Two identical metal plates are given charges | vedclass

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Question

(C) When two charged plates with charges $q_1$ and $q_2$ are brought close to form a parallel plate capacitor,the charge on the inner faces of the pla

Vedclass · Accepted Answer

$\frac{q_1-q_2}{2C}$

Vedclass · Answer

(C) When two charged plates with charges $q_1$ and $q_2$ are brought close to form a parallel plate capacitor,the charge on the inner faces of the plates becomes $q_{inner} = \frac{q_1 - q_2}{2}$.This is because the total charge on the inner surfaces must be equal and opposite to create a uniform electric field between the plates.The potential difference $V$ across a capacitor is given by the formula $V = \frac{Q}{C}$,where $Q$ is the magnitude of the charge on the inner face of the capacitor plates.Substituting $Q = \frac{q_1 - q_2}{2}$,we get $V = \frac{q_1 - q_2}{2C}$.Therefore,the correct option is $C$.

Two identical metal plates are given charges $q_1$ and $q_2$ $(q_2 < q_1)$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance $C$,the potential difference $V$ between the plates is

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