Question

Two equally charged particles are held3.2$\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{m}$ apart and then released from rest. The initial acceleration of the first particle is observed to be 7.0m/s² and that of the second to be 9.0m/s². If the mass of the first particle is$\mathbf{6}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{kg}\mathbf{,}$, what are (a) the mass of the second particle and (b) the magnitude of the charge of each particle?

Short answer

1. The mass of the second particle is $4.9\times {10}^{-7}kg$
2. The magnitude of the charge of each particle is $7.1\times {10}^{-11}C$

Step-by-step solution

 Step 1: The given data

The charge on each particle is the same.

Initial separation between the particles, $r=3.2\times {10}^{-3}m$

The initial acceleration of the first particle, $a_1=7.0\hspace{0.33em}m/s^2$

The initial acceleration of the second particle,$a_2=9.0\hspace{0.33em}m/s^2$

Mass of the first particle,$m_1=6.3\times {10}^{-7}kg$

Understanding the concept of Coulomb’s law and Newton’s law

According to Coulomb's Law of electrostatic attraction or repulsion within particles, the force acting on them is given as being directly proportional to the product of the charges on the particles and being inversely proportional to the separation between them. Using this concept, we can find out the force acting on them. Again, from the concept of Newton's second law, we can get the mass of the particle.

Formulae:

The magnitude of the electrostatic force between any two particles,

$F=k\frac{\left|q_1\right|\left|q_2\right|}{r^2}$ (1)

The force value according to Newton’s second law of motion,

F=ma (2)

a) Calculation of the mass of the second particle

As the two particles were held and released at the same time, the force acting on them due to their acceleration is the same. Thus, using equation (2), we can get the mass of the second particle as given:

$$\begin{gathered} m_2{a}_2=m_1{a}_1 \\ {m}_2=\frac{(6.3\times {10}^{-7}kg)(7.0\hspace{0.33em}m/s^2)}{(9.0\hspace{0.33em}m/s^2)} \\ {m}_2=4.9\times {10}^{-7}kg \end{gathered}$$

Hence, the value of the mass of the particle is$4.9\times {10}^{-7}kg$

b) Calculation of the charge on each particle

If we consider the force acting on the particle, then due to conservation law, the force acting on particle 1 due to acceleration is the same as the force acting on it due to its electrostatic charge. Thus, using equations (1) and (2), we can get the magnitude of the charge on particle 1 (or particle 2, as both have the same charge) as follows

$$\begin{gathered} m_1{a}_1=k\left|q_1\right|\left|q_2\right|/r^2 \\ (6.3\times {10}^{-7}kg)(7.0\hspace{0.33em}m/s^2)=(9\times {10}^9\hspace{0.33em}(N\cdot m^2)/C^2)\frac{|q{|}^2}{(0.0032\hspace{0.33em}m{)}^2} \\ \left|q\right|=7.1\times {10}^{-11}C \end{gathered}$$

Hence, the value of charge on each particle is $7.1\times {10}^{-11}C$