Suppose two similar charged particles are placed 2.3 × 10-2 m apart and then released from
rest. The acceleration of the 1st charge particle is observed to be 5.0 m/s2 and that of the second
to be 7.0 m/s2. 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?

Question

Dissection Guru · Answer

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, M 1 a 1 +M 2 a 2 = 0 (Initial acceleration is zero) Substituting values, you get M 2 = 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. 1 and from Coulomb’s law, F = K q 1 q 2 /r 2 – Eqn. 2 Equating Eqn.1 and Eqn.2 and setting q 1 = q 2 (Since they have equal charges) You will get q 1 =q 2 = 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.

Vibekjyoti Sahoo · Answer

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, M 1 a 1 +M 2 a 2 = 0 (Initial acceleration is zero) Substituting values, you get M 2 = 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. 1 and from Coulomb’s law, F = K q 1 q 2 /r 2 – Eqn. 2 Equating Eqn.1 and Eqn.2 and setting q 1 = q 2 (Since they have equal charges) You will get q 1 =q 2 = 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. Thank You

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