Two equally charged particles are held 2


Two equally charged particles are held 3.2\times 10^{-3}  m apart and then released from rest. The initial acceleration of the first particle is observed to be 7.0 m/s^2 and that of the second to be 9.0 m/s^2. If the mass of the first particle is 6.3\times 10^{-7} kg, what are (a) the mass of the second particle and (b) the magnitude of the charge of each particle?

Solution:

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Let us look at particle one first. From Newton’s second law, we know that:

F=ma

(Where F is force, m is mass, and a is acceleration)

Let us substitute the values we know for particle one.

F=ma

F=(6.3\times 10^{-7})(7)

F=4.41\times 10^{-6} N

 

We can now apply this to Coulomb’s law. Coulomb’s law states:

F=k\dfrac{(q_1)(q_2)}{r^2}

(Where F is the electrostatic force, k is Coulomb’s constant, q_1 is the charge of the first particle, q_2 is the charge of the second particle, and r is the distance between the two charges)

 

We know the value of F, and so, let us again, substitute the values into this equation.

F=k\dfrac{(q_1)(q_2)}{r^2}

4.41\times 10^{-6}=(8.99\times 10^{9})\dfrac{q^2}{(3.2\times 10^{-3})^2}

(Remember, the question tells us that the particles are equally charged, thus q_1=q_2)

q=7.09\times 10^{-11} C

 

The mass of the second particle can be calculated by using Newton’s second law again. Since the force is the same for both particles, we can isolate for mass.

F=ma

4.41\times 10^{-6}=(m)(9)

m=4.9\times 10^{-7} kg

 

Final Answers:

a) m_2=4.9\times 10^{-7} kg

b) q=7.09\times 10^{-11} C

 

This question can be found in Fundamentals of Physics, 10th edition, chapter 21, question 6.

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