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 ${\displaystyle \frac{v_A}{v_B}}$ 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 ${\displaystyle \frac{v_A}{v_B}}$.
Formula used:
$K={\displaystyle \frac{1}{2}}m{v}^2$
$K=charge\times potential$

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,
$K={\displaystyle \frac{1}{2}}m{v}^2$ …(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,
$K=QV={\displaystyle \frac{1}{2}}m{v}^2$ …{3}
Using equation. {3}, kinetic energy of particle A is given by,
$qV={\displaystyle \frac{1}{2}}m{v_A}^2$ …(4)
Similarly, kinetic energy of particle B is given by,
$4qV={\displaystyle \frac{1}{2}}m{v_B}^2$ …(5)
Dividing equation. (4) by (5) we get,
${\displaystyle \frac{1}{4}}={\displaystyle \frac{{v_A}^2}{{v_A}^2}}$
$\Rightarrow {\displaystyle \frac{v_A}{v_B}}={\displaystyle \frac{1}{2}}$
Hence, the ratio of their speed ${\displaystyle \frac{v_A}{v_B}}$ 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,
$a={\displaystyle \frac{F}{m}}$ …(2)
We know, force is also given by,
$F=qE$
Substituting this value in the equation. (2) we get,
$a={\displaystyle \frac{qE}{m}}$
Acceleration due to particle A is given by.
$a_A={\displaystyle \frac{qE}{m}}$ …(3)
Similarly, acceleration due to particle B is given by,
$a_B={\displaystyle \frac{4qE}{m}}$ …(4)
Substituting equation. (3) in equation. (4) we get,
$a_B=4a_A$
$\to a_A={\displaystyle \frac{1}{4}}{a}_B$
Equation of motion is given by,
$v^2=u^2+2as$
For particle A above equation becomes,
${v_A}^2=0+2a_{A}s$ …(3)
Similarly for particle B,
${v_B}^2=0+2a_{B}s$ …(4)
Dividing equation. (3) by (4) we get,
${\displaystyle \frac{{v_A}^2}{{v_A}^2}}={\displaystyle \frac{a_A}{a_B}}$
Substituting value in above equation we get,
${\displaystyle \frac{{v_A}^2}{{v_A}^2}}={\displaystyle \frac{a_B}{4a_B}}$
$\Rightarrow {\displaystyle \frac{v_A}{v_B}}={\displaystyle \frac{1}{2}}$
Hence, the ratio of their speed ${\displaystyle \frac{v_A}{v_B}}$ will become 1:2.
So, the correct answer is option B i.e. 1:2.