Question

Charge of uniform density (80 nC/m3) is distributed throughout a hollow cylindrical region formed by two coaxial cylindrical surfaces of radii 1.0 mm and 3.0 mm. Determine the magnitude of the electric field at a point which is 2.0 mm from the symmetry axis.

Answer 1

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
`nC/m^(3)` (1 n =
`10^(-9)` m) =
`80 * 10^(-9) C/m^(2)`

`r_(1)` = 1.0 mm (1 mm = 0.001 m) = 0.001 m

`r_(2)` = 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.

`\lambda = \rho \pi (r^(2)_(2) - r^(2)_(1))`

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`

Therefore, electric field at r = 0.002 m from the symmetry axis is calculated as follows.

`E = (\lambda)/(2 \pi r \epsilon_(o))`

Substitute the values into above formula as follows.

`E = (\lambda)/(2 \pi r \epsilon_(o))\\= (2.01 * 10^(-12) C/m)/(2 * 3.14 * 0.002 m * 8.85 * 10^(-12))\\= 18.08 N/C`

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.