Two identical parallel-plate capacitors, each with capacitance 11.0 F, are charg

Question

Two identical parallel-plate capacitors, each with capacitance 11.0 F, are charged to potential difference 47.0 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.

(a) Find the total energy of the system of two capacitors before the plate separation is doubled.
J

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

(c) Find the total energy of the system after the plate separation is doubled.
J

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

Positive work is done by the agent pulling the plates apart.Negative work is done by the agent pulling the plates apart.    No work is done by pulling the agent pulling the plates apart.

Explanation / Answer

Parallel plate cap
C = r(A/d) in Farads
is 8.8542e-12 F/m
r 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

(a)

Initial energy is E = ½CV² x 2 = E = CV²

E = 11.0*10-6*472 = 0.0243 J

(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 = CV x2 = 2CV

after change, capacitance is C + ½C = 1.5C
charge stays the same, voltage changes.
2CV = V*(1.5C)
V = (2/1.5)V = 1.33V

V = 1.33*47V = 62.51 V

(c) E = ½CV²
E = ½(1.5C)(1.33V)²
E = 1.33CV²

E = 1.33*11*10-6*472

E = 0.0323 J

(d)

work is done by an external person when pulling a plates apart

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