Question

The total electrostatic energy stored in both the capacitors (in $μJ$ ) is

Answer 1

According to circuit diagram, capacitors are connected in series combination.
So, equivalent capacitance $ \, \left(C_{e q}\right)=\frac{3 \times 6}{3 + 6}$
$=\frac{18}{9}=2 \, \mu F$
We know that,
Total electrostatic energy $\left(U\right)=\frac{1}{2}C_{e q}V^{2}$
$=\frac{1}{2}\times 2\times \left(- 3\right)^{2}$
$=9\,\mu J$