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(A) The energy stored in a capacitor is given by $U = \frac{Q^2}{2C}$.Since the plates are isolated,the charge $Q$ remains constant.The force $F$ on t

Vedclass · Accepted Answer

$x^{-2}$

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(A) The energy stored in a capacitor is given by $U = \frac{Q^2}{2C}$.Since the plates are isolated,the charge $Q$ remains constant.The force $F$ on the movable plate is given by the negative gradient of the potential energy with respect to the displacement $x$:$F = -\frac{dU}{dx} = -\frac{d}{dx} \left( \frac{Q^2}{2C} \right) = \frac{Q^2}{2C^2} \frac{dC}{dx}$.The capacitance of a parallel-plate capacitor with overlap length $x$ and width $b$ is $C = \frac{\epsilon_0 A}{d} = \frac{\epsilon_0 (bx)}{d}$,where $d$ is the fixed separation between the plates.Thus,$\frac{dC}{dx} = \frac{\epsilon_0 b}{d}$.Substituting this into the force equation:$F = \frac{Q^2}{2 \left( \frac{\epsilon_0 bx}{d} \right)^2} \left( \frac{\epsilon_0 b}{d} \right) = \frac{Q^2 d}{2 \epsilon_0 b x^2}$.Therefore,the force $F$ is proportional to $x^{-2}$.

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