Physics - Electrostatics Question with Solution | TestHub

According to circuit diagram, capacitors are connected in series combination.

So, equivalent capacitance$\mathrm{ }\left({\mathrm{C}}_{\mathrm{e}\mathrm{q}}\right)=\frac{3\times 6}{3+6}$

$=\frac{18}{9}=2\mathrm{ }\mathrm{\mu }\mathrm{F}$

We know that,

Total electrostatic energy$\left(U\right)=\frac{1}{2}{\mathrm{C}}_{eq}{V}^2$

$=\frac{1}{2}\times 2\times {\left(-3\right)}^2$

$=\mathrm{9 }\mu \mathrm{J}$