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?
1 answer:
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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