A variable capacitor consists of two stacks of plates that can be adjusted so a

Question

A variable capacitor consists of two stacks of plates that can be adjusted so a variable amount of area of the plates faces each other. Suppose a potential difference is applied across a variable capacitor and it stores of energy with its plates in one position. Then it is adjusted so as to increase its capacitance, and with its plates in the new position, it stores of energy. How much does its capacitance change as its plates adjust from the first to the second position? 0.048 F 0.012 F 0.0096 F 0.024 F

Explanation / Answer

E1 = initial energy stored = 5 mJ

E2 = energy stored after adjuctment = 11 mJ

C1 = initial capacitance

C2 = capacitance after adjustment

V = Voltage = 500 volts

energy stored is given as

E1 = (0.5) C1 V2

0.005 = (0.5) C1 (500)2

C1 = 4 x 10-8 F

energy after adjustment is given as

E2 = (0.5) C2 V2

0.011 = (0.5) C2 (500)2

C2 = 8.8 x 10-8 F

change = C2 - C1 = 8.8 x 10-8 - 4 x 10-8 = 0.048 F

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