The force between two charged objects can be analyzed using Coulomb's Law, which states that the magnitude of the electrostatic force \( F \) between two charges is given by the formula:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
where:
- \( k \) is the electrostatic constant,
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
- \( r \) is the distance between the charges.
Let's evaluate each interaction based on the charge magnitudes and distance:
-
Interaction A: \( q_1 = +2 \), \( q_2 = -2 \), \( r = 1 \) \[ F_A = k \frac{|2 \cdot (-2)|}{1^2} = k \frac{4}{1} = 4k \]
-
Interaction B: \( q_1 = +1 \), \( q_2 = -1 \), \( r = 1 \) \[ F_B = k \frac{|1 \cdot (-1)|}{1^2} = k \frac{1}{1} = 1k \]
-
Interaction C: \( q_1 = +2 \), \( q_2 = -2 \), \( r = 4 \) \[ F_C = k \frac{|2 \cdot (-2)|}{4^2} = k \frac{4}{16} = \frac{1}{4}k \]
-
Interaction D: \( q_1 = +1 \), \( q_2 = -1 \), \( r = 4 \) \[ F_D = k \frac{|1 \cdot (-1)|}{4^2} = k \frac{1}{16} = \frac{1}{16}k \]
Now, let's compare the calculated forces:
- \( F_A = 4k \)
- \( F_B = 1k \)
- \( F_C = \frac{1}{4}k \)
- \( F_D = \frac{1}{16}k \)
Clearly, the strongest force occurs in Interaction A.
So the correct answer is:
Interaction A.