JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Electrostatics

Picture the hollow as the full sphere superposed on a smaller sphere of equal but opposite charge density filling the cavity. At $A$, the centre of the cavity and a distance $\frac{R}{2}$ from the centre of the big sphere, the full sphere gives $\frac{\rho (R/2)}{3{\epsilon }_0}=\frac{\rho R}{6{\epsilon }_0}$ while the carved sphere contributes nothing at its own centre, so $\left|{\vec{E}}_A\right|=\frac{\rho R}{6{\epsilon }_0}$. At $B$ on the surface, a distance $R$ from the centre and $\frac{3R}{2}$ from the cavity centre, the full sphere gives $\frac{\rho R}{3{\epsilon }_0}$, and the carved sphere now acts as a point charge of magnitude $\frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3\rho$ at distance $\frac{3R}{2}$, subtracting $\frac{\rho R}{54{\epsilon }_0}$, so $\left|{\vec{E}}_B\right|=\frac{17\rho R}{54{\epsilon }_0}$. The ratio is $\frac{\rho R/6}{17\rho R/54}=\frac{54}{6\times 17}=\frac{9}{17}=\frac{18}{34}$.