Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 2 / 6A (Electromagnetism)

In SI units, Gauss's law relates the outward electric flux through a closed surface to its enclosed charge:The outward unit normal specifies the sign of the electric flux.

The uniform charge density is ${\rho }_e=3Q/(4\pi R^3)$. Spherical symmetry makes the electric field radial and constant in magnitude on each concentric sphere. A sphere of radius $r<R$ encloses charge $Q(r/R{)}^3$. Gauss's law therefore gives $4\pi r^2{E}_r=Q{r}^3/({\epsilon }_0{R}^3)$, soThis also defines the zero electric field at the centre by continuity.

Take the upper hemisphere to have $z>0$. Integrate the force $\mathbf{E}dq$ using the total interior electric field from part (a). This computes the net external force on the upper hemisphere, since the forces between pairs of its own charge elements cancel by antisymmetry of the Coulomb force. Horizontal components cancel by symmetry. The vertical component isA horizontal section at height $z$ has area $\pi (R^2-z^2)$, soThus the lower hemisphere repels the upper one vertically upward, withThe result is the same for either sign of $Q$, as expected for the force between hemispheres of a uniformly charged ball.