Two equally charged particles are held

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}

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}

Final Answers:

m_2=4.9\times 10^{-7}

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

Two equally charged particles are held 2

Solution:

Final Answers:

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

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