--- title: "SHM of charge between two positive charges" subject: "Engineering Physics" chapter: 7 key: "semester_1/engineering_physics:two-positive-charges-shm-period" source: https://www.bctengineer.com/q/semester_1/engineering_physics/two-positive-charges-shm-period site: BCT Engineering --- # SHM of charge between two positive charges Engineering Physics · Chapter 7: Electrostatics - Asked in: 2071 Shrawan, 5 marks; 2069 Asar, 5 marks; 2068 Shrawan, 5 marks (asked 3 times) - Video: [SHM between two charges and position of maximum force](https://www.youtube.com/watch?v=ytYOGZeXix8) — physics with yogendra - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/engineering_physics/two-positive-charges-shm-period - Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7 ## Question A particle of charge −q and mass m is placed midway between two equal positive charges q0 of separation d. If the negative charge −q is displaced in a perpendicular direction to the line joining them and released, show that the particle describes a SHM with a period T = √(ε0 m π³ d³ / (q q0)). ## Short answer For small perpendicular displacement y, the net restoring force is F = −(4 q q₀/(π ε₀ d³)) y, giving ω² = 4 q q₀/(π ε₀ m d³). The period is T = 2π/ω = √(ε₀ m π³ d³/(q q₀)). ## Answer **Setup:** two fixed positive charges q₀ are a distance d apart. A particle of charge −q and mass m starts at the midpoint O. It is displaced by a small distance y perpendicular to the line joining the charges. ```text -q (y) o /|\ / | \ F (towards each q0) q0 o---/--O--\---o q0 <---- d ----> ``` #### Derivation 1. The distance of −q from each fixed charge is s = √(d²/4 + y²). 2. Each positive charge attracts −q with force F = q q₀ / (4πε₀ s²), along the line to that charge. 3. The components of the two forces along the line joining the charges cancel by symmetry. The components along y (towards O) add: F_y = 2 F (y/s) = 2 q q₀ y / [4πε₀ (d²/4 + y²)^(3/2)] 4. For small displacement, y ≪ d, neglect y² compared to d²/4: F_y ≈ 2 q q₀ y / [4πε₀ (d/2)³] = 16 q q₀ y / (4πε₀ d³) = 4 q q₀ y / (π ε₀ d³) 5. This force is directed towards O, opposite to the displacement, and proportional to y. So the equation of motion is m d²y/dt² = − (4 q q₀ / π ε₀ d³) y, which is the equation of **simple harmonic motion**, d²y/dt² = −ω² y. 6. Angular frequency: ω² = 4 q q₀ / (π ε₀ m d³). 7. Period: T = 2π/ω = 2π √[π ε₀ m d³ / (4 q q₀)] = √[4π² · π ε₀ m d³ / (4 q q₀)] **T = √[ ε₀ m π³ d³ / (q q₀) ]** This is the required result. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When -q is displaced by y perpendicular to the line joining the charges, what happens to the components of the two attractive forces? A. Components along y cancel; those along the line add B. All components add C. All components cancel D. Components along the joining line cancel; components along y add toward O Answer: D. By symmetry the sideways parts cancel while the parts toward O add: F_y = 2F(y/s). (Key point: Force components) 2. Why must y be small to get SHM? A. So that the charges stay fixed B. So that the force becomes constant C. So that (d^2/4 + y^2)^(3/2) ≈ (d/2)^3, making F proportional to y D. So that the charge -q is not attracted Answer: C. Neglecting y^2 relative to d^2/4 gives F ≈ (4 q q0 / π ε0 d^3) y. (Key point: Small displacement) 3. Why does the motion qualify as SHM? A. The force is constant and directed toward O B. The force is proportional to displacement and directed toward O C. The force is proportional to y^2 and directed away from O D. The force is proportional to displacement and directed away from O Answer: B. m d^2y/dt^2 = -k y is the SHM equation. (Key point: SHM condition) 4. With ω^2 = 4 q q0/(π ε0 m d^3), what is T = 2π/ω? A. T = sqrt(ε0 m π^3 d^3/(q q0)) B. T = sqrt(ε0 m π^2 d^3/(q q0)) C. T = sqrt(ε0 m π d^3/(4 q q0)) D. T = 2π sqrt(q q0/(π ε0 m d^3)) Answer: A. T^2 = 4π^2 π ε0 m d^3/(4 q q0) = π^3 ε0 m d^3/(q q0). 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