Force of repulsion between 2 hemisphere of different charge density 
Consider two hemispheres of radius r which are given a uniform surface charge density of $\sigma_1$ and $\sigma_2$. Find the force of repulsion when these two hemispheres are grouped together to complete a sphere .  

In this problem I could not get started with how I to find the repulsion because of the different charge densities. 
If they had same charge density then the question would have been easy as it would just have been the electrostatic pressure into area  $F = \frac{\sigma^2}{2\epsilon_0} \pi r^2$.
How can I compute this problem as finding the pressure on a single element due to this non uniform distribution is very difficult. Help please..
 A: When you calculate the total force acting on one hempisphere, you can neglect the contributions from self-interatcion, as from the 3-rd Newton's Law, all such contributions will cancel out. You only need to consider only the contributions from the interaction between elements of one hemisphere with elements of the second hemisphere, asnd all such contributions are to be proportional to both $\sigma_1$ and $\sigma_2$, Because of this the total force also has to be proportional to both $\sigma_1$ and $\sigma_2$ that is $$ F = \alpha \sigma_1 \sigma_2$$
Knowing the result for $\sigma_1=\sigma_2$ allows you to find the value of constant $\alpha$:
$$ F = \frac{\pi r^2}{2\epsilon_0}\sigma_1\sigma_2$$
A: Alright then:
The total charge $q_1$ on $hs_1$ is the surface of $hs_1$ times $\sigma_1$:
$$q_1=2\pi r^2\sigma_1$$
Likewise, the total charge $q_2$ on $hs_2$ is:
$$q_2=2\pi r^2\sigma_2$$
I use the procedure to place the two charges $q_1$ and $q_2$ in the two centers of mass of the two hemispheres to find the force between the two hemispheres.
The center of mass of each hemisphere ($CM_{hs}$) with radius $r$ is (see here):
$$CM_{hs}=\frac{r}{2},$$
which tells us that the distance between the two charges in the two $CM_{hs}$'s is equal to $r$. Of course, we use

to find the force. So the force between the hemispheres is:
$$F=\frac {1}{4\pi\epsilon_0}4{\pi}^2 r^4 r^{-2} \sigma_1 \sigma_2=\frac{\pi}{\epsilon_0} r^2 \sigma_1 \sigma_2(N),$$
which differs a factor $2$ with the answer given in the question (if $\sigma_1=\sigma_2=\sigma$), but I have the suspicion that this has to do with this procedure of finding the answer to the question asked. 
