A positive point charge is released from rest at a distance `r _ 0 ` from a positive line charge with uniform density. The speed (v) of the point charge, as a function of instantaneous distance r from line charge, is proportional to:

Answer

`0.5`

Explanation

At any point P the force
`( f _(p))` experienced by change `= (2k lamda q)/(x)`
Using newton.s second law of motion,
`mv (dv)/(dx) = (2 k lamda q)/(x)....`
`. (because (dv)/(dt) = (dv)/(dx) . (dx)/( dt) = v (dv)/(dx))`
`therefore int _(0) ^(v) mvdv = 2 k lamda q int _(r _(0)) ^(r) (dx)/(x)`
`therefore ( mv ^(2))/( 2 ) | _(0) ^(v) = 2 klamda q ln k | _(r _(0)) ^(r)`
`therefore v = sqrt (( 2 k lamda q)/( m )) sqrt( ln ((r)/(r _(0))))`
`therefore v prop sqrt (ln ((r)/( r_(0))))`
` therefore N = 1implies ( 1)/( 2 n ) = 0.5`