--- title: "Position of third charge for zero force" subject: "Engineering Physics" chapter: 7 key: "semester_1/engineering_physics:third-charge-no-net-force" source: https://www.bctengineer.com/q/semester_1/engineering_physics/third-charge-no-net-force site: BCT Engineering --- # Position of third charge for zero force Engineering Physics · Chapter 7: Electrostatics - Asked in: 2071 Chaitra, 5 marks - Video: [Find where a third charge would be in equilibrium | Electric charges & fields | Khan Academy](https://www.youtube.com/watch?v=Lv8Tt2QyHTA) — Khan Academy India - English - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/engineering_physics/third-charge-no-net-force - Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7 ## Question Two fixed charges 1.07 μC and −3.28 μC are 61.8 cm apart, where may a third charge be located so that no net force acts on it? ## Short answer Place the third charge on the line outside the two charges, near the smaller magnitude (+1.07 μC). Balance condition: 1.07/x² = 3.28/(x+61.8)² gives x ≈ 82.3 cm from +1.07 μC, 144.1 cm from −3.28 μC. ## Answer Given q₁ = +1.07 μC, q₂ = −3.28 μC, separation d = 61.8 cm. #### Where can the net force be zero? - Between the charges: the forces on a third charge from opposite charges point in the same direction, so they cannot cancel. - Outside the charges: the forces are opposite in direction. They can balance only on the side of the smaller-magnitude charge (q₁): there the third charge is closer to the weaker charge, so the shorter distance makes up for its smaller size. ```text third charge q1 (+1.07) q2 (-3.28) o---------------(+)----------------(-) <------ x ------><-------- d --------> ``` Let the third charge be at distance x from q₁, on the side away from q₂. Its distance from q₂ is (x + d). #### Condition |q₁| / x² = |q₂| / (x + d)² (x + d)/x = √(|q₂|/|q₁|) = √(3.28/1.07) = √3.065 = 1.751 1 + d/x = 1.751 → d/x = 0.751 → x = 61.8 / 0.751 = 82.3 cm Check: 1.07/(0.823)² = 1.58 and 3.28/(1.441)² = 1.58 (μC/m²). ✓ **Answer: the third charge should be placed on the line joining the charges, outside them, 82.3 cm from the +1.07 μC charge on the side away from the −3.28 μC charge (144.1 cm from the −3.28 μC charge). The result is independent of the third charge's sign and magnitude.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can the net force on a third charge not be zero between +1.07 μC and −3.28 μC? A. Both forces point the same way, so they add B. The forces are equal in size but cancel only at the midpoint C. Opposite charges screen each other's field D. The forces are perpendicular there Answer: A. Between opposite charges, one attracts and the other repels the third charge in the same direction, so the forces add. (Key point: Region between charges) 2. Outside the charges, on which side does the net force vanish? A. Beside the larger charge q2 B. Beside the smaller-magnitude charge q1 C. At the midpoint outside the line D. On either side equally Answer: B. Near the smaller charge, its shorter distance compensates for its weaker magnitude, allowing balance. (Key point: Side of smaller charge) 3. What condition gives the position x from q1 (distance x+d from q2)? A. q1/x = q2/(x+d) B. q1 x^2 = q2 (x+d)^2 C. |q1|/x^2 = |q2|/(x+d)^2 D. q1 x = q2 (x+d) Answer: C. Equal force magnitudes require kq1q/x^2 = kq2q/(x+d)^2, so the charge q cancels. (Key point: Balance condition) 4. Where is the third charge placed, and does it depend on its own charge? A. 82.3 cm from q1 away from q2; independent of its sign and size B. 30.9 cm from q1 between the charges; depends on its sign C. 82.3 cm from q2 beyond q1; depends on its magnitude D. 61.8 cm from q1; independent of its sign Answer: A. x = 61.8/0.751 = 82.3 cm on the far side of q1; the third charge cancels from the equation. 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