Two charged spherical conductors of radius R_ | vedclass

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Question

$\mathrm{Q}_{1}=\frac{\sum \mathrm{Q}}{\mathrm{R}_{1}+\mathrm{R}_{2}} 	imes \mathrm{R}_{1}$

$\mathrm{Q}_{2}=\frac{\sum Q}{\mathrm{R}_{1}+\mathrm{R

Vedclass · Accepted Answer

$\frac{R_{2}}{R_{1}}$

Vedclass · Answer

$\mathrm{Q}_{1}=\frac{\sum \mathrm{Q}}{\mathrm{R}_{1}+\mathrm{R}_{2}} 	imes \mathrm{R}_{1}$

$\mathrm{Q}_{2}=\frac{\sum Q}{\mathrm{R}_{1}+\mathrm{R}_{2}} 	imes \mathrm{R}_{2}$

$\sigma_{1}=\frac{\mathrm{Q}_{1}}{4 \pi \mathrm{R}_{1}^{2}}=\frac{\sum Q}{\mathrm{R}_{1}+\mathrm{R}_{2}} 	imes \frac{\mathrm{R}_{1}}{4 \pi \mathrm{R}_{1}^{2}} \propto \frac{1}{\mathrm{R}_{1}}$

$\sigma_{2}=\frac{\mathrm{Q}_{2}}{4 \pi \mathrm{R}_{2}^{2}}=\frac{\sum Q}{\mathrm{R}_{1}+\mathrm{R}_{2}} 	imes \frac{\mathrm{R}_{2}}{4 \pi \mathrm{R}_{2}^{2}} \propto \frac{1}{\mathrm{R}_{2}}$

$\frac{\sigma_{1}}{\sigma_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}$

Two charged spherical conductors of radius $R_{1}$ and $\mathrm{R}_{2}$ are connected by a wire. Then the ratio of surface charge densities of the spheres $\left(\sigma_{1} / \sigma_{2}\right)$ is :

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