Answer to Question #159634 in Electric Circuits for Ariel

Question

A parallel-plate capacitor is charged and then is disconnected from the battery. By what factor does the stored energy change when the plate separation is then doubled?
(a) It becomes 4 times larger
(b) It becomes 2 times larger
(c) It stays the same
(d) It becomes one-half as large.

Answer 1

The energy stored in capacitor can be written as follows:

$$E=\dfrac{Q^2}{2C}.$$

The capacitance of the capacitor can be found as follows:

$$C=\epsilon_0\dfrac{A}{d}.$$

If the separation between the plates doubled the new capacitance will be:

$$C=\epsilon_0\dfrac{A}{2d}=\dfrac{1}{2}\epsilon_0\dfrac{A}{d}=\dfrac{1}{2}C.$$

Therefore, when the plate separation is doubled the capacitance halved.

Let's substitute the new capacitance into the formula for the energy stored:

$$E_{new}=\dfrac{1}{2}\dfrac{Q^2}{\dfrac{C}{2}}=2\cdot\dfrac{Q^2}{2C}=2E.$$

As we can see from the calculations, when the plate separation is doubled the stored energy becomes 2 times larger. Therefore, the correct answer is (b).

Answer:

(b) It becomes 2 times larger.