--- title: "Line and point charges, find D" subject: "Electromagnetism" chapter: 2 key: "semester_3/electromagnetism:line-and-point-charges-find-d" source: https://www.bctengineer.com/q/semester_3/electromagnetism/line-and-point-charges-find-d site: BCT Engineering --- # Line and point charges, find D Electromagnetism · Chapter 2: Electric field - Asked in: 2076 Chaitra, 6 marks - Video: [Electric Field due to Point, Line & Surface charge | EMFT lecture](https://www.youtube.com/watch?v=xghSPIgqBmU) — CADDD Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electromagnetism/line-and-point-charges-find-d - Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2 ## Question An infinite uniform line charge ρ_L = 2 nC/m lies along the x-axis in free space, while point charges of 8 nC each are located at (0, 0, 1) and (0, 0, −1). Find D at (2, 3, −4). ## Short answer By superposition, D from the line charge is ρ_L/(2πρ) a_ρ; point charges contribute Qr/(4πr³). At P(2,3,−4), D = 0.0178 a_x + 0.0649 a_y − 0.0830 a_z nC/m². ## Answer **Given:** Line charge ρ_L = 2 nC/m along the x-axis; point charges 8 nC at (0, 0, 1) and (0, 0, −1). Find D at P(2, 3, −4). Use the superposition principle. #### 1. Line charge (along x-axis) The perpendicular distance from the x-axis to P is in the yz-plane: R_L = 3 a_y − 4 a_z, |R_L| = 5. ```text D_L = ρ_L / (2π ρ) · a_ρ = 2 / (2π × 5) × (3 a_y − 4 a_z)/5 nC/m² = 0.06366 (0.6 a_y − 0.8 a_z) = 0.0382 a_y − 0.0509 a_z nC/m² ``` #### 2. Point charge at (0, 0, 1) ```text R₁ = P − (0, 0, 1) = 2 a_x + 3 a_y − 5 a_z, |R₁| = √38 = 6.164 |R₁|³ = 234.2 D₁ = Q R₁ / (4π |R₁|³) = 8 (2, 3, −5) / (4π × 234.2) nC/m² = 2.718 × 10⁻³ (2, 3, −5) nC/m² ``` ```text D₁ = (0.00544 a_x + 0.00815 a_y − 0.01359 a_z) nC/m² ``` #### 3. Point charge at (0, 0, −1) ```text R₂ = P − (0, 0, −1) = 2 a_x + 3 a_y − 3 a_z, |R₂| = √22 = 4.690 |R₂|³ = 103.2 D₂ = 8 (2, 3, −3) / (4π × 103.2) = 6.17 × 10⁻³ × (2, 3, −3) nC/m² = (0.01234 a_x + 0.01851 a_y − 0.01851 a_z) nC/m² ``` #### 4. Total ```text D = D_L + D₁ + D₂ D_x = 0 + 0.00544 + 0.01234 = 0.01777 D_y = 0.0382 + 0.00815 + 0.01851 = 0.0649 D_z = −0.0509 − 0.01359 − 0.01851 = −0.0830 ``` **Answer: D = 0.0178 a_x + 0.0649 a_y − 0.0830 a_z nC/m²**, i.e. D = 17.8 a_x + 64.9 a_y − 83.0 a_z pC/m². (|D| ≈ 0.107 nC/m².) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What principle is used to find D at P from a line charge and two point charges? A. Use only the nearest charge B. Superposition: add the D from each source as vectors C. Add the magnitudes of D scalar-wise D. Apply Gauss's law with a single cube Answer: B. Fields of the individual sources add vectorially. (Key point: Method) 2. For the line along the x-axis, what is the perpendicular distance to P(2, 3, -4)? A. 5 m, since only y and z matter B. √29 C. 7 m D. √38 Answer: A. The x-coordinate does not matter; ρ = √(3² + 4²) = 5 m. (Key point: Line charge distance) 3. Which formula gives D due to a point charge Q at the position R from it? A. D = Q / (2π|R|) B. D = Q R / (2π|R|²) C. D = Q / (4π|R|) D. D = Q R / (4π|R|³) Answer: D. D = Q a_R/(4πR²) = Q R/(4π|R|³). (Key point: Point charge formula) 4. What is the resulting D at P? A. 0.0649 a_x + 0.0178 a_y - 0.0830 a_z nC/m² B. -0.0178 a_x - 0.0649 a_y + 0.0830 a_z nC/m² C. 0.0178 a_x + 0.0649 a_y - 0.0830 a_z nC/m² D. 0.0178 a_x - 0.0649 a_y + 0.0830 a_z nC/m² Answer: C. Summing the line charge and both point charge contributions gives this vector. 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