Question

The area of each plate of parallel plate air capacitor is 150 cm². The distance between its plates is 0.8 mm. It is charged to a potential difference of 1200 volt. What will be its energy? What will be the energy when it is filled with a medium of k = 3 and then charged. If it is charged first as an air capacitor and then filled with this dielectric, what will happen to energy?

Solution

Given,
Area of each plate of parallel plate capacitor, A= 150 cm²=

150 \times 10^{- 4} m^{2}

Distance between the plates, d= 0.8 mm =

0.8 \times 10^{- 4} m

Potential applied across the capacitor, V= 1200 V

Capacitance, before the dielectric is introduced is given by C_o ,

C_{0} = \frac{ε_{0} A}{d} = \frac{8.85 \times 10^{- 12} \times 150 \times 10^{- 4}}{8 \times 10^{- 4}} = 1.66 \times 10^{- 10} F

Energy stored in the capacitor is,

E_{0} = \frac{1}{2} C_{0} V_{0}^{2} = \frac{1}{2} \times 1.66 \times 10^{- 10} \times (1200)^{2} = 1.2 \times 10^{- 4} J .

When, the capacitor is filled with a medium K and is charged to the same portential V, capacitance is given by,
C = kC₀ = 3 x 1.66 x 10^–10 Farad

Energy stored in the capacitor is given by,

E = \frac{1}{2} {CV}^{2} = \frac{1}{2} ({kC}_{0}) V_{0}^{2} = k \frac{1}{2} C_{o} {V_{o}}^{2} = k (E_{0}) = 3 \times 1.2 \times 10^{- 4} = 3.6 \times 10^{- 4} J

On filling with the dielectric in between, charge remains same and capacitances incresaes thrice of initial value.
Therefore, the potential becomes,

V = \frac{V_{0}}{k} = \frac{1200}{3} = 400 volt

New energy of capacitor, U_fis

U_{f} = \frac{1}{2} {CV}^{2} = \frac{1}{2} ({kC}_{0}) {(\frac{V_{0}}{k})}^{2} = \frac{1}{2} \frac{C_{0} V_{0}^{2}}{k} = \frac{1.2 \times 10^{- 4}}{3} = 4 \times 10^{- 5} J .

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