Two identical parallel-plate capacitors, each with capacitance C, are charged to

Question

Two identical parallel-plate capacitors, each with capacitance C, are charged to potential difference ?V and then disconnected from the battery. They are then connected to each other in parallel with plates of like sign connected. Finally, the plate separation in one of the capacitors is doubled. (Use any variable or symbol stated above as necessary.)
(a) Find the total energy of the system of two capacitors before the plate separation is doubled.
Utotal = 1


(b) Find the potential difference across each capacitor after the plate separation is doubled.
?V ' = 2


(c) Find the total energy of the system after the plate separation is doubled.
U 'total = 3


(d) Reconcile the difference in the answers to parts (a) and (c) with the law of conservation of energy.
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Explanation / Answer

Parallel plate cap C = e0er(A/d) in Farads e0 is 8.8542e-12 F/m er is dielectric constant (vacuum = 1) A and d are area of plate in m² and separation in m E = ½CV² = ½QV Energy in a cap in Joules Initial energy is E = ½CV0² x 2 = E = CV0² (a) (b) double the spacing and the capacitance is cut in half. You don't say, but I assume the battery is removed before the spacing is changed. Energy is not conserved, but charge is. Initial charge is Q = CV0 x2 = 2CV0 after change, capacitance is C + ½C = 1.5C charge stays the same, voltage changes. 2CV0 = V*(1.5C) V = (2/1.5)V0 = 1.33V0 (c) E = ½CV² E = ½(1.5C)(1.33V0)² E = 1.33CV0² you can substitute ?V for V if you want, it seems needless.

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