Question

In which of the following interactions will the amount of force between the two objects be the strongest?

Interaction Charge on Object 1 Charge on Object 2 Distance Between Objects
A +2 -2 1
B +1 -1 1
C +2 -2 4
D +1 -1 4
(1 point)
Responses

Interaction B
Interaction B

Interaction D
Interaction D

Interaction A
Interaction A

Interaction C

Answer 1

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**