A long cylindrical conductor (radius = 1.0 mm) carries a charge density of 4.0 pC/m and is inside a coaxial, hollow, cylindrical conductor (inner radius = 3.0 mm, outer radius = 4.0 mm) that has a total charge of −8.0 pC/m. What is the magnitude of the electric field 2.0 mm from the axis of these conductors?

Question

Quizplus · Accepted Answer

Since the system is symmetric around the axis, we can use Gauss's Law to find the electric field.

Consider a cylindrical Gaussian surface of radius 2.0 mm and height L (which we will take to infinity). By symmetry, the electric field is perpendicular to the surface at every point on it, and has the same magnitude at every point on a given circle.

The electric field magnitude at a point distance r from the axis is given by:

E = (1/2ε₀)(q_enclosed / πr²),

where ε₀ is the permittivity of free space, and q_enclosed is the charge enclosed inside the Gaussian surface.

The charge enclosed by the Gaussian surface includes the charge inside the smaller cylinder and also the charge on the inner surface of the larger cylinder. The charge on the outer surface of the larger cylinder does not contribute since it is outside the Gaussian surface.

The total charge q_enclosed is:

q_enclosed = q_small + q_inner,

where

q_small = λ_small x L = λ_small x π(0.001 m)²,

q_inner = λ_inner x L = λ_inner x π[(0.004 m)² - (0.003 m)²].

Substituting in the values, we get:

q_small = (4 pC/m)(π(0.001 m)²) = 1.2566 x 10^-8 C

q_inner = (-8 pC/m)(π[(0.004 m)² - (0.003 m)²]) = -2.5132 x 10^-8 C

q_enclosed = -1.2566 x 10^-8 C

Therefore,

E = (1/2ε₀)(q_enclosed / πr²) = (1/2ε₀)(-1.2566 x 10^-8 C / π(0.002 m)²)

= 35.98 N/C ≈ 36 N/C

So the answer is D.

A Long Cylindrical Conductor (Radius = 1