Yahoo Answers: Answers and Comments for Derive a relation between the potential V and the magnitude of the field E at a radial distance, r, from the a? [Physics]
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From Destiny
enUS
Sat, 04 Aug 2012 20:55:33 +0000
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Yahoo Answers: Answers and Comments for Derive a relation between the potential V and the magnitude of the field E at a radial distance, r, from the a? [Physics]
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https://answers.yahoo.com/question/index?qid=20120804205532AAAt5MW
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From jeffrey: Answer ... V =  r E • ln [ r / a ] ...
Consid...
https://answers.yahoo.com/question/index?qid=20120804205532AAAt5MW
https://answers.yahoo.com/question/index?qid=20120804205532AAAt5MW
Wed, 08 Aug 2012 01:54:24 +0000
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 zaxis and that the field point lies on the xyplane 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 ] …