A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance (Fig.).
Let medium between the plates to be vacuum and A be the area of each plate and d the separation between them. The two plates have charges Q and –Q. Since d is much smaller than the linear dimension of the plates (d^2 << A).
Plate 1 has surface charge density σ = Q/A and plate 2 has a surface charge density –σ.
The electric field in different regions is: Outer region I (region above the plate 1),
E = σ/2εo-σ/2εo=0
Outer region II (region below the plate 2),
E = σ/2εo-σ/2εo=0
In the inner region between the plates 1 and 2, the electric fields due to the two charged plates add up, giving
E = σ/2εo+σ/2εo = σ/εo = Q/Aεo
The direction of electric field is from the positive to the negative plate. Thus, the electric field is localised between the two plates and is uniform throughout.
Now for uniform electric field, potential difference is simply the electric field times the distance between the plates, that is,
V = Ed = Qd/Aεo
The capacitance C of the parallel plate capacitor is then Q = CV = Aεo/d
which, as expected, depends only on the geometry of the system.
