A hollow glass tube A thin-walled hollow circular glass tube, open at both ends,

Question

A hollow glass tube A thin-walled hollow circular glass tube, open at both ends, has a radius R and length L (see the figure). The axis of the tube lies along the x axis, with the left end at the origin. The outer sides are rubbed with silk and acquire a net positive charge Q distributed uniformly. Determine the electric field at a location on the x axis, a distance w from the origin. Carry out all steps including checking your result. Explain each step. You may have to refer to a table of integrals. (Do this on paper. Your instructor may ask you to turn in this work.)

Step 1:
Since we know the electric field of a charged ring, divide the tube into rings, of thicknessdx. Consider a representative ring somewhere in the middle of the tube, with its center at <x,0,0>. Draw a diagram illustrating this situation.

What is the vector from source to observation location?

I'm not sure how to define the source as a vector I know its in the tube

Obs-Source <x,0,0>-<?,?,?>?

My first guess was the source is <0,Rsin0,-Rcos0> because of the book example-wronge! :( I guess because no angles are taken in concideration in the example.

my 2nd guess was <L/2,R,R> still wrong I think I'm on my last try, please help!

Explanation / Answer

sol:

point where the electric field is to be calculated is (w,0)

surface charge density==(Q/2piRL),

from symmetry E points along x-axis,

E=(1/4pik)(from x=0 to L) (1/((R/2)^2 +(w-x)^2))*((w-x)/(R/2)^2 +(w-x)^2)(Q/2piRL)(2piR) dx

=(1/4pik)(from x=0 to L) ((w-x)/((w-x)^2 +(R/2)^2)(3/2))*(Q/L) dx

=(Q/4pikL)(from x=0 to L) -(s/((s^2 +(R/2)^2)(3/2)) ds,

s=w-x

now rest you can calculate.

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