2. Electric Potential and Capacitance
hard

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

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.