Answer:
Explanation:
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.
