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Electrostatics

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

Profile image of sunny chawla
15 Years agoGrade
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1 Answer

Profile image of Chetan Mandayam Nayakar
15 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