Answer: Magnitude of the electric field at a point which is 2.0 mm from the symmetry axis is 18.08 N/C.
Step-by-step explanation:
Given: Density = 80
(1 n =
m) =

= 1.0 mm (1 mm = 0.001 m) = 0.001 m
= 3.0 mm = 0.003 m
r = 2.0 mm = 0.002 m (from the symmetry axis)
The charge per unit length of the cylinder is calculated as follows.

Substitute the values into above formula as follows.
![\lambda = \rho \pi (r^(2)_(2) - r^(2)_(1))\\= 80 * 10^(-9) * 3.14 * [(0.003)^(2) - (0.001)^(2)]\\= 2.01 * 10^(-12) C/m](https://img.qammunity.org/2022/formulas/physics/college/693nd1dtd18yo5t6v0nj4itpj6ia8xax8g.png)
Therefore, electric field at r = 0.002 m from the symmetry axis is calculated as follows.

Substitute the values into above formula as follows.

Thus, we can conclude that magnitude of the electric field at a point which is 2.0 mm from the symmetry axis is 18.08 N/C.