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Question

(A) Let $F$ be the force applied to separate the plates of a parallel plate capacitor by a small distance $\Delta x$. The work done by the force is $W

Vedclass · Accepted Answer

$\frac{1}{2} QE$

Vedclass · Answer

(A) Let $F$ be the force applied to separate the plates of a parallel plate capacitor by a small distance $\Delta x$. The work done by the force is $W = F \Delta x$.This work done is stored as an increase in the potential energy of the capacitor. The energy density $u$ between the plates is given by $u = \frac{1}{2} \epsilon_0 E^2$.The total energy stored in the volume $V = A \Delta x$ is $\Delta U = u 	imes (A \Delta x) = (\frac{1}{2} \epsilon_0 E^2) A \Delta x$.Equating work done to the change in potential energy: $F \Delta x = \frac{1}{2} \epsilon_0 E^2 A \Delta x$.Thus,$F = \frac{1}{2} \epsilon_0 A E^2$.Since the electric field between the plates is $E = \frac{\sigma}{\epsilon_0} = \frac{Q}{A \epsilon_0}$,we have $\epsilon_0 A = \frac{Q}{E}$.Substituting this into the force equation: $F = \frac{1}{2} (\frac{Q}{E}) E^2 = \frac{1}{2} QE$.

If the charge on each plate of a parallel plate capacitor is $Q$ and the magnitude of the electric field between the plates is $E$,then the force on each plate of the parallel plate capacitor will be:

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