Derive expression for the capacitance of the parallel plate capacitor,

Explanation

Arrangement of two large plane parallel conducting plates separated by a small distance is known as parallel plate capacitor. Non-conducting medium is kept between two planes of such capacitor. Free space will be taken as non-conducting medium.

According to figure, the charge on plate I and plate 2 Q and - Q respectively and d is the separation between them A is the area of each plate.
d is much smaller than the linear dimension of the plates `[ d ltltA]` so we can use the result on electric field ` E= (sigma)/(2 in_(0))` by an infinite plane sheet of uniform surface charge density where,
`sigma=` surface charge density & `pm sigma= (Q)/(A)`
`in_(0)` = permittivity of free space
A = area of plate
Q= Charge on plate Region below the plate 1,
`E= (sigma)/(2 in_(0))-(sigma)/(2 in_(0))`
`:. E =0`
Region above the plate
`E= (sigma)/(2in_(0))- (sigma)/(2 in_(0))`
`:. E=0`
The electric field between plate 1 and 2
` E= (sigma)/(2 in_(0))+(sigma)/(2in_(0))`
`= (sigma)/(in_(0))`
`:. E = (Q)/(in_(0)A)`
The direction of electric field is from the positive to the negative plate. The electric field is localised between the two plates but the field lines bend outward at the edges an effect called .fringing of the field.. Hence `E= (Q)/(in_(0)A)` will not be uniform on the entire plate.
However for `d^(2) lt lt A` these effects can be ignored in the region far from the edges. For uniform electric field,
`V= Ed `
`:. V = (Qd)/(in_(0)A)`
Capacitance of parallel plate capacitor
`C= (Q)/(V)= (Q)/(Qd//in_(0)A)`
`:. (in_(0)A)/(d)` is the capacitance of capacitor .
Capacitance of parallel plate capacitor depend on the foil owing factors :
(1) Area of plate `[C prop A]`
(2) Separation between two plates `[C prop (1)/(d)]`
(3) Permittivity of medium between two plates `C prop epsilon`