Question

If the charge on left plate of the $5\mu \text{F}$ capacitor in the circuit segment shown in the figure is $-20\mu \text{C}$, then the charge on the right plate of $3\mu \text{F}$ capacitor is:

• A. $+8.58\mu \text{C}$
• B. $-8.58\mu \text{C}$
• C. $+11.42\mu \text{C}$
• D. $-11.42\mu \text{C}$

Answer 1

The correct option is A $+8.58\mu \text{C}$

The charge on left plate of $5\mu \text{F}$ capacitor is $-20\mu \text{C}$ i.e. it is a negative plate and charge on this capacitor is $20\mu \text{C}(=q)$

From formula, $q=CV$

$20\mu \text{C}=\left(5\mu \text{F}\right)\times V$

$\therefore V=4\text{Volt}$

Effective capacitance of $3\mu \text{F}$ and $4\mu \text{F}$ capacitors is

$C_e=3+4=7\mu \text{F}$

Once again, we can redraw circuit as;

Thus, the potential difference across $7\mu \text{F}$ capacitor is,

$V^{\prime }=\frac{q}{C}=\frac{20\mu \text{C}}{7\mu \text{F}}=2.86\text{Volt}$

Now,


So, charge $\left(q^{\prime }\right)$ on $3\mu \text{F}$ is given by

$\Rightarrow q^{\prime }=\left(3\mu \text{F}\right)\times \left(2.86\text{V}\right)$

$\Rightarrow q^{\prime }=+8.58\mu \text{C}$

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

Why this question? Always try to simplify the circuit such that we can apply $q=CV$ for a group of parallel combination of capacitors. This will help us to find $q$ and potential difference accross each component capacitor.