The charges on two plates of a 10,mu F capaci | vedclass

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(A) When two plates of a capacitor have charges $Q_1$ and $Q_2$,the charge $Q$ that appears on the inner facing surfaces is given by $Q = \frac{Q_1 -

Vedclass · Accepted Answer

$0.5$

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(A) When two plates of a capacitor have charges $Q_1$ and $Q_2$,the charge $Q$ that appears on the inner facing surfaces is given by $Q = \frac{Q_1 - Q_2}{2}$.Here,$Q_1 = 15\,\mu C$ and $Q_2 = 5\,\mu C$.So,the charge on the inner surfaces is $Q = \frac{15\,\mu C - 5\,\mu C}{2} = \frac{10\,\mu C}{2} = 5\,\mu C$.The potential difference $V$ across the capacitor is given by $V = \frac{Q}{C}$.Given $C = 10\,\mu F = 10 	imes 10^{-6}\,F$ and $Q = 5\,\mu C = 5 	imes 10^{-6}\,C$.Therefore,$V = \frac{5 	imes 10^{-6}}{10 	imes 10^{-6}} = 0.5\,V$.

The charges on two plates of a $10\,\mu F$ capacitor are $5\,\mu C$ and $15\,\mu C$. The potential difference across the capacitor plates is........$V$.

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