a battery, the voltage (potential difference) across the capacitor is held constant by the battery. The capacitance \( C \) of a parallel plate capacitor is given by the formula:

\[ C = \frac{\varepsilon_0 A}{d} \]

where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates.

When the plates are pulled apart, the separation \( d \) increases, which means that the capacitance \( C \) will decrease (since \( C \) is inversely proportional to \( d \)).

Since the battery maintains a constant voltage \( V \), the charge \( Q \) on the plates can be described by the formula:

\[ Q = C \cdot V \]

With the capacitance \( C \) decreasing and the voltage \( V \) remaining constant, the charge \( Q \) will also decrease.

Thus, the correct statement is:

The amount of the charge decreases, because the capacitance decreases.