Question

What is dielectric? Why does the capacitance of a parallel plate capacitor increase on introduction of a dielectric between its two plates? Derive an expression for the capacitance of such a capacitor having two identical plates each of area A and separated by distance x. The space between the plates has a medium of dielectric constant K.

Answer 1

Dielectric is an insulting medium which transmits electric current without conducting.

When a dielectric is introduced between the plates of a charged capacitor the dielectric gets polarised and opposite electric field is developed in the dielectric due to polarisation. Thus the net electric field inside the dielectric decreases without bringing any change in the charge. Some charge and decrease in electric field produces decrease in electric potential and hence increase in capacity of the capacitor ( C = \(\frac{q}{V}\)):

Capacitance of a parallel plate capacitor

(a) Parallel plate capacitor. A parallel plate capacitor is a combination of two conducting plates held parallel to each other at suitable distance having some dielectric in between them. One of the plate is positively charged and the other is earthed.

Expression for capacitance of parallel plate capacitor

Let a parallel plate capacitor consist of two thin conducting plates A and B held parallel to each other at a certain distance d apart. Medium between plates is given to be vacuum. Plate A is insulated and B is earthed.

When charge +q is given to A, it induces -q on the nearer face of B and +q on the farther face of the plate B. Free +ve charge on B flows to the earth.

Electric field intensity between the plate is

E = \(\frac{\sigma}{\varepsilon_0}\)

where σ is the charge density and equal to q/A.

Also E = \(\frac{V}{d}\), where V is the P.D. between plates.

V = Ed = \(\frac{\sigma}{\varepsilon_0}d = \frac{qd}{A\varepsilon_0} \)          [∵\(\sigma= \frac{q}{A}\)]

If C is the capacitance of the parallel plate capacitor, then

\(\frac{q}{V} = \frac{q}{\frac{qd}{A\varepsilon_0}} = \frac{A\varepsilon_0}{d}\)

or C = \(\frac{A\varepsilon_0}{d}\)

This is the expression for the capacitance when the medium is air or vacuum.

If A is in m², d is in m, then C is taken in farad. If medium between plates is other than air, capacitance of capacitor is given by

C = \(\frac{\varepsilon_0KA}{d}\)

where K is dielectric constant.

(b) Definition of dielectric constant

In vacuum the capacitance of capacitor,

C₁ = \(\frac{\varepsilon_0A}{d}\)  .........................(i)

In dielectric medium, the capacitance of capacitor

C₂ = \(\frac{\varepsilon_0KA}{d}\) .........................(i)

Dividing (i) and (ii), we get

\(\frac{C_2}{C_1} = \frac{\varepsilon_0KA}{d} \times \frac{d}{\varepsilon_0A} = K\)

∴ Dielectric constant of a medium is defined as the ratio of capacitance of capacitor with dielectric as medium to the capacitance of the same capacitor with air as medium.