Question

Two parallel plate capacitors X and Y, have the same area of plates and same separation between plates. X has air and Y with dielectric of constant 2 , between its plates. They are connected in series to a battery of 12 V. The ratio of electrostatic energy stored in X and Y is-
a) 4:1
b) 1:4
c) 2:1
d) 1:2​

Answer 1

Answer:

the answer is option c 2.1

Answer 2

The correct option to the above question is (c), which is 2:1

Given: Dielectric constant of the plate X = 1

           The Dielectric constant of the plate Y = 2

            Potential difference = 12V

To find: Ratio of electrostatic energy stored in X and Y.

Solution:

• When a capacitor is connected in a circuit of potential difference V. The capacitor begins to charge with charge Q.

• A charged capacitor stores energy in the electrical field between its plates, and it increases as the electric field builds up.

• The capacitance of the capacitor when a dielectric is between the plates = KC ( where K is the dielectric constant, C is the capacitance of the capacitor when there is no dielectric in between the plates.)

Electrostatic energy stored in a capacitor is equal to Q²/2C

In capacitor X the energy stored U₁ = Q²/2C

In capacitor Y the energy stored U₂ = Q²/2KC  (Where K is the dielectric constant.)

                                                      = Q²/4C

Ratio of electrostatic energy stored in X and Y = U₁/U₂

                                                                             = (Q²/2C)/(Q²/4C)

                                                                             = 2/1

Therefore, the ratio of  electrostatic energy stored in X and Y is 2:1