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Two very small particles `1` and `2` each of mass `0.5kg` and of charge `q_(1)=1mc` and `q_(2)=1muC` resepectivley are connected by a mass less spring of spring constant `200Nm^(-1)` and placed on a horizontal rough surface.The particle `1` is fixed and `2` is free to move ,Initially the spring is in its natural length `(2m)` where `q_(2)` is released.If coefficient of static friction between `2` and horizontal surface is `0.5` the separation between the charges when they are in equilibrium will be (Neglect gravitational interactions between `1` and `2g=10ms^(-2)` |
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Answer» Correct Answer - 2 Based on friction and Coulombus Law `F=(9xx10^(9)xx10^(-9))/(4)=2.25N` which is less `mumg` acting on `2` rArr final separation =2m only. |
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