A long nonconducting cylinder (radius = 12 cm) has a charge of uniform density (5.0 nC/m³) distributed throughout its column. Determine the magnitude of the electric field 5.0 cm from the axis of the cylinder.

Question

Quizplus · Accepted Answer

Assuming the cylinder is infinitely long, we can use Gauss's Law to find the electric field. We construct a cylindrical Gaussian surface with radius 5.0 cm and length L that surrounds the cylinder. The electric field is radial, and has the same magnitude at every point on the surface due to the symmetry of the charge distribution. The electric flux through the curved surface of the cylinder is zero since the electric field is perpendicular to it. The electric flux through the top and bottom surfaces is also zero since the electric field is parallel to them. Therefore, the electric flux through the Gaussian surface is simply the product of the electric field and the area of the two flat ends of the cylinder:
flux = 2 * E * π(0.05 m)^2
Applying Gauss's Law, the electric flux through the Gaussian surface is also equal to the total charge q enclosed by the surface divided by the permittivity of free space ε0:
flux = q/ε0
Solving for the electric field, we get:
E = q/(2*πε0rL) = (ρL/2ε0)r
where ρ is the charge density, which is given as 5.0 nC/m, and L is the length of the cylinder (which doesn't affect the electric field at points far from the ends, so we can take L to be any convenient value, such as 1 m). Plugging in the values and solving, we get:
E = (5.0 nC/m * 0.05 m)/(2 * π * 8.85×10^-12 C^2/N·m^2 * 1 m) = 14 N/C.

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