Charge q experiences a repulsive force of 0.50 mN when placed at a distance of 25 cm from charge q₁, and it experiences an attractive force of 0.25 mN when placed a distance of 75 cm from charge q₂. The ratio q₁/ q₂ is

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The force between two charges is given by Coulomb's law:$$F = k\frac{q_1q_2}{d^2}$$where k is Coulomb's constant, q1 and q2 are the charges, and d is the distance between them.In this problem, we have two different forces and two different distances. We can set up two equations:$$0.50	ext{ mN} = k\frac{q_1q_2}{(0.25	ext{ m})^2}$$$$0.25	ext{ mN} = k\frac{q_1q_2}{(0.75	ext{ m})^2}$$We can divide the first equation by the second equation to eliminate k and q2:$$\frac{0.50	ext{ mN}}{0.25	ext{ mN}} = \frac{(0.25	ext{ m})^2}{(0.75	ext{ m})^2}$$$$\frac{q_1}{q_2} = \frac{2}{9}$$Therefore, the ratio q1/q2 is 2/9.

Charge Q Experiences a Repulsive Force of 0