Question

(a) Discuss the polarisation of dielectric slab placed in an electric field. What happens to the applied electric field?

(b) Define dielectric constant in reference to applied and reduced electric field.

(c) Define polarisation density. Show that it is equal to reduced surface charge density.

(d) Define electric susceptibility and prove that

(i) K = 1 + ✗, (ii) K = 1 + N α.

Answer 1

(a) Polarisation of a dielectric slab

Consider a charged parallel plate capacitor with vacuum between the two plates such that E₀ is the electric field between the plates

E₀ = \(\frac{\sigma}{\varepsilon_0}\)  ............................(i)

where σ is the magnitude of surface charge densities on each plate.

Suppose a dielectric slab of non-polar atom is introduced between the plates. The polarization of the slab takes place. On the left face, a net negative charge -qi and on the right face, a positive charge +qi appears. There is no net charge in the interior of the dielectric marked by dotted boundary in the Fig. The charge -qi and +qi on the two surfaces of the dielectric slab are called induced charges. The induced charges set up an electric field E_p inside the dielectric. It is called electric field due to polarisation. The direction of electric field due to polarisation is from positive surface to the negative surface of the slab i.e. along a direction opposite to the direction of applied electric field E₀. Therefore, the resultant electric field is given by

E = E₀ - E_p

Since, magnitude of E is less than E₀, E is called the reduced value of electric field. In other words, on placing a dielectric slab inside an electric field, the strength of the electric field gets reduced. It can be understood in another way also. If -σ' and +σ' are surface charge densities of the left and right faces of the dielectric slab (due to induced charges developed), then electric field between the two plates of the capacitor will be

E = \(\frac{\sigma}{\varepsilon_0} + (\frac{-\sigma'}{\varepsilon_0})\)

or E = \(\frac{\sigma-\sigma'}{\varepsilon_0}\) ...........................(2)

Obviously, the value of E is less than E₀.

(b) Dielectric constant: The ratio of the strength of applied electric field to the strength of reduced value of electric field on placing the dielectric between the plates of the capacitor is called the dielectric constant of the dielectric medium.

It is also known as relative permittivity or specific inductive capacity and is denoted by K (or ε_r). Therefore, dielectric constant of a dielectric is given by

K = \(\frac{E_0}{E}\) ...........................(3)

The value of K is always greater than 1.

(c) Polarisation density: The induced dipole moment development per unit volume in a dielectric slab on placing it inside the electric field is called polarisation density. It is denoted by P. If p is induced dipole moment of an atom acquired by the dielectric and N is number of atoms per unit volume, then

P = Np ........................(4)

Since, p = α ε₀ E,

∴ P = N α ε₀ E, ............................(5)

If A is the area of each plate of the capacitor and d, the distance between them, then the volume of the dielectric slab = Ad. Since induced charges -qi and +qi are developed on the two faces of the dielectric, the slab possesses total dipole moment equal to qid. Therefore, from definition of polarisation density,

P = \(\frac{qid}{Add} = \frac{qi}{A}\)

Since \(\frac{qi}{A}\) = σ', induced surface charge density, we have

P = σ' ..................... (6)

Therefore, reduced value of electric field on placing dielectric between the two plates, a capacitor may be expressed as:

E = \(\)\(\frac{\sigma-\sigma'}{\varepsilon_0} = \frac{\sigma}{\varepsilon_0} - \frac{\sigma'}{\varepsilon_0}\)

or E = E₀ - \(\frac{P}{\varepsilon_0}\)  ..................... (7)

(d) Electric susceptibility. The polarisation density of a dielectric slab is directly proportional to the reduced value of the electric field and may be expressed as P ∝ ε₀ E

P = ✗ε₀E ....................(8)

The constant of proportionality ✗ is called electric susceptibility of the dielectric slab. It is a dimensionless constant.

Setting the value of P in equation (7), we have

E = E₀ - \(\frac{X\varepsilon_0E}{\varepsilon_0}\)  = E₀ - ✗_E

or E₀ = E (1 + ✗) .....................(9)

or \(\frac{E_0}{E}\)  = 1 + ✗ ...............................(10)

From equation (3) and (10), we have

K = 1 + ✗ ........................(11)

Also, from equations (5) and (8), we have

✗ ε₀E = Naε₀E

or ✗ = Na

and K = 1 + Na