# Homework Help: Coulomb's Law dielectric problem

1. Mar 15, 2015

### Tanya Sharma

1. The problem statement, all variables and given/known data

2. Relevant equations

Coulomb's law :

F=q1q2/4πεr2

where ε=εoK

3. The attempt at a solution

Suppose there is a dielectric slab of thickness 't' and dielectric constant 'K' inserted between the charges separated by a distance 'r' ,then if we convert the slab thickness to its effective air separation ,then it turns out to be equal to x= √(K)t .

So the effective separation is r-t+x .

The force turns out to be $$F = \frac{1}{4\piε_0}\frac{q_1q_2}{[(r-t)+\sqrt{K}t]^2}$$

This means , for K>0 ,the force decreases .

Now in Q 1 the answer should be b) i.e force decreases and in Q 2 the answer should be d) i.e force becomes zero as K= ∞ (since brass is a conductor).

I some how feel this is not correct . May be I am missing something very obvious .

Would be grateful if somebody could help me with the problem.

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2. Mar 15, 2015

### Shawnyboy

I think it usually doesn't serve to think of conductors in terms of dielectric constants as the two things behave completely differently.

What happens when you put a conductor between two charges is that the charges will pull the electrons in the conductor to the edge of the conductor (or push them away leaving positive ions) . The resulting induced charge will contribute to pulling the charges in towards the conductor because the sides facing the charges will have opposite charge.

However I think that if by calling it a plate, instead of a slab, the question meant to say the thickness of the plate was negligible, the two (now charged) sides of the plate are basically on top of each other so they would serve to cancel each other's effect.

3. Mar 15, 2015

### Delta²

When one puts a conductor between the two charges, then the electric field everywhere inside the conductor will become zero, however at the surface of the conductor where the two point charges lie it will not be zero, it will have direction normal to the surface of conductor. Its magnitude will be the same as before because the conductor has zero net electric charge, therefore the force will remain the same in this case.

4. Mar 15, 2015

### Delta²

Allow me to correct my self, the electric field will be a bit bigger in the case of placing a conductor in between because opposite charge will be induced in the surface of the conductor in the place near the original charge. This opposite charge will increase locally the electric field between this charge and the original charge hence the force will be abit bigger. The total force to the conductor from the original charges will be zero, however the force that the conductor exerts to each of the charges will not be zero due to this anisotropy in the distribution of charge density in the conductor.

5. Mar 15, 2015

### Tanya Sharma

What is the flaw in the reasoning I have given in the OP ?

6. Mar 15, 2015

### Delta²

By treating the conductor as a dielectric with K=infinite you essentially end up that the electric field is zero everywhere, while it is zero only inside the conductor. If the two original charges were put inside the conductor then your reasoning would be true, but we know this cannot be true the charges are either at the surface of the conductor plate or somewhere further of the surface.

7. Mar 15, 2015

### Tanya Sharma

Suppose the two point charges in the question are +q and -q .Let us call the surface facing +q be A and that facing B be -q .

Would the charge induced on A be -q and that on B be +q ? If yes , what is the justification ?

Last edited: Mar 16, 2015
8. Mar 16, 2015

### Tanya Sharma

What is the basic difference in the two cases given in the two questions ?

Aren't charges induced in both the cases ?

9. Mar 16, 2015

### Tanya Sharma

Would any mentor like to give his/her opinion on this problem ?

10. Mar 16, 2015

### TSny

I don’t believe the formula given in the OP is correct, in general. If you have two positive point charges and a thin dielectric wall (t << r) with $\kappa$ just a little greater than 1, then I believe the wall would cause the force on each charge to decrease a little. But I have never seen the formula given in the OP discussed in any textbook of EM. So, I don’t know if it applies to this case.

If the wall is thick so that r is just a little larger than t, then I don’t believe the formula is applicable at all. We know that a charged balloon will stick to a wall and that you could have a balloon on each side of the wall as shown below. The balloons now feel a net force toward each other and it is possible for the magnitude of the net force felt by a balloon to be greater with the wall than without the wall. This is not at all what the formula predicts.

Likewise, with a conducting wall, the like-charged balloons can “attract each other” with a force greater than the repulsion without the wall.

