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:

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

• The capacitance is defined as

• 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 

• The potential difference across the capacitor is 

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

• The capacitance of this capacitor is

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