Question

Two parallel plate capacitors $\text{A}$ and $\text{B}$ having capacitances $4\mu \text{F}$ and $6\mu \text{F}$ are seperately charged upto $10\text{V}$ and $5\text{V}$ respectively. Now positive plate of $\text{A}$ is connected to the positive plate of $\text{B}$ and negative plate of $\text{A}$ is connected with negative plate of $\text{B}$. Find the final charge on each capacitor ?

• A. $35\mu \text{C}$, $35\mu \text{C}$
• B. $20\mu \text{C}$, $50\mu \text{C}$
• C. $40\mu \text{C}$, $30\mu \text{C}$
• D. $28\mu \text{C}$, $42\mu \text{C}$

Answer 1

The correct option is D $28\mu \text{C}$, $42\mu \text{C}$

$\text{Before connection}:$

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

So,
$Q_1=4\mu \text{F}\times 10\text{V}=40\mu \text{C}$
$Q_2=6\mu \text{F}\times 5\text{V}=30\mu \text{C}$

$\text{After Connection:}$


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

$Q_1^{\prime }=4V$ and $Q_2^{\prime }=6V$

Using principle of conservation of charge,

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

$\Rightarrow 40+30=4V+6V$

$\therefore V=7\text{V}$

Therefore,

$Q_1^{\prime }=4V=4\times 7=28\mu \text{C}$

$Q_2^{\prime }=6V=6\times 7=42\mu \text{C}$

Hence, option (d) is the correct answer.