Best Answer:
Answer ... V = - r E • ln [ r / a ] ...

Consider the rod as a very long uniformly charged cylinder, with constant

volume charge density ρ₀ and radius a₀ … assume that the axis of the cylinder

lies along the z-axis and that the field point lies on the xy-plane at a

perpendicular distance r from the cylinder axis … by taking an imaginary

cylindrical Gaussian surface S with radius r and height h containing the field

point on its lateral side, we find that the total charge Q enclosed by S is …

…Q = ρ₀ V = ρ₀ ( π a ² ) h … then, according to Gauss law, the flux ϕ of the

electric field through S is … ϕ = Q / ϵ₀ = ρ₀ π a ² h / ϵ₀ …

Now, since the electric field E due to the cylindrical charge distribution is

radially away from the axis of the cylinder, the flux of E is through the lateral

side of S only … there’s no flux of E through the two circular sides at both ends

of S since E is parallel to them … also, E is parallel to the outward unit normal

vectors at the lateral side of S, so that in accordance with the definition of flux

… ϕ = ∫∫ E ( dS ) cos θ = E cos θ ∫∫ dS = E cos 0° ∫∫ dS = E ∫∫ dS … since the

magnitude of E is the same everywhere at all points on the lateral side of S for

a fixed radius r … it follows that … ϕ = E ∫∫ dS = E ( 2 π r h )

…………………………………………. = Q / ϵ₀ = ρ₀ π a ² h / ϵ₀ … and solving for

E, we get … E = ρ₀ π a ² h / [ ϵ₀ ( 2 π r h ) ] = ρ₀ π a ² / [ 2 π ϵ₀ r ]

……………….. = 2 π ρ₀ a ² / [ 4 π ϵ₀ r ] …

Now, since E = - ∇ V —> V = - ∫ E dr = - { 2 π ρ₀ a ² / [ 4 π ϵ₀ ] } ∫ dr / r

……………………………….. = - { 2 π ρ₀ a ² / [ 4 π ϵ₀ ] } ln r evaluated from r = a

to r, we get … V = - { 2 π ρ₀ a ² / [ 4 π ϵ₀ ] } ln [ r / a ] … and using the expression

for E, we find that … 2 π ρ₀ a ² / [ 4 π ϵ₀ ] = r E … which, when substituted in V

gives … V = - r E • ln [ r / a ] …

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