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|                      

insects!

explanation:

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards