Three particles are fixed on an x axis Particle 1 of charge q1 is at x a and particle 2 of charge q2 is at x a If their net electrostatic force on particle 3 of charge Q is to be zero what must be the ratio q1 q2 when particle 3 is at x 0.560a Tries 0 10 If their net electrostatic force on particle 3 of charge Q is to be zero what must be the ratio q1 q2 when particle 3 is at x 1.68a?

a) q₁ / q₂ = 1.62, With both charges of the same sign

b) q₁ / q₂ = 6.1, With the two charges of opposite sign

Explanation:

a) The electric force is given by Coulomb's law

F = k q₁q₂ / r²

If all three charges have the same sign the force is repulsive, therefore they are subtracted

F₁₃ - F₂₃ = 0

F₁₃ = F₂₃

K q₁ Q / r₁₃² = k q₂ Q / r₂₃²

q₁ / q₂ = r₁₃² / r₂₃²

The position of the second charge at x = a, the third charge at x = 0.560a and the first face is not clear it could be at x = 0 or at x = - a, let's make the callus assuming it is at x = 0

q₁ / q₂ = (0.560a-0) ² / (0.560a - a) ²

q₁ / q₂ = (0.560 / 0.44) ²

q₁ / q₂ = 1.62

With both charges of the same sign

b) The distance of the 3 particle is x = 1.68 a

In this case the third particle is outside the other two, so that we have a subtraction on the outside one of the one of charges 1 or 2 must have an opposite sign

q₁ / q₂ = (1.68 a - 0) ² / (1.68a - a) ²

q₁ / q₂ = (1.68 / 0.68) ²

q₁ / q₂ = 6.1

With the two charges of opposite sign