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

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Xema8 Xema8 · Answer 1 · 2020-06-05T06:00:31+02:00

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.