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(B) Let the plates be numbered $1, 2, 3, 4, 5$ from left to right. Plate $2$ is $M$ and plate $4$ is $N$.Since plates $1$ and $5$ are grounded,their p

Vedclass · Accepted Answer

$\frac{d(Q_1-Q_2)}{2 \epsilon_0 A}$

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(B) Let the plates be numbered $1, 2, 3, 4, 5$ from left to right. Plate $2$ is $M$ and plate $4$ is $N$.Since plates $1$ and $5$ are grounded,their potentials are zero. The electric field in the regions outside the system is zero.Let the charges on the inner and outer surfaces of the plates be $q_1, q_2, q_3, q_4, q_5, q_6, q_7, q_8, q_9, q_{10}$.Due to grounding,the outer surfaces of plates $1$ and $5$ have zero charge. The total charge on plate $M$ is $Q_1$ and on plate $N$ is $Q_2$.Using the property that the electric field inside a conductor is zero,the potential difference between plates $M$ and $N$ is given by $V_M - V_N = E \cdot d$,where $E$ is the electric field between them.For this configuration,the electric field between plates $M$ and $N$ is $E = \frac{Q_1 - Q_2}{2 \epsilon_0 A}$.Thus,the potential difference is $\Delta V = E \cdot d = \frac{d(Q_1 - Q_2)}{2 \epsilon_0 A}$.Therefore,the correct option is $B$.

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