Question

Two charged particles of masses m' and 2m have charges +2q and +q respectively. They are kept in uniform electric field and allowed to move for the same time. The ratio of their kinetic energies.

Answer 1

We know that kinetic energy is equal to mv2/2

We know both of their mass so im guessing you are unable to find their velocities.

V = u + at

u = 0

a = qE/m

V1 = 2qEt/m

V2 = qEt/m

Now simply plug these values into the kinetic energy equation and viola there’s your answer

Please mark as brainliest and feel free to ask for any clarification.

Answer 2

Answer:

Required ratio = 2 : 1

Explanation:

Let the electric field be $E$.

Force due to this electric field on a charge $Q$ is, $F=QE$ ----(1)

Also, force is given by, $F=ma$ ----(2)

where, $m$ is mass,

and $a$ is the acceleration.

Equating (1) and (2),

$QE=ma$

$\implies a=\frac{QE}{m}$----(3)

Let's say that the time for which charges remain in the electric field is $t.$

using newton's equations of motion,

$v= u + at$    

$\implies v=at$             [$u = 0$]

$\implies v=\frac{QE}{m}t$

Kinetic energy is, K.E.= $\frac{mv^2}{2}$

K.E. $=\frac{1}{2} *m*{(\frac{QE}{m}t})^2$

For particle 1:

$Q=+2q;$     mass= $m$

$K.E._1=\frac{1}{2} *m*{(\frac{2qE}{m}t})^2=2m{(\frac{qE}{m}t})^2$

For particle 2:

$Q=+q;$     mass= $2m$

$K.E._2==\frac{1}{2} *2m*{(\frac{qE}{m}t})^2=m{(\frac{qE}{m}t})^2$

Required ratio = $\frac{K.E._1}{K.E._2} =\frac{2m{(\frac{qE}{m}t})^2}{m{(\frac{qE}{m}t})^2}=\frac{2}{1}$

Required ratio = 2 : 1

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