Categories > Electrostatics > electric-field-potential > 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

Electrostatics

Question: 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


1) $2:1$
2) $1:2$
3) $1:4$
4) $4:1$
Solution: Explanation: Explanation
electric-field-potential

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