Question

The total electrostatic energy stored in both the capacitor is:

• A. $9\mu J$
• B. $40.5\mu J$
• C. $13.5\mu J$
• D. $18\mu J$

Answer 1

The correct option is A $9\mu J$

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