- 2075 Chaitra · 7 marks
Point charge and four line charges, D
Short answer
Superposition: the four line charges are symmetrically placed about y=−3, so their contributions cancel. Only the 12 nC point charge at the origin contributes: D = −61.1 a_y + 40.8 a_z pC/m².
A point charge of 12 nC is located at the origin. Four uniform line charges are located in the x = 0 plane as follows: 80 nC/m at y = −1 and −5 m, −50 nC/m at y = −2 and −4 m. Find the electric flux density D at P(0, −3, 2).
Given: Point charge 12 nC at the origin. Four infinite line charges lie in the x = 0 plane, each at a fixed y, so they run parallel to the z-axis: 80 nC/m at y = −1 and y = −5 m; −50 nC/m at y = −2 and y = −4 m. Find D at P(0, −3, 2).
Use superposition. For a line charge, D = ρ_L/(2πρ) a_ρ, where ρ is the perpendicular distance. A line at (x, y) = (0, y₀) is parallel to the z-axis, so the perpendicular vector from it to P is (0, −3 − y₀, 0), which lies along a_y only.
Point charge
R = P − 0 = −3 a_y + 2 a_z, |R| = √13 = 3.606
D_pt = Q R / (4π |R|³) = (12 × 10⁻⁹)(−3 a_y + 2 a_z) / (4π × 46.87)
= (−0.0611 a_y + 0.0408 a_z) nC/m²
Line charges
| Line | ρ_L (nC/m) | Vector to P | ρ (m) | D (nC/m²) |
|---|---|---|---|---|
| y = −1 | 80 | −2 a_y | 2 | −6.366 a_y |
| y = −5 | 80 | +2 a_y | 2 | +6.366 a_y |
| y = −2 | −50 | −1 a_y | 1 | +7.958 a_y |
| y = −4 | −50 | +1 a_y | 1 | −7.958 a_y |
For example, line 1: D = 80/(2π × 2) × (−a_y) = −6.366 a_y nC/m². Line 3: D = −50/(2π × 1) × (−a_y) = +7.958 a_y nC/m².
Sum of the four line contributions:
D_y: −6.366 + 6.366 + 7.958 − 7.958 = 0
The line charges are placed symmetrically about y = −3, so they cancel in pairs at P.
Total
D = D_pt = −0.0611 a_y + 0.0408 a_z nC/m²
Answer: D = −61.1 a_y + 40.8 a_z pC/m² (|D| ≈ 73.4 pC/m²).
✓ Answer checked by Claude Opus 5.5 (high effort)
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