# Potential sign question

Let's consider a positive test charge $$q$$ and a positive source charge $$q_o$$

If we take direction of $$\vec{r}$$ from $$q_o$$ to $$q$$, then potential energy of $$q_o$$ due to $$q$$ will be:

$$P=k\dfrac{q_oq}{r}$$

However if we reverse the direction of $$\vec{r}$$ from $$q$$ to $$q_o$$, will there be any change in the sign of potential energy? If yes/no , why?

Edit @Utkarshfutous and @ Demosthene:

$$P=-\int_{\infty}^{r}\vec{F}.\vec{dr}=-kqq_o\int_{\infty}^{r}\dfrac{1}{r^2}dr\cos\theta$$

Now,

if $$\cos\theta$$ is negative when we take direction of $$\vec{r}$$ from $$q_o$$ to $$q$$

then $$\cos\theta$$ will be positive when we take direction of $$\vec{r}$$ from $$q$$ to $$q_o$$

Thus the sign of potential changes. Is there something I am missing?

• I am so confused as to why you'd take anything other than a straight line path for calculating potential – Buraian Nov 6 '20 at 4:40

As you write it, the potential is a scalar and is thus independent of orientation. Notice that it's a function of $r=\lvert\vec{r}\rvert$ only.
If you "reverse the direction of $\vec{r}$", then you're simply looking at the potential of $q$ due to $q_0$.