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Suppose two similar charged particles are placed 2.3 × 10 -2 m apart and then released from rest. The acceleration of the 1 st charge particle is observed to be 5.0 m/s 2 and that of the second to be 7.0 m/s 2 . If the mass of the first particle is 3.6 × 10 -5 kg, what are (a) the mass of the second particle and (b) the magnitude of the charge of each particle?
Basically, two charges are placed at some distance away from each other. (A) So, if you notice, the net force on the charge system (Charge 1 + Charge 2) is zero.Therefore, Linear Momentum must be conserved. In other words, M1a1+M2a2 = 0 (Initial acceleration is zero)Substituting values, you get M2 = 2.57 × 10-5. (b) In the second part, you can use Newton’s Second Law and Coulomb’s Law.From Newton’s Law, we know F = Ma – Eqn. 1and from Coulomb’s law, F = K q1q2/r2 – Eqn. 2Equating Eqn.1 and Eqn.2 and setting q1= q2 (Since they have equal charges) You will get q1=q2 = 1.058 × 10-17. Note:- You can use Eqn. 1 and Eqn. 2 for any of these two particles since electrostatic force is same for both.
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