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An infinitely long cylinder of radius R has linear charge density λ. The potential on the surface of the cylinder is V0, and the electric field outside the cylinder is Er=λ2πϵ0r. Part A Find the potential relative to the surface at a point that is distance r from the axis, assuming r>R. Express your answer in terms of the variables λ, r, R, V0, and appropriate constants. Vr = nothing

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zuwairahdahir

Answer:

V = V_0 - (lamda)/(2pi(epsilon_0))*ln(R/r)

Explanation:

Attached is the full solution

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Cricetus

The expression will be "[tex]V = V_0-\frac{\lambda}{2 \pi \epsilon_0} ln(\frac{R}{r} )[/tex]".

Electric field:

Whenever charge seems to be present in whatsoever configuration or manner, an electric property has been linked among each point in space.

This same electric field might well be conceived of as the force per unit positively charged ions that would have been exerted well before appearance of the charged particle disturbs the field.

→ [tex]E = \frac{\lambda}{2 \pi \epsilon_0R}[/tex]

now,

→       [tex]\int\limits^V_{V_b} = \int\limits^R_{r} E.dr[/tex]

 [tex]V -V_0 = \int\limits^R_r {\frac{\lambda}{2 \pi \epsilon_0 R} }[/tex]

             [tex]= \frac{\lambda}{2 \pi \epsilon_0} (GR)^R_r[/tex]

         [tex]V = V_0 - \frac{\lambda}{2 \pi \epsilon_0} ln (\frac{R}{r} )[/tex]  

Find out more information about electric potential here:

https://brainly.com/question/4185099

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