Electrostatics: Common Mistakes and Fixes (9)

Question

A charge $+Q$ is placed at the centre of a hollow conducting sphere of inner radius $a$ and outer radius $b$. Find the electric field at distances (i) $r < a$, (ii) $a < r < b$, (iii) $r > b$. State what charge sits on each surface.

Solution — Step by Step

Why This Works

A conductor cannot tolerate a non-zero field inside it — free charges would migrate until the field cancels. This redistribution creates exactly the right surface charges to cancel the field of the enclosed charge within the metal. The outer surface, however, “knows” only that the conductor as a whole is neutral and that some charge is sealed inside; it spreads $+Q$ uniformly over itself, and the field outside looks identical to a point charge $Q$ at the centre.

This is a JEE Advanced staple. The cleanest way to see it: $E_{\text{inside metal}} = 0$ forces the induced charges by Gauss’s law alone.

Alternative Method

Use superposition. The field at a point is the sum of contributions from the central $+Q$ plus the induced surface charges. By symmetry, induced charges arrange uniformly. Solve for them by demanding $E = 0$ inside the metal — same answer, more algebra.

Common Mistake

If we now ground the outer surface, the $+Q$ on the outside drains away to earth and the field outside becomes zero. Don’t forget grounding kills only what’s electrically connected to the ground; the inner $-Q$ stays put.