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Two fixed charge A and B of 5 mu C each ...

Two fixed charge `A` and `B` of `5 mu C` each are separated by a distance of `6m` .C is the mid point of the line joining `A` and `B` .A charge `Q` of `-5 mu C` is shot perpendicular 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

A

3m

B

`sqrt3m`

C

`3sqrt3m`

D

4m

Text Solution

Verified by Experts

The correct Answer is:
D
The charge will stop when the increase in PE will be same as initial KE.

When charge Q is at C
`U_2 = 2xx1/(4piepsilon_0).(qQ)/r +1/(4piepsilon_0).q^2/6`
Now, `KE = U_1 - U_2`
`implies (2qQ)/(4piepsilon_0)[1/3-1/r]=0.06`
`implies (-1)/r+1/3=(0.06)/(2xx25 xx10^(-12)xx9xx10^9)`
`implies (-1)/r+1/3 = (0.06)/(0.45)`
`impliesr = 5m, CD =sqrt(5^2-3^2) =4m`
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