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Suppose two point charges, Q1 = -6.25 x 10-9 C and Q2 = -6.25 x 10-9 C, are separated by a distance d = 0.617 m.

Using Coulomb's Law, find the repulsive force between the charges.


A)
F = 9.23 x 10-7 N


B)
F = 9.67 x 10-7 N


C)
F = -9.67 x 107 N


D)
F = -9.23 x 10-7 N

Respuesta :

joseaaronlara

Answer:

The answer to your question is the letter A) F =  9.23 x 10⁻⁷ N

Explanation:

Data

q₁ = -6.25 x 10⁻⁹ C

q₂ = -6.25 x 10⁻⁹ C

d = 0.617 m

k = 9 x 10⁹ Nm²/C²

F = ?

Formula

              F = k q₁q₂ /r²

-Substitution

              F = (9 x 10⁹)(-6.25 x 10⁻⁹)(-6.25 x 10⁻⁹) / (0.617)²

-Simplification

              F = 3.512 x 10⁻⁷ / 0.381

-Result

              F = 9.227 x 10⁻⁷ N  ≈ 9.23 x 10⁻⁷ N

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