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` is released from rest at `y=4.0m`.Find the speed of the particle as it reaches the origin.

Options

A. `4.2m//s`
B. `6.2m//s`
C. `7.2m//s`
D. `8.2m//s`

Answer

B

Explanation

Let v be the speed of particle at origin.From conservation mechanical energy,
`U_i+K_i=U_f+K_(f)` or `1/(4piepsilon_0) [(q_3q_2)/((r_32)_i)+(q_3q_1)/((r_31)_i)+(q_2 q_1)/((r_21)_i)]+0=1/(4piepsilon_0)[(q_3q_2)/((r_32)_f)+(q_3q_1)/((r_31)_f)+(q_3q_1)/((r_31)_f)+(q_2q_1)/((r_21)_f)]+1/2mv^2`
Here `(r_21)_i=(r_21)_f`
Substituting the proper values, we have
`(9.0xx10^9)[((-4)(2))/((5.0))+((-4)(2))/((5.0))]xx10^-12=(9.0xx10^9)[((-4)(2))/((3.0))=((-4)(2))/(3.0)]xx10^-12=1/2xx10^-3xxv^2`
`:. (9xx10^-3)(-16/5)=(9xx10^-3)(-16/3)+1/2xx10^-3xxv^2`
`(9xx10^-3)(16)(2/15)=1/2xx10^-3xxv^2`
`v=6.2m//s`