Exams › JEE Advanced › Physics
Two fixed charges A and B, each +5 microcoulomb, are 6 m apart. C is the midpoint of AB. A charge Q = -5 microcoulomb is projected from C, perpendicular to AB, with kinetic energy 0.06 J. It momentarily comes to rest at a point D on this perpendicular line. What is the distance CD?
- 4 m
- 3 m
- sqrt(3) m
- 3*sqrt(3) m
Correct answer: 4 m
Solution
The negative charge is attracted, so it slows as it moves away. Total mechanical energy (KE + PE) is conserved; setting KE = 0 at D and solving for the distance gives CD.
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