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

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.

Ez = 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?

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SammyS
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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.

Ez = 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 !

Ez = 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.​