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Grade: 12th pass

                        

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?

one month ago

Answers : (1)

Dissection Guru
39 Points
							
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. 1
and from Coulomb’s law, F = K q1q2/r2  – Eqn. 2
Equating 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. 
one month ago
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