Electric Dipole - Oppositely charged hemispheres

In summary, we are asked to find the leading order term of the potential produced by a sphere with uniform charge density in one hemisphere and opposite charge density in the other hemisphere. The potential, to leading order, is given by a formula of the form g_n(\theta,\phi)/r^n. We need to determine the value of n and the function g_n(\theta,\phi). To find the gravitational force exerted by this body at large distances, we can use the formula for the field at large distances from a dipole. However, for a more accurate approximation, we can use the formula for the electric field around a dipole at all distances. By expanding this formula in spherical coordinates, we can get the expression for the potential
  • #1
ModusPwnd
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Homework Statement


A sphere of radius [itex] a[/itex] carries a uniform charge density [itex] \rho_0 [/itex] in one hemisphere and the opposite charge density [itex] -\rho_0 [/itex] in the other hemisphere. The charged sphere produces an electric potential [itex]\Phi (r,\theta,\phi)[/itex]. For [itex] r>>a[/itex], the potential to leading order is of the form,

[tex]
\Phi(r,\theta,\phi) \approx \frac{g_n(\theta,\phi)}{r^n}
[/tex]

Find [itex] g_n(\theta,\phi)[/itex] and the numerical value for n.


Homework Equations



??
I presume the multiple expansion could be used, but this was a test question so I think I could do it without having the expansion memorized.

Any ideas?

Thanks!
 
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  • #2
If you had to get an expression for gravitational force exerted by this body at large distances, except that one-half would consist of "negative" matter, what would you do?
 
  • #3
If I wanted the gravitational force, I would set up the integral over the region, taylor expand it and then integrate the leading term. Does that sound right?

Plus, since I need potential and not force, it will be easier since its a scalar.
 
  • #4
That sounds good overall. You want to find the c.m., pretend all the charge is there, now you have a standard dipole. Potentials may be the way to go, but I'm not sure. There are probably formulas floating around out there in cyberspace giving the c.m. of a hemisphere ... I would forget about Taylor series expansions. The formula for the field at large distances from a dipole is very standard textbook stuff.
 
  • #5
The question clearly asks for the leading order term of the potential though.
 
  • #6
I interpreted that to mean the far-field formula. The far-field formula IS a first-order ("leading") approximation. That's why it's called far-field. Does not apply when near the dipole, gets too inaccurate. It's 100% accurate only at infinity.

See http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dipole.html

Look at "dipole moment" part. Not that z has to be >> r, the distance between the spehers' c.m. in your case.

However, I just noticed they want an expression for everywhere around the dipole (albeit at far distances). The formula in the link only givs the field along the line z, perpendicular to r.

This is a bigger deal than I thought. May come back to it, might not be too bad.

Here's the best I've found on your problem:

http://phys.columbia.edu/~nicolis/Dipole_electric_field.pdf

If the last equation is expanded in spherical coordinates, that should be the answer. I would set up the dipole along the z axis, with the origin half-way between the two effective point charges.
 
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Related to Electric Dipole - Oppositely charged hemispheres

1. What is an electric dipole?

An electric dipole is a pair of equal and opposite charges separated by a small distance. This results in a dipole moment, which is a measure of the strength of the dipole and the distance between the charges.

2. How are electric dipoles created?

Electric dipoles can be created by separating positive and negative charges, such as in the case of oppositely charged hemispheres.

3. What is the significance of oppositely charged hemispheres in an electric dipole?

Oppositely charged hemispheres are essential for creating an electric dipole. They allow for a separation of charges, resulting in a dipole moment.

4. What is the relationship between the distance between charges and the strength of an electric dipole?

The strength of an electric dipole is directly proportional to the distance between the charges. This means that as the distance between the charges increases, the dipole moment decreases.

5. How are electric dipoles used in real-world applications?

Electric dipoles are used in various technologies, such as antennas, capacitors, and molecular interactions. They also play a significant role in understanding the behavior of molecules and atoms in chemistry and physics.

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