Question 427: Electrostats & Capacitance

Question

Two isolated metallic solid spheres of radii $$\mathrm{R}$$ and $$2 \mathrm{R}$$ are charged such that both have same charge density $$\sigma$$. The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is $$\sigma^{\prime}$$. The ratio $$\frac{\sigma^{\prime}}{\sigma}$$ is :

Accepted Answer

$$\sigma  = {{{Q_1}} \over {4\pi {R^2}}} = {{{Q_2}} \over {4\pi {{(2R)}^2}}}$$
Now $$Q{'_2} = {Q_{total}}\left[ {{{{R_2}} \over {{R_1} + {R_2}}}} ight]$$
$$ = ({Q_1} + {Q_2})\left[ {{{2R} \over {3R}}} ight]$$
$$ = \sigma (20\pi {R^2}){2 \over 3}$$
$$	herefore$$ $$\sigma {'_2} = {{Q{'_2}} \over {4\pi {{(2R)}^2}}} = {{\sigma (20\pi {R^2}){2 \over 3}} \over {16\pi {R^2}}}$$
$$ = {5 \over 4} 	imes {2 \over 3}\sigma $$
$$ = {5 \over 6}\sigma $$

Answer

$$\frac{5}{3}$$

Answer

$$\frac{5}{6}$$

Answer

$$\frac{9}{4}$$

Answer

$$\frac{4}{3}$$

Electrostats & Capacitance

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