Question

A parallel-plate capacitor is connected to a battery and becomes fully charged. The capacitor is then disconnected, and the separation between the plates is increased in such a way that no charge leaks off. As the plates are being separated, the energy stored in this capacitor:_____________
a) decreases
b) becomes zero
c) increases
d) does not change

Answer 1

Answer:

Increases

Explanation:

The expression for the capacitance is as follows as;

$C=\frac{\epsilon _{0}A}{d}$

Here, C is the capacitance, $\epsilon _{0}$ is the permittivity of free space, A is the area and d is the distance between the parallel plate capacitor.

It can be concluded from the above expression, the capacitance is inversely proportional to the distance. According to the given problem, the capacitor is disconnected from the battery and the distance between the plates is increased. Then, the capacitance of the given capacitor will decrease in this case.

The expression for the energy stored in the parallel plate capacitor is as follows;

$E=\frac{Q^{2}}{2C}$

Here, E is the energy stored in the capacitor, C is the capacitance and Q is the charge.

Energy stored in the given capacitor is inversely proportional to the capacitor. The charge on the capacitor is constant. In the given problem, as the distance between the parallel plates is being separated, the energy stored in this capacitor increases.

Therefore, the option (c) is correct.