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A proton is fixed at origin and another proton is released
from rest from a point at distance from the origin. Its
speed at 2r from origin is (m - mass of proton, e - charge of
proton)
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Explanation:
ANSWER
ANSWERinitial energy of system = 4πϵ11q1q2
ANSWERinitial energy of system = 4πϵ11q1q2when q1 reached x=0.5, energy of system = 4πϵ10.5q1q2+21mv2
ANSWERinitial energy of system = 4πϵ11q1q2when q1 reached x=0.5, energy of system = 4πϵ10.5q1q2+21mv2equating the two energy,
ANSWERinitial energy of system = 4πϵ11q1q2when q1 reached x=0.5, energy of system = 4πϵ10.5q1q2+21mv2equating the two energy,4πϵ1q1q2(11−0.51)=21mv2⇒v=18.9m/s
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