What is a parallel plate capacitor? Obtain the formula for the capacitance of such a parallel plate capacitor and state the factors on which the value of capacitance depends.

(N/A) parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance $d$.
$A$ non-conducting medium (dielectric) is typically kept between the two plates.
Let the charges on plate $1$ and plate $2$ be $+Q$ and $-Q$ respectively,$A$ be the area of each plate,and $d$ be the separation between them.
Since $d$ is much smaller than the linear dimensions of the plates $(d^2 << A)$,we can use the result for the electric field $E = \frac{\sigma}{2\epsilon_0}$ produced by an infinite plane sheet of uniform surface charge density $\sigma = \frac{Q}{A}$.
In the region between the plates,the electric fields due to both plates point in the same direction (from positive to negative plate):
$E = E_1 + E_2 = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0}$.
The potential difference $V$ between the plates is given by $V = E \cdot d = \frac{Qd}{A\epsilon_0}$.
The capacitance $C$ is defined as $C = \frac{Q}{V} = \frac{Q}{(Qd / A\epsilon_0)} = \frac{\epsilon_0 A}{d}$.
The capacitance depends on:
$1$. The area of the plates $(A)$.
$2$. The distance between the plates $(d)$.
$3$. The permittivity of the medium between the plates ($\epsilon_0$ or $\epsilon = k\epsilon_0$).