In the given circuit, the potential difference across $3\,\mu F$ capacitor will be:

A) $12\,V$
B) $10\,V$
C) $6\,V$
D) $4\,V$

Hint: The potential difference of the capacitor circuit is determined by using the potential difference formula, in this circuit the capacitance are connected in series, so the total capacitance is determined by using the capacitance in series formula, then the charge is determined, then the potential difference across $3\,\mu F$ capacitor can be determined.

Formula used:
The capacitance in the series can be given by,
$\dfrac{1}{C} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}$
Where, $C$ is the total capacitance, ${C_1}$ and ${C_2}$ are the capacitance of the capacitor which are connected in series.
The potential difference formula in the capacitor is given by,
$C = \dfrac{q}{V}$
Where, $C$ is the capacitance of the capacitor, $q$ is the charge in the capacitor circuit and $V$ is the potential difference in the capacitor circuit.

Complete step by step solution:
Given that,
The voltage across the first batter is, $V = 20\,V$,
The voltage across the second battery is, $V = 4\,V$,
The capacitance of the first capacitor is, ${C_1} = 3\,\mu F$,
The capacitance of the second capacitor is, ${C_2} = 5\,\mu F$
Now, the equivalent voltage is given by,
$V = 20\,V - 4\,V$
By subtracting the above equation, then the above equation is written as,
$V = 16\,V$
Now,
The capacitance in the series can be given by,
$\dfrac{1}{C} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}\,...............\left( 1 \right)$
By substituting the capacitance values in the above equation, then the above equation is written as,
$\dfrac{1}{C} = \dfrac{1}{3} + \dfrac{1}{5}$
By cross multiplying the terms in the above equation, then
$\dfrac{1}{C} = \dfrac{{5 + 3}}{{3 \times 5}}$
On further simplification in the above equation, then
$\dfrac{1}{C} = \dfrac{8}{{15}}$
By taking reciprocal on both sides in the above equation, then
$C = \dfrac{{15}}{8}$
Now, the chare in the capacitor circuit is given by,
$q = C \times V$
By substituting the capacitance and the voltage in the above equation then,
$q = \dfrac{{15}}{8} \times 16$
On further simplification in the above equation, then
$q = 30\,C$
Now, the potential difference across the $3\,\mu F$ capacitor is given by,
$C = \dfrac{q}{V}$
From the above equation, the potential difference is given by,
$V = \dfrac{q}{C}$
By substituting the charge and the capacitance in the above equation, then
$V = \dfrac{{30}}{3}$
By dividing the terms in the above equation, then the above equation is written as,
$V = 10\,V$

Hence, the option (B) is the correct answer.

Note: The capacitance of the capacitor is directly proportional to the charge in the capacitor circuit and the capacitance of the capacitor is inversely proportional to the voltage or the potential difference. As the charge increases, the capacitance of the capacitor also increases.