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

The magnitude of the electric field is 8.47 N/C

Step-by-step 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 = (V)/(A) = (3*10^(-6)*\pi*l)/(4\pi *10^(-3) l) = 75 *10^(-5) \ m`

the magnitude of the electric field

`E = (\lambda *d)/(\epsilon_o) = (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