InterviewSolution
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InterviewSolution
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A particle of mass `m_(2)` carrying a charge `Q_(2)` is fixed on the surface of the earth .Another particle of mass `m_(1)` and charge `Q_(1)` is positioned right above the first one at an altitude `h( ltlt R)`.R is radius of earth ,the charge `Q_(1)` and `Q_(2)` are of same sign ,then the magnitude of charge `Q_(2)` at which the velocity of `m_(1)` at an altitude `h_(2)` is zero is given byA. `Q_(2)=(piepsilon_(0)m_(1)ghh_(2))/(Q_(1))`B. `Q_(2)=(4piepsilon_(0)m_(1)ghh_(2))/(Q_(1))`C. `Q_(2)=(2piepsilon_(0)m_(1)ghh_(2))/(Q_(1))`D. `Q_(2)=(2piepsilon_(0)m_(1)ghh_(2))/(4Q_(1))` |
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Answer» Correct Answer - B `(1)/(4piepsilon_(0))(q_(1)q_(2))/(h)+m_(1)gh=m_(1)gh_(2)+(1)/(4piepsilon_(0))(q_(1)q_(2))/(h_(2))` `(:.KE` is zero at both positions) `rArr (1)/(4pepsilon_(0))q_(1)ql2((1)/(h_(2))-(1)/(h))=m_(1)g(h-h_(2))` `(1)/(4piepsilon_(0))q_(1)q_(2)((h-h_(2))/(hh_(2)))=m_(1)g(h-h_(2))` `rArr q_(2)=(4piepsilon_(0)m_(1)ghh_(2))/(q_(1))` |
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