Two point charges each `50 mu C` are fixed on `y`-axis at `y = + 4 m` and `y = - 4m`. Another charged particle having charge `-50 mu C` and mass `20 g` is moving along the positive `x`-axis. When it is at `x = -3 m` , its speed is `20 m//sec`. Find the speed of charged particle when it reaches to origin. Also , find distance of charged particle from origin , when its kinetic energy becomes zero.

Explanation

Let the speed of particle at origin be `v`. Applying energy conservation between `A` and `O` , we get
`K_(A) + U_(A) = K_(o) + U_(o)`
`K_(A) + (-Q) V_(A) = K_(o) + (-Q) V_(o)`
`(1)/(2) mv_(A)^(2) + (-Q) (1)/(4 pi in_(0)) . (2Q)/(sqrt(3^(2) + 4^(2))) = (1)/(2) mv_(o)^(2)`
`+ (-Q) (1)/(4 pi in_(0)) ((1)/(4) - (1)/(5))`
`= 2(50 xx 10^(-6))^(2) xx 9 xx 10^(9) xx (1)/(20)`
`(1)/(2) xx 20 xx 10^(-3) (v_(o)^(2) - 20^(2)) = 2.25`
`rArr v_(o)^(2) - 400 = 225 rArr v_(o)^(2) = 625`
`v_(o) = 25 m//sec`
Applying energy conservation between `A` and `B` , we have
`K_(A) + U_(A) = K_(B) + U_(B)`
`K_(A) + (-Q) V_(A) = 0 + (- Q) V_(B)`
`(1)/(2) mv_(A)^(2) + (-Q) (1)/(4 pi in_(0)) .(2Q)/(5) = (-Q) (1)/(4 pi in_(0)).(1)/(sqrt(4^(2) + x^(2)))`
`(1)/(2) mv_(A)^(2) = (1)/( 4 pi in_(0)) 2 Q^(2) [ (1)/(5) - (1)/(sqrt(4^(2) + x^(2))]`
`(1)/(2) xx 20 xx 10^(-3) xx (20)^(2) = 9 xx 10^(9) xx 2 xx (50 xx 10^(-6))^(2) [(1)/(5) - (1)/(sqrt(4^(2) + x^(2)))]`
`4 = 45 ((1)/(5) - (1)/(sqrt(16 + x^(2))))`
`rArr (1)/(sqrt(16 + x^(2))) = (1)/(5) - (4)/(45) = (5)/(45) = (1)/(9)`
`x = sqrt(65) m`