The capacitance of a variable radio capacitor, shown here, can be changed from 5

Question

The capacitance of a variable radio capacitor, shown here, can be changed from 50.0 pF to 950.0 pF by turning the dial from 0.00° to 180°. With the dial set at 180° the capacitor is connected to a 400.0 V battery. After charging, the capacitor is disconnected from the battery and the dial is turned to 0.00°.

     (a) What is the charge on the capacitor?                       

     (b) What is the potential difference across

            the capacitor when the dial reads 0°?

     (c) What is the energy of the capacitor at 0°?

     (d) How much work was required to turn the dial

to 0°, if friction is neglected?

Explanation / Answer

After charging the capacitor will have q = CV = 950 * 10 -12 * 400 = 3.8 * 10-7 C

a) After disconnecting the capacitor isn't attached to anything hence the charge will remain same. Hence current charge on capacitor is 3.8 * 10-7 C.

b) When dial reads Zero capacitance is 50 pF hence Voltage V = q/C =  3.8 * 10-7 / (50 * 10-12) = 7600 V

c) Energy is given by E = 1/2*C*V2 = 0.5*50 * 10-12 * (7600)2 = 1.44*10-3 Joules

d) Initial energy at 180o E = 1/2*C*V2 = 0.5*950 * 10 -12 * (400)2 = 7.6 * 10-5 Joules

Energy required to turn the dial = final energy - Initial ebergy = 1.44*10-3 - 7.6 * 10-5 = 1.364 * 10-3 Joules.

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