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 = Q/A∈0, you might think the force is F = QE = Q2/A∈0. 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 = Q2/2A∈0. Suggestion: Let C = ∈0A/x for an arbitrary plate separation x and note that the work done in separating the two charged plates is W = ∫ F dx.

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Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.