Derivation: Capacitance of a Parallel Plate Capacitor , Factors Affecting Capacitance - Physicskund

Derivation: Capacitance of a Parallel Plate Capacitor

Consider a parallel plate capacitor consisting of two parallel plates separated by a distance ( d ).  

- Each plate has an area ( A ).  

- Plate R carries charge ( +q ), and Plate S carries charge ( -q ).  

- The surface charge densities are ( $+\sigma$ ) and ( $-\sigma$) respectively.  

The electric field between the plates is:

$E = \frac{\sigma}{\varepsilon_0} \quad ...(i)$

where ( $\varepsilon_0$ ) is the permittivity of free space (assuming the medium is air or vacuum).  

The direction of the uniform electric field is from the positive plate to the negative plate.

Derivation : 

- Potential difference between the plates:

$V = E \cdot d$

- Since ( $\sigma = \frac{q}{A}$ ), substituting into equation (i):

$E = \frac{q}{A \varepsilon_0}$

So,

$V = \frac{q}{A \varepsilon_0} \cdot d$

- Capacitance is defined as:

$C = \frac{q}{V}$

Substituting ( V):

$C = \frac{q}{\frac{q d}{A \varepsilon_0}} = \frac{\varepsilon_0 A}{d}$

Final Expression : 

$C = \frac{\varepsilon_0 A}{d} \quad ...(ii)$

This is the expression for the capacitance of a parallel plate capacitor filled with air.

If a dielectric of permittivity \(\varepsilon\) occupies the space between the conducting plates, then capacitance is:

$C = \frac{\varepsilon A}{d}, \quad \text{where } \varepsilon = \varepsilon_0 \varepsilon_r \quad ...(iii)$

Factors on which capacitance of a parallel plate capacitor depends:

1. Area of the plates of the capacitor.  

2. Distance between the plates of the capacitor.  

3. Nature of the dielectric medium or insulator between the plates of the capacitor.