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Flux through sphere, two line charges

Short answer

By Gauss’s law, flux equals enclosed charge. Each line charge cuts the sphere; chord length L=2√(R²−d²)=2√3 m. Total enclosed charge = 2 × 20 nC/m × 2√3 m ≈ 138.6 nC.

Two uniform line charges, each 20 nC/m, are located at y = 1, z = ±1 m. Find the total electric flux leaving the surface of a sphere having a radius of 2 m, if it is centered at A(3, 1, 0).

Given: Two uniform line charges, ρ_L = 20 nC/m each, at y = 1, z = ±1 m (running parallel to the x-axis). Sphere of radius 2 m centred at A(3, 1, 0). Find the total flux leaving the sphere.

Method: By Gauss's law, flux Ψ = charge enclosed by the sphere = ρ_L × (length of line inside the sphere), summed for both lines.

Distance of each line from the centre

The lines are parallel to the x-axis, so the perpendicular distance from the centre A(3, 1, 0) to the line at (y, z) = (1, ±1) is

d = √( (1 − 1)² + (±1 − 0)² ) = 1 m

Since d = 1 < R = 2, each line cuts through the sphere.

Length of the line inside the sphere

The line passes at distance d from the centre, so the chord length is

L = 2 √(R² − d²) = 2 √(4 − 1) = 2√3 = 3.464 m

          ╭───────╮
         ╱    d    ╲        chord of length L
        │  ●────┼───│───    ← line charge
         ╲  A      ╱
          ╰───────╯

Charge enclosed and flux

Q_one line = ρ_L L = 20 × 10⁻⁹ × 3.464 = 69.28 nC
Q_total = 2 × 69.28 = 138.6 nC

Answer: Ψ = 138.6 nC ≈ 139 nC (outward flux, since charges are positive).

✓ Answer checked by Claude Opus 5.5 (high effort)

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