A metal sphere of radius $R$ carries a total charge $Q$. What is the force of repulsion between the northern hemisphere and southern hemisphere?
In this problem you consider the force per unit area on the surface of the northern hemisphere and then add it up basically: You get the average electric field as$$ \frac{1}{2}(E_{\text{outside}} + E_{\text{inside}}) ~=~ {E_{\text{average}}} ~=~ \frac{1}{2}\frac{1}{4 \pi \epsilon_0}\frac{Q}{R^2}\hat{r} \,,$$because of the symmetry the horizontal components cancel for the force per unit area you get $f_{z} = {\sigma \left(E_{\text{average}}\right)}_{z}$ where $\sigma = \frac{Q}{4 \pi R^2}$. Then the total force on the northern hemisphere is$$ F_{z} ~=~ \int f_{z} \, \mathrm{d}a ~=~ \int_{0}^{2 \pi} \int_{0}^{\frac{\pi}{2}} \left(\frac{Q}{4 \pi R^2}\right)\frac{1}{2}\left(\frac{Q}{4 \pi \epsilon_0 R^2}\right) \cos{\theta} \, R^2 \sin{\theta} \, \mathrm{d} \theta \, \mathrm{d} \phi ~=~ \frac{Q^2}{32 \pi R^2 \epsilon_0}. $$
Question: Are the forces of repulsion on the northern hemisphere a result of only the charges from the southern hemisphere while there is symmetric cancellation from charges in the northern hemisphere, or is the force on each unit area in the northern hemisphere due to the whole sphere?
The way the question is worded seems to suggest that the answer is the first case, i.e. that the force on the northern hemisphere is due completely to the southern hemisphere.
