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 (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{ (Fr^2)/(k_e) }= \sqrt{ ((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 = (F)/(a_2)= (1.67 \cdot 10^(-2)N)/(8.47 \cdot 10^3 m/s^2)=1.97 \cdot 10^(-6)kg`