Derive a relation between the potential V and the magnitude of the field E at a radial distance, r, from the a?

Derive a relation between the potential V and the magnitude of the field E at a radial distance, r, from the axis of a very long uniformly charged rod of radius a (r>a)

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  • 7 years ago
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    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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