RJC 2001 Common Test 1 Q 17
The capacitance of a certain variable capacitor may be varied between limits of 120 pF and 600 pF by turning a knob attached to the movable plates. The capacitor is set to 600 pF, and is charged by connecting it to a battery of e.m.f. 250 V.
(a) What is the charge on the plate?
The battery is then disconnected and the capacitance changed to 120 pF.
(b) Assuming that no charge is lost from the plates, what is now the potential difference between them?
(c) Calculate the initial and final energy stored in the capacitor during this change.
(d) Account for the change in energy.
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Answer:
(a) Q = CiVi
= 600 * 10-12 * 250
= 1.50 * 10-7 C (3 s.f.)
(b) Vf = Q / Cf
= 1.50 * 10-7 C (3 s.f.) / 120 * 10-12
= 1250 V (3 s.f.)
(c) initial energy = ½CiVi2
= ½ (600 * 10-12) (250)2
= 1.88 * 10-5 J
Final energy = ½CfVf2
= ½ (120 * 10-12) (1250)2
= 9.38 * 10-5 J
(d) Work is done by an external force in turning the knob to decrease capacitance. As a result, more energy is stored in the capacitor.
