# Properties of this electric field

1. Jul 23, 2013

### Gavroy

I have the following vector field:$E=E_0(-sin(\phi),cos(\phi),0)^T$ and [/itex]E_0[/itex] is some constant. Does anybody here have an idea what this electric field does to a metallic sphere that is in it? Thereby I mean, does it create a dipole moment or does it create a current and so on?

Last edited: Jul 23, 2013
2. Jul 23, 2013

### Staff: Mentor

What is ϕ? Is it a coordinate (in which way?), is it a parameter, does it depend on anything?

In general, yes.
Not in equilibrium.

3. Jul 23, 2013

### Gavroy

it is the 0 to 2pi angle of the spherical coordinates

4. Jul 23, 2013

### Staff: Mentor

So (1,0,0) has ϕ=0?
In other words, E is rotating around the z-axis?
This cannot be the electric field of a static setup.

Where is the metallic sphere?

5. Jul 23, 2013

### Gavroy

The origin of the sphere is at (0,0,0). Yeah, probably it is a rotation field around the z-axis. I do not understand what you mean by (1,0,0) ?

Do you know how to calculate the dipole moment that is induced by this field?

6. Jul 23, 2013

### Gavroy

Frankly, it would be even better if you could give me a general idea of how to calculate the dipole moment of a sphere in an electric field.

7. Jul 23, 2013

### Staff: Mentor

Due to the symmetry of the problem, you don't get a dipole moment if the sphere is in the origin.

In general, a conducting sphere in an electric field is hard to analyze in an analytic way. A numerical simulation is possible.

8. Jul 23, 2013

### Gavroy

So basically this field does nothing to our sphere or can you think of some effect that will happen due to this field?

9. Jul 24, 2013

### Staff: Mentor

Well, the sphere has to influence the field in some way, as an electric field perpendicular to the surface is not a stable configuration. Where does your field come from? A natural source would be a time-dependent current in z-direction*, but that current would have to go through your sphere (and through everything else) - and the current would grow without bound, which is not a very physical setup.

*edit: no, that would give this field configuration as magnetic field. You need something more complicated

Last edited: Jul 25, 2013
10. Jul 24, 2013

### TSny

Looks like you just have a uniform electric field (before the conducting sphere is brought in). The field is parallel to the xy plane and points in a direction that makes an angle $\phi$ to the y-axis.

This problem is worked out in many EM texts (with the initial uniform field usually along one of the axes). Placing the sphere in the electric field causes a nonuniform surface charge density to be induced on the sphere. This surface charge produces an electric field outside the sphere that has the same form as the electric field of a dipole. So, the total electric field outside the sphere will be the superposition of the uniform field and the dipole field of the induced charge.

This solution is usually obtained by first solving Laplace's equation for the electric potential and then using the potential to get the field. See for example here.

An alternate approach is to use the method of images. See here.

Last edited: Jul 24, 2013
11. Jul 25, 2013

### Staff: Mentor

ϕ is a coordinate, it depends on the position.

12. Jul 25, 2013

### TSny

Hello, mfb.
I guess I misinterpreted the notation. I thought $E = E_0(-sin(\phi), cos(\phi), 0)^T$ was just a way of saying that $E_x = -E_0sin(\phi)$, $E_y = E_0cos(\phi)$, and $E_z = 0$ where $\phi$ is a some constant.

I see now that earlier Gavroy stated that $\phi$ is apparently the azimuthal angle in spherical coordinates.

So, I'm confused now as to what the components in $E = E_0(-sin(\phi), cos(\phi), 0)^T$ represent. Are they Cartesian components $E_x, E_y, E_z$ or spherical components $E_r, E_\theta, E_\phi$, or something else?

13. Jul 25, 2013

### TSny

OK, does this picture represent the field and the sphere, (i.e., a cross-section through the sphere parallel to the xy plane)? This assumes the components given in the OP are Cartesian and $\phi$ is azimuthal angle.

Clearly this field doesn't represent an electrostatic field, as $\vec{\nabla} \times \vec{E} \neq 0$. It would have to be generated with a particular time-varying magnetic field such that $\vec{\nabla} \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$. Don't know if this is possible. [EDIT: I guess you could have a magnetic field in the negative z direction that increases proportional to time and has a magnitude that falls off inversely as the distance from the z-axis.]

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Last edited: Jul 25, 2013