Physics - Elecrostatics Question with Solution | TestHub

Field due to small sphere
$=\frac{\text{K}{\rho }_1\frac{4}{3}\pi {\text{R}}^3}{{\left(2\text{R}\right)}^2}$
Field due to bigger sphere
$=\frac{\text{K}{\rho }_2\frac{4}{3}\mathrm{\pi }{\left(2\text{R}\right)}^3\times \text{R}}{{\left(2\text{R}\right)}^3}$
The net electric field at point $P$ will be

${\overrightarrow{E}}_{\mathrm{net}}=0$

Equating above two, we get $\frac{{\rho }_1}{{\rho }_2}=4$
[Charges cannot be of an opposite sign otherwise field will not zero]

Similarly, the net electric field at point $Q$ becomes zero. Then $\frac{{\rho }_1}{{\rho }_2}=-\frac{32}{25}$ here both spheres should have opposite polarity.