1. Derive the expression for the capacitance of a parallel plate … - Vidyadip

Question

Derive the expression for the capacitance of a parallel plate area A and plate separation d.
Two charged spherical conductors of radii $R_1$ and $R_2$ when conducting wire acquire charges $q_1$ and $q_2$ respectively. surface charge densities in terms of their radii.

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Answer

Electric field between the plates of capacitor E = $\frac{\sigma}{\varepsilon_{0}} =\frac{\text{Q}}{\text{A}\varepsilon_{0}}$

$\therefore$ potential difference

$\text{V} = \text{Ed} = \frac{\text{Qd}}{\text{A}\varepsilon_{0}}$

Capacitance

$\text{C} = \frac{\text{Q}}{\text{V}} = \frac{\varepsilon_{0}\text{A}}{\text{d}}$

When the two charged spherical conductors are connected by a conducting wire , they acquire the same potential

i.e $\frac{\text{Kq}_{1}}{\text{R}_{1}} = \frac{\text{Kq}_{2}}{\text{R}_{2}}\Rightarrow\frac{\text{q}_{1}}{\text{q}_{2}} = \frac{\text{R}_{1}}{\text{R}_{2}}$

Hence, ratio of surface charge densities

$\frac{\sigma_{1}}{\sigma_{2}} = \frac{\text{q}_{1}/4\pi\text{R}_{1}^{2}}{\text{q}_{2}/4\pi\text{R}_{2}^{2}}$

$ = \frac{\text{q}_{1}\text{R}_{2}^{2}}{\text{q}_{2}\text{R}_{1}^{2}}$

$ = \frac{\text{R}_{1}}{\text{R}_{2}}\times\frac{\text{R}_{2}^{2}}{\text{R}_{1}^{2}} = \frac{\text{R}_{2}}{\text{R}_{1}}.$

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