Question

A capacitor $C_1$ is charged by a potential difference $v_0$ as shown in the figure. This charging battery is then removed and the capacitor is connected as into an uncharged capacitor $C_2$. What is the final potential difference $V$ across the combination?

• A. $V_{0} \frac{\left(C_{1}-C_{2}\right)}{C_{1}}$
• B. $V_{0} \frac{C_{1}}{\left(C_{1}-C_{2}\right)}$
• C. $V_{0} \frac{C_{1}}{\left(C_{1}+C_{2}\right)}$
• D. $V_{0} \frac{\left(C_{1}+C_{2}\right)}{C_{1}}$

Answer 1

Answer: (C)

Total amount of charge in an isolated system remains constant.
When a conductor is given a charge, its potential rises in proportion to the' charge given. Thus, if a charge $Q$ raises the potential of the conductor by $V_{0}$, then
$Q=C_{1} V_{0}$
where $C_{1}$ is capacitance of first capacitor.
When the charging battery is removed and capacitor is connected to an uncharged capacitor $C_{2}$, this charge $C_{1} V_{0}$ is now shared across $C_{1}$ and $C_{2}$. From law of conservation of charge, we have
$C_{1} V_{0} =C_{1} V+C_{2} V $
$\Rightarrow V =V_{0} \frac{C_{1}}{C_{1}+C_{2}} $