Derive the expression for capacitance of parallel plate capacitor.

To derive the expression for the capacitance of a parallel plate capacitor, we can use the concept of electric field and electric potential. Let's consider a parallel plate capacitor with two plates of area A, separated by a distance d. The plate on the left is connected to the positive terminal of a battery, while the plate on the right is connected to the negative terminal.

Electric Field:
When a potential difference is applied across the plates, an electric field is created between them. The magnitude of the electric field (E) between the plates is given by:
E = V/d

Where V is the potential difference applied across the plates.

Electric Flux:
The electric field between the plates causes electric flux to pass through the area A. The electric flux (Φ) passing through an area A is given by:
Φ = E * A

Gauss's Law:
According to Gauss's law, the electric flux passing through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of the medium. In this case, the electric field is constant between the plates, so the electric flux is also constant.
Φ = Q / ε₀

Where Q is the charge on either plate, and ε₀ is the permittivity of free space.

Capacitance:
The capacitance (C) of a capacitor is defined as the ratio of the charge on either plate to the potential difference across the plates.
C = Q / V

Using the relationships established earlier, we can substitute Q/ε₀ for Φ and V/d for E:

Φ = Q / ε₀ = E * A
Q / ε₀ = (V/d) * A

Now, rearranging the equation for capacitance:

C = Q / V = (ε₀ * A) / (V/d) = ε₀ * (A/d)

Therefore, the expression for the capacitance of a parallel plate capacitor is:

C = ε₀ * (A/d)

where ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.