Electric Potential energy Is the potential energy due to negatively charged sphere positive or negative?
I thought it would be negative as if we consider a test charge at infinity, in order to move it to the sphere no external work is needed
 A: It depends. The potential energy $U(\mathbf{x})$ of a charge* in an electric field at a position $\mathbf{x}$ is defined as the work needed to move that charge from infinity to $\mathbf{x}$. This work depends on the charge of the sphere, but it also depends on the test charge. That includes whether the two charges have the same or opposite polarity.
If the polarities are the same, i.e. they are both positive or negative, they will repel each other due to the Coulomb force. Thus we need to do positive work to the charge to move it from infinity to a point near the sphere.
The opposite is the case if the polarities are different, then the potential energy is negative.
It is often useful to define a potential $V$ as the potential energy per unit charge. That way, we can talk about the potential due to a charge, e.g. the charge on a sphere, independent of whatever other charges are present.
* This isn't entirely accurate. The potential energy is attributed to the whole system, and it is not a property of each charge separately. In fact, if we let the second charge be stationary and moved the sphere from infinity, the potential energy would be the same.
A: Well, a sphere, if you are far enough, can be well approximated as a point charge. Eviidently the value will change but the sign should be the same. Now think, what's the Potential Energy due to a negative charge?
According to the usual standards, (with reference in $\infty$), it is
$E_p(\infty)-E_p(r)= -\int_{\infty}^{r} \vec{F}\cdot d\vec{r} $
so the first term vanishes, and the vectors are antiparallel, so you have
$+E_p(r)= +\int_{\infty}^{r} \dfrac{\ominus KQq}{r^2}\cdot d{r} $
The integral is as easy as 
$ \ominus KQq \cdot \frac{1}{r}$
So, if the sphere is negatively charged, the first minus sign cancels and so the $E_p$ has got the same sign as the test-charge $q$. 
A: Electric potential at a certain point in space is defined as the amount of work done to move a unit positive charge from infinity to that point. Now, when you move the charge you need to make sure that you are able to move it at a near zero relative velocity, which means, if you have a positive charge at infinity which you want to move to a certain predefined spot, the positive charge will want to get attracted to your negetively charged sphere. But also, since you want a zero relative velocity on your test charge, you would have to apply a force along the line joining the sphere and the charge, but towards the direction of the charge. But, when you try to bring it from infinity to a non-infinite point $\vec{r}$, then the direction of motion of the charge is opposite to the direction of motion of the force you need to apply to balance forces on the charge. Hence the work done is negative, thus so is the potential. 
