Question

harge of uniform density (100 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. Group of answer choices

Answer 1

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

$Volume =\pi (R^2 -r^2)l\\\\Volume =\pi ((2*10^{-3})^2 -(1*10^{-3})^2)l\\\\Volume =\pi (4*10^{-6} - 1*10^{-6})l\\\\Volume = 3*10^{-6} \pi l \ m^3$

Area = 2πrl

Area =2π(2 x 10⁻)l

Volume = A x d

d = Volume / Area

$d = \frac{V}{A} = \frac{3*10^{-6}*\pi*l}{4\pi *10^{-3} l} = 75 *10^{-5} \ m$

the magnitude of the electric field

$E = \frac{\lambda *d}{\epsilon_o} = \frac{100*10^{-9} *75*10^{-5}}{8.854*10^{-12}} \\\\E = 8.47 \ N/C$

Therefore, the magnitude of the electric field is 8.47 N/C