# Determining if a field is electrostatic or not

If i have an arbitrary vector field, say F = constant / (x^2 + y^2 +z^2)
I want to determine if it is electrostatic or not. For this I am thinking I should first determine if it is a conservative field, but not 100% sure, any guidance would be appreciated

:)

mitch

Meir Achuz
Homework Helper
Gold Member
An electrostatic field has curl F=0, but the F you right is not a vector, unless "constant" is a vector.

but the F you right is not a vector, unless "constant" is a vector.
Yea i think its a trick question, the F is bold in the example I'm looking at, but then there are no vector directions on the right side of the = so in that case, you can't take the curl can you?

I think the OP refers to the modulus of a vector.Notice x^2+y^2+z^2 equals r^2.
My guess is that F is the modulus of a gravitational or electrical field. In such a case we can write the vector field using the direction cosines and, upon taking the curl of that vector, we find it's conservative.

vanhees71
Gold Member
If you have only the modulus of a field, you cannot determine whether it's electrostatic or not.

Of course, any time-independent vector field, which is curl free, can be an electrostatic field. The source (charge distribution) is given by

$$\rho=\vec{\nabla} \cdot \vec{E},$$

where I'm using Heaviside-Lorentz (rationalized Gauß) units.

Unless there's more hidden in the problem than stated, F = constant / (x2 + y2 +z2) is just F = constant / r2. You have not stated the direction of this vector, but I assume it is radial:
F = Fr = constant / r2
This is just the static field produced by a point source if the field is central and inverse-square-law in nature. For instance this would be the electrostatic field due to a point charge, or the gravitational field due to a point mass.

vanhees71
Gold Member
The radius vector, $\vec{e}_r=\vec{r}/r$ (with $r=|\vec{r}|$) is not a constant. Of course, the Coulomb field is potential field,

$$\vec{E}=\frac{q}{4 \pi r^2} \vec{e}_r=-\vec{\nabla} \frac{q}{4 \pi r},$$

and thus an electrostatic field. The charge distribution is

$$\vec{\nabla} \cdot \vec{E}=-\Delta \frac{q}{4 \pi r}=q \delta^{(3)}(\vec{r}).$$

Actually, I think the question is a bit more simple. It is asked whether or not a field is electrostatic. In other terms, if it is time-independent. By looking at the expression, there is no time variable in there. So there we have the "static" part. Now is it electrostatic? Well, from Faraday:
$$\vec{\nabla}\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}$$

Where the right side becomes zero (time-independent). Thus, in electrostatic, we would need the electric field E to be conservative also.

vanhees71