Four point `+ 8 mu C, -1 mu C, -1 mu C`, and `+ 8 mu C` 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 `6 xx 10^-4 kg` and charge `+ 0.1 mu C` moves along the x- direction. Its speed at `x = + oo` is `V_0`. Find the least value of `V_0` for which the particle will cross the origin. Find also the kinetic energy of the particle at the origin. Find also the kinetic energy of the particle at the origin. Assume that space is gravity free. Given `(1// 4 pi epsilon_0) = 9 xx 10^9 Nm^2 C^-2`.

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`.