A particle A has charge +q and particle B has charge +4q with each of them having the same mass m. when allowed to fall from rest through the same electric potential difference, the ratio of their speeds Va/Vb will become
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Ka/Kb=q×v/4q×v =1/2mVa^2/1/2mVb^2
Va^2/Vb^2=1/4
Va/VB=1/2
Va:Vb=1:2
It is the ratio between the kinectic energies between the two particles....
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