Combining charged caps

Consider the case when whole of the sphere has charge density $\rho$ . We will find the force of interaction in this case and then we will replace $\rho^2$ by $\rho_{1}\rho_{2}$ to get the final answer(because the force between any two charge components would change in this way).

Let the electric fields due to th caps be $E_{1}$ and $E_{2}$ respectively.

Clearly , the force of interaction is

$F = \displaystyle \int_{q \in \text{cap 1}} \vec{E_{2}} dq$

Now as the sum of internal forces is $0$ , hence,

$\displaystyle \int_{q \in \text{cap 1}} \vec{E_{1}} dq = 0$

Hence,

$F = \displaystyle \int_{q \in \text{cap 1}} (\vec{E_{1}} +\vec{E_{2}}) dq$

We know that $\vec{E_{1}} + \vec{E_{2}} = \dfrac{\rho \vec{r}}{3 \epsilon_{0}}$

Now,

Consider a small ring of radius $y$ at height x above the center subtending an angle $\theta$ at the center. Clearly, $\cos \theta = \dfrac{x}{r}$

Clearly, $dF = \dfrac{\rho r}{3 \epsilon_{0}} \times \rho \times 2 \pi y \cos \theta dy dx$

= $\dfrac{2\rho^2 \pi}{3\epsilon_{0}} xy dx dy$

Clearly, x varies from $h$ to $R$ , and $y$ varies from $0$ to $\sqrt{R^2-x^2}$

Hence,

$F = \displaystyle \dfrac{2\rho^2 \pi}{3\epsilon_{0}} \int_{h}^{R}x \bigg(\int_{0}^{\sqrt{R^2-x^2}} y dy\bigg) dx$

= $\dfrac{2\rho^2 \pi}{3\epsilon_{0}} \displaystyle \int_{h}^{R} x(R^2-x^2) dx$

= $\dfrac{\rho^2 \pi(R^2-h^2)^2}{12 \epsilon_{0}}$

Replacing $\rho^2$ by $\rho_{1} \rho_{2}$ , the force of interaction is :

$\boxed{\dfrac{\pi \rho_{1} \rho_{2} (R^2-h^2)^2}{12 \epsilon_{0}}}$