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Sagot :
Answer:
0.3 N
Explanation:
Electromagnetic force is F= Kq1q2/r^2, where r is the distance between charges. If r is doubled then the force will be 1/4F which is 0.3 N.
The magnitude of the force exerted on Q by q when the distance between them is doubled is 0.3 N
Coulomb's law equation
F = Kq₁q₂ / r²
Where
- F is the force of attraction
- K is the electrical constant
- q₁ and q₂ are two point charges
- r is the distance apart
Data obtained from the question
- Initial distance apart (r₁) = r
- Initial force (F₁) = 1.2 N
- Final distance apart (r₂) = 2r
- Final force (F₂) =?
How to determine the final force
From Coulomb's law,
F = Kq₁q₂ / r²
Cross multiply
Fr² = Kq₁q₂
Kq₁q₂ = constant
F₁r₁² = F₂r₂²
With the above formula, we can obtain the final force as follow:
F₁r₁² = F₂r₂²
1.2 × r² = F₂ × (2r)²
1.2r² = F₂ × 4r²
Divide both side by 4r²
F₂ = 1.2r² / 4r²
F₂ = 0.3 N
Learn more about Coulomb's law:
https://brainly.com/question/506926
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