Plz ans this question how and why we need to find the potential difference across the capacitor???

$$\begin{gathered} Applying Kirchhoff\text{'}s law to loop GABFG and EDCBFE, \\ we will get 4i_1+4i_2=6...\left(1\right) \\ 2i_2+3i_2+4i_1+4i_2=3+2\Rightarrow 4i_1+9i_2=5..\left(2\right) \\ \left(1\right)-\left(2\right), \\ -5i_2=1 in magnitude i_2=0.2A \\ Potential difference E and F is V=E2+i_{2}R2=2+(0.2)2=2.4V \\ As no current passes through R_1 so the potential at H and F are same. \\ Thus the potential difference across capacitor is V=2.4V. \\ Energy stored in capacitor is U=\frac{1}{2}C{V}^2=0.5\times 5\times {10}^{-6}\times 2.42 \\ =1.44\times {10}^{-5}J \end{gathered}$$

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