Question

Derive an expression for capacitance of a capacitor with dielectric medium and Effect of Dielectric on Capacitance.

Answer 1

If we consider a capacitor made of two large plates, each of area A, separated by a distance d. The charge on the plates is ±Q, corresponding to the charge density ± σ (with σ = Q/A). When there is vacuum between the plates, 

and the potential difference V₀ is 

V₀ = E₀d 

The capacitance C₀ in this case is 

Consider next a dielectric inserted between the plates fully occupying the intervening region. The dielectric is polarised by the field and the effect is equivalent to two charged sheets (at the surfaces of the dielectric normal to the field) with surface charge densities σ_p and – σ_p. The electric field in the dielectric then corresponds to the case when the net surface charge density on the plates is ±( σ – σ_p). 

That is,

so that the potential difference across the plates is 

 

For linear dielectrics, we expect σp to be proportional to E₀, i.e., to σ. Thus, ( σ – σ_p) is proportional to σ and we can write  

where K is a constant characteristic of the dielectric. Clearly, K > 1. We then have 

The capacitance C, with dielectric between the plates, is then 

The product ε₀K is called the permittivity of the medium and is denoted by ε

ε = ε₀K

For vacuum K = 1 and ε = ε₀; ε₀ is called the permittivity of the vacuum.  

 Dielectric constant : 

The dimensionless ratio 

is called the dielectric constant of the substance. From Eq. (3), it is clear that K is greater than 1. From Eqs. (1) and (5),

Thus, the dielectric constant of a substance is the factor (>1) by which the capacitance increases from its vacuum value, when the dielectric is inserted fully between the plates of a capacitor.