Question

Two charged particles of masses $m$ and $3m$ have charges $3q$ and $q$ respectively. They are kept in a uniform electric field and allowed to move for the same time. The ratio of their kinetic energies is:

• A. $1:9$
• B. $1:27$
• C. $9:1$
• D. $27:1$

Answer 1

The correct option is D $27:1$

A $\to$ $m,3q$ B$\to$ $3m,q$

Force acting on charge $Q$ when placed in an electric field $E$ is $F=EQ$

$\therefore$ acceleration of Ais $\frac{3qE}{m}$
Velocity of A at time $t$ is given as: $\frac{3qE}{m}t$

So, K.E of A, $K_A$$=\frac{1}{2}m\frac{9q^2{E}^2{t}^2}{m^2}$

Similarly, acceleration of B: $\frac{qE}{3m}$
Velocity of B at time $t$ is given by: $\frac{qE}{3m}t$
So, K.E of B, $K_B=\frac{1}{2}3m\frac{q^2{E}^2{t}^2}{9m^3}$
$\therefore$ $K_A:K_B=27:1$