Thread: Electric Field between concentric cylinders

I wanted the electric field between concentric cylinders, and got anexpression using Gauss' Law that considers only charge enclosed. Thus, there was no influence due to the outer cylinder.
But, if we consider the limiting case of the inner cylinder having very large curvature and the outer cylinder not too far off from it..i.e a is very large and b is just larger than a.
I assume that then the electric field between the cylinders can be considered as a flat plate analogy, which gives some other answer of the field between(for flat plate analogy, the outer cylinder also supplies electric field but it had no influence watsoever when we considered the normal cases)...Please help. I cant figure out what's wrong.
**
And suppose now, I consider both the cylinders to be dielectrics. Does that affect the analysis? Will the Gauss' Law still be applicable? Will the field inside an infinitely long uniformly charged cylinder still be zero, and have no affect on the electric field .

I will try to get back to you on this later today.

-Dan

Originally Posted by samarthvk

I wanted the electric field between concentric cylinders, and got anexpression using Gauss' Law that considers only charge enclosed. Thus, there was no influence due to the outer cylinder.
But, if we consider the limiting case of the inner cylinder having very large curvature and the outer cylinder not too far off from it..i.e a is very large and b is just larger than a.
I assume that then the electric field between the cylinders can be considered as a flat plate analogy, which gives some other answer of the field between(for flat plate analogy, the outer cylinder also supplies electric field but it had no influence watsoever when we considered the normal cases)...Please help. I cant figure out what's wrong.
**
And suppose now, I consider both the cylinders to be dielectrics. Does that affect the analysis? Will the Gauss' Law still be applicable? Will the field inside an infinitely long uniformly charged cylinder still be zero, and have no affect on the electric field .

I'm probably being way simple here but ....:

Parallel-plate capacitor:

It's easy to show using Gauss's Law that where q is the total charge on a plate and A is the area of a plate.


Cylindrical capacitor:

It's easy to show using Gauss's Law that where q is the total charge on the inner cylinder and is the radius of a cylindrical Gaussian surface between both cylinders.

Note that the surface area of the cylindrical Gaussian surface is . Substitute in the expression for E for the cylindrical capacitor: .


As the radii of the cylinders get really really large, A can be considered approximately flat ......

If I am understanding the problem it's a case of "Why is it when we find the E field for a parallel plate capacitor we need to add the fields from both plates, but when we find the E field for a cylindrical capacitor we don't."

This question has been kicking my butt a bit and I thank you for asking it, just for that reason.

The best answer I can give is that it is related to the fact that the electric field adds as a vector quantity, but the flux adds as a scalar, so we have some "weird" things happening occasionally.

For instance, in the case of the parallel plate capacitor we have to add the E fields due to two planes of charge. Why is that? Because we cannot create a single Gaussian surface to do the job. I can't give a good reason why this must be so in a case of such high symmetry in the charge distribution, but I liken it to the case where we have two spheres of charge, offset from one another. In this case we have to find the electric field due to each charge distribution and add them to get the total, else our flux integral has to practically be done numerically.

However we can do the cylindrical case in the same manner as the parallel plate case if you like. The difference here is that the outer "plate" of the capacitor bends around, so the field due to it inside the cylinder is 0. So when we add 0 to the field from the inner "plate" we just get the term for the inner plate. (If you want to get really advanced about this we are dealing with a non-Riemannian geometry here. That's going to foul up the mathematical form for Gauss' law, though not the principle of it.)

The difference between the case of conductors vs. dielectrics is only for inside the inner cylinder: now the inner cylinder will have an internal E field whereas in the case of the conductors there was no E field inside the inner cylinder.

-Dan

So, can we suggest that electric field inside an infinite hollow cylinder with charge on the inner surface is 0 ?
I tried to this problem analytically assuming the cylinder to be made of many infinite lines of thickness R*d(theta) , of which i know the contribution to the electric field. Adding the vector components and relating them through the projected triangle, and using cosine law, i get an integral that is very complex to solve..Solving it online, gave me 2 different answers . I just cant figure out the method to do this.

Simply put, my question is what is the electric field inside an infinite hollow cylinder with charge on inner surface?

Originally Posted by samarthvk

So, can we suggest that electric field inside an infinite hollow cylinder with charge on the inner surface is 0 ?
I tried to this problem analytically assuming the cylinder to be made of many infinite lines of thickness R*d(theta) , of which i know the contribution to the electric field. Adding the vector components and relating them through the projected triangle, and using cosine law, i get an integral that is very complex to solve..Solving it online, gave me 2 different answers . I just cant figure out the method to do this.

Simply put, my question is what is the electric field inside an infinite hollow cylinder with charge on inner surface?

It's a real to do find the field that way, isn't it? In this case I'd just trust Gauss' Law: There's no charge inside any Gaussian surface inside of the cylinder, so there is no electric field there.

FYI: One way to approach the problem of integrating the field "by hand" is to use some symmetry. The field, if it existed, would have to be radial. So only consider the radial components of the field due to your infinitesimal source points.

-Dan