The capacitance of the variable capacitor of a radio can be changed from 50 to 7

Question

The capacitance of the variable capacitor of a radio can be changed from 50 to 760 pF by turning the dial from 0 degrees to 180 degrees. With the dial set at 180 degrees the capacitor is connected to a 430 V battery. After charging, the capacitor is disconnected from the battery and the dial is turned to 0 degrees. What is the charge on the capacitor now? b) What is the potential difference between the plates of the capacitor when the dial reads 130 degrees assuming the dial is linearly calibrated? c) What is the energy stored in the capacitor in this position? d)How much work is required to turn the dial from 0 degrees to 130 degrees, if friction is neglected? e) select true or false: The voltage across the capacitor varied when the capacitance was varied. The energy stored in the capacitor varied when the capacitance was varied. Work was done on the capacitor to turn the dial from 180 to 0 degrees. The charge on the capacitor increased when the capacitance was varied. More charge would have accumulated on the capacitor if it had been charged with the dial in the 0 degree position.

Explanation / Answer

a) at 180 deg, C = 760 p. For V= 430 V, charge stored Q= CV = 760 x 10-12 * 430 = 3.268 x 10 -7 C

The capacitor is linearly calibrated, hence if we plot theta in x-axis, and Potential difference (PD) in the y, the slope of the curve (st. line) = (760-50)/ (180-0) = 3.94, with a positve cut in y axis, C0=50pF

so, for 130 deg, capacitance C1= (3.94 x 130) + 50 = 562.2 pF

b) at 130 deg, if same charge is stored. then PD = Q/C1 = 3.268 x 10 -7/ 562.2 x 10 -12 = 581.28 Volt.

c) energy stored in this position = 0.5 * C * V2 = 9.49 x 10 -5 Joule.

d) PD at 0 deg = V0 = 3.268 x 10 -7 / 50 x 10-12 = 6536 Volt

energy stored at 0 deg = 0.5 * C0 * V02 = 1.067 x 10 -3 Joule

work to be done to dial from 0 deg to 130 deg = (1.067 x 10 -3 - 9.49 x 10 -5 ) = 97.21 x 10 -3 Joule

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