A thin spherical shell radius of r has a charge Q uniformly distributed on it. At the centre of the shell, a
negative point charge –q is placed. If the shell is cut into two identical hemispheres, still equilibrium is maintained. Then find the relation between Q and q?

Here the outward electric pressure at every point on the shell due to its own charge is
${\mathrm{P}}_1=\frac{{\mathrm{\sigma }}^2}{2{\mathrm{\epsilon }}_0}=\frac{1}{2{\mathrm{\epsilon }}_0}{\left(\frac{\mathrm{Q}}{4{\mathrm{\pi r}}^2}\right)}^2\Rightarrow {\mathrm{P}}_1=\frac{{\mathrm{Q}}^2}{32{\mathrm{\pi }}^2{\mathrm{\epsilon }}_0{\mathrm{r}}^4}$
Due to –q, the electric field on the surface of the shell is $\mathrm{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}$
This electric field pulls every point of the shell in inward direction. The inward pressure on the surface of the shell due to the negative charge is ${\mathrm{P}}_2=\mathrm{\sigma E}$
$=\left(\frac{\mathrm{Q}}{4{\mathrm{\pi r}}^2}\right)\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}\right)=\frac{\mathrm{Qq}}{16{\mathrm{\pi }}^2{\mathrm{\epsilon }}_0{\mathrm{r}}^4}$
For equilibrium of the hemispherical shells ${\mathrm{P}}_2\ge {\mathrm{P}}_1\text{ or }\frac{\mathrm{Qq}}{16{\mathrm{\pi }}^2{\mathrm{\epsilon }}_0{\mathrm{r}}^4}\ge \frac{{\mathrm{Q}}^2}{32{\mathrm{\pi }}^2{\mathrm{\epsilon }}_0{\mathrm{r}}^4};\mathrm{q}\ge \frac{\mathrm{Q}}{2}$