Answer:
The magnitude of the electric field is 8.47 N/C
Explanation:
Given;
uniform charge density, λ = 100 nC/m³
inner radii of the cylinder, r = 1.0 mm and 3.0 mm
distance from the symmetry axis, R = 2.0 mm
[tex]Volume =\pi (R^2 -r^2)l\\\\Volume =\pi ((2*10^{-3})^2 -(1*10^{-3})^2)l\\\\Volume =\pi (4*10^{-6} - 1*10^{-6})l\\\\Volume = 3*10^{-6} \pi l \ m^3[/tex]
Area = 2πrl
Area =2π(2 x 10⁻)l
Volume = A x d
d = Volume / Area
[tex]d = \frac{V}{A} = \frac{3*10^{-6}*\pi*l}{4\pi *10^{-3} l} = 75 *10^{-5} \ m[/tex]
the magnitude of the electric field
[tex]E = \frac{\lambda *d}{\epsilon_o} = \frac{100*10^{-9} *75*10^{-5}}{8.854*10^{-12}} \\\\E = 8.47 \ N/C[/tex]
Therefore, the magnitude of the electric field is 8.47 N/C
