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