Coulomb's Law
Three point charges of (+2) μC, (-4) μC, and (+2) μC are placed in a straight line with equal spacing of 20 cm. What is the net force on the middle charge?
Select the correct option:
Solution
zero, since the geometry is symmetric
The superposition principle in electrostatics states that the net force on any charge is the vector sum of individual Coulomb forces due to all other charges. This principle is essential for multi-charge problems in JEE. The middle charge is (-4) μC. The left (+2) μC charge exerts an attractive force on the middle charge directed towards the left. The right (+2) μC charge exerts an attractive force on the middle charge directed towards the right. Since both (+2) μC charges are at equal distances of 20 cm and have equal magnitude, the two forces are equal in magnitude: (F = 9 \times 10^9 \times \frac{2 \times 10^{-6} \times 4 \times 10^{-6}}{(0.20)^2} = 1.8) N each. Being exactly opposite, they cancel, giving a net force of zero. Option 1.8 N is incorrect because it counts only one of the two forces. Option 3.6 N is incorrect because it sums the forces without accounting for their opposite directions. Option 0.9 N is incorrect because it halves one of the individual force values without physical justification. Symmetry is the critical insight here, directly tested in JEE Main problems. Plausibility check: by symmetry, the middle charge lies on an axis of symmetry, so the net force must vanish.
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About This Question
- Subject
- physics
- Chapter
- electrostatics
- Topic
- coulomb's law
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
zero, since the geometry is symmetric
The superposition principle in electrostatics states that the net force on any charge is the vector sum of individual Coulomb forces due to all other charges. This principle is essential for multi-charge problems in JEE. The middle charge is (-4) μC. The left (+2) μC charge exerts an attractive force on the middle charge directed towards the left. The right (+2) μC charge exerts an attractive force on the middle charge directed towards the right. Since both (+2) μC charges are at equal distances of 20 cm and have equal magnitude, the two forces are equal in magnitude: (F = 9 \times 10^9 \times \frac{2 \times 10^{-6} \times 4 \times 10^{-6}}{(0.20)^2} = 1.8) N each. Being exactly opposite, they cancel, giving a net force of zero. Option 1.8 N is incorrect because it counts only one of the two forces. Option 3.6 N is incorrect because it sums the forces without accounting for their opposite directions. Option 0.9 N is incorrect because it halves one of the individual force values without physical justification. Symmetry is the critical insight here, directly tested in JEE Main problems. Plausibility check: by symmetry, the middle charge lies on an axis of symmetry, so the net force must vanish.
This medium difficulty physics question is from the chapter electrostatics, covering the topic of coulomb's law. It appeared in the 2025 exam.
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