A particle A has charge +q and particle B has charge +4q with each of them having the same mass ‘m’. When allowed to fall from rest through the same electrical potential difference. The ratio of their speeds v_A: v_B will be

Complete Solution:

When a charged particle falls through an electrical potential difference V, its potential energy is converted into kinetic energy. The change in potential energy is given by:

ΔU = q ⋅ V

where q is the charge of the particle and V is the potential difference.

The kinetic energy gained is:

KE = (1/2) m v²

For each particle, the electrical potential energy is entirely converted to kinetic energy:

q ⋅ V = (1/2) m v²

Solving for v:

v = √(2 q V / m)

For particle A (q = +q):

v_A = √(2 q V / m)

For particle B (q = +4q):

v_B = √(2 (4q) V / m) = √4 ⋅ √(2 q V / m) = 2 ⋅ √(2 q V / m)

v_A : v_B = √(2 q V / m) : 2 ⋅ √(2 q V / m)

Simplifying:

v_A : v_B = 1 : 2

Final Answer

The ratio of their speeds is: 1 : 2