# 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:

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

a) $m_2=4.9\times 10^{-7}$ kg
b) $q=7.09\times 10^{-11}$ C