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.

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|                      

(75°f − 32) × 5/9 = 23.889°c

pe = weightxheight = 2000x1000=2,000,000joules