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Four identical metallic plates are arranged very close and parallel to each other as shown in the figure. Plates A and D are connected across a voltage source, which maintains potential of plate A higher than that of D by an amount $V_{0}$ . Plates B and C are connected by a short conducting wire for a while and then the wire is removed. Now the voltage source is removed and then plates A and D are connected by another short conducting wire. Then choose the correct options.

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The potential of B is higher than that of D

Potential of C is $V_{0} / 3$ higher than the potential of B

Potential difference between A and B has magnitude $V_{0} / 6$ , with A at higher potential

$V_{C} - V_{A}$ is equal to $V_{D} - V_{B}$

answer is A, B, D.

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Detailed Solution

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$- \frac{C V}{2} - x + \frac{C V}{2}$

$(\frac{C V}{2} - x) + x + (\frac{C V}{2} - x) = 0$

$x = \frac{C V}{3}$

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