A 600pF capacitor is charged by a 200V supply. It is then disconnected form the supply and is connected to another uncharged 600pF capacitor. How much electrostatic energy is lost in the process?

Explanation

When an uncharged capacitor is connected across a charged capacitor, the charged capacitor discharges through the uncharged capacitor, till both the capacitors attain equal potential difference between their plates.
Given: `C_(1)=C_(2)=600pF, V_(1)=200V, V_(2)=0`
Net charge on both the capacitors will be:
`q=C_(1)V_(1)+C_(2)V_(2)=600xx10^(-12)xx200+0=12xx10^(-8)C`
As both the capacitors are connected in parallel, net capacitance of the combination will be:
`C=C_(1)+C_(2)=600+600=1,200pF=1,200xx10^(-12)F`
After charge sharing, the common potential attained by both the capacitors will be:
`V=(q)/(C )=(12xx10^(-8)C)/(1200xx10^(-12))=100V`
Final energy stored in combination of both the capacitors will be:
`U_(2)=(1)/(2)CV^(2)=(1)/(2)xx1200xx10^(-12)xx(100)^(2)= 6xx10^(-6)J`
Energy initially stored in the capacitor,
`U_(1)=(1)/(2)C_(1)V_(1)^(2)=(1)/(2)xx600xx10^(-12)xx(200)^(2)=12xx10^(-6)J`
While sharing the charges between both the capacitors, energy lost can be calculated as:
`DeltaU=U_(1)-U_(2)=12xx10^(-6)-6xx10^(-6)=6xx10^(-6)J`