Question

Two particles A and B having the same mass have charges +q and +4q, respectively. When they are allowed to fall from rest through the same electric potential difference the ratio of their speeds v_A to v_B will become ______.

Options

• 2 : 1

• 1 : 4

• 4 : 1

• 1 : 2

Answer 1

Two particles A and B having the same mass have charges +q and +4q, respectively. When they are allowed to fall from rest through the same electric potential difference the ratio of their speeds v_A to v_B will become 1 : 2.

Explanation:

Given, the mass of both particles = m

Charge of particle, A = +q

Charge of particle, B = +4q

Potential difference = V

Kinetic energy is given by K = `1/2m"v"^2` ...........(i)

This energy is equal to electrostatic potential energy is K = V × Q .............(ii)

From Eqs. (i) and (ii), we have

`1/2m"v"^2 =  V xx Q`

For particle A, `qV = 1/2m"v"_A^2` .........(iii)

For particle B, `4qV = 1/2m"v"_B^2` ................(iv)

Dividing Eq. (iii) by Eq. (iv), we get

`1/4 = "v"_A^2/"v"_B^2 ⇒ "v"_A/"v"_B = 1/2`

Hence, the ratio of their speed `"v"_A/"v"_B = 1/2`.