A spherical conducting shell of inner radius r₁ and outer radius r₂ has a charge Q. A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell.

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$\frac{- q}{4 {πr}_{1}^{2}}, \frac{(Q + q)}{4 {πr}_{2}^{2}}$

$\frac{q}{4 {πr}_{1}^{2}}, \frac{(Q)}{4 {πr}_{2}^{2}}$

$\frac{Q}{4 {πr}_{1}^{2}}, \frac{(Q - q)}{4 {πr}_{2}^{2}}$

$\frac{q}{4 {πr}_{1}^{2}}, - \frac{(Q + q)}{4 {πr}_{2}^{2}}$

answer is A.

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Detailed Solution

The charge of magnitude +q present inside the spherical shell induces a charge of magnitude −q.On the inner part of the shell. Therefore, for the charge on the inner part, the surface charge density will be due to −q.

It is given by:

σ₁=Total charge/surface area of inner part

⇒ $\frac{- q}{4 π r_{1}^{2}}$

The charge on the outer shell will be +q and the surface charge density at the outer part will be due to sum of q and Q charges present on the outer shell.

σ₂=Total charge/Outer surface area= $\frac{Q + q}{4 {πr}_{2}^{2}}$

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