Question

In the figure shown below, the charge on the left plate of the $10\mu \text{F}$ capacitor is $-30\mu \text{C}$. The charge on the right place of the $6\mu \text{F}$ capacitor is:

• A. $-12\mu \text{C}$
• B. $+12\mu \text{C}$
• C. $-18\mu \text{C}$
• D. $+18\mu \text{C}$

Answer 1

The correct option is D $+18\mu \text{C}$


As given in the figure, $6\mu F\text{and}4\mu F$ are in parallel.

The combination of $6\mu F\text{and}4\mu F$ will be having a charge of $30\mu \text{C}$ as they are in series with $10\mu \text{F}$

$\Rightarrow Q_6+Q_4=30\mu \text{C}$

As the potential difference is same in $6\mu F\text{and}4\mu F$,

$\frac{Q_6}{Q_4}=\frac{6\mu \text{F}}{4\mu \text{F}}$

Charge on $6\mu F$ capacitor

$Q_6=\frac{6}{6+4}\times 30$

$\Rightarrow Q_6=18\mu \text{C}$

Since charge is asked on right plate, therefore it is $+18\mu C$

Hence, option $\left(D\right)$ is correct.