Derive an expression for the capacitance of a parallel plate capacitor.

Explanation

Expression for the capacitance of a parallel plate capacitor.
1. P and Q are two parallel plates of a capacitor separated by a distance of d.
2. The area of each plate is A. The plate P is charged and Q is earth connected.
3. The charge on P is `+q` and surface charge density of charge `=sigma`
`therefore q=Asigma`
4. The electric intensity at point `x,E=(|sigma|)/(epsi_0)`
5. Potential difference between the plates P and Q.
`V=int dV =int_(d)^(0)-Edx=int_(d)^(0)(-sigma)/(epsi_0)dx=(sigmad)/(epsi_0)`
6.Capacitance of the capacitor `C=(Q)/(V)=(Asigma)/((sigmad)/(epsi_0))=(epsi_0A)/(d)` Farads (In air)
Note : Capacity of a capacitor with dielectric medium is `C=(epsi_0A)/([d-t+(t)/(k)])` Farads.