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(C) Let the charges on the outer surfaces of the system be $q_1$ and $q_2$. For a system of parallel plates, the outer charges are equal to half the t

Vedclass · Accepted Answer

The potential difference between $A$ and $C$ is one-third of the potential difference between $B$ and $C$.

Vedclass · Answer

(C) Let the charges on the outer surfaces of the system be $q_1$ and $q_2$. For a system of parallel plates, the outer charges are equal to half the total charge: $q_1 = q_2 = (Q - Q + Q)/2 = Q/2$.Let the charge on the inner face of $A$ be $x$. Then the charge on the inner face of $C$ (facing $A$) is $Q - x$. The charge on the other face of $C$ is $Q - (Q - x) = x$. The charge on the inner face of $B$ is $-Q - Q/2 = -3Q/2$.Using the property that the electric field inside a conductor is zero, the field between $A$ and $C$ is $E_1 = x/(\epsilon_0 A)$ and between $C$ and $B$ is $E_2 = (x - Q)/(\epsilon_0 A)$.Since the potential difference $V = E \cdot d$, and the distance between plates is $d/2$, we find $V_{AC} = (x \cdot d)/(2\epsilon_0 A)$ and $V_{CB} = ((x - Q) \cdot d)/(2\epsilon_0 A)$.Solving for the charge distribution, we find $x = Q/4$. Thus, $V_{AC} = Q d / (8\epsilon_0 A)$ and $V_{CB} = 3Q d / (8\epsilon_0 A)$.Therefore, $V_{AC} = (1/3) V_{CB}$, making option $C$ correct.

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