Question

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 velocity of `m_(1)` at a point `P` very close to `Q_(2)` at a distance `h_(1)` from the surface of the earth ( if the initial velocity of `m_(1)` was zero ,air drag and earths magnitude field being ignored) is
A. `sqrt(2gh-(Q_(1)Q_(2))/(2piepsilon_(0)h_(1)m_(1)))`
B. `sqrt(2gh-(Q_(1)Q_(2))/(4piepsilon_(0)h_(1)m_(1)))`
C. `sqrt(2gh)`
D. `sqrt(2gh-(Q_(1)Q_(2))/(4piepsilon_(0)h_(1)m_(1)))`

Answer 1

Correct Answer - A

`h ltltR` where `R` is radius of earth.
Applying mechanical energy conservation
`m_(1)gh+(1)/(4piepsilon_(0))(Q_(1)Q_(2))/(h)=m_(1)gh_(1)+(1)/(4piepsilon_(0))(Q_(1)Q_(2))/(h_(1))+(1)/(2)m_(1)v^(2)`
`(1)/(2)mv^(2)=m_(1)g(h-h_(1))+(Q_(1)Q_(2))/(4piepsilon_(0))[(1)/(h)-(1)/(h_(1))]`
`v^(2)=2g(h-h_(1))+(2Q_(1)Q_(2))/(4piepsilon_(0))[(h_(1)-h)/(hh_(1))]`
``
`v^(2)=2g(h-h_(1))-(2Q_(1)Q_(2))/(4piepsilon_(0)m_(1))[(h-h_(1))/(hh_(1))]`
`v^(2)=2gh(1-(h_(1))/(h))-(2Q_(1)Q_(2))/(4piepsilon_(0)m_(1))[(h(1-(h_(1))/(h)))/(hh_(1))]`
Since `h_(1) ltltlth` then `(h_(1))/(h)` can be neglecteed
`v^(2)=2gh-(Q_(1)Q_(2))/(2piepsilon_(0)m_(1)h_(1))v=sqrt(2gh-(Q_(1)Q_(2))/(2piepsilon_(0)m_(1)h_(1)))`