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Fynjy0 [20]
3 years ago
12

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?
Physics
1 answer:
Fynjy0 [20]3 years ago
7 0

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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I hope it helps you!                        

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