# Electric field due to Half Cylinder

1. Oct 22, 2013

### ChowPuppy

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

We have an infinitely long half-cylindrical shell of radius r and charge density σ as shown below

I am supposed to find the electric field at a point on the cylindrical axis, as seen in the diagram. Consider a coordinate system where the cylinder extends infinitely along the z-axis and is in the upper half of the x-y plane, centered at the origin. Based on the symmetry of the problem, I know that the electric field must point in the -y direction. However I can't find a good method to figure out what the magnitude of the field is.

Direct integration is a no go, because this is an infinite half cylinder.
Gauss' Law doesn't apply here.
I am not sure about the method of images, but I don't see any clear way to match boundary conditions.
Possibly Laplace's Equation would apply? But again the boundary conditions would be odd.

Thanks for any help.

2. Oct 23, 2013

### TSny

Did you give it a try?

[EDIT: If you want to avoid direct integration, you can try the following. Imagine the z-axis is uniformly charged with a linear charge density. It is easy to find the E-field produced by the line charge (Gauss' law). Find the force that the infinite line charge exerts on a unit length of the half-cylinder. This will require an integration, but it will be simple. Then, think about how you can use this to determine the E-field produced by the half cylinder at the z-axis.]

Last edited: Oct 23, 2013
3. Dec 20, 2013

### Saitama

Hi TSny! :)

I recently came along a similar problem and I am very interested to know how you can determine the E-field without integration.

As per your suggestion, I found the force due to infinite line charge on the unit length of half-cylinder.

It comes out be:
$$\frac{\lambda \sigma}{\pi \epsilon_o}$$
where $\lambda$ is the linear charge density of infinite wire.

I am not sure how I can find the E-field of half-cylinder using the above. I am making a guess, do I have to use Newton's third law?

Many thanks!

4. Dec 20, 2013

### TSny

Hi Pranav-Arora.

That looks good.

Yes, think about the implication of Newton's 3rd law for the force per unit length on the infinite cylinder compared to the force per unit length on the infinite line of charge.

5. Dec 20, 2013

### Saitama

From Newton's third law, the force acting on the unit length of infinite line charge is also $\dfrac{\lambda \sigma}{\pi \epsilon_0}$. Hence, the E-field due to cylinder is $\dfrac{\sigma}{\pi \epsilon_0}$, is this correct?

But can you please explain why this method works? Why only an infinite line charge, why not some finite line charge or something else?

6. Dec 20, 2013

### TSny

If you had an infinite half-cylinder but a finite line charge, then the 3rd law would tell you that the magnitude of the net force on the cylinder would equal the net force on the line charge. But the 3rd law would not be able to tell you very much about the force per unit length (i.e., the force density) on the cylinder. The force density would not be uniform on the cylinder.

However, when both the half-cylinder and the line charge are infinite, then you can use symmetry along with the 3rd law to reason that the force densities must be uniform and of equal magnitude for each object.

7. Dec 20, 2013

### TSny

Yes.

8. Dec 20, 2013

### Saitama

Umm....I have trouble understanding why are we concerned with the force density and not the net force itself. I am sorry if this is a dumb question.

9. Dec 20, 2013

### TSny

When both the half-cylinder and line charge are infinitely long, then the net force on each is infinite. But the force densities are finite.

If you use a finite line charge with an infinite half-cylinder, then the net force on each will be finite. But then you cannot use the simple expression for the electric field of an infinite line charge in your analysis.

10. Dec 21, 2013

### Saitama

I am not sure if I understand it correctly but if I use a finite line charge, would that still give me the right answer?

11. Dec 21, 2013

### TSny

Yes, it would give you the right answer if you use a finite line charge. But it seems to me that you would then have to do essentially the same integration as you would if you just integrate over the half-cylinder directly to find the E field on the axis produced by the half-cylinder.

For example, suppose you shrunk the finite line charge down to a point charge on the axis of the cylinder. Would there be any advantage to using the point charge at all?

12. Dec 21, 2013

### Saitama

I don't think there would be any advantage, that would be similar to finding E-field of cylinder through direct integration.

I was wondering that instead, if we had a solid half-cylinder, can the same method be used?

13. Dec 21, 2013

### TSny

Yes, you could use the infinite line charge for the infinitely long, solid half-cylinder.

14. Dec 21, 2013

### Saitama

Thanks a lot TSny!