Two spheres of radii $R_{1}$ and $R_{2}$ joined by a fine wire are raised to a potential V. Let the surface charge densities at these two spheres be $\sigma_{1}$ and $\sigma_{2}$ respectively. Then the ratio $\frac{\sigma_{2}}{\sigma_{1}}$ has a value

Question

Dr. Abubacker · Accepted Answer

(a) Since the two spheres are joined by a wire, their potential are equal $ie,$
$\frac{q_{1}}{4\pi\varepsilon_{0}R_{1}} = \frac{q_{2}}{4\pi\varepsilon_{0}R_{2}} \Rightarrow \frac{q_{1}}{q_{2}} = \frac{R_{1}}{R_{2}}$
Now, $\sigma_{1} = \frac{q_{1}}{4\pi\varepsilon_{0}R_{1}^{2}}$
And $\sigma_{2} = \frac{q_{2}}{4\pi\varepsilon_{0}R_{2}^{2}}\ ,$
Hence $\frac{\mathbf{\sigma}_{\mathbf{2}}}{\mathbf{\sigma}_{\mathbf{1}}}$=$\frac{\mathbf{\sigma}_{\mathbf{2}}}{\mathbf{\sigma}_{\mathbf{1}}}\mathbf{	imes}\frac{R_{1}^{2}}{R_{2}^{2}} = \left( \frac{R_{2}}{R_{1}} ight)\left( \frac{R_{1}}{R_{2}} ight)^{2}$
$\Rightarrow \ \ \ \frac{\sigma_{2}}{\sigma_{1}}$=$\frac{R_{1}}{R_{2}}$

Q. Two spheres of radii $R_{1}$ and $R_{2}$ joined by a fine wire are raised to a potential V. Let the surface charge densities at these two spheres be $\sigma_{1}$ and $\sigma_{2}$ respectively. Then the ratio $\frac{\sigma_{2}}{\sigma_{1}}$ has a value

Explanation

Reviewed by Dr. Abubacker

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