Question

A charge q is placed at O in the cavity in a spherical uncharged conductor. Point S is outside the conductor. If the charge is displaced from O towards S, still remaining within the cavity,

• A. electric field at S will increase
• B. electric field at S will decrease
• C. electric field at S will first increase and then decrease
• D. electric field at S will not change

Answer 1

Answer: (D)

By Gauss's law, the electric field at $S$ is $E.4\pi r^2=\frac{q_{en}}{{ϵ}_0}$
$\Rightarrow E=\frac{q_{en}}{4\pi {ϵ}_0{r}^2}$
where $r=$ distance from center to $S.$
When we move charge q towards $S$ within cavity,
the enclosed charge $q_{en}$ will remain same i.e, $q_{eq}=q$.
Thus, the field at $S$ will not change.