Question: A parallel-plate capacitor of plate area A and separation x is given a charge Q and is then removed from the charging source. (a) Find
A parallel-plate capacitor of plate area A and separation x is given a charge Q and is then removed from the charging source.
(a) Find the stored electrostatic energy as a function of x.
(b) Find the increase in energy dU due to an increase in plate separation dx from dU = (dU/dx) dx.
(c) If F is the force exerted by one plate on the other, the work needed to move one plate a distance dx is F dx = dU. Show that F = Q2/2є0A.
(d) Show that the force in part (c) equals ½EQ, where Q is the charge on one plate and E is the electric field between the plates. Discuss the reason for the factor ½ in this result.
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a Neglecting end effects U 12 Q 2 C U 12 Q 2 x e 0 A b dU dU dx dx 12 Q 2 e ... View full answer
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