Two particles, with identical positive charges and a separation of 3.41 × 10-2 m, are released from rest. Immediately after the release, particle 1 has an acceleration a1 whose magnitude is 3.68 × 103 m/s2, while particle 2 has an acceleration a2 whose magnitude is 10.8 × 103 m/s2. Particle 1 has a mass of 4.66 × 10-6 kg. Find (a) the charge on each particle and (b) the mass of particle 2.

(a) the charge on each particle is 4.71 x 10⁻⁸ C

(b) the mass of particle 2 is 1.59 x 10⁻⁶ kg

Explanation:

given information:

distance of both particle, r = 3.41 × 10⁻² m

acceleration of the particle 1, a₁ = 3.68 × 10³ m/s²

acceleration of the particle 1, a₂ = 10.8 × 10³ m/s²

the mass of particle 1, m₁ = 4.66 × 10⁻⁶ kg

(a) the charge on each particle

the both particle has positive identical charge, thus q₁=q₂=q. to find the charge we can use the electrostatic force equation

F = k q₁q₂/r²

where

F = electrostatic force (N)

k = coulomb's constant (8.99 x 10⁹ N.m²/C²)

q = the charge (C)

r = then distance of the charge (m)

F = k q₁q₂/r²

  =  k q²/r²

  = 8.99 x 10⁹ q²/(3.41 × 10⁻²)²

  = 7.73 x 10¹² q² N

according to newton's second law

F = ma

F = m₁a₁

7.73 x 10¹² q² = (4.66 × 10⁻⁶) (3.68 × 10³)

q² = (4.66 × 10⁻⁶) (3.68 × 10³)/7.73 x 10¹²

q = 4.71 x 10⁻⁸ C

(b) the mass of particle 2

F₁=F₂, the force of both particle is the same

m₁a₁ = m₂a₂

m₂ = m₁a₁/a₂

     = (4.66 × 10⁻⁶) (3.68 × 10³)/(10.8 x 10³)

     = 1.59 x 10⁻⁶ kg

answer; /// the above statement is (true);

no. dot product   =0 when vecs at right angles