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A parallel-plate capacitor is connected to a cell. Its positive plate A and its negative plate B have charges +Q and -Q respectively and then the cell is disconnected from the capacitor . A third plate C, identical to A and B, with charge +Q is now introduced midway between A and B, parallel to them. Which of the following are correct?

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The charge on the inner face of B now is $\frac{- 3 Q}{2}$

There is no change in the potential difference between A and B

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

The charge on the inner face of A now is $\frac{Q}{2} .$

answer is A, B, C, D.

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Detailed Solution

Charge distribution on the three plates will be as shown

This will ensure that the electric field is zero inside the metal plates. Potential difference between A and B is

$P_{A B} = E_{1} x + E_{2} x = \frac{Q}{2 A ε_{0}} x + \frac{3 Q}{2 A ε_{0}} x = \frac{2 Q}{A ε_{0}} x$

Before insertion of plate C the potential difference was same.

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