A very long, solid cylinder with radius R has positive charge uniformly distributed throughout it, with charge per unit volume \rhorho.
(a) Derive the expression for the electric field inside the volume at a distance r from the axis of the cylinder in terms of the charge density \rhorho.
(b) What is the electric field at a point outside the volume in terms of the charge per unit length \lambdaλ in the cylinder?
(c) Compare the answers to parts (a) and (b) for r = R.
(d) Graph the electric-field magnitude as a function of r from r = 0 to r = 3R.

Applying Gauss's law which states that the net electric field in an enclosed surface is directly ∝ to the charge found in the enclosed surface.

A ) The expression for the electric field inside the volume at a distance r

Gauss law : E. A = [tex]\frac{q}{e_{0} }[/tex] ----- ( 1 )

where : A = surface area = 2πrL , q = p(πr²L)

back to equation ( 1 )

E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]

B) Electric field at point Outside the volume in terms of charge per unit length λ

Given that: linear charge density = area * volume charge density

λ = πR²P

from Gauss's law : E ( 2πrL) = [tex]\frac{q}{e_{0} }[/tex]

∴ E = [tex]\frac{\pi R^{2}P }{2e_{0}r\pi }[/tex] ----- ( 2 )

where : πR²P = λ

Back to equation ( 2 )

E = λ / 2e₀π*r where : k = 1 / 4πe₀

∴ The electric field ( E ) at point outside the volume in terms of charge per unit Length λ

E = 2K λ / r

C) Comparing answers A and B

Answers to part A and B are similar

Hence we can conclude that Applying Gauss law to the long solid cylinder

E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex], E = 2K λ / r also Answers from parts a and b are the same.

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