Calculating the Coulomb Force between 2 charges in space

In summary, the conversation discusses two problems related to charges and dielectric materials. Problem #1 involves finding the force between two charges when a dielectric slab is placed between them. Problem #2 extends this to finding the force between two charges when two dielectric slabs of different thicknesses and dielectric constants are placed between them. The conversation also mentions a related problem of finding the induced charge on a conducting plate near a point charge, which can be solved using the method of images.
  • #1
maverick280857
1,789
4
Hi

Here is an interesting problem which I have been trying to solve. For convienience of discussion, I have broken it into parts to show where I get stuck:

Two charges [itex]Q_{1}[/itex] and [itex]Q_{2}[/itex] are kept in space at a distance r from each other in a medium of dielectric constant K. The force between them is given by

[tex]F = \frac{1}{4\pi\epsilon_{0}K}\frac{Q_{1}Q_{2}}{r^2}[/tex]

Problem #1: Now if instead of the original dielectric medium we have a slab of dielectric constant K and width r/2 placed somewhere between the two charges, then find the force betwen the two charges (this is not homework so I guess its in the right place).

Problem #2 (This is what I could not do): The space between the two charges (i.e. r) is filled with two dielectric slabs of thickness [itex]d_{1}[/itex] and [itex]d_{2}[/itex] and dielectric constants [itex]K_{1}[/itex] and [itex]K_{2}[/itex] such that [itex]d_{1}+d_{2} = d[/itex].

My reasoning (so far) for the first problem: If two charges are kept at a distance r in a medium of dielectric constant K then they must be kept in air at a separation = [itex]r\sqrt{K}[/itex] in air to keep the force of interaction constant. This gives the effective distance they must be placed at in air, as [itex]\frac{r}{2} + \frac{r\sqrt{K}}{k}[/itex]. How do I use this for the second problem?

Advice/help is greatly appreciated.

Thanks and cheers
vivek
 
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  • #2
TO add to my previous post:

Is it possible to find an effective dielectric constant in such a case?

I read somewhere that I could do this using:

[tex]\frac{d_{1}+d_{2}}{K} = \frac{d_1}{k_1}+\frac{d_2}{k_2}[/tex]

Obvious as this equation seems, I can't seem to extract anything out of it let alone derive it.

Thanks,
vivek
 
  • #3
To everyone: a reminder...

To the Moderator: please relocate this if you feel the same.
 
  • #4
maverick280857 said:
TO add to my previous post:

Is it possible to find an effective dielectric constant in such a case?

I read somewhere that I could do this using:

[tex]\frac{d_{1}+d_{2}}{K} = \frac{d_1}{k_1}+\frac{d_2}{k_2}[/tex]

Obvious as this equation seems, I can't seem to extract anything out of it let alone derive it.
To derive this equation for the effective dielectric constant of the two slabs, consider two parallel plate capacitors in series:
[tex]1/C = 1/C_1 + 1/C_2[/tex]
[tex]1/(k_{eff}/d) = 1/(k_1/d_1) + 1/(k_2/d_2)[/tex]
... etc ([itex]d = d_1 + d_2[/itex])
 
  • #5
Are the slabs of dielectric stacked on top of one another or is it that the capacitor has a dielectric [itex]K_1[/itex] on the left half and a dielectric [itex]K_2[/itex] on the right half?
 
  • #6
Maverick... Both problems 1 and 2 are difficult, and require either an infinite series of images, or direct solutions of the boundary value problem in terms of Fourier Bessel integrals. Prob 1 can be found in Symthe's Static and Dynamic Electricity, Chapter 5, Section 5.303 Point Charge and Dialectric Plate. Prob 2. is, or should be, a relatively straightforward extension of prob 1.
Regards,
Reilly Atkinson
 
  • #7
Thanks a lot Doc Al, Corneo and reilly...

I didn't realize that problem 2 would be so tough (but that's because I already knew the governing equation...maybe?). This is because they were posed as prospective problems for an engineering entrance test (called the Joint Entrance Examination in India) which requires a background of very basic physics and electrodynamics (no Fourier-Bessel Integrals, Maxwell's Equations, PDEs, images, Laplace/Poisson equations, etc--just total derivatives and an extension of the theory mentioned in books like Crane,Zemansky/Sears, etc.) I suppose I need to know much more physics/mathematics to be able to derive these equations.

Doc, I could do the effective dielectric derivation for a parallel plate capacitor but I am not sure if I can use it for two point charges. I mean, you have two point charges in space and the space between them is filled with two dielectric slabs (only the width of which is given). I am not sure if we can use the idea of induced charge in much the same way as we set up equations for a parallel plate capacitor using Gauss's Theorem.

If however, I assume that the effective dielectric equation mentioned in my third post is correct then the problem is solved (though I don't think so as I don't know where the equation came from, for the point charges). :smile:

A related problem is (which I think can be done using the images method) to find the induced charge on a conducting plate when placed near a point charge. I would be grateful if you folks could point me to some book or internet resource where I can learn some basic stuff about the method of images to be able to apply it to relatively simple situations. I have been reading Cheng but as I have little time left for my exams, I cannot really start from scratch and understand all the mathematics required for a proper treatment of the subject.

Thanks and cheers,
Vivek
 

1. How is the Coulomb force calculated between two charges in space?

The Coulomb force between two charges in space is calculated using the formula F = (k * q1 * q2)/r^2, where F is the force, k is the Coulomb's constant (8.99 * 10^9 N*m^2/C^2), q1 and q2 are the magnitudes of the two charges, and r is the distance between the two charges.

2. What is the unit of measurement for Coulomb force?

The unit of measurement for Coulomb force is Newtons (N).

3. Does the Coulomb force between two charges depend on the type of charges?

Yes, the Coulomb force between two charges depends on the type of charges. Like charges (positive-positive or negative-negative) repel each other, while opposite charges (positive-negative) attract each other.

4. Is the Coulomb force affected by the distance between the two charges?

Yes, the Coulomb force between two charges is inversely proportional to the square of the distance between them. This means that as the distance increases, the force decreases, and vice versa.

5. Can the Coulomb force between two charges in space be zero?

Yes, the Coulomb force between two charges in space can be zero if the charges have the same magnitude and are located at an infinite distance from each other. This is known as the "zero-force" condition.

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