Question

A particle A has charge +q and a particle B has charge +4q with each of them having the same mass m. When allowed to fall from rest through the same electric potential difference, the ratio of their speed $\dfrac {{v}_{A}}{{v}_{B}}$ will become.
A.2:1
B.1:2
C.1:4
D.4:1

Answer 1

Hint: To solve this problem, use the formula for kinetic energy of a particle. And then use the formula for kinetic energy in terms of potential and charge. Equate both the equations. Using these two equations, find the kinetic energy for particle A and particle B. Then, divide these two equations. Dividing the equations will give the ratio of their speed $\dfrac {{v}_{A}}{{v}_{B}}$.
Formula used:
$K= \dfrac {1}{2}m{v}^{2}$
$K= charge \times potential$

Complete step by step answer:
Given: Charge of particle A = +q
            Charge of particle B= +4q
            Mass of both the particles= m
             Electric potential difference of both the particles= V
Kinetic energy of a particle is given by,
$K= \dfrac {1}{2}m{v}^{2}$ …(1)
Kinetic energy in terms of charge and potential is given by,
$K= charge \times potential$
$\Rightarrow K= QV$ …(2)
Equating equation. (1) and (2) we get,
$K=QV= \dfrac {1}{2} m{v}^{2}$ …{3}
Using equation. {3}, kinetic energy of particle A is given by,
$qV= \dfrac {1}{2} m {{v}_{A}}^{2}$ …(4)
Similarly, kinetic energy of particle B is given by,
$4qV= \dfrac {1}{2} m {{v}_{B}}^{2}$ …(5)
Dividing equation. (4) by (5) we get,
$\dfrac {1}{4} = \dfrac {{{v}_{A}}^{2}} {{{v}_{A}}^{2}}$
$\Rightarrow \dfrac {{v}_{A}}{{v}_{B}} = \dfrac {1}{2}$
Hence, the ratio of their speed $\dfrac {{v}_{A}}{{v}_{B}}$ will become 1:2.

So, the correct answer is option B i.e. 1:2.

Note:
This problem can also be solved using an alternate method. The alternate method is given below:
We know, force is given by.
$F= ma$ …(1)
Where, m is the mass of the particle
               a is the acceleration
Rearranging equation. (1) we get,
$a = \dfrac {F}{m}$ …(2)
We know, force is also given by,
$F= qE$
Substituting this value in the equation. (2) we get,
$a = \dfrac {qE}{m}$
Acceleration due to particle A is given by.
${a}_{A} = \dfrac {qE}{m}$ …(3)
Similarly, acceleration due to particle B is given by,
${a}_{B} = \dfrac {4qE}{m}$ …(4)
Substituting equation. (3) in equation. (4) we get,
${a}_{B}= 4{a}_{A}$
$\rightarrow {a}_{A}= \dfrac {1}{4}{a}_{B}$
Equation of motion is given by,
$ {v}^{2} = {u}^{2} +2as$
For particle A above equation becomes,
${{v}_{A}}^{2} = 0 + 2{a}_{A}s$ …(3)
Similarly for particle B,
${{v}_{B}}^{2} = 0 + 2{a}_{B}s$ …(4)
Dividing equation. (3) by (4) we get,
$\dfrac {{{v}_{A}}^{2}} {{{v}_{A}}^{2}} = \dfrac {{a}_{A}}{{a}_{B}}$
Substituting value in above equation we get,
$\dfrac {{{v}_{A}}^{2}} {{{v}_{A}}^{2}} =\dfrac {{a}_{B}}{4{a}_{B}}$
$\Rightarrow \dfrac {{v}_{A}}{{v}_{B}} = \dfrac {1}{2}$
Hence, the ratio of their speed $\dfrac {{v}_{A}}{{v}_{B}}$ will become 1:2.
So, the correct answer is option B i.e. 1:2.