Question

The cross section of a long coaxial cable is shown in the figure, with radii as given. The linear charge density on the inner coductor is -20 nC/m and the linear charge density on the outer conductor is -60 nC/m The inner and outer cylindrical surfaces are respectively denoted by A, B, C, and D, as shown. (-o-885: 10-12 c2N.m2) The radial component of the electric field at a point that 35 mm from the axis is closest to D B 28 mm 1 30 mm 49 mm -51 mm O A) -31,000 N/C. B) -10,000 N/C. C) +31,000 N/C. D) zero E) +10,000 N/C.

Answer 1

The radial component of the electric field at a point 35 mm from the axis of the coaxial cable is approximately +31,000 N/C.

Step-by-step explanation:

The radial component of the electric field at a point that is 35 mm from the axis of the long coaxial cable can be found using Gauss' law. In this case, since the electric field is directed radially, we can consider a cylindrical Gaussian surface with its axis aligned with the axis of the cable. The electric field at a distance r from the axis can be calculated as:

E = (λ_in - λ_out) / (2πε₀r)

where λ_in is the linear charge density on the inner conductor, λ_out is the linear charge density on the outer conductor, ε₀ is the permittivity of free space (8.85 x 10-12 C2/(N.m2)), and r is the distance from the axis. Plugging in the given values:

E = (-20 x 10-9 C/m - (-60 x 10-9 C/m)) / (2π(8.85 x 10-12 C2/(N.m2))(0.035 m)

E ≈ +31,000 N/C

Therefore, the closest answer choice is C) +31,000 N/C.

Answer 2

The radial component of the electric field at a point that is 35 mm from the axis, closest to D, is approximately -31,000 N/C.

The correct answer is A) -31,000 N/C.

Step-by-step explanation:

To determine the radial component of the electric field at a point that is 35 mm from the axis, we need to consider the electric field contributions from both the inner and outer conductors.

1. Inner Conductor (A and B):

- The linear charge density on the inner conductor is -20 nC/m.

- The radius of the inner conductor is 28 mm.

- The point of interest is 35 mm from the axis, which is within the inner conductor.

- Since the point is inside the inner conductor, the electric field contribution from the inner conductor will be zero.

2. Outer Conductor (C and D):

- The linear charge density on the outer conductor is -60 nC/m.

- The radius of the outer conductor is 49 mm.

- The point of interest is 35 mm from the axis, which is within the outer conductor.

- To calculate the electric field contribution from the outer conductor, we can use the formula for the electric field due to a charged cylindrical shell:

E = (k * λ) / r

where E is the electric field, k is the electrostatic constant (8.85 x 10⁽⁻¹²⁾C²/N.m²), λ is the linear charge density, and r is the distance from the axis.

- Plugging in the values, we get:

E = (8.85 x 10⁽⁻¹²⁾ * (-60 x 10⁽⁻⁹⁾) / 0.035

E ≈ -31,000 N/C

Therefore, the radial component of the electric field at a point that is 35 mm from the axis, closest to D, is approximately -31,000 N/C.

The correct answer is A) -31,000 N/C.