# Intensity of electric field and electric potential [duplicate]

This question already has an answer here:

I know that the vector of the intensity of the electric field has the opposite direction of the gradient of the electric potential. But what baffles me is that the magnitude of the intensity and the electric potential both get smaller with distance. Is that because we consider the charge creating the field to be positive? If so, if it's negative charge creating the field the potential should get bigger with distance and the intensity gets smaller?

P.S. sorry if my question doesn't use the right terminology. also for not being styled right.

## marked as duplicate by sammy gerbil, glS, Michael Seifert, Bill N, MitchellJan 27 '18 at 5:02

• I hope I understand you but am not sure. Anyhow, think of the function $f(x)=1/x$, when $x=1$ for $f$ to be half as much you need to move only one unit $\Delta x =1$ for a drop by a factor of two, but if $x=100$ it takes $\Delta x = 100$ distance to achieve the same fractional drop. And so is with the field intensity that is (a directional) rate of change of the potential. – hyportnex Jan 24 '18 at 15:17
If you are beginning with electrostatics, you must have been introduced to Coulomb's Law for force between two point charges $q_1$ and $q_2$: $$F \propto \frac{q_1 q_2}{r^2}$$ where $r$ is the distance between them, and the force acts such that unlike charges repel and like charges attract.