The problem asks to calculate the force on one part due to other. But the electric field I'm using to do this calculation is the result E of complete solid non-conducting sphere. **So why are we considering E of the northern hemisphere to calculate force it experience from the southern hemisphere?**

This is a very good question.

One way of finding the total electrostatic force on the N hemisphere is to calculate the vector force $F_{ij}=k q_i q_j / r_{ij}^2$ on every charge $q_i$ in the N hemisphere due to every charge $q_j$ in the S hemisphere, then take the vector sum of all forces $F_{ij}$ :

$$F_N=\sum_{i \in N} \sum_{j \in S} F_{ij}$$

This double sum or integral is very difficult to calculate, for two reasons :

the general expression for the force $F_{ij}$ between every pair of 2 charges is a complicated function using the Cartesian co-ordinate system. even more so when using spherical co-ordinates ;

each of the 2 charges $q_i, q_j$ in every pair has 3 co-ordinates, so in total there will be a sum ranging over 6 possible co-ordinates.

An alternative method of solution (used in Jorge Daniel's answer) is to sum the **total** electrical force $F_i$ on each charge $q_i$ in the N hemisphere due to **all** of the charges in **both** the S and N hemispheres :

$$F_N=\sum_{i \in N} F_i=\sum_{i \in N} \sum_{j \in N,S} F_{ij}$$

Whereas $F_{ij}$ is a very complicated expression, the total force $F_i$ on each charge is very much simpler because it is known to be radial and depends only on the radial distance of the charge $q_i$ from the centre of the sphere.

As you rightly point out, this means that within the expression for $F_i$ we will include the force $F_{ij}$ on charge $q_i$ due to other charges $q_j$ which are **also in the N hemisphere**.

However, if charge $q_j$ is also within the N hemisphere ($j \in N$), then when we take the sum over all charges $q_i$ in the N hemisphere ($i \in N$), this sum will include not only $F_{ij}$ but also the equal and opposite force $F_{ji}$ acting on $q_j$ due to charge $q_i$. These 2 contributions to the total force on all particles in the N hemisphere will cancel out because $F_{ij}=-F_{ji}$.

On the other hand, if charge $q_j$ is in the S hemisphere ($j \in S$) then the force $F_{ji}$ will **not** be included in the first summation over $i \in N$.

See Find the net force the southern hemisphere of a uniformly charged sphere exerts.

To make the above explanation clearer, suppose we have a system of only 4 charges $q_i$ with $i=1 \to 4$. Between every pair of charges there is an electrical force $F_{ij}$ meaning the force that charge $j$ exerts on charge $i$.

Suppose that charges $i=1 , 2$ constitute one object (which we call "the N hemisphere") and charges $i=3, 4$ constitute a second object ("the S hemisphere"). Then the total force on the N hemisphere due to the S hemisphere is $$F_N=F_{13}+F_{14}+F_{23}+F_{24}$$ The total forces on charges $q_1, q_2$ due to **all** other charges in **both** hemispheres (ie summing over $j \in N, S$) are $$F_1=F_{12}+F_{13}+F_{14}$$ $$F_2=F_{21}+F_{23}+F_{24}$$ If we add the forces on charges $q_1$ and $q_2$ (summing over $i \in N$) we get $$F_N'=F_1+F_2=(F_{12}+F_{21})+F_{13}+F_{14}+F_{23}+F_{24}=F_N$$ because $F_{12}=-F_{21}$.

Thus both methods give the same answer.