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A particle of charge `-q` and mass m moves in a circle of radius r around an infinitely long line charge of linear charge density `+lamda`. Then, time period will be where , `k=1/4piepsilon_0)`A. `T=2pi rsqrt((m)/(2k lambdaq))`B. `T^(2)=(4pi^(2)m)/(2k lambda q)r^(3)`C. `T=(1)/(2 pi r) sqrt((2k lambdaq)/(m))`D. `T=(1)/(2pi r) sqrt((m)/(2k lambdaq))` |
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Answer» Correct Answer - A We have centripetal force equation `q((2k gamma)/(r))=(mv^(2))/(r)` so `v=sqrt((2kq gamma)/(m))` Now `T=(2pi r)/(v)=2pirsqrt((m)/(2 k gamma q))` where, `k=(1)/(4pi epsilon_(0))` |
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