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# Suppose two similar charged particles are placed 2.3 × 10-2 m apart and then released fromrest. The acceleration of the 1st charge particle is observed to be 5.0 m/s2 and that of the secondto 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 secondparticle and (b) the magnitude of the charge of each particle?

Dissection Guru
39 Points
one year ago
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.
Vibekjyoti Sahoo
145 Points
2 months ago
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.
Thank You