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What must be the distance between point charge 91 = 20 MC and point charge 02 -45,0 AC for the electrostatic force between them to have a magnitude of 7.0 N?

Respuesta :

Answer:

Explanation:

Q₁ = 20 X 10⁻⁶ C

Q₂ = -45 X 10⁻⁶ C

d is required distance.

F = Force between the = 7 N

F = k x Q₁ X Q₂ / d²

d² = k x Q₁ X Q₂ / F = 20 X 45 X 10⁻¹² X 9 X 10⁹  /7

=1.157

d = 1.075 m

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