A long pipe of outer radius R 1 = 3.70 cm and inner radius R 2 = 3.15 cm carries a uniform charge density of 1.22 mC/m 3. Assuming that the pipe is sufficiently long to consider it infinitely long, use Gauss's law calculate the electric field E at a distance r = 7.77 cm from the centerline of the pipe. Use ε 0 = 8.85 * 10 - 12 C / N⋅m 2 for the the permittivity of free space.

E = 4.72 * 10⁻⁶ Nm²

Explanation:

Parameters given:

Outer radius, R = 3.70cm = 0.037m

Inner radius, r = 3.15cm = 0.0315m

Permittivity of free space, ε₀ = 8.85 * 10⁻¹² C/Nm²

Charge density: 1.22 * 10⁻³ C/m³

The question requires that we solve using Gauss law which states that the net electric field through a closed surface is proportional to the enclosed electric charge.

Hence,

E = Q/Aε₀

Charge Q is given as

Q = ρπ (R² ⁻ r²) L

A = 2π (R - r) L

E = [ρπ (R² ⁻ r²) L]/[2π (R - r) ε₀L]

Using difference of two squares,

(R² ⁻ r²) = (R + r) (R - r)

E = [ρ (R + r) ] / (2ε₀)

E = [1.22 * 10⁻³ * (0.0370 + 0.0315) ] / (2 * 8.85 * 10⁻¹²)

E = 4.72 * 10⁻⁶ Nm²