A particle A has a charge +q and particle B has a 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 speed VA : VB will become _____?

4:1

2:1

1:2

None of these

The kinetic energy gained by a charged particle in an electric field is given by KE = qV, where q is charge and V is the potential difference.

Since kinetic energy is also KE = ½ mV², equating both expressions gives V = √(2qV/m).

For particle A (q) and particle B (4q), their velocities will be VA = √(2qV/m) and VB = √(8qV/m).

Taking the ratio: VA/VB = √(2qV/m) / √(8qV/m) = 1/2.