Determining if a field is electrostatic or not

[mitch_1211]

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]

An electrostatic field has curl F=0, but the F you right is not a vector, unless "constant" is a vector.

 

[mitch_1211]

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?

 

[Gordianus]

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]

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.

 

[chrisbaird]

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 = F_r = 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]

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}).

 

[Mbert]

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]

Sure, any conservative time-independent field can be an electrostatic field. I've only shown this for the Coulomb field.