# What is the connection between electric field and electric potential

If a charged body is brought from infinity to a point in space where the electric field is zero, what will be it's potential then? There is no electric field in the region other than that of the field of the charged body itself.

• ...What? Please explain a little clearer – user95137 Apr 5 '17 at 1:18
• The potential of a charged body due to itself is not defined. – garyp Apr 5 '17 at 1:19
• Well, no, it's not. Think about it like this, if you were to drop a point charge into the centre of a radial electric field, where would it move? – user95137 Apr 5 '17 at 1:24
• It shouldn't move anywhere – user150960 Apr 5 '17 at 1:26
• Yup, it wouldn't. – user95137 Apr 5 '17 at 1:31

The potential can be anything.

Think of the potential as the height of the land in a mountainous region, and the electric field as the slope of the land. Zero potential might be the level of the plain far away from the mountains, or sea level.

There is no necessary relation between the height and slope at any particular point. You can have steep slopes below sea level, and plateaus at the top of the highest mountains. If the land is flat (zero electric field) it could be at any height (the potential could have any value).

We can only measure a potential difference, not the absolute value of the potential. That's because the potential has a gauge symmetry. This has recently been discussed in Potential is with respect to what, which is nearly but quite a duplicate of this question.

The potential difference is just the work required to bring a unit charge between the two locations. So in this case take an infinitesimal test charge $q$, move it from infinity to a point in space where the electric field is zero and measure/calculate the work done. Divide this work by the charge $q$ to get the work per unit charge and that gives you the potential difference between infinity and its current position.

In the absence of a time-dependent magnetic field. The electric field $E$ is irrotational and can be represented by the negative gradient of a potential $\Phi$.

$$E = -\nabla \Phi$$

It is clear from this equation that the new potential $\Phi'=\Phi+f(t)$ can be used as a potential for the electric field, as $\nabla f(t)=0$. Hence, the electric potential for a given electric field is not unique.