Question

In a vacuum, two particles have charges of q1 and q2, where q1 = +4.1C. They are separated by a distance of 0.23 m, and particle 1 experiences an attractive force of 4.6 N. What is the value of q2, with its sign?

Answer 1

According to coulomb's law, we have

F = k_e q₁ q₂ / r²

where, q₁ = charge of particle 1 = +4.1 C

r = separation distance between them = 0.23 m

k_e = proportionality constant = 9 x 10⁹ Nm²/C²

F = attractive force = 4.6 N

then, we get

(4.6 N) = (9 x 10⁹ Nm²/C²) (+4.1 C) q₂ / (0.23 m)

q₂ = [(4.6 N) (0.23 m)] / [(9 x 10⁹ Nm²/C²) (+4.1 C)]

q₂ = [(1.058 N.m) / (3.69 x 10¹⁰ Nm²/C)]

q₂ = 2.86 x 10^-11 C

{ Since the force between the charges is attractive and q₁ is positive, then naturally q₂ must be negative. So there has to be a minus sign in front of our calculated number }

Then, we get

q₂ = - 2.86 x 10^-11 C