X and Y are large,parallel conducting plates | vedclass

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(D) When two large parallel conducting plates are placed close to each other,the charges on the outer faces are equal to half the sum of the total cha

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All of the above

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(D) When two large parallel conducting plates are placed close to each other,the charges on the outer faces are equal to half the sum of the total charges on the plates. Let the charges on the four faces be $q_1, q_2, q_3$,and $q_4$ from left to right.$q_1 = q_4 = \frac{Q + 0}{2} = \frac{Q}{2}$Since plate $X$ has total charge $Q$,$q_1 + q_2 = Q \implies q_2 = Q - \frac{Q}{2} = \frac{Q}{2}$.Since plate $Y$ has total charge $0$,$q_3 + q_4 = 0 \implies q_3 = -q_4 = -\frac{Q}{2}$.The electric field due to a single charged sheet is $E = \frac{\sigma}{2\varepsilon_0} = \frac{q}{2A\varepsilon_0}$.At point $A$ (to the left of both plates): The fields from all four faces point to the left. $E_A = \frac{q_1 + q_2 + q_3 + q_4}{2A\varepsilon_0} = \frac{Q/2 + Q/2 - Q/2 + Q/2}{2A\varepsilon_0} = \frac{Q}{2A\varepsilon_0}$ (towards the left).At point $B$ (between the plates): The fields from $q_1, q_2$ point to the right,and from $q_3, q_4$ point to the left. $E_B = \frac{q_1 + q_2 - q_3 - q_4}{2A\varepsilon_0} = \frac{Q/2 + Q/2 - (-Q/2) - Q/2}{2A\varepsilon_0} = \frac{Q}{2A\varepsilon_0}$ (towards the right).At point $C$ (to the right of both plates): The fields from all four faces point to the right. $E_C = \frac{q_1 + q_2 + q_3 + q_4}{2A\varepsilon_0} = \frac{Q}{2A\varepsilon_0}$ (towards the right).Thus,the magnitude of the electric field at $A, B$,and $C$ is $\frac{Q}{2A\varepsilon_0}$. Therefore,all statements are correct.

$X$ and $Y$ are large,parallel conducting plates close to each other. Each face has an area $A$. $X$ is given a charge $Q$. $Y$ is without any charge. Points $A, B$,and $C$ are as shown in the figure.

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