Electric field due to a FINITE cylinder of charge - Tricky binomial expansion

[nnj3k]

1. Homework Statement

(a) Calculate the electric field at an axial point z of a thin, uniformly charged cylinder of charge density ρ , radius R, and length 2L. z is the distance measured from the center of the cylinder. (b) What becomes of your result in the event z >> L ?

2. Homework Equations

I found the answer to (a) by doing a triple integral and then double checked by integrating the equation for a disk of charge given in my textbook over the z-axis and came up with the same exact result. So I am quite confident this expression is correct. The image linked is my work to find the answer if this doesn't look right.

E_z = kq/R²L { 2L + √[ R² + (z-L)² ] - √[ R² + (z+L)² }

(As a 2nd check...i wrote the expression as kq/z² { 2L + √[ R² + (z-L)² ] - √[ R² + (z+L)² } [ z²/R²L ] and in my calculator wrote a quick program to calculate the limit and sure enough...all that garbage on the right goes to 1 as z >> L and R.)

3. The Attempt at a Solution

This expression should reduce to kq/z² because any charge distribution should mimic a point charge at large distances. The answer to the problem is some type of application of the binomial expansion which I cannot seem to figure out. Everything I try just leads to everything in the bracket except 2L to be zero. Any help is greatly appreciated.

Side Question: This expression is only valid for z outside the cylinder. To calculate z inside the cylinder...is it valid to use the previous integral but changing the bounds from -L to z, and then adding the second integral of the remaining part of the cylinder that is above the charge from z to L and adjusting the radius of dq accordingly, or is there a more simple way to do it?

 

[SammyS]

1. Homework Statement

(a) Calculate the electric field at an axial point z of a thin, uniformly charged cylinder of charge density ρ , radius R, and length 2L. z is the distance measured from the center of the cylinder. (b) What becomes of your result in the event z >> L ?

2. Homework Equations

I found the answer to (a) by doing a triple integral and then double checked by integrating the equation for a disk of charge given in my textbook over the z-axis and came up with the same exact result. So I am quite confident this expression is correct. The image linked is my work to find the answer if this doesn't look right.

E_z = kq/R²L { 2L + √[ R² + (z-L)² ] - √[ R² + (z+L)² }

(As a 2nd check...i wrote the expression as kq/z² { 2L + √[ R² + (z-L)² ] - √[ R² + (z+L)² } [ z²/R²L ] and in my calculator wrote a quick program to calculate the limit and sure enough...all that garbage on the right goes to 1 as z >> L and R.)

3. The Attempt at a Solution

This expression should reduce to kq/z² because any charge distribution should mimic a point charge at large distances. The answer to the problem is some type of application of the binomial expansion which I cannot seem to figure out. Everything I try just leads to everything in the bracket except 2L to be zero. Any help is greatly appreciated.

Side Question: This expression is only valid for z outside the cylinder. To calculate z inside the cylinder...is it valid to use the previous integral but changing the bounds from -L to z, and then adding the second integral of the remaining part of the cylinder that is above the charge from z to L and adjusting the radius of dq accordingly, or is there a more simple way to do it?

Hello nnj3k. Welcome to PF !

Your expression,

E_z = kq/(R²L) { 2L + √[ R² + (z-L)² ] - √[ R² + (z+L)² ] } ,​

should include the parentheses I added to clearly indicate what's in the denominator.

One thing that may help you is to write everything in terms of z/L, since if z >> L, then z/L >> 1 .

For your Side Question: If 0 ≤ z < L,

Break the rod into two pieces.

One piece has a length of 2(L-z) and extends from L-2(L-z) = 2z-L to L. The point, z, at which you're finding the E field is centered in this piece, so the E field, at z, due to this piece is zero.

The other piece extends from -L to 2z-L. Calculate the E field, at z, due to this piece.​

Symmetry is your friend!​

 

[Mindscrape]

I imagine what you should get for z>>L is something like a point charge with a total charge equal to the charge over the cylinder.

 

[nnj3k]

I finally figured out what to do for the limit (it is far too much work for me to be arsed to explain though :P)...and thanks for the tip about symmetry!

 

[asteriatic]

As a fellow scientist, I would like to commend you on your thorough work and double checking of your answer. It is always important to make sure our calculations are correct, especially in the field of science where precision and accuracy are crucial.

Regarding the question at hand, you are correct in your assumption that the expression should reduce to kq/z^2 at large distances. This can be seen by using the binomial expansion for the square root terms in your expression. As z >> L, the terms inside the square root can be approximated as R^2/z^2. By using the binomial expansion, we can simplify the expression to kq/z^2 [1 + (1/2)(R^2/z^2) - (1/8)(R^4/z^4) + ...]. As z >> L, the higher order terms become negligible and we are left with kq/z^2, as expected.

As for the side question, your approach seems valid. Another way to approach it would be to use the method of images, where you replace the cylinder with a line of charge along the z-axis and a point charge at the center of the cylinder. This would require you to integrate over the entire length of the cylinder, but it may simplify the calculation and make it more intuitive.

I hope this helps and keep up the good work in your scientific endeavors!