If you could help me with 10 it would be lovely. Thank you. The capacitance of t

Question

If you could help me with 10 it would be lovely.

Thank you.

The capacitance of the variable capacitor of a radio can be changed from 30 to 130 pF by turning the dial from 0 degree to 180 degree. With the dial set at 180 degree the capacitor is connected to a 170 V battery. After charging, the capacitor is disconnected from the battery and the dial is turned to 0 degree. What is the charge on the capacitor now? You are correct. Computers answer now shown above. Your receipt is 157-5745 What is the potential difference between the plates of the capacitor when the dial reads 100 degree assuming the dial is linearly calibrated? Here linear means the change in angle is proportional to the change in capacitance. What is the energy stored in the capacitor at this position? How much work is required to turn the dial from 0 degree to 100 degree if friction is neglected? Select True or False. Less charge would have accumulated on the capacitor if it had been charged with the dial in the 0 degree position. The electrostatic energy stored in the capacitor decreased when the dial was turned from 108 to 0 degrees. The voltage across the capacitor was constant even when the capacitance was varied. The energy stored in the capacitor remained constant when the capacitance was varied. The charge on the capacitor increased when the capacitance was varied.

Explanation / Answer

When battery is disconnected the charge remains the same

Let V2 be the required potential

At 100o

Capacitance = (130-30)pF*100/180 = 55.55pF

Hence Using

q = C2V2

2.21*10-8 = 55.55*10-12V2

V2 = 397.8 V

Energy =0.5CV2 = 0.5*55.55*10-12*397.8 = 1.105*10-8 J

SImilarly calculating energy at 0o

E = 0.5Q2/C = 0.5*(2.21*10-8)2/30*10-12 = 8.14*10-6 J

So energy required to turn the dial = Gain in energy of capacitor = 8.14*10-6 - 1.105*10-8 = 8.13*10-6 J

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