A metallic spherical shell has an inner radius $R_{1}$ and outer radius $R_{2}$ . A charge is placed at the centre of the spherical cavity. The surface charge density on the outer surface is

\frac{- q}{4 π R_{1}^{2}}

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\frac{q}{4 π R_{2}^{2}}

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\frac{q^{2}}{4 π R_{1}^{2}}

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\frac{2 q}{4 π R_{2}^{2}}

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Solution

The correct option is B $\frac{q}{4 π R_{2}^{2}}$

Inner radius $= R_{1}$
outer radius $= R_{2}$
The surface density depends on only outer surface. It does depend on inner surface.
Surface charge density $= \frac{q}{4 π R_{2}^{2}}$
The surface area $= 4 π R_{2}^{2}$
$ρ = \frac{q}{A}$ ( $∵$ $A =$ area)

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