Question

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

• A. $2:1$
• B. $1:2$
• C. $1:4$
• D. $4:1$

Answer 1

The correct option is B $1:2$

Since kinetic energy imparted to a system because of accelerated voltage V

$KE=qV$____(1) and $Kineticenergy\left(KE\right)=\frac{1}{2}m{v}^2$ ____(2)

From (1) and (2)

$\frac{K_A}{K_B}=\frac{qV}{4qV}=\frac{1/2m{V}_A^2}{1/2m{V}_B^2}$

$\frac{V_A^2}{V_B^2}=\frac{1}{4}=\frac{1}{2}$