Deduce expression for capacitance of parallel plate capacitor. Discuss the action of dielectric between the plates.

Capacitance Long Question Solution

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Deduce expression for capacitance of parallel plate capacitor. Discuss the action of dielectric between the plates.

Capacitance - Long Question Answer 3

Question: Deduce the expression for capacitance of a parallel plate capacitor. Discuss the action of dielectric between the plates.

Answer: Parallel plate capacitors are is a combination of two parallel plates each of equal area and paced very close to each other. One plate is given positive charge while another is connected to the earth.

Let two plate X and Y with equal area A separated by distance d. The plate X is given a charge +q and the plate Y is earthed. When positive charge is given to plate X, negative charge is itself induces in another plate at the side facing plate X and positive chargeat the other side.

Since, the plate Y is earthed the positive charge flows to the earth and the positive and negative charges in plate X and Y produces an electric field between them.  If each plate has a uniform surface charge density σ, then electric field E between the plates is uniform and is given as;

E = σ/Є0

Where Є0 is the permittivity of the free space between the plate as it is kept in the air. Then, the potential difference between the plates is

V = Ed = σ d/Є0

Since σ = q/A, so

V = q. d/Є0A

q/V = Є0A/d

But q/V is the capacitance of the capacitor, so,

C = Є0A/d

If the space between the plates is filled with a medium of absolute permittivity Є, then the capacitance of the capacitor is given by,

C = ЄA/d   is the required expression for the capacitance of a capacitor.

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