--- title: "Point charge and four line charges, D" subject: "Electromagnetism" chapter: 2 key: "semester_3/electromagnetism:point-and-four-line-charges-d" source: https://www.bctengineer.com/q/semester_3/electromagnetism/point-and-four-line-charges-d site: BCT Engineering --- # Point charge and four line charges, D Electromagnetism · Chapter 2: Electric field - Asked in: 2075 Chaitra, 7 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electromagnetism/point-and-four-line-charges-d - Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2 ## Question 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). ## 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. ## Answer **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 ```text 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: ```text 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 ```text 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). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is D of an infinite line charge at perpendicular distance ρ? A. ρ_L / (2πρ) a_ρ B. ρ_L / (4πρ²) a_ρ C. ρ_L / (2π ρ²) a_ρ D. ρ_L ρ / (2π) a_ρ Answer: A. D = ρ_L/(2πρ) in the radial direction. (Key point: Line charge formula) 2. For the lines in the x = 0 plane parallel to the x-axis, the perpendicular vector from the line at y_0 to P(0, -3, 2) is: A. (−3, 0, 2) B. (0, -3, 2 - y₀) C. (0, 3 + y₀, 2) D. (0, -3 - y₀, 2) Answer: D. Each line is described by its y position, so the vector is (-3 - y₀) a_y + 2 a_z. (Key point: Using perpendicular vector) 3. What happens to the four line charge contributions at P? A. They add to give a large field B. Only the 80 nC/m lines cancel C. They cancel in pairs, giving zero net D. Only D_z cancels Answer: C. The symmetric positions about y = -3 make the y- and z-components cancel. (Key point: Cancellation) 4. Therefore D at P equals: A. 0 B. -61.1 a_y + 40.8 a_z pC/m² from the point charge only C. +61.1 a_y - 40.8 a_z pC/m² D. -3.18 a_y + 3.18 a_z nC/m² Answer: B. With the lines cancelling, D equals the point charge field Q R/(4π|R|³). 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