Question

Particle \(\mathrm{A}\) and \(\mathrm{B}\) have electric charge \(+\mathrm{q}\) and \(+4 \mathrm{q} .\) Both have mass \(\mathrm{m}\). If both are allowed to fall under the same p.d., ratio of velocities $\left(\mathrm{V}_{\mathrm{A}} / \mathrm{V}_{\mathrm{B}}\right)=\ldots \ldots \ldots \ldots \ldots$ (A) \(2: 1\) (B) \(1: 2\) (C) \(1: 4\) (D) \(4: 1\)

Short answer

The ratio of velocities \(\frac{V_{A}}{V_{B}}\) for Particle A and Particle B is \(1:2\).

Step-by-step solution

Consider the work-energy principle

The work-energy principle states that the work done on an object is equal to its change in kinetic energy. The work done on a charged particle in an electric field is determined by the charge, the potential difference, and the potential energy of the particle. We can find the kinetic energy of each particle after they fall through the potential difference and use it to calculate their velocities.

Write the equations for work done and kinetic energy

For particle A (charge +q): Work done, \(W_{A} = qV\) Change in kinetic energy, \(\Delta K_{A} = \frac{1}{2}mV_{A}^2\) For particle B (charge +4q): Work done, \(W_{B} = 4qV\) Change in kinetic energy, \(\Delta K_{B} = \frac{1}{2}mV_{B}^2\) Since work done is equal to the change in kinetic energy for each particle: \(qV = \frac{1}{2}mV_{A}^2\) and \(4qV = \frac{1}{2}mV_{B}^2\)

Solve for velocities

Now we need to solve for \(V_{A}\) and \(V_{B}\): For particle A: \(qV = \frac{1}{2}mV_{A}^2\) \(V_{A}^2 = \frac{2qV}{m}\) \(V_{A} = \sqrt{\frac{2qV}{m}}\) For particle B: \(4qV = \frac{1}{2}mV_{B}^2\) \(V_{B}^2 = \frac{8qV}{m}\) \(V_{B} = \sqrt{\frac{8qV}{m}}\)

Find the ratio of velocities

Now that we have expressions for \(V_{A}\) and \(V_{B}\), we can find the ratio between them: \(\frac{V_{A}}{V_{B}}=\frac{\sqrt{\frac{2qV}{m}}}{\sqrt{\frac{8qV}{m}}}= \sqrt{\frac{2}{8}}=\sqrt{\frac{1}{4}}\) \(\frac{V_{A}}{V_{B}}= \frac{1}{2}\) So the final answer is \(\frac{V_{A}}{V_{B}} = 1:2\) which corresponds to Option (B).