A charge q1 of - 5.00 * 10-⁹ C and a charge q2 of - 2.00 * 10-⁹ C are separated by a distance of 40.0 cm. Find the equilibrium position for a third charge of + 15.0 * 10-⁹ C.

P. S. 10-⁹ is 10^-9

The equilibrium position will be in between q₁ and q₂.

Let this position of third charge be x distance from q₁

q₁ will pull the third charge towards it with force F₁

F₁ = 9 x 10⁹x 5 x 10⁻⁹x15 x 10⁻⁹ / x²

= 675 x 10⁻⁹ / x²

q₂ will pull the third charge towards it with force F₂

F₂ = 9 x 10⁹x 2 x 10⁻⁹x15 x 10⁻⁹ / (.40-x) ²

= 270 x 10⁻⁹ / (.40-x) ²

For equilibrium

675 x 10⁻⁹ / x² = 270 x 10⁻⁹ / (.40-x) ²

5 / x² = 2 / (.40-x) ²

(.40-x) ² / x² = 2/5 =.4

.4 - x / x =.632

.4 - x =.632x

.4 = 1.632 x

x =.245.

24.5 cm

so third charge must be placed at 24.5 cm away from q₁ charge.