Question

Two particles, with identical positive charges and a separation of 2.12 × 10-2 m, are released from rest. immediately after the release, particle 1 has an acceleration a1 whose magnitude is 3.64 × 103 m/s2, while particle 2 has an acceleration a2 whose magnitude is 8.47 × 103 m/s2. particle 1 has a mass of 4.60 × 10-6 kg. find (a) the charge on each particle and (b) the mass of particle 2.

Answer 1

1) Particle 1 has an acceleration of $3.64 \cdot 10^3 m/s^2$, and this acceleration is caused by the electric force between the two particles, F.
For Newton's second law:
$F=m_1 a_1$
and using the mass of particle 1, we find the magnitude of the electric force:
$F=(4.60 \cdot 10^{-6} kg)(3.64 \cdot 10^3 m/s^2)=1.67 \cdot 10^{-2} N$

The electric force is also given by:
$F=k_e \frac{q_1 q_2}{r^2}$
where 
$k_e = 8.99 \cdot 10^9 N m^2 C^{-2}$ is the Coulomb's constant
$q_1 = q_2 =q$ is the charge of the two particles (they are identical, so they have same charge)
$r=2.12 \cdot 10^{-2} m$ 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:
$q= \sqrt{ \frac{Fr^2}{k_e} }= \sqrt{ \frac{(1.67 \cdot 10^{-2} N)(2.12 \cdot 10^{-2} m)^2}{8.99\cdot 10^9 N m^2 C^{-2}} }=2.89 \cdot 10^{-9} C$

(b) We can calculate the mass of particle 2 by applying Newton's second law:
$F=m_2 a_2$
where F is again the electric force. So, we find
$m_2 = \frac{F}{a_2}= \frac{1.67 \cdot 10^{-2}N}{8.47 \cdot 10^3 m/s^2}=1.97 \cdot 10^{-6}kg$