Question

Two particles, one with charge −2.93×10−6 C and the other with charge 4.39×10−6 C , are 0.0209 m apart. What is the magnitude of the force that one particle exerts on the other?

magnitude of force:

Two new particles with identical positive charge q3 are placed the same 0.0209 m apart. The force between them is measured to be the same as that between the original particles. What is q3 ?

Answer 1

Given,

The charges, q₁ = - 2.93 * 10^-6 C and q₂ = 4.39 * 10^-6 C

The distance, r = 0.0209 m

a)

The electrostatic force, F_e = ( k q₁q₂) / r²

F_e = ( 9 * 10⁹ * 2.93 * 10^-6 * 4.39 * 10^-6 ) / (0.0209)²

     = 265.022 N

b)

Let the charge of a new particle be q₃

The distance is, r = 0.0209 m

F_e = ( k q₁q₂) / r²   ( q₁ = q₂)

F_e = k (q₃)² / r²

(q₃)² = F_e r² / k

(q₃)² =  265.022 * (0.0209)² / 9 * 10⁹

(q₃)²  = 1.29 * 10^-11 C

q₃ = 3.59 * 10^-6 C