To determine which interaction has the strongest force between two charged objects, we can use Coulomb's law, which states that the electric force \( F \) between two charges is given by:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
where:
- \( F \) is the force,
- \( k \) is Coulomb's constant,
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
- \( r \) is the distance between the charges.
We need to calculate the force for each interaction:
Interaction A:
- Charge 1 = +2
- Charge 2 = -2
- Distance = 1 \[ F_A = k \frac{|2 \times -2|}{1^2} = k \frac{4}{1} = 4k \]
Interaction B:
- Charge 1 = +1
- Charge 2 = -1
- Distance = 1 \[ F_B = k \frac{|1 \times -1|}{1^2} = k \frac{1}{1} = 1k \]
Interaction C:
- Charge 1 = +2
- Charge 2 = -2
- Distance = 4 \[ F_C = k \frac{|2 \times -2|}{4^2} = k \frac{4}{16} = \frac{1}{4}k \]
Interaction D:
- Charge 1 = +1
- Charge 2 = -1
- Distance = 4 \[ F_D = k \frac{|1 \times -1|}{4^2} = k \frac{1}{16} = \frac{1}{16}k \]
Now, we can summarize the calculations:
- \( F_A = 4k \)
- \( F_B = 1k \)
- \( F_C = \frac{1}{4}k \)
- \( F_D = \frac{1}{16}k \)
From the computed forces, we see that the force is strongest in Interaction A.
Therefore, the correct answer is:
Interaction A