Two fixed charges A and B of `5 mu C ` each are separated by a distance of 6 m . O is the mid point of the ine joining A and B . A charge Q is of ` - 5 mu C` is shot perpendicualr to the line joining A and B through C with a kinetic energy of 0.06 J . The charge Q comes to rest at a point D . The distance CD is :

Options

A. 4m
B. 3 m
C. `sqrt(3)` m
D. `3sqrt(3)` m

Answer

A

Explanation

By conservation of energy ,
Loss of PE = Gain in KE
` (2a^(2))/(4piepsilon_(0)) (1/r - 1/(sqrt(r^(2)+x^(2)) ))= K `

`or 2xx9 xx10^(9) xx ( 5xx 10^(-6))^(2) (1/3 - 1/(sqrt(3^(2)+x^(2))))`
` = 0.06`
` or 1/3 - 1/(sqrt(9+x^(2))) = 2/15`
or x = 4 m