# Should the electric potential of a positively charged sphere be negative?

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

$V_f - V_i = \int_i^f \vec{E} \cdot d \vec{r}$. 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

You are missing the minus sign itself!
The electric potential at the surface of the sphere is the work done by you on unit positive charge moving it against the forces of the field to bring that charge from place with zero potential (this place is $R\rightarrow\infty$) to the surface of the sphere.
Work done by moving charge against the forces of the field is like you are holding the charge with a force exactly opposite and infinitesimally larger then that of the field, thus when the charge is moving under your control you are doing positive work (you are loosing energy giving it to the charge), thus the charge is gaining energy (the potential) moving opposite to the forces of the field.
Now the potential should be the energy stored in the charge during its motion and if the charge gains energy then the potential is defined as positive, but if the charge looses energy the potential is defined as negative.
So to conclude you move the charge against the forces of the field and it gains energy (your energy), so the potential must be positive and equal to the magnitude of the work done by the electric forces (opposite to the work done by you), but this work of the electric forces is negative and thus it must be that:
$$V=-\int_{∞}^{R}\textbf{E}\cdot\textbf{dr}=-\int_{∞}^{R}\frac{q}{4πϵ_0r^2} dr=\frac{q}{4πϵ_0R}$$ Here the vectors $\textbf{E}$ and $\textbf{dr}$ are in the same direction because radius vector $\textbf{r}$ points away from the charge just like the field vector $\textbf{E}$. It is important to realize that it is not $\textbf{dr}$ vector that shows the direction you are moving but the limits of integration. Note that the final result is positive since the minus sign in the formula is compensated by the path of integration, which expresses the fact that you are moving the charge opposite to the forces of the field, i.e. form infinity to the source.

One more thing:
Your force is infinitesimally larger then that of the electric field in order to displace the charge but not to induce any inertia on it since we have the static case. Off course to do this displacement (displacement with zero velocity) means to take infinite amount of time ($t=s/v$) but this should not bother you since in the statics every process takes such amount of time.

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Please do use the facility of mathjax here. – user36790 Jan 9 at 12:14
To extend user36790's comment, you can read about MathJax at this link. – Kyle Kanos Jan 9 at 12:34