With the understanding of the behaviour of dielectrics in an external field developed in Section 2.10, let us see how the capacitance of a parallel plate capacitor is modified when a dielectric is present. As before, we have 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=G/A). When there is vacuum between the plates,
E0=σε0
and the potential difference V0 is
V0=E0d
The capacitance C0 in this case is
C0=BV0=ε0Ad…(2.46)
Consider next a dielectric inserted between the plates fully occupying the intervening region. The dielectric is polarised by the field and, as explained in Section 2.10, the effect is equivalent to two charged sheets (at the surfaces of the dielectric normal to the field) with surface charge densities σp and −σp. The electric field in the dielectric then corresponds to the case when the net surface charge density on the plates is ±(σ−σp). That is,
E=σ−σpε0…(2.47)
so that the potential difference across the plates is
V=Ed=σ−σpε0d…(2.48)
For linear dielectrics, we expect σp to be proportional to E0, i.e., to σ. Thus, (σ−σp) is proportional to σ and we can write
σ−σp=σK…(2.49)
where K is a constant characteristic of the dielectric. Clearly, K>1. We then have
V=σdε0K=QdAε0K…(2.50)
The capacitance C , with dielectric between the plates, is then
C=QV=ε0KAd…(2.51)
The product ε0K is called the permittivity of the medium and is denoted by ε
ε=ε0K…(2.52)
For vacuum K=1 and ε=ε0;ε0 is called the permittivity of the vacuum. The dimensionless ratio
K=εε0…(2.53)
is called the dielectric constant of the substance. As remarked before, from Eq. (2.49), it is clear that K is greater than 1. From Eqs. (2.46) and (2.51)
K=CC0…(2.54)
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. Though we arrived at Eq. (2.54) for the case of a parallel plate capacitor, it holds good for any type of capacitor and can, in fact, be viewed in general as a definition of the dielectric constant of a substance.
Example 2.8: A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness (3/4)d, where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?
Solution: Let E0=V0/d be the electric field between the plates when there is no dielectric and the potential difference is V0. If the dielectric is now inserted, the electric field in the dielectric will be E=E0/K. The potential difference will then be
V=E0(14d)+E0K(34d)=E0d(14+34K)=V0K+34K
The potential difference decreases by the factor (K+3)/4K while the free charge Q0 on the plates remains unchanged. The capacitance thus increases
C=Q0V=4KK+3Q0V0=4KK+3C0
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