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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?

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?

Grade:12th pass

1 Answers

Eshan
askIITians Faculty 2095 Points
5 years ago
Dear student,

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

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