A particle of mass 40 mg and carrying a charge `5 xx 10^(-9) C` is moving directly towards a fixed positive point charge on magnitude `10^(-8) C`. When it is at a distance of 10 cm from the fixed positive point charge it has a velocity of `50 cm s^(-1)` at what distance from the fixed point charge will the particle come momentarily to rest ? Is the acceleration constant during motion?

Explanation

If the particle comes to rest momentarily at a distance r from the fixed charge, then from .conservation of energy. we have
`(1)/(2) mv^(2) + (1)/(4pi in_(0)) (Qq)/(a) = (1)/(4pi in_(0)) (Qq)/(r )`
Substituting the given data, we get
`(1)/(2) xx 40 xx 10^(-6) xx (1)/(2) xx (1)/(2)`
`=9 xx 10^(9) xx 10^(-8) xx 5 xx 10^(-9) [(1)/(r ) - 10]`
or `(1)/(r ) - 10 = (5 xx 10^(-4))/(9 xx 5 xx 10^(-8)) = (100)/(9) or, (1)/(r ) = (100)/(9) + 10 =(190)/(9)m`
i.e, `r = 4.7 xx 10^(-2)m`
As here, `F = (1)/(4pi in_(0)) (qQ)/(r^(2)), so a = (F)/(m) prop (1)/(r^(2))`
i.e., acceleration is not constant during motion.