An electric potential (also called the electric field potential or the electrostatic potential) is the amount of work needed to move a unit positive charge from a reference point to a specific point inside the field without producing any acceleration. Typically, the reference point is Earth or a point at Infinity, although any point beyond the influence of the electric field charge can be used.

I can understand why Electric field is a vector field but I can't understand why Electric potential is a scalar quantity and why the electric potential/electric potential energy produced by multiple number of point charges equal to the sum of the point charges' individual potentials/potential energies.

And what is the meaning behind electric potential produced by multiple number of point charges. enter image description here

In the image above, if I change the position of Q1 but keep r1 the same (or rotate Q1 around the point), the electric field at that point will change since the vector sum change, but how can the electric potential keeps the same, it makes no sense.


2 Answers 2


1) The existence of the electric potential, and why it is a scalar field is due to the fact that the electric field is a conservative field.

2) If you rotate $Q_1$ around then point the potential field will not remain the same. The only thing that remains the same is the value of the potential (only at that point), but the gradient of the field will change.

For example take a positive point charge located at $(0,0,0)$. The scalar field $\phi( \boldsymbol{r})$ is given by:

$$ \phi_1 ( \boldsymbol{r} ) = \frac{q}{4 \pi \epsilon _0 \sqrt{x^2+y^2+z^2}} $$

If you now locate your positive charge in any other place $(x_0,y_0,z_0)$ then this other potential would be given by:

$$ \phi_2 ( \boldsymbol{r} ) = \frac{q}{4 \pi \epsilon _0 \sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}} $$

Note that if $\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}=r$ then this would be the case you ask, and the potential would have the same value at $(x,y,z)$, but the gradient at this point (and at every other point) would have changed.

  • $\begingroup$ Thanks for your answer, I want to check both of yours as the answer but this site only accept one. Your answer address the question from the formula point of view and Neil_UK's answer give me a clear picture to conceptualize $\endgroup$
    – Snoob
    Commented Oct 31, 2017 at 2:04

The electric field E is not just a vector field, but it's a conservative vector field (at least in the statics case, and after a straightforward modification for magnetic effects, in the dynamics case as well).

Because the field is conservative, a line integral from a reference point to a test point produces a unique scalar, regardless of the path of that integral. This is the work that is done taking a unit charge between those two points.

Conversely, and equivalently, the electric field can be obtained as the gradient of the potential.

The potential finds use because it's so much easier to work with sums of scalar fields than vector fields. In an electrostatics problem, it's much easier to place charges around, sum the potentials, and differentiate to find the electric field. In particular, a mirror symmetric arrangement of opposite charges results in a zero potential along the plane of symmetry. This construction is frequently used to model ground planes, which of course have zero voltage on them.

How does the electric potential staying the same make sense under charge movement? Making sense is just a luxury, few people understand stuff, you just get used to how things behave. The total energy of the test charge is the sum of the component energies due to its interaction with each point charge. That's the way energy adds up. If you move a point charge to keep the same distance from the test charge, that energy component stays the same, and with it, the total potential. It's what happens.

Electric potential's existence, uniqueness and scalar form can be explained by the fact that the E field is conservative. Why the E field is conservative should be the subject of a separate question.


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