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.

Question

Quizplus · Accepted Answer

We can use Gauss's law to find the electric field at a distance of 15 cm from the axis of the cylinder. We choose a Gaussian cylinder with radius 15 cm and height L. The electric field will be uniform and perpendicular to the curved surface of the cylinder, so we can simplify the calculation to:
ϕ = EAd = qenc/ε0
where ϕ is the electric flux through the surface, E is the electric field, A is the area of the surface, d is the distance between the surfaces, qenc is the charge enclosed by the surface, and ε0 is the permittivity of free space.
The charge enclosed by the Gaussian cylinder is:
qenc = ρV = ρπr²L = (5.0 nC/m³)(π(0.12 m)²(0.15 m)) = 0.0324 nC
The area of the curved surface of the Gaussian cylinder is:
A = 2πrh = 2π(0.12 m)(0.15 m) = 0.113 m²
Using d = 0.15 m and ε0 = 8.85 x 10^-12 F/m, we can solve for the electric field:
E = ϕ/Ad = qenc/ε0Ad = (0.0324 nC)/(8.85 x 10^-12 F/m)(0.113 m²) = 27 N/C. 
Therefore, the correct answer is B.

A Long Nonconducting Cylinder (Radius = 12 Cm) Has a Charge