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Particle A carries charge +q and particle B carries charge +4q; both have the same mass m. Each is released from rest and accelerated through the same potential difference. What is the ratio of their final speeds v_A: v_B?
- 2: 1
- 1: 2
- 1: 4
- 4: 1
Correct answer: 1: 2
Solution
Energy conservation: qV = (1/2)mv², so v = sqrt(2qV/m). With equal m and V, v is proportional to sqrt(q). For A, q; for B, 4q. So v_A/v_B = sqrt(q)/sqrt(4q) = 1/2.
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