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A long nonconducting cylinder (radius = 12 cm) has a charge of uniform density (5.0 nC/m3) distributed throughout its volume. Determine the magnitude of the electric field 15 cm from the axis of the cylinder. Group of answer choices 12 N/C 20 N/C 27 N/C 16 N/C 54 N/C

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onyebuchinnaji

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

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