Physics - Electrostatics Question with Solution | TestHub

In a capacitor, energy is stored in the electric field between the plates.

Increase in energy

$∆\mathrm{U}={\mathrm{U}}_{\mathrm{f}}-{\mathrm{U}}_{\mathrm{i}}$

$=\frac{1}{2}{\mathrm{CV}}_{\mathrm{f}}^2-\frac{1}{2}{\mathrm{CV}}_{\mathrm{i}}^2=\frac{1}{2}\mathrm{C}({\mathrm{V}}_{\mathrm{f}}^2-{\mathrm{V}}_{\mathrm{i}}^2)$

Given, $\mathrm{C}=\mathrm{6 }\mathrm{\mu F}=6\times {10}^{-6}$, ${\mathrm{V}}_{\mathrm{i}}=10 \mathrm{V}$

${\mathrm{V}}_{\mathrm{f}} =20 \mathrm{V}$

$\therefore ∆\mathrm{U}=\frac{1}{2}\times 6\times {10}^{-6}[{\left(20\right)}^2-{\left(10\right)}^2]$

$=3\times {10}^{-6}\times 300=9\times {10}^{-4}\mathrm{ }\mathrm{J}$