I understand that this question has been asked multiple times before but my question is regarding something specific.
I came across the following solution on the web:
Since $\vec{E}$ is the field at the said point due to the entire sphere and not the southern hemisphere alone, why is this solution correct? I feel that the integral should be much more complicated than this.
It turns out that both points of view are equivalent. The force exerted by the total sphere on the north hemisphere is the same as the force exerted by the southern hemisphere on the northern hemisphere.
Since in the former case, the expression of the field is simpler, it is a better way for calculating the force. Both methods agree because the self force of a localized stationary distribution of charge is null. In your case, the force of the northern hemisphere on itself is zero.
This is general fact can be proven as follows. Consider a stationary charge density $\rho$ generating a field $E$ according to the laws of electrostatic (setting $\epsilon_0=1$):
$$
\nabla\cdot E=\rho\\
\nabla\times E=0
$$
The self force is:
$$
F=\int \rho E d^3x
$$
The idea is to express this bulk integral into a boundary one (which will vanish due to the localisation of the charges). This naturally leads to Maxwell’s stress tensor. Mathematically, you can use the analogue of Stokes’ formula for vector integrands. If you’re not familiar with that, you can project everything along a direction $u$:
$$
\begin{align}
u\cdot F &= \int \rho E\cdot u d^3x \\
&= \int (\nabla \cdot E )(E\cdot u )d^3x \\
&=\oint (E\cdot u )E\cdot d^2x -\int (E \cdot \nabla) (E\cdot u)d^3x \\
(E \cdot \nabla) (E\cdot u) &= (E\times u)(\nabla \times E)+E\cdot ((u\cdot \nabla) E) \\
&= \nabla\cdot \frac{u}{2}E^2 \\
u\cdot F &= \oint \left[(E\cdot u )E- \frac{u}{2}E^2\right]\cdot d^2x
\end{align}
$$
For large distance $r$, $d^2x=O(r^2)$ and for a localized charge, the field falls off at most like the monopole contribution $E=O(r^{-2})$ so the integral is 0 generically:
$$
u\cdot F=0
$$
Actually, this suffices for you since you are interested only in the z component. But the result is more general, since the projection under any direction is 0, then:
$$
F=0
$$
Btw, as advertised, the integrand is Maxwell’s stress tensor: $$ T = E\otimes E-\frac{1}{2}E^2 1\\ u\cdot F = \oint u\cdot T\cdot d^2x \\ F = \oint T\cdot d^2x $$ This gives an alternative method to calculate your force, using the Maxwell stress tensor and integrating only at the boundary of the hemisphere (2 integrals instead of 3, so it is an improvement). This is a classic exercise, you can find it for example in Griffiths’ Introduction to Electrodynamics.
Hope this helps
