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A particle A of charge 1 mu C is held f...

A particle A of charge ` 1 mu C` is held fixed at a point P in free space . Another particle B of same charge and mass `4 mu g` is kept at a distance of 1mm from P. If B is released then its velocity at a distance of 9 mm from P is (Take `(1)/( 4pi epsilon_(0)) =9 xx 10^(9)Nm^(2) C^(-2))`

A

`1.5xx10^(2)m//s`

B

`1.0m//s`

C

`3.0xx10^(4)m//s`

D

`2.0xx10^(3)m//s`

Text Solution

Verified by Experts

The correct Answer is:
D
Loss in P.E. = Gain in K.E.
`Kxx10^(-6)xx10^(-6)[(1)/(10^(-3))-(1)/(9xx10^(-3))]=(1)/(2)mv^(2)`
`9xx10^(9)xx(10^(-6)xx10^(-6))/(10^(-3))xx(8)/(9)=(1)/(2)xx4xx10^(-6)xxv^(2)`
`v=2xx10^(3)m//s`.
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