So this is problem 563 from the Physics Problem Solver. Thequestion is as follow

Question

So this is problem 563 from the Physics Problem Solver. Thequestion is as follows:

Prove the following theorem: The electric field outside of aninfinite cylindrically symmetrical charge distribution isequivalent to the field due to an infinite line charge of equallinear charge density.

Solution: First, we compute the field due to a cylindrical shell ofcharge. We note that the field must be radial since no onedirection is preferred. We erect a Gaussian surface, a cylinder, orradius r (r>R) and height l, concentric to the shell. By Gauss'law (since E is constant along any cylinder concentric to theaxis):
E * dA =4q, 2 r l E = 4 (sRl)

I understand all the work after this preliminary step. I am havingdifficulties with the above step.

I know that the surface area will be 2rl, and we willthus have E d2rl=   2rlE.

But where does the 4q come from, and why? And why do they thenget =4 (sRl) ? I assume then that the sRl= q?

Thanks much

Explanation / Answer

r =R      flux at surface =flux thru imaginary surface of radius R R = ER A = ( /0) A = ( / 0) 2 LR       we know the field at thesurface of the conductor r = Er 2 L r Er = R / (0 r) = Q / (2 L R)     surfacecharge density on cylinder Er = (Q / L) / ( 2 0 r) =   / ( 2 0 r)     where is the linear charge density on the cylinder

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