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|