Question

Two parallel plate capacitors $\text{A}$ and $\text{B}$ having capacitances $2\mu \text{F}$ and $10\mu \text{F}$ are charged seperately upto $10\text{V}$ and $20\text{V}$ respectively. Now positive plate of $\text{A}$ is connected to the negative plate of $\text{B}$ and vice-versa. Find the final charge on each capacitor?

• A. $200\mu \text{C}$, $20\mu \text{C}$
• B. $30\mu \text{C}$, $150\mu \text{C}$
• C. $20\mu \text{C}$, $200\mu \text{C}$
• D. $180\mu \text{C}$, $20\mu \text{C}$

Answer 1

The correct option is B $30\mu \text{C}$, $150\mu \text{C}$

$\text{Before connection}:$

Let, $Q_1$ and $Q_2$ be initial charges on capacitors $\text{A}$ and $\text{B}$ respectively.

So,
$Q_1=2\mu \text{F}\times 10\text{V}=20\mu \text{C}$

$Q_2=10\mu \text{F}\times 20\text{V}=200\mu \text{C}$

$\text{After Connection:}$

Let, due to the charge distribution both the capacitors attain a common potential $V$ and their charges become $Q_1^{\prime }$ and $Q_2^{\prime }$ respectively. So, using $Q=CV$, we get

$Q_1^{\prime }=2V$ and $Q_2^{\prime }=10V$

Using principle of conservation of charge,

$Q_1+Q_2=Q_1^{\prime }+Q_2^{\prime }$

$\Rightarrow -20+200=2V+10V$

$\therefore V=15\text{V}$

Therefore,

$Q_1^{\prime }=2V=2\times 15=30\mu \text{C}$

$Q_2^{\prime }=10V=10\times 15=150\mu \text{C}$

Hence, option (b) is the correct answer.