Question

Consider a long cylindrical charge distribution of radius R = 14 cm with a uniform charge density of ? = 20 C/m³. Find the electric field at a distance r = 25 cm from the axis.

Elecltrical Field = ?

Answer 1

Your equation states that the E field * surface area = enclosed charge / e_0

translate lambda from rho

lambda = 2*Pi*(rho)*(14cm). That is the "linear" charge density of the cylinder!

Now simply treat this as a uniformly charged wire and you are done.

Source(s):

http://farside.ph.utexas.edu/teaching/316/lectures/node26.html

Answer 2

Let us draw a cylindrical gaussian surface, co-axial with the wire, of radius and length . The above symmetry arguments imply that the electric field generated by the wire is everywhere perpendicular to the curved surface of the cylinder. Thus, according to Gauss' law,

$\begin{displaymath} E(R)\,2\pi \,R\, L = \frac{\lambda\, L}{\epsilon_0}, \end{displaymath}$

where
E(R)
is the electric field-strength a perpendicular distance from the wire. Here, the left-hand side represents the electric flux through the gaussian surface. Note that there is no contribution from the two flat ends of the cylinder, since the field is parallel to the surface there. The right-hand side represents the total charge enclosed by the cylinder, divided by $$\epsilon_0$$. It follows that

$\begin{displaymath} E(R) = \frac{\lambda}{2\pi\epsilon_0\,R}. \end{displaymath}$ now , translate lambda from rho lambda = 2*Pi*(rho)*(14cm). That is the "linear" charge density of the cylinder! Now simply treat this as a uniformly charged wire and you are done. lambda = 2*(22/7)*20*14/100 = 17.6 E(25) = 17.6/(2*pi*8.85*10^-12*25/100) = 1.26*10^12 Vm^-1