The plates of a parallel plate capacitor are at a distance of 1 mm. The space between the plates is inserted with a dielectric of dielectric constant 2. What should be the area of a plate of the parallel plate capacitor to generate the capacitance of 8.85 nF

CONCEPT:

• The capacitance of a capacitor (C): The capacitance of a conductor is the ratio of charge (Q) to it by a rise in its potential (V), i.e.

C = Q/V

For a Parallel Plate Capacitor:

• A parallel plate capacitor consists of two large plane parallel conducting plates of area A and separated by a small distance d.

• Mathematical expression for the capacitance of the parallel plate capacitor is given by:

\(\Rightarrow C = \frac{{{\epsilon_o}A}}{d}\)

Where C = capacitance, A = area of the two plates, ε = dielectric constant (simplified!), d = separation between the plates.

• The unit of capacitance is the farad, (symbol F ).
• The electric field between the plates, from Gauss law, is

\(⇒ E=\frac{\sigma }{ε_0}\)

• The capacitance is defined as

\(⇒ C= \frac {Q}{V} = \frac {\sigma A}{Ed} =\frac {ε_0 A}{d}\)

• Capacitance is independent of charge Q on the plates and potential difference V across the plates.

EXPLANATION:

Given C = 8.85 nF, d = 1mm, K = 2 where C is the capacitance, d is the distance between the two plates of the capacitor, K is the dielectric constant.

• If a dielectric of dielectric constant K is inserted between the plates of a parallel plate capacitor, the potential difference changes due to induced charges on the edges of the dielectric.

• The electric field inside the capacitor is 

\(⇒ E = \frac{\sigma - \sigma _p}{ε _0}\)

• The potential difference across the capacitor is 

\(⇒ V= Ed = \frac{\sigma - \sigma _p}{ε _0}d\)

• The ratio of the total electric field with dielectric 'E' and the total electric field without the dielectric 'E0' can be represented as

\(⇒ \frac {E}{E_0}= \frac{\sigma - \sigma _p}{\sigma} = \frac{1}{K}\)

• The capacitance of this capacitor is

\(⇒ C= \frac {Q}{V} = \frac {\sigma A}{Ed} =\frac {ε_0 \sigma A}{(\sigma-\sigma_p)d} = \frac {Kε_0A}{d}=KC_0\)

• The area of a plate of the parallel plate capacitor to generate the capacitance of 8.85 nF is

\(⇒ A= \frac {Cd}{Kε_0} =\frac {8.85 \times 10 ^{-9} \times 10 ^{-3}}{ 2 \times 8.85 \times 10^{-12} } =0.5 m^2\)