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(B) The work done by the electric field on a charge $q$ accelerated through a potential difference $V$ is equal to its kinetic energy.$W = qV = \frac{

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$1 : 3$

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(B) The work done by the electric field on a charge $q$ accelerated through a potential difference $V$ is equal to its kinetic energy.$W = qV = \frac{1}{2}mv^2$Since the mass $m$ and potential difference $V$ are the same for both particles,we have $v^2 \propto q$,which implies $v \propto \sqrt{q}$.Therefore,the ratio of their speeds is:$\frac{v_A}{v_B} = \sqrt{\frac{q_A}{q_B}} = \sqrt{\frac{q}{9q}} = \sqrt{\frac{1}{9}} = \frac{1}{3}$.Thus,the ratio of their speeds is $1 : 3$.

$A$ particle $A$ has a charge of $+q$ and a particle $B$ has a charge of $+9q$. Both particles have the same mass $m$. If both particles are released from rest through the same potential difference,the ratio of their speeds will be .......

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