Mention any two factors on which the capacitance of a parallel plate capacitor depends.

Hint: Capacitor is a device used to store charges in an electric circuit. Capacitor has two plates separated by a distance having equal and opposite charges. Derive the expression for capacitance of a parallel plate capacitor using the equation for potential difference and electric field between two plates. Then, using this obtained expression, find the factors on which it’s capacitance is dependent.

Complete answer:
Electric field between the plates of parallel plate capacitors given by,
$E= \dfrac {\sigma}{K{\epsilon}_{0}}$ ...(1)
Potential difference between two plates is given by,
$V=Ed$
Substituting equation. (1) in above equation we get,
$V= \dfrac {\sigma d}{K{\epsilon}_{0}}$ ...(2)
But, $\sigma = \dfrac {Q}{A} $
Substituting this value in the equation. (2) we get,
$V= \dfrac {({Q}/{A})d}{K {\epsilon}_{0}}$
$\therefore V= \dfrac {Qd}{K {\epsilon}_{0}A}$ ...(3)
Now, capacitance of a capacitor is given by,
$C= \dfrac {Q}{V}$ ...(4)
Substituting equation. (4) in equation. (3) we get,
$C= \dfrac {Q} {\dfrac {Qd}{K {\epsilon}_{0}A}}$
$\therefore C= \dfrac {K {\epsilon}_{0}A}{d}$
This is the expression for capacitance of a parallel plate capacitor.
From the above equation, it can be understood that the capacitance depends on the dielectric constant of the medium between the plates, area of each plate and distance between the plates.
Hence, capacitance of a parallel plate capacitor depends on the dielectric constant of the medium between the plates, area of each plate and distance between the plates.

Note:
Amount of charge a capacitor can store depends on two factors. First is the voltage and second is the physical characteristics of the capacitors. When a dielectric medium is placed between the two parallel plates of the capacitor, it’s capacity of storing the energy increases. It’s energy increases by a factor K which is called a dielectric constant. If you want to increase the capacitance of parallel plate capacitors then increase the area, decrease the separation between two plates and use a dielectric medium.