The charges on two plates of a $10\,\mu f$ capacitor are $5\,\mu C$ and $15\,\mu C$ then the potential difference across the capacitor plates is........$V$
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The charge of the capacitor in expression $Q=C V$ is the magnitude of charge which appears on facing surfaces of plates of capacitor.
The distribution of charge on various faces for present case are shown in figure.
So, here $Q=5\, \mu C$
$\therefore \quad \mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}}=\frac{5 \times 10^{-6}}{10 \times 14^{-6}}=0.5 \mathrm{\,volt}$
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