Determining if a field is electrostatic or not

  • Thread starter mitch_1211
  • Start date
  • #1
99
0

Main Question or Discussion Point

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
 

Answers and Replies

  • #2
clem
Science Advisor
1,308
15
An electrostatic field has curl F=0, but the F you right is not a vector, unless "constant" is a vector.
 
  • #3
99
0
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?
 
  • #4
261
14
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.
 
  • #5
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
14,409
5,985
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

[tex]\rho=\vec{\nabla} \cdot \vec{E},[/tex]

where I'm using Heaviside-Lorentz (rationalized Gauß) units.
 
  • #6
610
5
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.
 
  • #7
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
14,409
5,985
The radius vector, [itex]\vec{e}_r=\vec{r}/r[/itex] (with [itex]r=|\vec{r}|[/itex]) is not a constant. Of course, the Coulomb field is potential field,

[tex]\vec{E}=\frac{q}{4 \pi r^2} \vec{e}_r=-\vec{\nabla} \frac{q}{4 \pi r},[/tex]

and thus an electrostatic field. The charge distribution is

[tex]\vec{\nabla} \cdot \vec{E}=-\Delta \frac{q}{4 \pi r}=q \delta^{(3)}(\vec{r}).[/tex]
 
  • #8
64
0
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:
[tex]\vec{\nabla}\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}[/tex]

Where the right side becomes zero (time-independent). Thus, in electrostatic, we would need the electric field E to be conservative also.
 
  • #9
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
14,409
5,985
Sure, any conservative time-independent field can be an electrostatic field. I've only shown this for the Coulomb field.
 

Related Threads on Determining if a field is electrostatic or not

Replies
10
Views
2K
  • Last Post
Replies
12
Views
5K
  • Last Post
Replies
1
Views
901
Replies
8
Views
2K
Replies
6
Views
2K
Replies
4
Views
377
Replies
1
Views
2K
Replies
2
Views
823
Replies
11
Views
5K
Top