Question

Two particles of masses $m$ and $2m$, having same charges $q$ each, are placed in a uniform electric field $E$ and allowed to move for the same time. Find the ratio of their kinetic energies :

Answer 1

If $a_1$ and $a_2$ are the acceleration of $m$ and $2m$ respectively.

thus, $F_1=qE\Rightarrow m{a}_1=qE$ and $F_2=qE\Rightarrow 2m{a}_2=qE$

$\therefore a_1=\frac{qE}{m}$ and $a_2=\frac{qE}{2m}$

using $v=u+at$, we will get

$v_1=0+a_{1}t\Rightarrow v_1=a_{1}t=\frac{qE}{m}t$

$v_2=0+a_{2}t\Rightarrow v_2=a_{2}t=\frac{qE}{2m}t$

Kinetic energy of $m$ is $K_1=\frac{1}{2}m{v}_1^2$

Kinetic energy of $2m$ is $K_2=\frac{1}{2}2m{v}_2^2$

$\therefore \frac{K_1}{K_2}=\frac{m}{2m}{\left(\frac{v_1}{v_2}\right)}^2=\frac{1}{2}{\left(\frac{qEt}{m}\right)}^2{\left(\frac{2m}{qEt}\right)}^2=2$