Electrostatics Ques 124
- In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance $C$. The switch $S_{1}$ is pressed first to fully charge the capacitor $C_{1}$ and then released. The switch $S_{2}$ is then pressed to charge the capacitor $C_{2}$. After some time, $S_{2}$ is released and then $S_{3}$ is pressed. After some time
(2013 Adv.)
(a) the charge on the upper plate of $C_{1}$ is $2 C V_{0}$
(b) the charge on the upper plate of $C_{1}$ is $C_{1} V_{0}$
(c) the charge on the upper plate of $C_{2}$ is $Q$
(d) the charge on the upper plate of $\mathrm{C}_ {2}$ is $-C V_{0}$
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Answer:
Correct Answer: 124.(b,d)
Solution:
Formula:
Distribution of Charges on Connecting Two Charged Capacitors:
- After pressing $S_{1}$ charge on upper plate of $C_{1}$ is $+2 C V_{0}$.
After pressing $S_{2}$ this charge equally distributes in two capacitors. Therefore, charge on the upper plates of both capacitors will be $+C V_{0}$.
When $S_{2}$ is released and $S_{3}$ is pressed, charge on upper plate of $C_{1}$ remains unchanged $\left(=+C V_{0}\right)$ but charge on upper plate of $C_{2}$ is according to new battery $\left(=+C V_{0}\right)$.
BETA
