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# A particle of mass 2 g and charge 1 microcoulomb is held at rest on a frictionless surface at a distance of 1 m from the fixed charge of 2 mC. if the particle is released, it will be repelled. the speed of the particle when it is at a distance of 10 m from the fixed charge is?

Khimraj
3007 Points
4 years ago
Just apply conservation of energy as intial energy is equal to final energy.K(2*10^-3)(1*10^-3)/1 + 0 = K(2*30^-3)(1*10^-3)/10 + 1/2mv². So 16200 = 1/2(2*10^-3)v² then v =√(16200000) = 100√(1620 and it is approx 4000 m/s.Hope it clears. If u have doubts then please clearify. And if u like my answer then please approve answer
Ramansh Parmar
36 Points
4 years ago
first use the coloumb’s formula i.e F= k*q1*q2/r, and then compare it with f=ma because both of them are forces. after finding acceleration we have inital velocity =0 .and we can use v2-u2=2as and s=9m we have the answer [about 400m/s]

Niranjan Pant
13 Points
3 years ago
No,we can't apply force equation in this case because acceleration varies as separation changes thus best is to apply energy conservation  as no force is exerted on the system. Just equate final and initial energy and the answer coming is 180ms^-1