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Two plates, each of area A, are placed parallel to each other at a distance d. One plate is connected to a battery of emf E and its negative is earthed. The other plate is also earthed. The charge drawn by plate is

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$\frac{2 ε_{0} A E}{d}$

$\frac{ε_{0} A E}{d}$

$\frac{ε_{0} A E}{2 d}$

$\frac{3 ε_{0} A E}{d}$

answer is B.

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Detailed Solution

The circuit can be redrawn as shown in figure. Charge on capacitor is

$Q = C E = \frac{ε_{0} A E}{d}$

Options 1 and 3 and 4 are incorrect.

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