The force between two charged objects can be calculated using Coulomb's Law, which states that the force \( F \) between two charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by:
\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \]
where \( k \) is Coulomb's constant.
To compare the interactions, we need to look at both the product of the charges and the distance between them:
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Interaction A:
- Charges: \( +2 \) and \( -2 \)
- Distance: \( 1 \)
- \( F = k \frac{|2 \cdot -2|}{1^2} = k \frac{4}{1} = 4k \)
-
Interaction B:
- Charges: \( +1 \) and \( -1 \)
- Distance: \( 1 \)
- \( F = k \frac{|1 \cdot -1|}{1^2} = k \frac{1}{1} = 1k \)
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Interaction C:
- Charges: \( +2 \) and \( -2 \)
- Distance: \( 4 \)
- \( F = k \frac{|2 \cdot -2|}{4^2} = k \frac{4}{16} = \frac{1}{4}k \)
-
Interaction D:
- Charges: \( +1 \) and \( -1 \)
- Distance: \( 4 \)
- \( F = k \frac{|1 \cdot -1|}{4^2} = k \frac{1}{16} = \frac{1}{16}k \)
Now we can compare the forces:
- Interaction A: \( 4k \)
- Interaction B: \( 1k \)
- Interaction C: \( \frac{1}{4}k \)
- Interaction D: \( \frac{1}{16}k \)
The strongest force is in Interaction A.
Thus, the answer is: Interaction A.