A spherical conducting shell of inner radius \( r_1 \) and outer radius \( r_2 \) has a charge Q. A charge q is placed at the centre of the shell. The surface charge density on outer surface of the shell is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{Q+q}{4\pi r_2^{2}} \)
In electrostatic equilibrium the electric field inside the material of a conductor is zero, and any charge on it lies on its surfaces. Surface charge density is the charge per unit area of the surface.
Take a Gaussian sphere inside the material of the shell. The field on it is zero, so by Gauss's law the total charge enclosed is zero:
\( \displaystyle q + q_{\mathrm{inner}} = 0 \;\Rightarrow\; q_{\mathrm{inner}} = -q \)
The shell carries a total charge \( Q \), so by conservation of charge the outer surface carries
\( \displaystyle q_{\mathrm{outer}} = Q - (-q) = Q + q \)
This charge spreads over the outer surface of area \( 4\pi r_2^{2} \):
\( \displaystyle \sigma_{\mathrm{outer}} = \frac{Q + q}{4\pi r_2^{2}} \)