Charges of 2Q and -Q are placed on two plates | vedclass

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(B) For a parallel plate capacitor with charges $Q_1$ and $Q_2$ on the two plates,the charge on the inner surfaces is given by $q = \frac{Q_1 - Q_2}{2

Vedclass · Accepted Answer

$V = \frac{3Q}{2C}$

Vedclass · Answer

(B) For a parallel plate capacitor with charges $Q_1$ and $Q_2$ on the two plates,the charge on the inner surfaces is given by $q = \frac{Q_1 - Q_2}{2}$.Here,$Q_1 = 2Q$ and $Q_2 = -Q$.Therefore,the charge on the inner surface of the first plate is $q = \frac{2Q - (-Q)}{2} = \frac{3Q}{2}$.The charge on the inner surface of the second plate is $-q = -\frac{3Q}{2}$.The potential difference $V$ between the plates is given by the relation $q = CV$,where $q$ is the magnitude of the charge on the inner surface.Substituting the values,we get $\frac{3Q}{2} = CV$.Thus,$V = \frac{3Q}{2C}$.

Charges of $2Q$ and $-Q$ are placed on two plates of a parallel plate capacitor. If the capacitance of the capacitor is $C$,find the potential difference between the plates.

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