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

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Q. A capacitor of capacitance $50\, pF$ is charged by $100\, V$ source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is _______$nJ$.

JEE MainJEE Main 2022Electrostatic Potential and Capacitance

Solution:

Energy loss $=\frac{1}{2} \frac{C_{1} C_{2}}{C_{1}+C_{2}}\left(V_{1}-V_{2}\right)^{2}$
$=\frac{1}{2} \frac{50 \times 50 \times 10^{-12} \times 10^{-12}}{(50+50) 10^{-12}}(100-0)^{2}=125 \,n J$