Electric Potential Particle Charge Problem

[Lancelot59]

I'm given the following problem:

A particle with charge q is located at x=-R and a particle with charge -2q is located at the origin. Prove that the equipotential surface that has zero potential is a sphere centered at (-4R/3,0,0) with radius r=2/3R.

I started by dividing the length between the two particles into two lengths: a and b. a is between the particle of charge q, and region b. b is between region a and the origin.

I then set up my potential equation like so:

$$V=k(\frac{q}{-R-b}+\frac{-2q}{-R-a})$$

I also had this equation to work with:
$$R=(a+b)$$

I set it equal to zero, and after some simplification I got:
$$b=\frac{2}{3}R$$$$a=\frac{1}{3}R$$

So I solved for b, but I'm not sure what to do now. I've proved that that point on the x-axis has a potential of zero. How do I prove that it is a sphere?

 

[vela]

The assumptions you made reduced the problem to a one-dimensional problem, but you're looking to solve the three-dimensional case. You're going to have to start over.

What does the cylindrical symmetry of the problem give you?

 

[Lancelot59]

The assumptions you made reduced the problem to a one-dimensional problem, but you're looking to solve the three-dimensional case. You're going to have to start over.

What does the cylindrical symmetry of the problem give you?

I'm not entirely sure. For any given position on the line, the potential only varies with respect to the radius? That would sweep out a cylinder.

EDIT: I think I see it now. I need to draw my radius out to an arbitrary point (x,y,z) and then using that in my potential equation set it equal to zero and solve after converting it to spherical coordinates?

 

[vela]

Yes, but you can just work in the xy-plane because of the cylindrical symmetry.

 

[jambaugh]

There's another possible approach. One can use the http://en.wikipedia.org/wiki/Method_of_image_charges" to solve for the fields (and hence potentials) for charges and conductors.

Given your conclusion you should get the equivalent field (outside the sphere) if you replace the inner charge with a spherical shell (grounded to zero potential).

The method of images says:
1.) Use geometric optics to find the virtual image of a charged point by treating the conductor as a mirror.
2.) Assign the image a charge of opposite sign times the magnification times the original charge.
3.) Remove the conductor and work out the field and potential for the real plus image charges.

I'm suggesting doing this in reverse. A conductor (in electrostatics) by its nature defines an equipotential surface. So one can always replace an equipotential surface with a conductor held at that voltage.

Maybe my suggestion is "driving tacks with a sledgehammer" but it has a certain elegance. I'd wager cash that the person who thought up the problem was thinking in terms of the method of images.

 

[Lancelot59]

Yes, but you can just work in the xy-plane because of the cylindrical symmetry.

I see, I'm not yet sure how I could go about setting it up. I'll try it out and see if I can come up with anything.

 

[ehild]

I then set up my potential equation like so:

$$V=k(\frac{q}{-R-b}+\frac{-2q}{-R-a})$$

The potential function is wrong.
The potential is a scalar function but its argument is the position vector, r.
The potential of a point charge q situated at R is

$$V(\vec r)=k \frac{q}{|\vec r-\vec R|}$$

So your potential function is

$$V(\vec r)=k \frac{-2q}{|\vec r|}+k\frac{q}{|\vec r+R \vec i|}$$

ehild