A capacitor consists of two closely spaced metal conductors of large area, separ
ID: 1586457 • Letter: A
Question
A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electrical capacity of 3200.0 F and is charged to a potential difference of 63.0 V. Calculate the amount of energy stored in the capacitor.
Calculate the charge on this capacitor when the electrical energy stored in the capacitor is 12.11 J.
If the two plates of the capacitor have their separation increased by a factor of 2 while the charge on the plates remains constant, by what factor is the energy stored in the capacitor increased?
Explanation / Answer
C = 3200 F
V = 63.0 V
(a)
Energy stored in the cpacitor, E = 1/2 * CV^2
E = 1/2 *3200 * 10^-6 * 63.0^2
E = 6.35 J
Energy stored in the capacitor, E = 6.35 J
(b)
Energy = Q^2/(2*C)
Q^2 = 12.11 * 2*3200 * 10^-6
Q = 0.278 C
Charge on the Capacitor, Q = 0.278 C
(C)
As the separation is increased by a factor of 2.
C = a*eo/d
So Capacitace would decrease by a factor of 2.
E = Q^2/2*C
As Energy is Inversely Proportional to Capacitance therefore,
Energy would increase by a factor of 2.
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