This problems presents a symmetric (cylindrical) charge distribution. The electric field of such distribution can be found using Gauss's Law, which states that the electric flux through a given closed surface equals the net electric charge enclosed by that surface, divided by `epsilon_0`

The surface of choice here is a cylinder of height *h* and radius
*r* = 1 cm, coaxial with the given cylinder. Since the given cylinder is
infinitely long, and the charge density is uniform, there is symmetry: the
electric field is horizontal (assuming the cylinder is vertical) and spreads
out radially away from the axis of the cylinder. The flux of this field through
the top and bottom of the cylinder is zero. Also, the electric field is
constant on the cylindrical surface itself, and perpendicular to it at every
point.

Then, the flux through this surface is field times the area of the surface `E*2*pi*r*h`

The charge enclosed by this surface equals the volume charge density times the volume of enclosed by the surface:

`rho*pi*r^2*h`

According to Gauss's Law,

`E*2*pi*r*h = (rho*pi*r^2*h)/epsilon_0`

From here,

`E = (rho*r)/(2*epsilon_0)`

Notice that we found the electric field as a function of the distance away
from the axis of the cylinder for all points—as long as they are inside the
given cylinder. Notice also that the field is independent on the radius
*R* of the cylinder.

Plugging in the given values and converting units results in

`E = (18*10^(-6)*0.01)/(2*8.85*10^(-12)) = 1.02*10^4 N/C`

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