The plates S and T of an uncharged parallel p | vedclass

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(A) The electrostatic force of attraction between the two plates of a parallel plate capacitor with charge $Q$ and area $A$ is given by $F_e = \frac{Q

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$\sqrt {2mgA{ \in _0}} $

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(A) The electrostatic force of attraction between the two plates of a parallel plate capacitor with charge $Q$ and area $A$ is given by $F_e = \frac{Q^2}{2A\epsilon_0}$.In the given system,the plate $S$ is connected to a block of mass $m$ via a spring of constant $k$ and a pulley. For the system to be in equilibrium,the tension in the string must balance the weight of the block,so $T = mg$.Since the spring is massless and in equilibrium,the force exerted by the spring on plate $S$ is equal to the tension in the string,which is $mg$.Equating the electrostatic force to the spring force for equilibrium of plate $S$:$\frac{Q^2}{2A\epsilon_0} = mg$Solving for $Q$:$Q^2 = 2mgA\epsilon_0$$Q = \sqrt{2mgA\epsilon_0}$

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