6.3 Applying gauss’s law  (Page 3/13)

Uniformly charged cylindrical shell

Strategy

Solution

1. Electric field at a point outside the shell. For a point outside the cylindrical shell, the Gaussian surface is the surface of a cylinder of radius $r>R$ and length L , as shown in [link] . The charge enclosed by the Gaussian cylinder is equal to the charge on the cylindrical shell of length L . Therefore, ${\lambda }_{\text{enc}}$ is given by
   
   ${\lambda }_{\text{enc}}=\frac{{\sigma }_{0}2\pi RL}{L}=2\pi R{\sigma }_0.$
   
   

   

   
   Hence, the electric field at a point P outside the shell at a distance r away from the axis is
   
   $\overrightarrow{\text{E}}=\frac{2\pi R{\sigma }_0}{2\pi {\epsilon }_o}\frac{1}{r}\widehat{\text{r}}=\frac{R{\sigma }_0}{{\epsilon }_o}\frac{1}{r}\widehat{\text{r}}(r>R)$
   
   where $\widehat{\text{r}}$ is a unit vector, perpendicular to the axis and pointing away from it, as shown in the figure. The electric field at P points in the direction of $\widehat{\text{r}}$ given in [link] if ${\sigma }_0>0$ and in the opposite direction to $\widehat{\text{r}}$ if ${\sigma }_0<0$ .
2. Electric field at a point inside the shell. For a point inside the cylindrical shell, the Gaussian surface is a cylinder whose radius r is less than R ( [link] ). This means no charges are included inside the Gaussian surface:
   
   ${\lambda }_{\text{enc}}=0.$
   
   

   

   
   This gives the following equation for the magnitude of the electric field $E_{\text{in}}$ at a point whose r is less than R of the shell of charges.
   
   $E_{\text{in}}2\pi rL=0(r<R),$
   
   This gives us
   
   $E_{\text{in}}=0(r<R).$

Significance