The force between two charged objects can be determined using Coulomb's Law, which states that the magnitude of the force \(F\) between two point charges is given by:
\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \]
where:
- \(F\) is the force,
- \(k\) is Coulomb's constant,
- \(q_1\) and \(q_2\) are the amounts of the charges, and
- \(r\) is the distance between the charges.
We must examine each interaction.
Interaction A:
- Charges: +2 and -2
- Distance: (not provided; will assume distance is 1 for calculation)
- Force: \(F_A = k \frac{|2 \cdot -2|}{1^2} = k \cdot 4\)
Interaction B:
- Charges: +1 and -1
- Distance: 1
- Force: \(F_B = k \frac{|1 \cdot -1|}{1^2} = k \cdot 1\)
Interaction C:
- Charges: +1 and -2
- Distance: 4
- Force: \(F_C = k \frac{|1 \cdot -2|}{4^2} = k \cdot \frac{2}{16} = k \cdot \frac{1}{8}\)
Interaction D:
- Charges: +1 and -1
- Distance: 4
- Force: \(F_D = k \frac{|1 \cdot -1|}{4^2} = k \cdot \frac{1}{16}\)
Now, let's compare the forces:
- \(F_A \text{ (force from Interaction A)} = k \cdot 4\)
- \(F_B \text{ (force from Interaction B)} = k \cdot 1\)
- \(F_C \text{ (force from Interaction C)} = k \cdot \frac{1}{8}\)
- \(F_D \text{ (force from Interaction D)} = k \cdot \frac{1}{16}\)
Based on the calculations, Interaction A has the strongest force: \(F_A = k \cdot 4\).
Thus, the answer is: • Interaction A
1 answer