One dimensional medium with 2 different dielectric constants

[Clever-Name]

I just got this question on a classical mechanics assignment... and unfortunately I know NOTHING about dielectric materials (never been introduced to me for some reason >.<). I don't even know why this is on my classical mechanics assignment but it is anyway. And btw there is a disclaimer on the assignment stating that I CAN collaborate with others to find a solution.

Homework Statement

A one dimensional medium has its dielectric property changed halfway. The dielectric constant is $\varepsilon_{0}$ for $0 \leq x \lt L$ (medium 1) and has a different value $\varepsilon$ for $L \leq x \leq 2L$ (medium 2). What is the potential energy of two charges q1 and q1 placed at a distance d apart when (a) both are in medium 1, and (b) both are in medium 2. Would it be possible to hold them at the same distance in medium 2 as in medium 1 without doing any work? (Note that the relative displacement between the charges remains zero, with the two charges maintained at the same relative distance between each other. Also, remember that $\partial W = \vec{F} \bullet \vec{\partial s}$

Homework Equations

No idea really..

I suspect Coulomb's Force equation might be used.

The Attempt at a Solution

Don't even know where to begin. As I said at the beginning I have never encountered dielectrics before. Also, after some textbook reading and online reading I should note that as I have written it is the question, there is no mention of an external field or that the material is wedged within a capacitor.

I have to head out now I just wanted to post this in case anyone might be able to help. I won't be able to respond until sometime tomorrow morning. Thanks in advance for any help you might be able to give!

 

[Clever-Name]

Sorry to bump but I still have no idea how to approach this. Any suggestions?

 

[ehild]

In a medium of dielectric constant ε, the magnitude of the Coulomb force between two point charges q1 and q2, r distance apart is

$$F=\frac{1}{4\pi \epsilon}\frac{q_1 g_2}{r^2}$$.

And you know, that no work is needed to keep two charges in rest.

ehild

 

[Clever-Name]

Is it really that simple? I feel like there's something more complicated with this question.
So the potential between the particles would be:

$$U = \frac{1}{4\pi\varepsilon}\frac{q_{1}q_{2}}{d}$$

Now if we consider the second part:

Would is be possible to hold them at the same distance in medium 2 as in medium 1 without doing any work?

This confuses me a little. If I understand it properly then we have the pair of particles at separation d in medium 1, and then we take those 2 particles with same separation d and move them to medium 2. I realize that the force on the particles will vary between the mediums due to the dielectric property changing; but I don't quite understand what this question is requiring me to do. We would have to apply some sort of resistive force to the particles to keep them stationary at separation d, otherwise they would move closer together or farther apart based on the charges of course. But since we wouldn't be applying a force over any distance then there wouldn't be any work done? Therefore yes it's possible to hold them at the same distance without doing any work?

 

[ehild]

I do not understand the question either. It might ask about the work when you move the charges from one medium to the other. In this case, you might do work on the medium as the charges polarize it. But just holding the charges at a given distance apart does not mean any work.
ehild

 

[Clever-Name]

Hm, alright I'll speak to my professor on Monday about it then. Thanks for the help

 

[ehild]

Hm, alright I'll speak to my professor on Monday about it then. Thanks for the help

And let me know please, what he said.


ehild

 

[Clever-Name]

So I actually emailed him tonight and he responded right away, here is what he said:

The 2 charges are to be held at the same distance in the new medium as was between them in the first medium.
Since the dielectric constants of the two media are different, the energy of the system for a given distance cannot be the same, and extra forces will need to have been applied to maintain them at the same distance. Beyond this, there is just one little point that completes the answer. I will let you discover it, or feel free to ask again.

 

[ehild]

Well, that is the same badly worded text it was before, only "hold" has been changed to "maintain" .
The forces are different in the different media, so are the potential energies. But no work is needed to keep the charges at the same position. Maybe your teacher wanted to ask work while transferring the charges from one medium to the other one. But he did not say that. ehild

 

[Clever-Name]

So apparently enough people complained about this assignment that he has posted the solutions prior to the due date (just for this assignment) so people can get an idea as to what he means when he says certain things. I figured you'd be interested in knowing the answer to this problem:

Note that work will have to be done to merely maintain the charges at the same distance in the two media. This requires application of an external force that would make the momentum change at a rate that is equal to the force that must be applied. The momentum will then not be conserved, and the non-conservation of the momentum is connected here to the fact that our 'space' in this problem is NOT homogeneous; translational invariance is lost!

I don't know if it's just me but that description doesn't seem to make sense.