Parallel plate capacitor consists of two conducting plates parallel to each other and separated by a distance which is small compared with the dimensions of the plate.

Let A be the area of each plate separated by a distance d in vacuum and +q and -q are the charges given to the plates.

The charges on each plate are attracted by the charges on the other plate. So the charges spread out uniformly on the inner surfaces of the plates and produce an electric field E between the two plates.

Consider a Gaussian surface PQRS whose walls are PS and QR (try to make the figure yourself).

The electric flux through the PS is zero (because inside the conductor, E = 0). Also the electric flux through the end PQ and RS is zero because E is parallel to both PQ and RS

i.e. E is perpendicular to dS for PQ and RS.

Thus flux φ = Eds cos 90 = 0

Now the electric flux through the Gaussian surface is equal to the electric flux through the surface QR only, that is

Φ = EdS cos 0 = Eds

Or φ = EA where A is area of each plate.

Now according to Gauss’s law, the φ through the Gaussian surface is equal to the 1/ε_{0} times the charge contained within the surface

Therefore φ = EA = q/ ε_{0}

Or E = q/ ε_{0}A (1)

As the potential difference V between the plates is defined as the work required to carry a unit charge from one plate to the other. Therefore,

V = Ed

Putting the value of E from equation (1), we get

V = qd/ ε_{0}A

But the capacitance of a capacitor is

C= q/V

Substitute the value of V in above equation, we get

C= ε_{0}A/d

This is the relation for capacitance of parallel plate capacitor.

If a dielectric of dielectric constant k is placed between the plates, then the capacitance will be

C = k ε_{0}A/d