A particle$\ $A has a charge q and particle$\ $B has charge+4q with each of them having the mass m. When they are allowed to fall from rest through same potential difference, the ratio of their speeds $v_{A}:v_{B}$ will be

Question

Dr. Abubacker · Accepted Answer

(c) Speed obtained by the particle after falling through a potential difference of $V$ volt is
$v_{A} = \sqrt{\frac{2Vq}{m}}\ \ \ldots(i)$
And $v_{B} = \ \sqrt{\frac{2V 	imes 4q}{m}}$ …(ii)
Now dividing Eq. (i) by Eq. (ii), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{v_{A}}{v_{B}} = \sqrt{\frac{1}{4}}$=$\frac{1}{2}$
So, $v_{A}:v_{B} = 1:2$

Q. A particle$\ $A has a charge q and particle$\ $B has charge+4q with each of them having the mass m. When they are allowed to fall from rest through same potential difference, the ratio of their speeds $v_{A}:v_{B}$ will be

Explanation

Reviewed by Dr. Abubacker

Want more practice?