= Force between hemispheres of a uniformly charged ball
{title2=$F=3Q^2/(64\pi\varepsilon_0R^2)$}
For uniform <charge density> $\rho_e=3Q/(4\pi R^3)$, <Gauss's law> gives the interior <electric field> $E=\rho_e r/(3\varepsilon_0)$. Integrate its vertical force density $\rho_e^2z/(3\varepsilon_0)$ over the upper hemisphere: $\int z\,dV=\pi R^4/4$. 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 $3Q^2/(64\pi\varepsilon_0R^2)$.
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