So I don’t like the formula! The only way I can see to “arrive at it” is by a very loose (and incorrect) argument. If two point charges are embedded well inside a dielectric material, the force between the charges is $F = \frac{q_1 q_2}{4 \pi\varepsilon_0 \kappa r^2}$. This is the same as if you replaced the distance $r$ by an “effective distance” $\sqrt{\kappa}r$ and then ignore the dielectric. So far so good (I think).

Now consider two point charges in air separated by a distance $r$ and a dielectric slab of thickness $t$ between the charges. To get the formula in the OP, it appears that someone has argued that you can take into account the slab by just saying that the effective thickness of the slab is $\sqrt{\kappa}t$ so that the effective distance between the charges becomes $r – t + \sqrt{\kappa}t$. This is unjustified. The effective distance argument only works if the point charges are both embedded inside the dielectric medium and the boundaries of the dielectric are far from the charges.

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11. Mar 16, 2015

### ehild

I have seen problems like that, with similar solution, but only in posts of Indian students.
The problem of electric forces in presence of dielectrics is rather difficult.
Coulomb force is defined between point charges in vacuum or between point-like free charges embedded in an infinite dielectrics as TSny pointed out.
A dielectric slab consists of its particles, either having charge, or having dipole momentum, or able to be polarised. So you have the original two charges and all charged particles and/or dipoles of the dielectrics. The original charges interact with the same force as before, but they also feel the forces from all the charged particles and dipoles of the dielectric slab.

I think, this is a useful link: http://energythic.com/view.php?node=209

12. Mar 16, 2015

### Tanya Sharma

Thanks TSny and ehild for replying .

Please leave this formula aside for a while . I will come back to this formula afterwards.The problems have raised a few basic fundamental doubts . Kindly help me in understanding these issues first .

1) What is the basic difference in the two cases given in the two problems (in the context of answering the question i.e effect on force between the two point charges )?

2) Suppose the two point charges in the question are +q and -q .Let us call the surface facing +q be A and that facing -q be B .Would the charge induced on A be -q and that on B be +q ,in case of metal plate ?

3) Aren't charges induced on the two facing surfaces of the plates in both the cases (glass in case 1 and metal in case 2) ? Shouldn't the answer to both the questions remain same ( i.e if we justify that force increases in the first case ,then it should also increase in the second case ) ?

Last edited: Mar 16, 2015
13. Mar 16, 2015

### SammyS

Staff Emeritus
Yes.
In the case of the dielectric, the induced charges on surfaces A and B will be somewhat less than q in magnitude.

Last edited: Mar 17, 2015
14. Mar 16, 2015

### Tanya Sharma

15. Mar 17, 2015

### Delta²

If the thickness t (and the breadth z) of the conductor is negligible compare to the distance r of the two original point charges then yes you can say that (so that the conductor appears like an idealized line segment between the two original point charges). The justification is that the induced charges must cancel out the electric field (which electric field is due to the original point charges +q and -q) inside the conductor.

Last edited: Mar 17, 2015
16. Mar 17, 2015

### ehild

In case you have a big metal plate and a point charge q+ at distance d from it (the lateral sizes of the plate much bigger than d) you can find the electric field with the method of mirror charges: the metal surface has to be equipotential and the electric force lines are perpendicular to it, as if an equal and opposite charge was at distance d behind the surface. From that, you find the surface charge density on the surface, and can determine the force exerted by the surface charges on the original charge. Note, that the surface charge density is not constant.

If the plate is not grounded, the other surface has the same charge but distributed evenly. The surface charge density can be considered zero if the metal plate can be considered infinity. The q- charge on the other side does not feel anything from the presence of the q+ charge on the other side. There is no sense to ask the force between them. Remember, a volume separated by metal from the surroundings does not feel the outside electric field .

When the charges are at different distances from the metal, they feel different forces from it.

If the metal plate is small, you have to consider the electric field at the edges, too. Still, the whole surface is at the same potential. But the induced charges distribute along the finite surface and the charges feel the presence of the other charge.

In the case when the thin metal sheet is placed just halfway between the q+ and q- charges , it is at the equipotential surface of the electric field. Introducing it, does not change anything. You can even fill the volume enclosed by the metal sheet with metal (half space in this case), the electric field outside it remains the same, and the force one charge feels remains the same.

