Two fixed, equal, positive charges, each of magnitude `5xx10^-5` coul are located at points A and B separated by a distance of 6m. An equal and opposite charge moves towards them along the line COD, the perpendicular bisector of the line AB.
The moving charge, when it reaches the point C at a distance of 4m from O, has a kinetic energy of 4 joules. Calculate the distance of the farthest point D which the negative charge will reach before returning towards C.

Answer

D

Explanation

Total energy of the system of three charges when the charge `-q` is at C
`=P.E.+K.E.`
`=[(Kqxxq)/(6)+(K(q)(-q))/(5)+(Kq(-q))/(5)]+4` …(i)

Final energy of the system of three charges when `-q` is at D and momentarily at rest
`=[(Kqxxq)/(6)+(Kq(-q))/(sqrt(x^2+3^2))+(Kq(-q))/(sqrt(x^2+3^2))]+0`
`=(Kqxxq)/(6)+(2Kq(-q))/(sqrt(x^2+3^2))` ...(ii)
By the principle of conservation of energy from (i) and (ii), we get
`(kqxxq)/(6)+(2kq(-q))/(5)+4=(kqxxq)/(6)+(2kq(-q))/(sqrt(x^2+3^2))`
`2=kq^2[1/5-(1)/(sqrt(x^2+3^2))]`
`:.` `x^2+9=81`
`:.` `x=8.48m`