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# A thin-walled hollow circular glass tube, open atboth ends, has a radius R and length L. Theaxis of the tube lies along the x axis, with theleft end at the origin. The outer sides arerubbed with silk and acquire a net positivecharge Q distributed uniformly. Determine theelectric field at a location on the x axis, adistance w from the origin. Carry out all steps,including checking your result. Explain eachstep. (You may look up the integral).Here is the diagram

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