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If a charge q is placed inside a conducting spherical cavity with inner radius of $R_{1}$ and outer radius of $R_{2}$ , the surface charge density on the outer surface is

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$\frac{q}{4 {πR}_{1}^{2}}$

$\frac{2 q}{4 {πR}_{2}^{2}}$

$\frac{q^{2}}{4 {πR}_{2}^{2}}$

$\frac{q}{4 {πR}_{2}^{2}}$

answer is B.

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Detailed Solution

Charge induced on the inner surface = -q

charge induced on the outer surface= +q

Surface charge density on the outer surface $= \frac{q}{4 π R_{2}^{2}}$

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