CGPET 2016 Solved Paper

When the capacitor of 600 pF is charged by 200V supply, the electrostatic energy stored =1.2CV2
=

1

2

×600×10−12×(200)2
=

1

2

×600×10−12×200×200
=12×10−6J
When it is connected to 600 pF uncharged capacitor, the electrostatic energy will be equally divided in the both the capacitors. So, the energy loss in first capacitor
=

1

2

×12×10−6J=6×10−6J