91Ó°ÊÓ

Increased stored energy,

$$\begin{gathered} ∆U/U=10\% \\ =0.1 \end{gathered}$$

In the capacitor, electrostatic potential energy is stored. Therefore, it has to do with the charge and voltage between the capacitor's plates. Here, we need to use the formula of energy stored in the capacitor to find the percentage of the required increase in potential difference.

Formula:

The energy stored between the capacitors,$U=\frac{1}{2}C{V}^2$ …(¾±)

If the increase in the stored energy is by small amount $∆U$, then we must have to increase the potential difference by small amount$∆V$. Thus, the increased equation is given using equation (i) as follows:

$$\begin{gathered} \left(U+∆U\right)=\frac{1}{2}C{\left(V+∆V\right)}^2 \\ U\left(1+\frac{∆U}{U}\right)=\frac{1}{2}C{V}^2{\left(1+\frac{∆V}{V}\right)}^2 \\ U\left(1+\frac{∆U}{U}\right)=U{\left(1+\frac{∆V}{V}\right)}^2\left(\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right)\right) \\ \sqrt{1+\frac{∆U}{U}}=1+\frac{∆V}{V} \\ \frac{∆V}{V}=\sqrt{1+\frac{∆U}{U}}-1 \\ =\sqrt{1+0.1}-1 \\ =\sqrt{1.1}-1 \\ =0.049 \\ =4.9\% \end{gathered}$$

Hence, the value of the increased percentage in the potential difference is 4.9%.