Explanation

(d) Let potential difference between the plates of the capacitors $C_{1},C_{2}\ and\ C_{3}\ be\ V_{1},\ V_{2}\ and\ V_{3}and\ q$ be the charge.
[image:2_image140.emf]
Then, $V = \frac{q}{C_{1}},\ V_{2} = \frac{q}{C_{2}},\ V_{3} = \frac{q}{C_{3}}$
The total potential difference $V = 11$volt
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V = V_{1} + V_{2} + V_{3}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V = \frac{q}{C_{1}} + \frac{q}{C_{2}} + \frac{q}{C_{3}} = 11$
Given, $C_{1} = 1\mu F,\ C_{2} = 2\mu F,C_{3} = 3\mu F$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 11 = q\left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} \right)$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 11 = \frac{11q}{6}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ q = 6\mu c$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V_{1}\ = \ \frac{q}{C_{1}} = \frac{6}{1} = 6$ V