Question

33) A 900 pF capacitor is charged by 100 V source. Calculate the electrostatic energy stored in the

capacitor? The capacitor is then disconnected from the source and connected to another

uncharged 900 pF capacitor. Find the common potential of the system?
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Answer 1

A

1.25×10−6J

B

5.25×10−6J

C

2.25×10−6J

Correct Answer

D

8.24×10−6J

Hard

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Solution

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Correct option is C)

We have the initial energy as 21CV2=21×9×10−10×1002=4.5×10−6J

Now the capacitor is disconnected and is connected to another 900 pF capacitor. 

In the steady situation, the two capacitors have their positive plates at the same potential. 

Let the common potential difference be V’. The charge on each capacitor is Q’=CV’.

By charge conservation, Q’=Q/2. This implies V’=V/2. 

The total energy of the system is = 2×(1/2)Q’V’=(1/4)QV=2.25×10−6J 

Thus we get the electrostatic energy stored by the system as 21(C1+C2)V2=2