Charge of $2Q$ and $-Q$ are placed on two plates of a parallel plate capacitor. If capacitance of capacitor is $C$, potential difference between the plates is
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$\therefore \frac{3}{2} \mathrm{Q}=\mathrm{CV} \Rightarrow \mathrm{V}=\frac{3 \mathrm{Q}}{2 \mathrm{C}}$
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