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2. Electric Potential and Capacitance
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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 $\dots \; nJ$.
A
$155$
B
$145$
C
$135$
D
$125$
(JEE MAIN-2022)
Solution
Energy loss $=\frac{1}{2} \frac{C_{1} C_{2}}{C_{1}+C_{2}}\left(V_{1}-V_{2}\right)^{2}$
$=\frac{1}{2} \frac{50 \times 50 \times 10^{-12} \times 10^{-12}}{(50+50) 10^{-12}}(100-0)^{2}=125 \; n J$
Standard 12
Physics
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