1) Particle 1 has an acceleration of [tex]3.64 \cdot 10^3 m/s^2[/tex], and this acceleration is caused by the electric force between the two particles, F.
For Newton's second law:
[tex]F=m_1 a_1[/tex]
and using the mass of particle 1, we find the magnitude of the electric force:
[tex]F=(4.60 \cdot 10^{-6} kg)(3.64 \cdot 10^3 m/s^2)=1.67 \cdot 10^{-2} N[/tex]
The electric force is also given by:
[tex]F=k_e \frac{q_1 q_2}{r^2} [/tex]
where
[tex]k_e = 8.99 \cdot 10^9 N m^2 C^{-2}[/tex] is the Coulomb's constant
[tex]q_1 = q_2 =q[/tex] is the charge of the two particles (they are identical, so they have same charge)
[tex]r=2.12 \cdot 10^{-2} m[/tex] is the initial distance between the two particles.
Using the value of the force we found previously, and by re-arranging this formula, we can calculate the value of q, the charge of each particle:
[tex]q= \sqrt{ \frac{Fr^2}{k_e} }= \sqrt{ \frac{(1.67 \cdot 10^{-2} N)(2.12 \cdot 10^{-2} m)^2}{8.99\cdot 10^9 N m^2 C^{-2}} }=2.89 \cdot 10^{-9} C [/tex]
(b) We can calculate the mass of particle 2 by applying Newton's second law:
[tex]F=m_2 a_2[/tex]
where F is again the electric force. So, we find
[tex]m_2 = \frac{F}{a_2}= \frac{1.67 \cdot 10^{-2}N}{8.47 \cdot 10^3 m/s^2}=1.97 \cdot 10^{-6}kg [/tex]
