Particles A and B have charges +q and +4q res | vedclass

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(B) The work done by the electric field on a charge $q$ accelerated through a potential difference $V$ is given by $W = qV$.According to the work-ener

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$1 : 2$

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(B) The work done by the electric field on a charge $q$ accelerated through a potential difference $V$ is given by $W = qV$.According to the work-energy theorem,this work is equal to the change in kinetic energy: $W = \frac{1}{2}mv^2$.Equating the two,we get $qV = \frac{1}{2}mv^2$,which implies $v = \sqrt{\frac{2qV}{m}}$.Since both particles have the same mass $m$ and are accelerated through the same potential difference $V$,the velocity $v$ is directly proportional to the square root of the charge: $v \propto \sqrt{q}$.Therefore,the ratio of their velocities 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 $v_A/v_B = 1:2$.

Particles $A$ and $B$ have charges $+q$ and $+4q$ respectively. Both have the same mass $m$. If they are released from rest and accelerated through the same potential difference $V$,the ratio of their velocities $v_A/v_B$ is:

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