Explanation

(c) [image:1_image340.emf]
Here, $2\mu F$ and $3\mu F$ capacitors are connected in series. The equivalent capacitance is
$\frac{1}{C_{S}} = \frac{1}{2} + \frac{1}{3}$ or $C_{S} = \frac{6}{5}\mu F$
Net voltage, $V = 16V - 6V = 10V$
The equivalent circuit diagram as shown in figure below
[image:1_image341.emf]
Charge on each capacitor,
$q = C_{S}V = \frac{6}{5} \times 10 = 12\mu C$
The potential difference between $A$ and $B$
$= - \frac{12\mu C}{2\mu F} = - 6V$