 Force between hemispheres of a uniformly charged ball

For uniform charge density ${\rho }_e=3Q/(4\pi R^3)$, Gauss's law gives the interior electric field $E={\rho }_{e}r/(3{\epsilon }_0)$. Integrate its vertical force density ${\rho }_e^{2}z/(3{\epsilon }_0)$ over the upper hemisphere: $\int zdV=\pi R^4/4$. Internal pair forces in that hemisphere cancel, so its net force is exactly the force from the lower hemisphere. The result is upward and has magnitude $3Q^2/(64\pi {\epsilon }_0{R}^2)$.