2.13 EFFECT OF DIELECTRIC ON CAPACITANCE

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 $\pm Q$ , corresponding to the charge density $\pm σ$ (with $σ = Q / A$ ). When there is vacuum between the plates, $E_{0} = \frac{σ}{ε_{0}}$ and the potential difference $V_{0}$ is $V_{0} = E_{0} d$ The capacitance $C_{0}$ in this case is

$C_{0} = \frac{Q}{V_{0}} = ε_{0} \frac{A}{d}$ (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 $σ_{p}$ and $- σ_{P}$ . The electric field in the dielectric then corresponds to the case when the net surface charge density on the plates is $\pm (σ - σ_{P})$ . That is,

$E = \frac{σ - σ_{P}}{ε_{0}}$ (2.47)

so that the potential difference across the plates is

$V = E d = \frac{σ - σ_{P}}{ε_{0}} d$ (2.48)

For linear dielectrics, we expect $σ_{P}$ to be proportional to $E_{0}$ , i.e., to $σ$ . Thus, ( $σ - σ_{P}$ ) is proportional to $σ$ and we can write

$σ - σ_{P} = \frac{σ}{K}$ (2.49)

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

$V = \frac{σ d}{ε_{0} K} = \frac{Q d}{A ε_{0} K}$ (2.50)

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

$C = \frac{Q}{V} = \frac{ε_{0} K A}{d}$ (2.51)

The product $ε_{0} K$ is called the permittivity of the medium and is denoted by $ε$

$ε = ε_{0} K$ (2.52)

For vacuum K = 1 and

ε = ε_{0}

;

ε_{0}

is called the permittivity of the vacuum.

The dimensionless ratio $K$ is called the dielectric constant of the substance.

$K = \frac{ε}{ε_{0}}$ (2.53)

As remarked before, from Eq. (2.49), it is clear that K is greater than 1. From Eqs. (2.46) and (2. 51)

$K = \frac{C}{C_{0}}$ (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.

Electric Displacement

We have introduced the notion of dielectric constant and arrived at Eq. (2.54), without giving the explicit relation between the induced charge density $σ_{P}$ and the polarisation $P$ .

We take without proof the result that $σ_{p} = P \cdot \hat{n}$ where $\hat{n}$ is a unit vector along the outward normal to the surface. Above equation is general, true for any shape of the dielectric. For the slab in Fig. 2.23, $P$ is along $\hat{n}$ at the right surface and opposite to $\hat{n}$ at the left surface. Thus at the right surface, induced charge density is positive and at the left surface, it is negative, as guessed already in our qualitative discussion before. Putting the equation for electric field in vector form $E \cdot \hat{n} = \frac{σ - P \cdot \hat{n}}{ε_{0}}$ Or $(ε_{0} E + P) \cdot \hat{n} = σ$

The quantity $ε_{0} E + P$ is called the electric displacement and is denoted by $D$ . It is a vector quantity. Thus, $D = ε_{0} E + P, D \cdot \hat{n} = σ$

The significance of $D$ is this : in vacuum, $E$ is related to the free charge density $σ$ . When a dielectric medium is present, the corresponding role is taken up by $D$ . For a dielectric medium, it is $D$ not $E$ that is directly related to free charge density $σ$ , as seen in above equation. Since $P$ is in the same direction as $E$ , all the three vectors $P$ , $E$ and $D$ are parallel.

The ratio of the magnitudes of $D$ and $E$ is $\frac{D}{E} = \frac{σ ε_{0}}{σ - σ_{P}} = ε_{0} K$ Thus, $D = ε_{0} K E$ and $P = D - ε_{0} E = ε_{0} (K - 1) E$ This gives for the electric susceptibility $χ_{e}$ defined in Eq. (2.37) $χ_{e} = ε_{0} (K - 1)$

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?

VIEW SOLUTION

Let $E_{0} = V_{0} / d$ be the electric field between the plates when there is no dielectric and the potential difference is $V_{0}$ . If the dielectric is now inserted, the electric field in the dielectric will be $E = E_{0} / K$ . The potential difference will then be $\begin{array}{l} V = E_{0} (\frac{1}{4} d) + \frac{E_{0}}{K} (\frac{3}{4} d) \\ = E_{0} d (\frac{1}{4} + \frac{3}{4 K}) = V_{0} \frac{K + 3}{4 K} \end{array}$

The potential difference decreases by the factor $(K + 3) / K$ while the free charge $Q_{0}$ on the plates remains unchanged. The capacitance thus increases $C = \frac{Q_{0}}{V} = \frac{4 K}{K + 3} \frac{Q_{0}}{V_{0}} = \frac{4 K}{K + 3} C_{0}$