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
$C_1{V}_0=C_{1}V+C_{2}V$
$\Rightarrow V=V_0\frac{C_1}{C_1+C_2}$