Let a total charge $2Q$ be distributed in a sphere of radius $R$, with the charge density given by $\rho(r) = kr$, where $r$ is the distance from the centre. Two charges A and B, of $-Q$ each, are placed on diametrically opposite points, at equal distance, $a$, from the centre. If A and B do not experience any force, then:
Answer: (D) $a = 8^{-1/4}R$
Charge within radius $r$: $q(r) = \int_0^rkr'\,4\pi r'^2dr' = \pi kr^4$. Since $q(R) = 2Q$, $q(a) = 2Q\dfrac{a^4}{R^4}$.
On A, the sphere's attraction (field of $q(a)$ at distance $a$) balances the repulsion from B at distance $2a$:
$$\frac{Q\,q(a)}{4\pi\varepsilon_0a^2} = \frac{Q^2}{4\pi\varepsilon_0(2a)^2} \Rightarrow q(a) = \frac Q4$$
$$2Q\frac{a^4}{R^4} = \frac Q4 \Rightarrow a = 8^{-1/4}R$$
Solution by Sreeraj P, M.Sc Physics