Four point charges `+8muC-1muC,-1muC and +8muC` are fixed at the points `-sqrt(27//2)m, -sqrt(3//2)m+sqrt(3//2)m and + sqrt(27//2)m` respectively on the Y-axis. A particle of mass `6xx10^-4` kg and charge `+0.1muC` moves along the -X direction. It speed at `x=+oo` is `v_0`. Find the least value of `v_0` for which the paticle will cross the origin. Find also the kinetic energy of the particle at the origin. Assume that space is gravity free.

Answer

`3 m//s, 3 xx 10^-4 J`

Explanation

Let electric field atv P due to all four charges be zero.
.
`2 E_1 cos alpha = 2 E_2 cos beta`
or `(2 k xx 8 xx 10^-6 x)/((x^2 + 27 // 2)^(3//2))= (2 k xx 1 xx 10^-6 x)/((x^2 + 3 // 2)^(3//2))` or `x = sqrt((5)/(2)) m`
Beyond P, E is toward right. To the left of `P, E` is toward left Once the charge crosses point P, attractive forces will pull the charge to origin. Applying conservation of energy between `oo` and P, we get
`(1)/(2) mv_0^2 = 2k [(8 xx 10^6)/(sqrt(27/(2) + (5)/(2)))- (1 xx 10^6)/(sqrt((3)/(2)+ (5)/(2)))] xx 0.1 xx 1 xx 10^6`
or `v_0 = 3 ms^-1`
To find K E at origin, apply conservation of energy between `oo` and origin.
`K E_(x = 0) = 2 k xx 0.1 xx 10^-12 [(8)/(sqrt(27 //2)) -(1)/(sqrt(3 // 2))] = (1)/(2) m v_0^2`
=`2.5 xx 10^-4 J`.