Infinity direction of potential Potential energy at a point is defined as: "Work done to move a charge from infinity to that point".
Now infinity is usually taken in the direction of $r$. Can we take the direction of infinity other than that of $r$. Why/Why not?

 A: You can take it in any direction you please.  You will find that the math works out to exactly the same value.  In fact, you can even do funny curved paths or loops!  This is the product of two features of the equations which are used in calculating energy.  The first is path independence -- the math behind the fields you are moving the particle through are such that the energy ends up being the same, no matter what path you took.  The second is that the effect of any forces in the system approaches zero as your particle approaches infinity.
A: Potential Energy is defined as $$U(r)=-\int_{ref}^{r} \vec{F}.\vec{dr}$$
In your case, let's consider the force on some test charge $q_o$.
$$\vec{F}=\frac{kq_oq}{r^2} \hat{r}$$ (where $\hat{r}$ is a unit vector pointing in any direction)
For simplicity, let's go towards infinity in a straight line. (The path doesn't matter, because the integral depends only on endpoints and not on the path itself. This happens to be the easiest case for evaluation.)
In this case, $\vec{dr}$ can be split into components perpendicular and parallel to $\hat{r}$.The relevant component here is $dr\hat{r}$ because the dot product of $\hat{r}$ with the perpendicular component is zero, and hence $U(r)$ is:
$$ U(r)=-\int_{\infty}^{r} \frac{kq_oq}{r^2} dr$$
$$U(r)=\frac{kq_oq}{r}\bigg\rvert^{r}_{\infty}$$
We did not take the direction of movement into account here, so $U(r)$ here is the same for "any infinity".
