Question

The plates of a parallel plate capacitor have an area of 90 cm² each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply. How much electrostatic energy is stored by the capacitor?

Answer 1

Answer:

$E =1.01\times 10^{-10} J$

Explanation:

We know that the energy stored in the capacitor is $\frac{CV^{2} }{2}$

Where C is the capacitance and V is the voltage.

For parallel plate capacitor capacitance is given by,

$C = \dfrac{\epsilon _{0}A }{d}$

$C = \dfrac{8.9\times 10^{-12}\times 90\times 10^{-4}}{2.5\times 10^{-3} } = 320.4\times 10^{-13}$

So, energy stored in capacitor is

$E= \dfrac{320.4\times 10^{-13}\times 400^{2}  }{2} =1.01\times 10^{-10} J$

Answer 2

The electrostatic energy is stored by the capacitor is $2.55\times10^-^6J$

Explanation:

Given area of parallel plate capacitor, A = 90 cm²

distance between the plates, d = 2.5 mm.

Voltage across capacitor, V = 400 V.

The capacitance is given as

$C=\epsilon_0\frac{A}{d}$

substitute the values, we get

$C=\frac{8.854\times10^-^1^2\times90\times10^-^4m^2}{2.5\times10^-^3m}$

Electrostatic energy stored in the capacitor,

$U=\frac{1}{2}CV^2$

substitute the values, we get

$U=\frac{1}{2}\frac{8.854\times10^-^1^2\times90\times10^-^4m^2\times(400V)^2}{2.5\times10^-^3m}$

$U=2.55\times10^-^6J$

Thus, the electrostatic energy is stored by the capacitor is $2.55\times10^-^6J$.

#Learn More: electrostatic energy

https://brainly.in/question/4646999