A particle A has charge +q and particle B has | vedclass

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Question

(B) The work done on a charge $q$ moving through a potential difference $\Delta V$ is equal to its change in kinetic energy.$W = q \Delta V = \Delta K

Vedclass · Accepted Answer

$1 : 2$

Vedclass · Answer

(B) The work done on a charge $q$ moving through a potential difference $\Delta V$ is equal to its change in kinetic energy.$W = q \Delta V = \Delta K = \frac{1}{2} m v^2 - 0$Since both particles start from rest and move through the same potential difference $\Delta V$,we have:$\frac{1}{2} m v_A^2 = q_A \Delta V$$\frac{1}{2} m v_B^2 = q_B \Delta V$Dividing the two equations:$\frac{v_A^2}{v_B^2} = \frac{q_A}{q_B} = \frac{q}{4q} = \frac{1}{4}$Taking the square root of both sides:$\frac{v_A}{v_B} = \sqrt{\frac{1}{4}} = \frac{1}{2}$Therefore,the ratio $V_A : V_B$ is $1 : 2$.

$A$ particle $A$ has charge $+q$ and particle $B$ has charge $+4q$,with each of them having the same mass $m$. When allowed to fall from rest through the same electrical potential difference,the ratio of their speeds $V_A : V_B$ will be:

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