Electric field due to a hollow, charged cylinder

[chipotleaway]

Homework Statement

Find the electric field at a point on the x-axis due to a uniformly charged, hollow cylinder. The x-axis is parallel to the cylinder and goes through it's centre. One end of the cylinder is located at the origin.

Homework Equations

R = radius of cylinder
Q = charge on cylinder
L = length of cylinder from the origin
D = distance from the origin to the point of interest
k = 1/4πε

The Attempt at a Solution

1) I found the charge density
\lambda=\frac{Q}{2\pi RL

2) So an infinitesimally thin slice of the cylinder 2πRdx would have charge
dQ=\frac{Q}{L}dx

3) The force due to one of these rings is
dF = k\frac{qdQ}{(D-L)^2+R^2}

4) We want only the horizontal component due to the symmetry of the cylinder and the cancellation of the vertical components of the forces. So since

cos(\theta) = \frac{D-L}{((D-L)^2+R^2)^{\frac{1}{2}}

Therefore dF_x=k\frac{qdQ(D-L)}{((D-L)^2+R^2)^{\frac{3}{2}}

I divided by q to get the electric field, replaced dQ with (Q/L)dx and then integrated with respect to x from 0 and L.

The formula I got is:

E=k\frac{Q(D-L)}{((D-L)^2+R^2)^\frac{3}{2} which doesn't seem right or 'behave' right. Based on what I know, if we used it to calculate the force on a test charge and we let the test charge be a very large distance away, then we should be left with Coulomb's law but that doesn't seem to happen with what I got.

 

[haruspex]

Fixing up the latex:

Homework Statement

Find the electric field at a point on the x-axis due to a uniformly charged, hollow cylinder. The x-axis is parallel to the cylinder and goes through it's centre. One end of the cylinder is located at the origin.

Homework Equations

R = radius of cylinder
Q = charge on cylinder
L = length of cylinder from the origin
D = distance from the origin to the point of interest
k = 1/4πε

The Attempt at a Solution

1) I found the charge density
\lambda=\frac{Q}{2\pi RL}

2) So an infinitesimally thin slice of the cylinder 2πRdx would have charge
dQ=\frac{Q}{L}dx

3) The force due to one of these rings is
dF = k\frac{qdQ}{(D-L)^2+R^2}

4) We want only the horizontal component due to the symmetry of the cylinder and the cancellation of the vertical components of the forces. So since

cos(\theta) = \frac{D-L}{((D-L)^2+R^2)^{\frac12}}

Therefore dF_x=k\frac{qdQ(D-L)}{((D-L)^2+R^2)^{\frac32}}

I divided by q to get the electric field, replaced dQ with (Q/L)dx and then integrated with respect to x from 0 and L.

The formula I got is:

E=k\frac{Q(D-L)}{((D-L)^2+R^2)^\frac32} which doesn't seem right or 'behave' right. Based on what I know, if we used it to calculate the force on a test charge and we let the test charge be a very large distance away, then we should be left with Coulomb's law but that doesn't seem to happen with what I got.

At a great distance, L and R are insignificant compared to D, so it tends to E=k\frac{QD}{|D|^3}. Isn't that reasonable?

 

[rude man]

Well, at D = L you have E = 0 wheras should be the negative of at D = 0, no? And at D = L/2 should the field not be zero?

When you set this up at some point 0 < D < L did you take into consideration the two opposing forces at your test charge q?

 

[chipotleaway]

Fixing up the latex:

At a great distance, L and R are insignificant compared to D, so it tends to E=k\frac{QD}{|D|^3}. Isn't that reasonable?

Oh yeah, I didn't think about R...

Well, at D = L you have E = 0 wheras should be the negative of at D = 0, no? And at D = L/2 should the field not be zero?

When you set this up at some point 0 < D < L did you take into consideration the two opposing forces at your test charge q?

In setting up the diagram, I only considered a test charge at a distance greater than the length of the cylinder and hoped that the resulting formula would be work even if L>D...hmmm...

 

[rude man]

Oh yeah, I didn't think about R...

In setting up the diagram, I only considered a test charge at a distance greater than the length of the cylinder and hoped that the resulting formula would be work even if L>D...hmmm...

The problem didn't really specify where the observation point is, so in a way you can't be blamed. You obviously know how to attack a problem like this.
If you do decide to analyze inside the cylinder, realize you only need to do it for half of it. The other half is a mirror image with the sign of E revered. Of course, that holds for ouside as well.

I was also thinking, would this be more easily done with potentials rather than fields? With potentials you don't have to worry about vectors until it's time to find grad U.

 

[chipotleaway]

The problem didn't really specify where the observation point is, so in a way you can't be blamed. You obviously know how to attack a problem like this.

Well it did, but they were randomly generated numbers (that weren't very nice) so it could be inside or outside the cylinder. I picked a point outside to work with because I assumed that L would be greater than D and that would work even if L<D. Guess I was wrong in assuming one formula would do the trick.

If you do decide to analyze inside the cylinder, realize you only need to do it for half of it. The other half is a mirror image with the sign of E revered. Of course, that holds for ouside as well.

Could you please elaborate?

I was also thinking, would this be more easily done with potentials rather than fields? With potentials you don't have to worry about vectors until it's time to find grad U.

We're not up to that yet and unfortunately I'm too far behind on my other courses to be doing extra physics! But doing it this way is kind of fun though :p

 

[rude man]

If you do decide to analyze inside the cylinder, realize you only need to do it for half of it. The other half is a mirror image with the sign of E revered. Of course, that holds for ouside as well.

Could you please elaborate?

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Sure. Supposing you're a distance D = L/3 inside the cylinder. Then you compute the field from that point. But that will be the same field at a distance L - D/3 except for reversal of sign.

In fact, you can do the integration in just one direction instead of two by canceling the symmetrically located charge about D. So for any D you integrate from 2D to L only to get the field at D.

In other words, at D the field due to 0 → D is canceled by the field due to D → 2D.