Mrthdrb

For finding electric potential energy of a uniformly charged sphere, we can assemble the sphere by brining charges from infinity to that point. So to make a uniformly charged sphere of radius $R$ and total charge $Q$, at some instant, charge will be assembled up to a certain radius $x$.
In order to find potential of this sphere at the surface, why is my approach giving different answers?

Approach 1:

$$rho = frac3Q4 pi R^3$$

$$q = frac43 pi x^3 rho = Q fracx^3R^3$$
Potential at the surface would be $$V = fracq4 pi epsilon_0 x = fracQ x^24 pi epsilon_0 R^3$$

Approach 2:
$$rho = frac3Q4 pi R^3$$
$$q = frac43 pi x^3 rho = Q fracx^3R^3$$
$$E = fracQ x4 pi epsilon_0 R^3$$ (From Gauss' Law)

Potential at the surface would be $$V = -intvecE cdot vecdx = -fracQ4 pi epsilon_0 R^3 int_0^xxdx = -fracQ x^28 pi epsilon_0 R^3$$

Why is the answer different in both the cases?

Two cases described are completely different. In first case you find the true potential of the sphere by taking the charge from infinity to the surface of the sphere. In another case you take the charge from the middle of the sphere or the centre of the sphere to the surface of the sphere which is not the potential of the sphere surface. The potential of the sphere surface can be described as the work needed to push a positive charge from infinity to a to the surface or the energy stored to push the charge from the the surface towards the infinity so you can see in your second case you are not calculating the potential of the surface of the sphere. SHORT NOTE:- You can find the potential at any point by finding the difference of potential at that point and any other point whose the potential is zero now at the centre of the the sphere you don't have the potential as 0. See this:http://physics.bu.edu/~duffy/semester2/d06_potential_spheres.html

Approach 2 is wrong. You didn't take into account the corresponding limits for potential. Potential at centre of sphere is not zero!! The expression is V(x)-V(0) instead of V(x).... Find potential at surface by integrating for electric field outside sphere from X to infinity V(infinity)=0. So Then if you wish you can find V(x) by integrating from x=x to any general x=y(

The first thing to note is that the electric potential at a point is entirely different to the electric potential energy of an assembly of charges.

I have assumed that you are finding the potential at a point and you have used two definitions of the zero of electric potential, one at infinity and the other at the centre of the charge distribution.

Using Gauss's law the graph of electric field strength $E(x)$ against distance from the centre of the charge distribution $x$ looks something like this.

The area under the graph $int E,dx$ is related to the change in potential.

In essence what you have done is found that areas $A$ and $B$ are not the same.

PS You may well have met a similar graph with $E(r)$ negative and labelled $g(r)$ when discussing the gravitational field due to the earth and the gravitational field strength inside the Earth?