The cross section of a long coaxial cable is shown in the figure,with radii as given.The linear charge density on the inner conductor is -30 nC/m and the linear charge density on the outer conductor is
-70 nC/m. The inner and outer cylindrical surfaces are respectively denoted by A,B,C,and D,as shown.(ε₀ = 8.85 × 10^-12 C²/N • m²) The radial component of the electric field at a point that 34 mm from the axis is closest to

Question

Quizplus · Accepted Answer

Using Gauss's Law, the electric field at a distance r from the center (where r is between r1 and r2) can be found by:
E = (λ1 - λ2) / ε
where λ1 is the linear charge density on the inner conductor (-30 nC/m), λ2 is the linear charge density on the outer conductor (-70 nC/m), and ε is the permittivity of free space (8.85 × 10^-12 N·m^2/C^2).

At a distance of 34 mm from the center, the electric field is given by:
E = (-30 × 10^-9 - (-70 × 10^-9)) / (8.85 × 10^-12)
E = 40 × 10^3 N/C (directed radially inward)

Since the question asks for the radial component of the electric field, which is the magnitude of the electric field at that point, the answer is:
|E| = 40,000 N/C or 16,000 N/C (closest choice)

The Cross Section of a Long Coaxial Cable Is Shown