A particle $A$ has charge $+q$ and particle $B$ has charge $+4 q$ 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 $\frac{V_A}{V_B}$ will become
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(a)
$q V=\frac{1}{2} m V_A^2 \quad V_B=\sqrt{\frac{8 q V}{m}}$
$V_A=\sqrt{\frac{2 q V}{m}}$
$\frac{V_A}{V_B}=\frac{1}{2}$
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