Two fixed, equal, positive charges, each of magnitude `5xx 10^-5` C are located at points A and B separated by a distance of 6 m. An equal and opposite charge moves towards them along the fine COD, the perpendicular bisector of the line AB. The moving charge when it reaches the Point C at a distance of 4 m from 0, has a kinetic energy of 4 J. Calculate the distance of the farthest point D which the negative charge will reach before returning towards C.

Answer

`8.48 m` from O

Explanation

COME: `K_(C)+U_(C)=K_(D)+U_(D)`
`4+2[(Kqxx(-q))/5]=0+((Kqxx-2)/r)`
`4-(2xx9xx10^(9)xx25xx10^(-10))/5`
`=(-2xx9xx10^(9)xx25xx10^(-10))/(r), r=9 m`
`OD=sqrt(AC^(2)-OA^(2))=sqrt(9^(2)-3^(2))=8.48m`