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Question

Two charged spherical conductors of radii R₁ and R₂ are connected by a wire. Then the ratio of surface charge densities of the spheres $σ_{1} / σ_{2}$ is:

Easy

A

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

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

$\sqrt{(\frac{R_{1}}{R_{2}})}$
D

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

Solution

Potential at the surface of spherical conductor of radius R carrying charge Q, $V_{S} = \frac{Q}{4 {πε}_{0} R}$
Let Q₁ and Q₂ are the charges on two spherical conductors of radii R₁ and R₂ respectively. When these two charged spherical conductors connected by a wire, the potential at their surfaces becomes equal.
$∴ V_{S_{1}} = V_{S_{2}}$
$\frac{Q_{1}}{4 {πε}_{0} R_{1}} = \frac{Q_{2}}{4 {πε}_{0} R_{2}} \Rightarrow \frac{Q_{1}}{Q_{2}} = \frac{R_{1}}{R_{2}} . . . . (i)$
Surface charge density, $σ = \frac{Charge}{Area}$
$∴ \frac{σ_{1}}{σ_{2}} = \frac{(\frac{Q_{1}}{4 {πR}_{1}^{2}})}{(\frac{Q_{2}}{4 {πR}_{2}^{2}})} = \frac{Q_{1}}{Q_{2}} \times \frac{R_{2}^{2}}{R_{1}^{2}} = \frac{R_{1}}{R_{2}} \times \frac{R_{2}^{2}}{R_{1}^{2}} = \frac{R_{2}}{R_{1}}$

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