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Figure shows two fixed positive point charges of magnitude `5muC` A negative point charge of `-10nC` and mass `1kg` is released as shown in figure (consider only electrostatic foces)

A

Initial acceleration fo `-10nC` charge is `0.216m//s^(2)`

B

initial acceleration of `-10nC` charge is `2.16m//s^(2)`

C

Kinetic energy of `-10nC` charge when it crosses the line joining two fixed charges is 4.5 mJ

D

Kinetic energy of `-10nC` charge when it crosses the line joining two fixed charges is 3 mJ.

Text Solution

Verified by Experts

The correct Answer is:
A, C

`a=(2Kq_(1)q_(2))/(r^(2))(cos53^(2))/(m)`
`=(2xx9xx10^(9)xx10xx10^(-9)xx5xx10^(-6))/((5xx10^(-2))xx1)(3)/(5)`
`=0.216m//s^(2)`
`K.E.=2xx9xx10^(9)xx5xx10^(-6)xx10xx10^(-9)[(1)/(4xx10^(-2))-(1)/(5xx10^(-2))]`
`=2xx9xx5xx10xx10^(9-6-9+2)[(5-4)/(20)]`
`=2xx9xx5xx(10)/(20)xx10^(-4)=4.5mJ`
Initial acceleration of `-10nC` charge is `0.216m//s^(2)`
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