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# Particle A has charge +q and particle B has charge +4q.Both have same masses....when allowed to fall from rest through the same electirc potential difference,then ratio of speed of A than that of B will be?2:1....1:2......4:1.....1:4?

Eshan
3 years ago
Dear student,

Loss in potential energy=qV
This is equal to gain in kinetic energy=$\dpi{80} \dfrac{1}{2}mv^2$
$\dpi{80} \implies \dfrac{1}{2}mv^2=qV$
$\dpi{80} \implies v\propto q^{1/2}$
Hence the speed of particle B is twice as that of A.