Question

. The plates of a parallel plate capacitor have an area of 100cm2
each and are
separated by 3mm. The capacitor is charged by connecting it to a 400V supply.
Calculate (a) the electrostatic energy stored in the capacitor, (b) if a dielectric
of constant 2.5 is introduced between the plates of the capacitor, then find
electrostatic energy stored and also change in the energy stored.

Answer 1

Answer:

(a) The electrostatic energy stored in capacitor is 2.36 × $10^{-6}$ J

(b) The electrostatic energy stored in capacitor is 5.9 × $10^{-6}$ J

Explanation:

Given as :

For parallel plate capacitor

Area of plats = A = 100 cm²

Distance between plate = d = 3 mm = 0.3 cm

The capacitor is charged to voltage = v = 400 v

now, Capacitance = C = $\dfrac{A\varepsilon _0}{d}$

where $\varepsilon _0$ = 8.854 × $10^{-12}$ F/m = 8.854 × $10^{-14}$ F/cm

According to question

(a) The electrostatic energy stored in capacitor = E = $\dfrac{1}{2}$ × C × v²

where C is capacitance,

and     v is voltage

So, C =  $\dfrac{A\varepsilon _0}{d}$

or, C = $\dfrac{100\times 8.854\times 10^{-14}}{0.3}$

∴   C = 29.513 × $10^{-12}$ F

Now,

E = $\dfrac{1}{2}$ × C × v²

or, E =  $\dfrac{1}{2}$ ×  29.513 × $10^{-12}$ × (400)²

∴   E = 2.36 × $10^{-6}$ J

Hence, The electrostatic energy stored in capacitor is 2.36 × $10^{-6}$ J . Answer

(b) A dielectric constant of K = 2.5 is placed between capacitor

So,

now, Capacitance = C' = KC

Or, C' = 2.5 × 29.513 × $10^{-12}$ F

or, C' = 73.78 × $10^{-12}$ F

Now,

E = $\dfrac{1}{2}$ × C × v²

or, E =  $\dfrac{1}{2}$ ×  73.78 × $10^{-12}$ × (400)²

∴   E = 5.9 × $10^{-6}$ J

Hence, The electrostatic energy stored in capacitor is 5.9 × $10^{-6}$ J . Answer