Question

The plates of a parallel plate capacitor have an area of $90c{m}^2$ each and are separated by $2.5mm$. The capacitor is charged by a $400$ volt supply. How much electrostatic energy is stored by the capacitor?

• A. $2.55\times {10}^{-6}J$
• B. $1.55\times {10}^{-6}J$
• C. $8.15\times {10}^{-6}J$
• D. $5.5\times {10}^{-6}J$

Answer 1

The correct option is A $2.55\times {10}^{-6}J$

Given, area $A=90c{m}^2=90\times {10}^{-4}m$, separation between plates $d=2.5mm=2.5\times {10}^{-3}m$,
and potential, $V=400$ volt.
The capacitance of parallel plate capacitor , $C=\frac{A{ϵ}_0}{d}=\frac{90\times {10}^{-4}\times 8.85\times {10}^{-12}}{2.5\times {10}^{-3}}=3.186\times {10}^{-11}F$
Energy stored, $U=\frac{1}{2}C{V}^2=\frac{1}{2}\times 3.186\times {10}^{-11}\times (400{)}^2=2.55\times {10}^{-6}J$