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  4. A parallel plate capacitor of capacitance 1 μF is charged on both the plates +2 μC and +4 μC respectively . The potential difference developed across the capacitor is:

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Q. A parallel plate capacitor of capacitance $1\,μF$ is charged on both the plates $+2\,μC$ and $+4\,μC$ respectively . The potential difference developed across the capacitor is:

NTA AbhyasNTA Abhyas 2022

Solution:

Charge on capacitor plate is given by, $Q=\frac{Q_{1} - Q_{2}}{2}$
$\Rightarrow Q=\frac{4 - 2}{2}=1μC$
$V=\frac{Q}{C}=\frac{1 μC}{1 μF}=1$ Volt