Electric Field at the end of a Half-Infinite Cylinder

In summary, the author used polar coordinates to solve the problem. Their approach was incorrect because it did not take into account the fact that the cylinder is hollow. The author will post a new answer later to get some critique.
  • #1
cwill53
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Homework Statement
(a) Consider a half-infinite hollow cylindrical shell (that is, one
that extends to infinity in one direction) with radius R and uniform surface charge density σ. What is the electric field at the
midpoint of the end face?
(b) Use your result to determine the field at the midpoint of a half-infinite solid cylinder with radius R and uniform volume charge density ρ, which can be considered to be built up from
many cylindrical shells.
Relevant Equations
$$\vec{E}=\frac{1}{4\pi \epsilon _0}\sum_{j=1}^{N}\frac{q_{j}}{r_{0j}^2}\vec{r_{0j}}\frac{1}{\left \| \vec{r_{0j}} \right \|}$$
The approach used in the book uses polar coordinates. I was wondering if my approach would still be correct. I set up the problem such that the midpoint of one face of the cylinder is at the origin while the midpoint of the other end's face is at the point (##l##,0).
The surface area of the cylinder is SA_{cylinder}=
$$2\pi Rl+2\pi R^2$$
$$q_{TOT}=\sigma (2\pi Rl+2\pi R^2)$$
So for the electric field at l, which is the midpoint of the face at the other end of the cylinder, I wrote
$$\vec{E}=\frac{1}{4\pi \epsilon _0}\int_{0}^{\infty }\frac{\sigma (2\pi Rl+2\pi R^2)}{l^3}\hat{x}dl$$

Does this make sense or do I have to use polar coordinates?
 
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  • #2
Hi. There are some problems.

The length the cylinder is infinite, not some value [itex]l[/itex]. The total surface charge will be infinite. If you are trying to consider the semi-infinite cylinder as (an infinite number of) finite cylinders, you have not made this clear.

From your equation[tex]\vec{E}=\frac{1}{4\pi \epsilon _0}\int_{0}^{\infty }\frac{\sigma (2\pi Rl+2\pi R^2)}{l^3}\hat{x}dl[/tex]it looks like:

1) your integral is summing the fields from an infinite number of overlapping cylinders of increasing length;

2) you are treating each cylinder as a point charge(applying the inverse square law) when finding its contribution to the total field.

Both of these are totally incorrect! I hope you can see why if you think about it.

Also, note that we are told that the cylinder is hollow, so it has no end faces (charge only on the curved surface).

To correct your approach you have to consider the semi-infinite hollow cylinder sliced-up into rings, each ring of width dz (if cylinder's axis is the z-axis).

You then sum (integrate) the fields from the rings. Note a ring of charge is not a point charge, so you can’t simply apply the inverse law. But each 'bit' of a ring is a point charge.

You will get exactly the same answer whatever coordinate system you use. But polar coordinates keep the amount of work to a minimum in this particular problem.
 
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  • #3
Steve4Physics said:
Hi. There are some problems.

The length the cylinder is infinite, not some value [itex]l[/itex]. The total surface charge will be infinite. If you are trying to consider the semi-infinite cylinder as (an infinite number of) finite cylinders, you have not made this clear.

From your equation[tex]\vec{E}=\frac{1}{4\pi \epsilon _0}\int_{0}^{\infty }\frac{\sigma (2\pi Rl+2\pi R^2)}{l^3}\hat{x}dl[/tex]it looks like:

1) your integral is summing the fields from an infinite number of overlapping cylinders of increasing length;

2) you are treating each cylinder as a point charge(applying the inverse square law) when finding its contribution to the total field.

Both of these are totally incorrect! I hope you can see why if you think about it.

Also, note that we are told that the cylinder is hollow, so it has no end faces (charge only on the curved surface).

To correct your approach you have to consider the semi-infinite hollow cylinder sliced-up into rings, each ring of width dz (if cylinder's axis is the z-axis).

You then sum (integrate) the fields from the rings. Note a ring of charge is not a point charge, so you can’t simply apply the inverse law. But each 'bit' of a ring is a point charge.

You will get exactly the same answer whatever coordinate system you use. But polar coordinates keep the amount of work to a minimum in this particular problem.
I had a feeling I was doing something wrong. Thanks for clearing this up. I see now that I was basically taking the E field of an infinite number of cylinders and not integrating the rings of a hollow cylinder with no faces and summing up the contributions of each ring to the E field felt at the point indicated. Thanks. My approach was TOTALLY off.
I will post my new answer later to get some critique.
 
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  • #4
cwill53 said:
I had a feeling I was doing something wrong. Thanks for clearing this up. I see now that I was basically taking the E field of an infinite number of cylinders and not integrating the rings of a hollow cylinder with no faces and summing up the contributions of each ring to the E field felt at the point indicated. Thanks. My approach was TOTALLY off.
I will post my new answer later to get some critique.
 
  • #5
I was curious if you ever completed the problem.
 
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1. What is an electric field at the end of a half-infinite cylinder?

An electric field at the end of a half-infinite cylinder refers to the strength and direction of the electric field that is present at the end of a cylindrical object that extends infinitely in one direction and has a finite length in the other direction.

2. How is the electric field at the end of a half-infinite cylinder calculated?

The electric field at the end of a half-infinite cylinder can be calculated using the formula E = λ/2πε₀r, where λ is the linear charge density of the cylinder, ε₀ is the permittivity of free space, and r is the distance from the end of the cylinder to the point where the electric field is being measured.

3. What factors affect the electric field at the end of a half-infinite cylinder?

The electric field at the end of a half-infinite cylinder is affected by the linear charge density of the cylinder, the distance from the end of the cylinder, and the permittivity of free space. Additionally, the presence of other nearby charged objects can also influence the electric field.

4. What is the direction of the electric field at the end of a half-infinite cylinder?

The direction of the electric field at the end of a half-infinite cylinder depends on the direction of the linear charge density. If the charge is positive, the electric field will point away from the end of the cylinder. If the charge is negative, the electric field will point towards the end of the cylinder.

5. Can the electric field at the end of a half-infinite cylinder be negative?

Yes, the electric field at the end of a half-infinite cylinder can be negative if the linear charge density is negative. This means that the electric field will point towards the end of the cylinder instead of away from it. However, the magnitude of the electric field will still be calculated using the same formula as for a positive linear charge density.

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