Electrostatics 5 Question 36
38. 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 $C V _0$
(c) the charge on the upper plate of $C _2$ is 0
(d) the charge on the upper plate of $C _2$ is $-C V _0$
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Solution:
- 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 an 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 $\left(=-C V _0\right)$.
