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12 December, 13:10

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.

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  1. 12 December, 13:34
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    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
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