Current density is given in cylindrical coordinates as J = −106z1.5az A/m2 in the region 0 ≤ rho ≤ 20 µm; for rho ≥ 20 µm, J = 0

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Current density is given in cylindrical coordinates as J = −106z1.5az A/m2 in the region 0 ≤ rho ≤ 20 µm; for rho ≥ 20 µm, J = 0

. (a) Find the total current crossing the surface z = 0.1 m in the az direction. (b) If the charge velocity is 2 × 106 m/s at z = 0.1 m, find rhoν there. (c) If the volume charge density at z = 0.15 m is −2000 C/m3, find the charge velocity there.

Engineering

1 answer:

Mrrafil [318] · Answer 1 · 2026-03-05T07:48:53+03:00

a. Total current yields -39.8μA. b. The value of ρν at z = 0.1 m is -15.81mC/m³. c. The charge velocity at z = 0.15 m is calculated to be 29.05m/s. To explain further: Total current is computed as follows: Using the formula J * ½((ρ1)² - (ρ0)²) * 2 π * φdza, where J represents Density = -10^6 * z^1.5, with ρ1 as 20 (upper limit for ρ), ρ0 as 0 (lower limit for ρ), π approximated as 22/7, and φdza as 10^-6, considering z = 0.1: The overall current then becomes: = -10^6 * z^1.5 * ½(20² - 0²) * 2 * 22/7 * 10^-6, = 10^6 * 0.1^1.5 * ½(20² - 0²) * 2 * 22/7 * 10^-6, which resolves to approximately -39.8μA. b. To derive the charge density velocity (ρv): Density (J) = ρv * V, where J remains the same as before, V is noted to be 2 * 10^6, within the context of z = 0.1: Substituting generates -10^6 * 0.1 ^1.5 = ρv * 2 * 10^6, from which it follows that ρv = (-10^6 * 0.1^1.5)/(2 * 10^6), leading to the final computation of ρv = -0.01581mC/m³. c. To evaluate velocity at z = 0.15: Velocity is computed as J/V: Given that the volume charge density amounts to -2000 C/m³, J at this height calculates to -10^6 * 0.15^1.5: Thus, Velocity = -58094.75019311125/-2000, which results in approximately 29.05m/s.