# Potential of charge inside a cylinder

1. Aug 17, 2011

### Beer-monster

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

A single positive point charge Q is located off-centre at radius R and height z = 0 inside a hollow cylindrical conducting shell of infinite length and inner radius a. The outer radius of the shell is b, and the axis of the cylinder is aligned with the z axis. What is the magnitude and direction of the electric field outside the shell, as a function of position?

2. Relevant equations

The principle of superposition and Gauss's Law.

3. The attempt at a solution

Can someone correct my thinking here. If you place a charge inside a conductor it will induce negative surface charges on this inner surface of this cylinder so as to cancel out the effect of the orginal charge and ensure the potential inside the cylinder is zero.

However, that would mean that the charge Q would be distributed across the surface of the cylinder and the problem would reduce to a charged uniform cylinder. The Electric field given by E= Q/2*pi*R*L.

That all seems a little too easy. Am I missing something in my thinking here?

2. Aug 17, 2011

### Bob at PC

How large is L? That will determine E.

3. Aug 19, 2011

### Beer-monster

If I'm right and the situation reduces to the problem of a cylinder were induced charge (equal to the original positive charge Q) is distributed uniformly along the surface of the cylinder the electric field as a function of L is.

$$E = \frac{+Q}{2\pi \epsilon_{0} b L)}$$

Which as a function of position in cylindrical coordinates (with the origin at the centre of the cylinder:

$$E = \frac {+Q}{2\pi \epsilon_{0} r z}$$

However, again this all seems too easy given this was supposed to be a graduate level question. Is there something wrong with my thinking that the charge inside the cylinder will be cancelled out by induced negative charge on the inner surface, leaving uniform positive charge on the outer surface of the shell.

Also, as a general matter, is there some rule of thumb or hint I can use to know whether to approach an electrostatic problem using the Method of Images or Laplace's equation.

Thanks again.

4. Aug 19, 2011

### Bob at PC

Your fundamental assumption that the fact that the field within the conductor = 0 leads to an induced charge of value -Q on the inside of the cylider and a charge Q on the outside is valid. This would hold regardless of the shape of the hollow object.

But here, the twist is that Q is some finite charge and the cylinder is infinitely long. What will be the surface charge density on the cylinder? Therefore, what is the field outside the cylinder?

5. Aug 19, 2011

### fleem

After equilibrium is reached, there is no electric field outside the shell. Think of it this way: If there were an electric field just above the outer surface of the shell, then electrons would adjust their position to counter it.

6. Aug 19, 2011

### Bob at PC

To Fleem:

If the cylinder were finite in length with a uniformly distributed surface charge Q, there would have to be a non-zero field outside (Gauss' Law).

I am assuming that we are not introducing any imposed external field not stated in the problem.

7. Aug 19, 2011

### Beer-monster

A cylinder of infinite length has infinite surface area. Therefore the charge on the surface would effectively be zero and there would be no field inside. Sort of like an inverted Faraday cage. Is that right still seems a bit too neat.

8. Aug 19, 2011

### Bob at PC

If the problem is exactly as you originally stated it - it is that neat!

There IS a field within the hollow - The field lines go from Q to the -Q induced on the inside of the conducting cylinder. The field in the conducting material that makes up the shell is zero because it is a conductor (see fleem's comment). The field on the outside is zero because the outer surface charge density is infinitesimal due to the infinite length of the conductor and the finite value of Q (dispersed evenly over the outer surface because no field from inside makes it through the conductor).

9. Aug 20, 2011

### fleem

Correct. But the problem states it has infinite length.

10. Aug 20, 2011

### Bob at PC

I think we are all in agreement (and probably have been right along). There cannot be a component of the electric field parallel to the outside surface regardless of the geometry because the charges will move until that component is zero. With an infinite surface, not only is the parallel component of field zero, the finite charge has spread itself out so much through mutual repulsion that all components of the field are zero.

11. Aug 20, 2011

### Beer-monster

It makes sense physically. I'm glad my intuitions were correct. I'm just surprised.

I copy/pasted that question from a past phd comprehensive exam, so I
expected to have to use mirror charges or Laplaces
equation and some messy
integration. I guess that was the trick.