Chapter 17, Problem 10Q

The parallel plates of an isolated capacitor carry opposite charges, Q. If the separation of the plates is increased, is a force required to do so? Is the potential difference changed? What happens to the work done in the pulling process?

To determine

If parallel plates of an isolated capacitor carry opposite charges, Q. Is a force required to increase the separation of the plates? Is the potential difference changed? What happens to the work done in the pulling process?

Answer to Problem 10Q

Solution:

Force is required, potential difference increases, and work is stored as potential energy.

Explanation of Solution

Since plates carry opposite charges there is force of attraction between them, so for increasing separation between them a force is necessary.

We have capacitance, $C=\frac{{\epsilon }_{0}A}{d}$, where A is the area of plates, d is the separation and ${\epsilon }_0$ is the dielectric constant. When distance between plates increases capacitance decreases.

We also have potential difference, $V=\frac{Q}{C}$, where Q is the charge. When C decreases potential difference increases.

Here the plates are in a force of attraction, so work is required to increase the separation between them. This work is stored as the potential energy between the plates.