Two point charges `q_1=q_2=2muC` are fixed at `x_1=+3 m` and `x_2=-3m` as shown in figure. A third particle of mass 1 g and charge `q_3=-4muC` are released from rest at `y=4.0m`.Find the speed of the particle as it reaches the origin.

Explanation

Here the charge `q_(3)` is attracted toward `q_(1)` and `q_(2)` both so the net force on `q_(3)` is toward the origin As only force acting is conservative we can use conservation of mechanical energy let v be the speed of particle at origin. From energy of conservation principle we get
`Ui+Ki=U+K`
or `(1)/(4piepsilon_(0))[(q_(3)q_(2))/((r_(32))_(i))+(q_(3)q_(1))/((r_(32))_(i))+(q_(2)q_(1))/((r_(21))_(i))]+0`
`=(1)/(4piepsilon_(0))[(q_(3)q_(2))/((r_(32))_(f))+(q_(3)q_(1))/((r_(31))_(f))+(q_(2)q_(1))/((r_(21))_(f))]+(1)/(2)mv^(2)`
Here `(r_(21))_(i)=(r_(21))_(f)` substituting the proper values
`9xx10^(9)[((-4)(2))/(5)+((-4)(2))/(3)]xx10^(-12)+(1)/(2)xx10^(-12)+(1)/(2)xx10^(-3)v^(2)`
or `(9xx10^(-3))((-16)/(5))=(xx10^(-3))((-16)/(3))+(1)/(2)xx10^(-3)v^(2)`
or `9xx10^(-3)(16)((2)/(15))=(1)/(2)xx10^(-3)v^(2)`
or `v=6.2=6ms^(-1)`