The distance between the plates of a parallel plate condenser is $\,4mm$ and potential difference is $60\;volts$. If the distance between the plates is increased to $12\,mm$, then
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(a) For capacitor $\frac{{{V_1}}}{{{V_2}}} = \frac{{{d_1}}}{{{d_2}}}$ $==>$ ${V_2} = \frac{{{V_1} \times {d_2}}}{{{d_1}}} = \frac{{60 \times 12}}{4} = 180\,V$
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