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Because the indefinite integral of the electric field results in a negative value? (As the function is proportional to $r^{-2}$?

I've got to be missing something... Help please!! Thanks!

Also, I took the integral from infinity to the radius of the charged sphere and got

$$V = \int_\infty^R{\frac{q*dr}{4*\pi*\epsilon_0*r^2}} = -\frac{q}{4*\pi*\epsilon_0*R}$$

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What does "dr" mean? A small increase in r. Which way are you traveling? –  DarenW Mar 15 '13 at 4:15

1 Answer 1

up vote 1 down vote accepted

$V_f - V_i = \int_i^f \vec{E} \cdot d \vec{r}$. The dot product has a sign depending on the relative orientation between the electric field $\vec{E}$ and infinitesimal displacement $d \vec{r}$. Also note that as you move radially inwards from infinity to some point, the displacement $d \vec{r} = \vec{r_f} - \vec{r_i}$ points radially inward, whereas the electric field of a positive point charge points radially outward.

Like @DarenW says in the comment, make sure you've accounted for that relative sign. That should take care of your sign mistake.

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It is, of course, vitally important to get the sign of $d\vec{r}$ right, but do not neglect that you can make a gauge transformation to get any particular point at any potential you care to name. –  dmckee Mar 15 '13 at 4:43
    
@dmckee Thanks for the comment. I've made my answer more explicit. –  Siva Mar 15 '13 at 5:14

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