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Question: Answered & Verified by Expert
A 600 pF capacitor is charged by 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. Electrostatic energy lost in the process is _____ μJ.
PhysicsCapacitanceJEE MainJEE Main 2023 (08 Apr Shift 2)
Solution:
1060 Upvotes Verified Answer
The correct answer is: 6

The formula to calculate the initial energy stored in the first capacitor is given by

Ui=12CV2   ...1

The formula to calculate the final energy stored when two capacitors are connected is given by 

Uf=12CV22×2=14CV2   ...2

Subtract equation (2) from equation (1) to obtain the energy loss U.

 U=12CV2-14CV2= CV24   ...3

Substitute the values of the known parameters into equation (3) to calculate the required loss.

U=600×10-12×(200)24 J= 6 μJ

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