The charges on the four surfaces of two isola | vedclass

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(C) For a parallel plate capacitor,the charges on the facing surfaces are equal in magnitude and opposite in sign. Thus,$Q_2 = -Q_3$.The potential dif

Vedclass · Accepted Answer

$\frac{Q_2 - Q_3}{2C}$

Vedclass · Answer

(C) For a parallel plate capacitor,the charges on the facing surfaces are equal in magnitude and opposite in sign. Thus,$Q_2 = -Q_3$.The potential difference $V$ between the plates of a capacitor is given by $V = \frac{Q}{C}$,where $Q$ is the charge on the inner surface of the positive plate.Here,the charge on the inner surface of the first plate is $Q_2$. Therefore,the potential difference is $V = \frac{Q_2}{C}$.Since $Q_2 = -Q_3$,we can write $Q_2 = \frac{Q_2 - Q_3}{2}$.Substituting this into the potential formula:$V = \frac{Q_2 - Q_3}{2C}$.

The charges on the four surfaces of two isolated plates with capacitance $C$ are $Q_1$,$Q_2$,$Q_3$,and $Q_4$ as shown in the figure. The potential difference between the plates is:

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