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

2

Explanation

From figure `AC=sqrt((AO)^(2)+(OC^(2)))`
`AC=sqrt(3^(2)+4^(2))=5m` similarly `BC=5m`
P.E at `C` is `U_(c)=2xx9xx10^(9)xx((q)xx(-q))/(AC)`
`K.E` at `C` is `4J`
Total energy at `C` is `E_(c) =P.E. +K.E rArr E_(C)=-9+4=-5J`
Potnetial energy at `D` is `U_(D)` is given by
`U_(D)=(-2xx9xx10^(9)(5xx10^(-5))^(2))/(r )`
where `AD=BD=r` (say)
`rArr U_(D)=(-45)/(r )=(-45)/(r )J`
By law of conservation of Energy
`U_(C)+K_(C)=U_(D)+K_(D)`
`rArr (-45)/(R )=-5 rArr r=9m`
Now `OD=sqrt((AD)^(2)-(AO)^(2))`
`rArr OD=sqrt(9^(2)-3^(2))=sqrt(72)m`
` rArr OD=6sqrt(2)rArr x=2`