Question

A parallel-plate capacitor has a charge Q and plates of area A. What force acts on one plate to attract it toward the other plate? Because the electric field between the plates is E 5 Q /Ae0, you might think the force is F 5 QE 5 Q2/Ae0. This conclusion is wrong because the field E includes contributions from both plates, and the field created by the positive plate cannot exert any force on the positive plate. Show that the force exerted on each plate is actually F 5 Q2/2Ae0. Suggestion: Let C 5 e0A/x for an arbitrary plate separation x and note that the work done in separating the two charged plates is W 5 e F dx.

Answer 1

Answer:

$F=\dfrac{Q^2x}{2\varepsilon _0A}$

Explanation:

Given that

Charge = Q

Area =A

Electric filed =E

We know that

$F=\dfrac{dU}{dx}$

U=Energy

We know that energy in capacitor given as

$U=\dfrac{Q^2}{2C}$

$C=\dfrac{\varepsilon _oA}{x}$

$U=\dfrac{Q^2x}{2\varepsilon _0A}$

So

Force F

$F=\dfrac{Q^2x}{2\varepsilon _0A}$