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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

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- Thread starter mitch_1211
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- #1

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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

- #2

Meir Achuz

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- #3

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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?but the F you right is not a vector, unless "constant" is a vector.

- #4

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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

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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

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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

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[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

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[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.

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