18) A particle 'A' has charge \(+q\) and a particle 'B' has charge \(+4q\). Each has the same mass \(m\). When they are allowed to fall from rest through the same potential difference, what is the ratio of their speeds (particle A to particle B)?
Solution:
When a particle falls through a potential difference \(V\), the change in its electric potential energy is converted into kinetic energy. For a particle with charge \(q\), the work done is: \[ qV. \]
This work becomes the kinetic energy: \[ \frac{1}{2}mv^2 = qV. \]
Therefore, the speed \(v\) of the particle is: \[ v = \sqrt{\frac{2qV}{m}}. \]
For particle A (charge \(+q\)): \[ v_A = \sqrt{\frac{2qV}{m}}. \]
For particle B (charge \(+4q\)): \[ v_B = \sqrt{\frac{2(4q)V}{m}} = \sqrt{\frac{8qV}{m}} = 2\sqrt{\frac{2qV}{m}} = 2v_A. \]
Thus, the ratio of their speeds (particle A to particle B) is: \[ v_A : v_B = 1 : 2. \]
Therefore, the correct answer is option B.
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