• question_answer  Determine the potential difference across the plates of the capacitor C i of the network shown in the figure. (Assume, ${{E}_{2}}>{{E}_{1}}$).   OR A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor, How much electrostatic energy is lost in the process?

Effective emf in the circuit, $E={{E}_{2}}-{{E}_{1}}$ The net capacitance in the circuit, $C=\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}$ Charge on each capacitor, $q=C\cdot E=\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}({{E}_{2}}-{{E}_{1}})$ The potential difference across the plates of ${{C}_{1}},$ $V=\frac{q}{{{C}_{1}}}=\frac{{{C}_{1}}{{C}_{2}}}{({{C}_{1}}+{{C}_{2}})}\frac{({{E}_{2}}-{{E}_{1}})}{{{C}_{1}}}$             $=\frac{{{C}_{2}}({{E}_{2}}-{{E}_{1}})}{({{C}_{1}}+{{C}_{2}})}$ Or Given,   ${{C}_{1}}={{C}_{2}}=600\times {{10}^{-12}}F$             ${{V}_{1}}=200V,$     ${{V}_{2}}=0$ $\therefore$ Loss of energy,$\Delta U=\frac{{{C}_{1}}{{C}_{2}}}{2({{C}_{1}}+{{C}_{2}})}{{({{V}_{1}}-{{V}_{2}})}^{2}}$             $=\frac{600\times {{10}^{-12}}\times 600\times {{10}^{-12}}}{2(600+600)\times {{10}^{-12}}}{{(200-0)}^{2}}$             $=\frac{36\times {{10}^{-8}}}{2400}\times 4\times {{10}^{4}}=6\times {{10}^{-6}}J$