A particle A has a charge q and particle B has charge 4 q with each of them having the mass m . When they are allowed to fall from rest through the same potential difference, the ratio of their speeds

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

TestHub · Accepted Answer

Speed obtained by the particle after falling through a potential difference of $V$ volt is

$v_{A} = \sqrt{\frac{2 V q}{m}} \dots (i)$

And $v_{B} = \sqrt{\frac{2 V \times 4 q}{m}}$ ...(ii)

Now dividing Eq. (i) by Eq. (ii), we get

$\frac{v_{A}}{v_{B}} = \sqrt{\frac{1}{4}}$ = $\frac{1}{2}$

So, $v_{A} : v_{B} = 1 : 2$

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