From the expression for EE obtained in Problem 22.42,22.42, find the expressions for the electric potential V as a function of r, both inside and outside the cylinder. Let V=0 at the surface of the cylinder. In each case, express your result in terms of the charge per unit length λ of the charge distribution. (b) Graph V and E as functions of rr from r=0 to r=3R.

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shahnoorazhar3 shahnoorazhar3 · Answer 1 · 2020-02-15T18:55:11+01:00

Answer:

a) V(r) = k*λ*r^2/R^2 r =< R

V(r) = -2*k*λ*Ln|r/R|

Explanation:

Given:

- The derived results for Electric Fields are:

r =< R, E(r) = 2*k*λ*r / R^2

r > R, E(r) = 2*k*λ/ r

Find:

-Expressions for the electric potential V as a function of r, both inside and outside the cylinder.

Solution:

- From definition we can establish the relation between E(r) and V(r) as follows:

E(r) = - dV / dr

- We will develop expression for each case as follows:

Case 1: r =< R

E(r) = 2*k*λ*r / R^2 = - dV / dr

Separate variables:

2*k*λ*r . dr / R^2 = -dV

Integrating both sides:

2*k*λ/R^2 integral(r).dr = -integral(dv)

k*λ*r^2/R^2 | = - ( 0 - V)

Put limits ( 0-r)

V(r) = k*λ*r^2/R^2

Case 2: r >= R

E(r) = 2*k*λ/ r = - dV / dr

Separate variables:

2*k*λ.dr/ r = -dV

Integrating both sides:

2*k*λ integral(1/r).dr = -integral(dv)

2*k*λ*Ln|r/R| = - ( V - 0)

Put limits ( R -> r)

V(r) = -2*k*λ*Ln|r/R|