Question

A parallel-plate capacitor (Homework #2) A parallel-plate capacitor is made by a pair of plates with area A and separation d. At first, the plates are charged up to +Q and -O respectively, and then a battery is disconnected. Express all the answers for a) through f) only by Q, d, A and 5 +Q 2d -Q Area A a) What is the electric field E inside of the capacitor? b) What is the potential difference V between the plates? c) What is the capacitance C? d) What is the energy U stored in the capacitor? e) What is the energy density inside of the capacitor? Show that it is equal to E

Answer 1

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Electric field = sigma/elementof_0 E = Q/A elementof_0 V = potential difference V = Ed = Qd/A elementof_0 Capacitance = Q/V = A elementof_0/d Energy stored in capacitor = 1/2 CV^2 = 1/2(A elementof_0/d) (Q^2d^2/A^2 elementof_0^2) V = Q^2 d/2 A elementof_0 Energy density = U/volume u = U/A d = 1/2 Q^2/A^2 elementof_0 = 1/2 elementof_0 E^2 E' = Q/A elementof_0 v' = 2V c' = C/2 E' = E U' = 2U work done U' - U = U