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.