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(B) The kinetic energy gained by a particle of charge $Q$ and mass $m$ when accelerated through a potential difference $V$ is given by $K = QV$.Since

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$1:2$

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(B) The kinetic energy gained by a particle of charge $Q$ and mass $m$ when accelerated through a potential difference $V$ is given by $K = QV$.Since the particles start from rest,$K = \frac{1}{2}mv^2$.Equating the two,we get $\frac{1}{2}mv^2 = QV$,which implies $v = \sqrt{\frac{2QV}{m}}$.Since $m$ and $V$ are the same for both particles,the speed $v$ is proportional to the square root of the charge: $v \propto \sqrt{Q}$.Therefore,the ratio of their speeds is $\frac{v_A}{v_B} = \sqrt{\frac{Q_A}{Q_B}}$.Substituting the given values $Q_A = q$ and $Q_B = 4q$,we get $\frac{v_A}{v_B} = \sqrt{\frac{q}{4q}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Thus,the ratio $V_A / V_B$ is $1:2$.

Particle $A$ has a charge of $+q$ and particle $B$ has a charge of $+4q$. Both particles have the same mass. If they are released from rest under the same electric potential difference,what is the ratio of their speeds $V_A / V_B$?

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