Say I have a 5 charges
and wanted to place them in vertices of a regular hexagon, and one asks me to find electric potential and electric field at the center, then
- Why we don't use electrostatic potential energy
The electrostatic potential energy of two charges is:
$$
U = \frac{q_1 q_2}{4\pi \epsilon_0 |\vec r_1 - \vec r_2|}\;,
$$
For more than two charges (e.g., $N$ charges), we sum over all unordered pairs of charges to calculate the electrostatic potential energy:
$$
U = \frac{1}{4\pi\epsilon_0}\sum_{i,j<i}^{N,N}\frac{q_iq_j}{|\vec r_i - \vec r_j|}
=\frac{1}{2}\frac{1}{4\pi\epsilon_0}\sum_{i=1}^N\sum_{j\neq i}^N\frac{q_i q_j}{|\vec r_i - \vec r_j|} \tag{1}
$$
concept to find electric potential just by dividing the source charge $Q$, Or in other words, bringing each of these 5 charges on the vertices of the hexagon one by one,
You could do it this way. But there would be components to this calculation that would not be relevant to the electrostatic potential calculation. This is because you may be assuming that all five charges (other than the "test" charge) are held fixed. Indeed, when this is the case, when you bring in that last "test" charge you only need to count up the contributions to the energy from the five pairs (the new sixth test charge paired with the five charges you already brought in and are holding fixed in their positions). This contribution will be proportional to the electrostatic potential of interest.
why is that electric potential is just $\frac{5q}{4\pi\epsilon_{0}r}$ at the center of the hexagon.
This is true by definition of the electrostatic potential (assuming the distance from the vertices to the middle is denoted by "$r$"). (Later in this answer, I will call this distance $L$ because I want to use $r$ for something else.)
i.e. if we bring one charge and then another one too, wont electric potential should get affected like electrostatic potential energy.
Sure. Yes, you could bring each charge in one at a time, and calculate the electrostatic potential energy of the whole system, but that would be a different calculation.
My guess is, If electric potential is not evaluated like electrostatic potential energy, then it means bringing charges from some point and just keep placing them one by one in a place without getting affected by the other charges, which is so as electric potential follows superposition theorem. Does that mean electrostatic potential energy do not follows superposition theorem.
The electrostatic potential follows a superposition rule (often called a linear superposition rule). The electrostatic potential energy due to assembling a system of charges is not linear (it is quadratic) in the charges.
Assuming the other five charges are held fixed then the potential energy due to bringing in one more "test" charge also follows linear superposition (because it is proportional to the potential). E.g., if you already brought in the five charges and hold them fixed, then when you bring in a sixth "test charge" (of charge $Q$ and location $\vec r$) then additional energy contribution is:
$$
U_{6th} = \sum_{j=1}^5\frac{q_j Q}{4\pi\epsilon_0 |\vec r - \vec r_j|}
=Q\phi_{5}(\vec r)\;,
$$
where $\phi_{5}$ is the electrostatic potential due to the other five charges, which are held fixed, and must be held fixed (or the test charge must be considered infinitesimal), otherwise the other charges would move and the problem statement would not hold.
Another way to think of this is that the electric potential at the location of a charge (e.g., $q_k$ at $\vec r_k$) is closely related to the derivative of the electric potential energy with respect to that charge. From Eq. (1) we have:
$$
\frac{\partial U}{\partial q_k} = \frac{1}{4\pi\epsilon_0}\sum_{j\neq k}\frac{q_j}{|\vec r_k - \vec r_j|} = \phi_k\;.
$$
The electric potential proper is the generalization to a field (evaluated at an arbitrary vector $\vec r$) of the above quantity:
$$
\phi(\vec r) = \frac{1}{4\pi\epsilon_0}\sum_{j}\frac{q_j}{|\vec r - \vec r_j|}
$$
- Now if we have electric potential, then why cant one use gradient of electric potential to find the electric field, rather the electric field simply just $\frac{q}{4\pi\epsilon_{0}r}$.
The electric field would not be $\frac{q}{4\pi\epsilon_{0}r}$. That expression has the wrong units and also it is not a vector.
The electric field is a vector and can be determined by:
$$
\vec E(\vec r) = -\vec \nabla\phi(\vec r)\;,
$$
where $\phi$ is the electrostatic potential.
In your case you will have:
$$
\phi(\vec r) =\sum_{i=1}^5\frac{q_i}{4\pi\epsilon_0|\vec r - \vec r_i|}\;,\tag{1}
$$
where the $\vec r_i$ are the location of the five vertices. If you assume all of the charges have equal magnitude $q_i = q$ and you evaluate this potential at $\vec r = 0$ you find
$$
\phi(0) = \frac{5q}{4\pi\epsilon_0 L}\;,\tag{2}
$$
where $L$ is the distance from the vertex to the center.
Note that Eq. (2) is not enough to determine the field, since the $\vec r$ dependence is gone. But from Eq. (1) we see that the field is:
$$
\vec E(\vec r) = -\vec \nabla \phi(\vec r) = \sum_{i=1}^5\frac{q_i(\vec r - \vec r_i)}{4\pi\epsilon_0|r - r_i|^3}
$$
