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Question

(A) The force $F$ between the plates of a parallel plate capacitor is given by $F = \frac{Q^2}{2\varepsilon_0 A}$.Since the charge $Q = CV$,we substit

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$\frac{CV^2}{2d}$

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(A) The force $F$ between the plates of a parallel plate capacitor is given by $F = \frac{Q^2}{2\varepsilon_0 A}$.Since the charge $Q = CV$,we substitute this into the formula:$F = \frac{(CV)^2}{2\varepsilon_0 A} = \frac{C^2V^2}{2\varepsilon_0 A}$.We know that the capacitance $C = \frac{\varepsilon_0 A}{d}$,which implies $\varepsilon_0 A = Cd$.Substituting $\varepsilon_0 A = Cd$ into the force equation:$F = \frac{C^2V^2}{2(Cd)} = \frac{CV^2}{2d}$.Thus,the force acting between the plates is $\frac{CV^2}{2d}$.

$A$ capacitor of capacitance $C$ with a distance $d$ between its plates is connected to a battery of $V$ volts. What is the force acting between the two plates?

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