If the distances are different, the metal sheet is not on an equipotential surface of the original electric field, but it will become that, changing the electric field.

Last edited: Mar 17, 2015
17. Mar 17, 2015

### Delta²

Oh dear, i had a ***slightly*** different geometric setup on my mind, namely that the plate is lying on the line connecting the two point charges and not perpendicular to it... Thanks ehild for providing the figure...Tanya you can safely ignore most of what i ve said, it is wrong...

18. Mar 17, 2015

### ehild

I do not know if the problem maker thought of that set-up, as it is not stated. It could be anything. It was a very ill-worded problem.

19. Mar 17, 2015

### ehild

As for the field and force between a dielectric and a charge,see https://archive.org/stream/Electrod...rodynamicsOfContinuousMedia#page/n49/mode/2up page 40, problem 1.

(Landau-Lifshitz: Electrodynamics of Continuous Media)

The figures are from a book of a Hungarian author(Simonyi)
The charge polarises the dielectrics (1) and the field in the upper half-space is as if produced by the original charge Q and an other charge Q' on the other side of the interface, and the field in the dielectric is as if a charge Q" would be instead of Q, and

$Q'=-\frac{ε_2-ε_1}{ε_1+ε_2}Q$,

$Q''=\frac{2ε_2}{ε_1+ε_2}Q$

And that is the final electric field.

Last edited: Mar 17, 2015
20. Mar 17, 2015

### Delta²

Thx ehild for trying to make me feel better and for the link as well. I have a good excuse also, i had to sleep 24 hours+ when i was posting in this thread

21. Mar 17, 2015

### TSny

I think the two cases (dielectric slab vs conducting slab) are similar in their qualitative effects.

Yes, ehild has discussed this. In the region of the surface closest to the one of the point charges the induced charge would be opposite to the point charge. In the case where the two point charges are both positive, the induced charge on the surfaces nearest to the charges would be negative. Assuming the slab is neutral overall, there would need to be some positive charge induced at other locations on the surfaces. But I think it will be the negative patches that will be more important in determining the net force on a point charge.

Yes, that's my understanding.

You can do some experimenting. You can easily charge pieces of sticky tape and use them as a rough substitution for point charges.

.

You can use a piece of paper or a metal cookie sheet as the slab of dielectric or conducing material. If you take two pieces of tape of the same sign of charge, you can see that the repulsion between the tapes becomes attractive as you move the charges closer to the slab (either dielectric or conducting). Tapes of opposite charge attract each other with or without the slab so the effect of the slab is harder to see. It helps to use a thicker slab here; say, a wooden door for the dielectric slab.

22. Mar 17, 2015

### Tanya Sharma

Tsny ,

Here is a simple reasoning for both the cases -

Under the influence of two point charges q and -q , positive and negative charges are induced on the facing surfaces of slab (metal/glass) . Now charge q will be under the influence of three charges i.e negative induced chared , positive induced charge and -q . But since negative induced charge will be closer to q , its attractive force would be more than that of repulsive force on q due to negative induce charge . So in addition to the force exerted by -q point charge , there would be an additional attractive force on point charge q due to the slab . Hence the overall attractive force on charge q increases due to the presence of the slab (metal/glass) .

So,the force increases in both the questions .

Does it make any sense ?

23. Mar 17, 2015

### TSny

Yes that makes sense for oppositely charged particles. The case of two like charges seems to me to be ambiguous. If the slab is thin and the charges relatively far apart, the repulsive force would decrease when the slab is inserted. But, if the slab is thick so that there is little space between the charges and the slab surfaces, then the repulsive force F1 (without the slab) switches to attractive force F2 (with the slab) and the magnitude of F2 could be larger than the magnitude of F1. (I think.)

24. Apr 18, 2018

### Rk classes

a dielectric slab of variable amount (k/2 to k) and thickness t is placed between two point charges separated by D. How will you calculate electric force between the two charges.

25. Apr 18, 2018

### Rk classes

a dielectric slab of variable amount (k/2 to k) and thickness t is placed between two point charges separated by D. How will you calculate electric force between the two charges.