A parallel plate capacitor has plate area $A$ and separation $d$. It is charged to a potential difference $V_o$. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is
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(d) Work done $W = {U_f} - {U_i}$
${U_i} = \frac{1}{2}C{V_0}^2{\rm{\, and \,}}{{\rm{U}}_{\rm{f}}} = \frac{1}{2}\frac{{(C)}}{3}.{(3{V_0})^2}$$ = 3 \times \frac{1}{2}C{V_0}^2$
So $W = \frac{{{\varepsilon _0}A{V_0}^2}}{d}$
${U_i} = \frac{1}{2}C{V_0}^2{\rm{\, and \,}}{{\rm{U}}_{\rm{f}}} = \frac{1}{2}\frac{{(C)}}{3}.{(3{V_0})^2}$$ = 3 \times \frac{1}{2}C{V_0}^2$
So $W = \frac{{{\varepsilon _0}A{V_0}^2}}{d}$
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