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A thin-walled hollow circular glass tube, open at both ends, has a radius R and length L. 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 look up the integral). Here is the diagram

A thin-walled hollow circular glass tube, open at
both ends, has a radius R and length L. 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 look up the integral).


 


Here is the diagram


 

Grade:

1 Answers

Chetan Mandayam Nayakar
312 Points
10 years ago

I am assuming that the point where the electric field is to be calculated is (w,0)

surface charge density=σ=(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

The rest of the solution is very easy, but if you want further details email me at ramsesthegreat2@gmail.com

 

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