Answer:
The magnitude of the electric field is 27 N/C
Explanation:
Given;
radius of the cylinder, r = 12 cm = 0.12 m
uniform density, λ = 5.0 nC/m³
distance from the axis, R = 15 cm = 0.15 m
Volume of a cylinder, V = πr²h
charge on the cylinder, q = V λ = πr²hλ = π (0.12)²h x 5.0 = 0.072πh nC
from Gauss law;
[tex]EA = \frac{q}{\epsilon_o}[/tex]
[tex]E * 2\pi Rh = \frac{q}{\epsilon_o}[/tex]
Substitute in the value of q into this equation and solve for E
q = 0.072πh nC
[tex]E * 2(0.15)\pi h = \frac{0.072\pi h*10^{-9}}{8.85*10^{-12}}\\\\E = \frac{0.072*10^{-9}}{0.3*8.85*10^{-12}} = 27.1 \ N/c[/tex]
Therefore, the magnitude of the electric field is 27 N/C
