Question

What happens to the energy stored in a capacitor if the separation between the plates is doubled after it is disconnected form the battery?

Answer 1

Step 1: Given and assumptions

Energy stored = $U$

Potential difference = $V$

Capacitance = $C$

Area = $A$

Distance between the plates = $d$

The permittivity of the air or vacuum = ${\epsilon }_0$

Step 2: Calculation of the energy stored in a capacitor

Potential difference, $V=\frac{Q}{C}$

Initially the capacitance,$C=\frac{{\epsilon }_{0}A}{d}$

Initially, energy is stored when the capacitor is disconnected from the battery

$$\begin{gathered} U=\frac{1}{2}C{V}^2 \\ U=\frac{1}{2}C{\left(\frac{Q}{C}\right)}^2 \\ U=\frac{1}{2}\times \frac{C{Q}^2}{C^2} \\ U=\frac{1}{2}\times \frac{Q^2}{C} \\ U=\frac{1}{2}\times \frac{Q^2}{{\displaystyle \frac{{\epsilon }_{0}A}{d}}} \\ U=\frac{1}{2}\frac{d{Q}^2}{{\epsilon }_{0}A} \end{gathered}$$

Step 3: Calculation of the energy stored in a capacitor if the separation between the plates is doubled

The distance will be $2d$

Finally the capacitance,$C=\frac{{\epsilon }_{0}A}{d}$

Finally, energy is stored when the capacitor is disconnected from the battery,

$$\begin{gathered} U\text{'}=\frac{1}{2}\frac{\left(2d\right)Q^2}{{\epsilon }_{0}A} \\ U\text{'}=2.\frac{1}{2}\frac{d{Q}^2}{{\epsilon }_{0}A} \\ U\text{'}=2U \end{gathered}$$

Hence, the energy stored in a capacitor will be doubled if the separation between the plates is doubled after it is disconnected from the battery.