Question

Two equally charged particles, held 3.0 × 10^-3 m apart, are released from rest. The initial acceleration of the first particle is observed to be 8.5 m/s² and that of the second to be 12 m/s². If the mass of the first particle is 8.7 × 10^-7 kg, what are

(a) the mass of the second particle and

(b) the magnitude of the charge of each particle?

Answer 1

Suppose q be the magnitude of the charge.

Distance between the charges, r = 3×10^-3m

Since the particles are of same charge, if they are released from rest, they will move apart with some initial acceleration

Given,

Initial acceleration of first particle, a₁ = 8.5 m/s²​​​​​​

Initial acceleration of the second particle, a₂ = 12 m/s²

The mass of first particle, m₁ = 8.7×10^-7 kg

(a) We have Newton’s third law which states for every action in nature there is an equal and opposite reaction. i.e, if an object A exerts a force on an another object B, then object B also exerts an equal and opposite force on object A.

Therefore the force on first particle exerted by second particle, say F₁₂ is equal to the force on second particle exerted by first particle F₂₁.

From Newton’s second law, Force is given by F = m×a

F₁₂ = F₂₁

m₁ × a₁ = m₂ × a₂

From this, mass of second particle can be determined

m₂ = 6.1625×10^-7 kg

(b) The magnitude of force from Newton’s law( F_g) equals the electrostatic force between the charged particles (Coulomb's law), say F_e​​​​​​.

F_g = m₁× a₁

  F_g = 8.7×10^-7 × 8.5

F_g = 73.95×10^-7 N

From Coulomb's law

Here q₁ = q₂ = q (since the particles are of equal charge)

r = 3×10^-3 m

F_e = F_g = 73.95 ×10^-7 N

k = Coulomb's constant = 9×10⁹ kg.m³/s².C²

  

  

  

q = 8.6×10^-11C

Therefore magnitude of charge of both particle is

8.6×10^-11C