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A particle of mass m(2) carrying a charg...

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 by

A

`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))`

Text Solution

Verified by Experts

The correct Answer is:
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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