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Equipotential Surfaces

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Question

Figure shows three plates A, B and C each of area S in which A and C are joined by a conducting wire. Plate A and B are given charge Q and 3Q respectively. Find the final charge on left side of plate B.

A

- 2Q

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B

2Q

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C

- Q

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D

\frac{3 Q}{2}

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Solution

The correct option is B 2Q

y + (- x) + {- (3Q - x)] + z = Q
y = z
$\int_{C}^{A} \to E . \to d r = 0$
solving these, x = 2Q and y = z = 2Q

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