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?

Answer 1

Answer:

0.5

Explanation:

The energy of a charged capacitor is given by $\frac{1}{2}QC$ where $Q$ is the charge on it and $C$ is its capacitance. The capacitance is defined, geometrically, as

$\dfrac{\epsilon A}{d}$

where $\epsilon$ is a constant that is determined by the material between the plates, $A$ is the area of the plates and $d$ is the distance between them. It is then that the capacitance is linearly inversely proportional to the separation distance; as the distance increases, the capacitance reduces.

Because it is linear, when the separation distance is doubled, other factors remaining constant, the capacitance is halved. Because the capacitance is halved, the energy is halved.

Let's see the mathematics.

Initial capacitance, $C_1$ at initial separation, $d_1$

$C_1=\dfrac{\epsilon A}{d_1}$

If separation is doubled, separation becomes $d_2=2d_1$. Then the capacitance becomes

$C_2=\dfrac{\epsilon A}{d_2}$

$C_2=\dfrac{\epsilon A}{2d_1}$

$C_2=0.5\dfrac{\epsilon A}{d_1}$

$C_2=0.5C_1$

Initial energy, $E_1=\frac{1}{2}QC_1$

Final energy, $E_2=\frac{1}{2}QC_2= \frac{1}{2}Q\times0.5C_1$

$E_2=0.5\times\frac{1}{2}QC_1=0.5E_1$

Hence, $\frac{E_2}{E_1}=0.5$