Respuesta :
Answer: Magnitude of the electric field at a point which is 2.0 mm from the symmetry axis is 18.08 N/C.
Explanation:
Given: Density = 80 [tex]nC/m^{3}[/tex] (1 n = [tex]10^{-9}[/tex] m) = [tex]80 \times 10^{-9} C/m^{2}[/tex]
[tex]r_{1}[/tex] = 1.0 mm (1 mm = 0.001 m) = 0.001 m
[tex]r_{2}[/tex] = 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.
[tex]\lambda = \rho \pi (r^{2}_{2} - r^{2}_{1})[/tex]
Substitute the values into above formula as follows.
[tex]\lambda = \rho \pi (r^{2}_{2} - r^{2}_{1})\\= 80 \times 10^{-9} \times 3.14 \times [(0.003)^{2} - (0.001)^{2}]\\= 2.01 \times 10^{-12} C/m[/tex]
Therefore, electric field at r = 0.002 m from the symmetry axis is calculated as follows.
[tex]E = \frac{\lambda}{2 \pi r \epsilon_{o}}[/tex]
Substitute the values into above formula as follows.
[tex]E = \frac{\lambda}{2 \pi r \epsilon_{o}}\\= \frac{2.01 \times 10^{-12} C/m}{2 \times 3.14 \times 0.002 m \times 8.85 \times 10^{-12}}\\= 18.08 N/C[/tex]
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.
