AP EAMCET · PHYSICS · Electrostatics
Two balls of charge \(q_1\) and \(q_2\) initially have a velocity of the same magnitude and direction. After a uniform electric field has been applied during a certain time, the direction of the velocity of the first ball changes by \(60^{\circ}\), and the velocity magnitude is reduced by half. The direction of the velocity of the second ball changes thereby by \(90^{\circ}\).
In what proportion will the velocity of the second ball change? Determine the magnitude of the charge-to-mass ratio for the second ball if it is equal to \(k_1\) for the first ball. The electrostatic interaction between the balls should be neglected.
- A \(\frac{k}{\sqrt{2}}\)
- B \(\frac{k}{\sqrt{3}}\)
- C \(\frac{k}{2}\)
- D \[\frac{4}{3} k_1\]
Answer & Solution
Correct Answer
(D) \[\frac{4}{3} k_1\]
Step-by-step Solution
Detailed explanation
Let \(v_1\) and \(v_2\) be the velocities of the first and second balls after the removal of the uniform electric field. By hypothesis, the angle between the velocity \(v_1\) and the initial velocity \(v\) is \(60^{\circ}\). Therefore, the change in the momentum of the first…
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