To determine which set has more coulombic force, we use Coulomb's law, which states that the electrostatic force (\(F_c\)) between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The formula is:
[tex]\[ F_c = \frac{k \cdot q_1 \cdot q_2}{d^2} \][/tex]
where:
- \(k\) is Coulomb's constant (\(8.9875517873681764 \times 10^9 \, \text{N·m}^2/\text{C}^2\)),
- \(q_1\) and \(q_2\) are the magnitudes of the charges,
- \(d\) is the distance between the charges.
First, let's calculate the Coulombic force for Set 1.
Set 1:
- \(q_1 = 2 \, \text{C}\)
- \(q_2 = 7 \, \text{C}\)
- \(d = 6 \, \text{Å} = 6 \times 10^{-10} \, \text{m}\)
Substitute these values into Coulomb's law formula:
[tex]\[ F_c^{\text{Set 1}} = \frac{8.9875517873681764 \times 10^9 \cdot (2 \cdot 7)}{(6 \times 10^{-10})^2} \][/tex]
The resulting coulombic force for Set 1 is:
[tex]\[ F_c^{\text{Set 1}} = 3,495,159,028.420957 \, \text{N} \][/tex]
Set 2:
- \(q_1 = 1 \, \text{C}\)
- \(q_2 = 7 \, \text{C}\)
- \(d = 6 \, \text{Å} = 6 \times 10^{-10} \, \text{m}\)
Substitute these values into Coulomb's law formula:
[tex]\[ F_c^{\text{Set 2}} = \frac{8.9875517873681764 \times 10^9 \cdot (1 \cdot 7)}{(6 \times 10^{-10})^2} \][/tex]
The resulting coulombic force for Set 2 is:
[tex]\[ F_c^{\text{Set 2}} = 1,747,579,514.2104785 \, \text{N} \][/tex]
Comparing the forces:
- The Coulombic force for Set 1 is \(3,495,159,028.420957 \, \text{N}\).
- The Coulombic force for Set 2 is \(1,747,579,514.2104785 \, \text{N}\).
Since \(3,495,159,028.420957 > 1,747,579,514.2104785\), Set 1 has more coulombic force.
Therefore, the answer is:
Set 1 has more coulombic force.
