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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?

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  • $\begingroup$ I am so confused as to why you'd take anything other than a straight line path for calculating potential $\endgroup$ – Buraian Nov 6 '20 at 4:40
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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 $q_0$.

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  • $\begingroup$ He is talking about potential "energy " though the answer is anyway correct but for energy you last statement is of no meaning $\endgroup$ – Utkarsh futous Apr 5 '17 at 14:11
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    $\begingroup$ @Utkarshfutous indeed :) $\endgroup$ – Demosthene Apr 5 '17 at 14:12
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No simply because potential energy is a scalar quantity

And that's because any form of energy is a scalar quantity (and scalar quantity have no direction)

Remember it depends on electric field and distance (both vectors) but their dot product leads again to a scalar quantity

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