A particle $A$ has charge $ + q$ and a particle $B$ has charge $ + \,4q$ 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 speed $\frac{{{v_A}}}{{{v_B}}}$ will become

  • A

    $2:1$

  • B

    $1:2$

  • C

    $1:4$

  • D

    $4:1$

Similar Questions

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