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 the inner and outer surface of the shell will be
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Electrostatic induction inside conductors always acts to mirror charges: a central charge of $-q$ requires an inner surface layout of exactly $+q$. This immediately eliminates option (C) and leaves option (A) as the correct choice!
$\frac{q}{4\pi r_1^2}$ and $\frac{Q - q}{4\pi r_2^2}$
$\frac{q}{4\pi r_1^2}$ and $\frac{Q}{4\pi r_2^2}$
$\frac{-q}{4\pi r_1^2}$ and $\frac{Q + q}{4\pi r_2^2}$
zero and $\frac{Q - q}{4\pi r_2^2}$
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The Correct Option isA
Solution and Explanation
Step 1: Inner surface induction.
The central charge $-q$ pulls an equal opposite charge $+q$ onto the inner surface of radius $r_1$. So inner density is $\frac{q}{4\pi r_1^2}$.
Step 2: Outer surface charge.
The metal stays neutral inside, so $-q$ moves to the outer surface. Adding the shell's own charge $Q$, the outer surface holds $Q-q$.
Step 3: Outer density.
Dividing by the outer area, the outer density is $\frac{Q-q}{4\pi r_2^2}$. \[ \boxed{\frac{q}{4\pi r_1^2}\ \text{and}\ \frac{Q-q}{4\pi r_2^2}} \]
