Problem #0098 Electricity & Magnetism Sub-menu Problem #0100 Chapters Chapters

369.

Problem 31.37 (RHK)

A parallel-plate capacitor has plates of area A and separation d and is charged to a potential difference V. The charging battery is then disconnected and the plates are pulled apart until their separation is 2d. We have to derive in terms of A, d, and V for (a) the new potential difference, (b) the initial and final stored energy, and (c) the work required for separating the plates.

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(a)

A parallel-plate capacitor has plates of area A and separation d and it is charged to a potential difference V.

The capacitance of a parallel plate capacitor of plate area A and separation d is

.

As the capacitor is charged by connecting it to a battery of voltage V, the magnitude of the charge on the plates of the capacitor will be

Now the capacitor is disconnected from the battery. The plates of the capacitor are pulled apart until their separation is 2d. Now the capacitance of the capacitor will become

.

As the charge on the capacitor plates remains unchanged when its plates are separated, the potential difference across the plates will change to

(b)

The stored energy in a capacitor of capacitance C with potential difference V is given by

.

Therefore, the initial stored energy is

and the final stored energy will be

.

(c)

The work required for separating the plates will be

.