A point charge Q is placed inside a cavity within a solid isolated conducting sphere. Consider points A, B and C as shown in the figure, where the magnitudes of the electric fields are $E_A$, $E_B$ and $E_C$, respectively. The points B and C are at the same distance from the center of the solid sphere. The correct option is :
Answer: (B) $E_A \neq 0,\ E_B = E_C$
Point A is inside the cavity, near the charge Q. The field there comes from Q and the induced charge on the cavity wall, and it is not zero: $E_A \neq 0$.
Charge $-Q$ is induced on the cavity wall and $+Q$ on the outer surface. The conductor shields the outside from the cavity, so the charge $+Q$ on the outer surface spreads uniformly, whatever the position of Q inside.
Outside, the field is therefore that of a point charge $Q$ at the centre of the sphere, $E = \dfrac{Q}{4\pi\epsilon_0 r^2}$. B and C are at equal distances from the centre, so $E_B = E_C$.
Solution by Sreeraj P, M.Sc Physics