Question

How will the (i) energy stored and (ii) the electric field inside the capacitor be affected when it is completely filled with a dielectric material of dielectric constant K?

Answer 1

When an additional amount of charge dq is transferred from negative to positive plate, the small work done is given by `dw =Vdq =q/Cdq`

The total work done in transferring total charge Q is given by

`w=int_0^Q q/c dq = 1/Cint_0^Q qdq = 1/c[q^2/2]_0^Q = 1/C [Q^2/2 -0]`

   =`Q^2/(2C)`

This work is stored as electrostatic potential energy U in the capacitor.

z`U =Q^2/(2C)`

`or  U =((CV)^2)/(2C)             [because Q = CV]`

`or U =1/2 CV^2`

`or U =1/2QV`

When dielectric material of dielectric constant ‘k’ is introduced inside the capacitor then

(ii) Electric field is reduced

`(E_0)/E =k`

`\text {But}  k>1 so E_0/E>1 or E_0>E`

When dielectric is introduced in capacitor opposite charge is induced on the plates of dielectric as a result of which an electric field is induced which is in opposite direction. Thus, Net electric field is reduced.

(i) Again, V₀ = E₀d ... (1)

Where V₀ is the potential when there is vacuum between the plates of the capacitor and d is the separation between the plates of the capacitor,

When dielectric is introduced, potential difference is given by

V = Ed ... (2)

Dividing (1) & (2)

`V_0/V = E_0/E =k`

But k > 1 ∴ V₀ > V

Thus potential difference also decreases.

We have energy stored as z``U=1/2 QV`Since V decreases, U also decreases.