Question

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

Answer 1

Step 1, Given data:

Area = 90cm290 cm^290cm2 = ${90\times 10}^{-4}m$

A=90cm2=90×10−4m2A=90{ cm }^{ 2 }=90 \times { 10 }^{ -4 }{ m }^{ 2 }Area(A)=90 cm2=90×10−4m2

Distance between the plate(d) = 2.5 mm = $2.5\times {10}^{-3}m$

d=2.5mm=2.5×10−3md=2.5 mm=2.5\times { 10 }^{ -3 }m

Potential difference(V)=400 V

Let C is capacitance.

Step 2: Formula used

$C=\frac{\epsilon \text{₀}A}{d}$

Energy of the capacitor =$\frac{1}{2}C{V}^2$ 12CV2\dfrac { 1 }{ 2 }C{ V }^{ 2 }21​CV2

Step 3: Calculating the capacitance

$C=\frac{\epsilon \text{₀}A}{d}$ = $\frac{8.85\times {10}^{-12}\times 90\times {10}^{-4}}{2.5\times {10}^{-3}}$ = $3.186\times {10}^{-11}F$

Step 4: Calculating the energy stored

Energy of the capacitor = $\frac{1}{2}C{V}^2$

Energy of the capacitor = $\frac{1}{2}\times 3.186\times {10}^{-11}\times {\left(400\right)}^2=2.55\times {10}^{-6}$ J

Hence, the energy stored in the capacitor is $2.55\times {10}^{-6}J$ .