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Question

Physics Question on electrostatic potential and capacitance

A capacitor of capacitance 50pF50pF is charged by 100V100 V source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is _____ nJ.

Answer

Loss in electrostatic energy
=12(12CV2)= \frac 12 (\frac 12CV^2)

=12×12×50×1012×(100)2= \frac 12 \times \frac 12 \times 50 \times 10^{-12} \times (100)^2

=5004 nJ= \frac {500}{4}\ nJ
=125 nJ= 125\ nJ

So, the answer is 125 nJ125\ nJ.