91Ó°ÊÓ

The total energy for a system of capacitors is given by the formula:

$$\begin{gathered} U=\frac{1}{2}{C}_1{V^2}_1+\frac{1}{2}{C}_2{V^2}_2 \\ =\frac{1}{2}C({V^2}_1+{V^2}_2) \end{gathered}$$

Now, we know that the voltage across the plates in a parallel connection is same.

Therefore, the equation becomes:

$$\begin{gathered} U_P=\frac{1}{2}{C}_1{V^2}_1+\frac{1}{2}{C}_2{V^2}_2 \\ =\frac{1}{2}C({V^2}_1+{V^2}_2) \\ =C{V}_2 \end{gathered}$$

We know that in a series connection the applied voltage is different than the voltage of each capacitor is different. The voltage is divided by both capacitors and the voltage one each capacitor will be given by:

$V_1=V_2=\frac{V}{2}$

Now, energy stored in each capacitor in series is given by:

$$\begin{gathered} U_s=\frac{1}{2}C({V^2}_1+{V^2}_2) \\ {U}_s=\frac{1}{2}C\left({\left[\frac{V}{2}\right]}^2+{\left[\frac{V}{2}\right]}^2\right) \\ =\frac{1}{4}C{V}_2 \end{gathered}$$

The ration between $U_p$and $U_s$is:

Charge Q is related to the stored energy, given by:

$$\begin{gathered} U=\frac{1}{2}QV \\ =\frac{2U}{V} \end{gathered}$$

For parallel:

$$\begin{gathered} Q_p=\frac{2U_p}{V_P} \\ =\frac{2\left(C{V}^2\right)}{V} \\ =2CV \end{gathered}$$

For series:

$$\begin{gathered} Q_p=\frac{2U_s}{V_s} \\ =\frac{2\left({\displaystyle \frac{1}{4}}C{V}^2\right)}{V/2} \\ =CV \end{gathered}$$

Now the ration between $Q_p$and $Q_s$is:

$$\begin{gathered} \frac{{\mathrm{Q}}_{\mathrm{p}}}{{\mathrm{Q}}_{\mathrm{S}}}=\frac{2\mathrm{CV}}{\mathrm{CV}} \\ {\mathrm{Q}}_{\mathrm{p}}=2{\mathrm{Q}}_{\mathrm{s}} \end{gathered}$$

The electric field between two parallel plates of capacitor is given by:

$E=\frac{V}{d}$

Now, putting the potential difference for both parallel and series connection we get the ratio of the electric field:

$$\begin{gathered} \frac{E_p}{E_s}=\frac{\left(v/d\right)}{\left(v/2d\right)} \\ =2 \\ {E}_p=2E_s \end{gathered}$$

Hence, the parallel connection shows higher energy.