Question

Derive an expression for capacitance of a parallel plate capacitor without a dielectric ​

Answer 1

Explanation:

First we have to find electric field in between the plates

Answer 2

$\huge \mathfrak \red{Answer}$

>> Consider a parallel plate capacitor without any dielectric medium (vacuum) between the plates. Let A be the area of the plates and d be the plate separation. The two plates have charges Q and – Q. Since d is much smaller than the linear dimension of the plates

(d2 <<A),

Plate 1 has surface charge density σ = Q/A and plate 2 has a surface charge density – σ. Using earlier result of the electric field in different regions is:

Outer region I (region above the plate 1),

(SEE THE ATTACHMENT 1)

Outer region II (region below the plate 2),

(SEE THE ATTACHMENT 2)

In the inner region between the plates 1 and 2, the electric fields due to the two charged plates add up, giving

(SEE THE ATTACHMENT 3)

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. For plates with finite area, this will not be true near the outer boundaries of the plates. The field lines bend outward at the edges – an effect called ‘fringing of the field’. Hence σ will not be strictly uniform on the entire plate. However, for

d2 << A,

these effects can be ignored in the regions sufficiently far from the edges and the field is given by Eq(1). Now for uniform electric field, potential difference is simply the electric field times the distance between the plates, then is,

(SEE THE ATTACHMENT 4)

The capacitance C of the parallel plate capacitor is then,

(SEE THE ATTACHMENT 5)

which, as expected, depends only on the geometry of the system.

:)