To determine which set has more Coulombic force, we need to use Coulomb's Law:
[tex]\[ F_c = \frac{k \left(q_1 q_2\right)}{d^2} \][/tex]
where:
- \( F_c \) is the Coulombic force
- \( k \) is Coulomb's constant, approximately \( 8.987551787 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
- \( q_1 \) and \( q_2 \) are the charges
- \( d \) is the distance between the charges
### For Set 1:
Given:
- \( q_1 = 30 \, \text{C} \)
- \( q_2 = 15 \, \text{C} \)
- \( d = 7 \, \text{\AA} = 7 \times 10^{-10} \, \text{m} \)
[tex]\[
F_{c, \text{Set 1}} = \frac{(8.987551787 \times 10^9) \left(30 \times 15\right)}{(7 \times 10^{-10})^2}
\][/tex]
From the numerical calculation, we get:
[tex]\[
F_{c, \text{Set 1}} \approx 82538740904.40161 \, \text{N}
\][/tex]
### For Set 2:
Given:
- \( q_1 = 45 \, \text{C} \)
- \( q_2 = 15 \, \text{C} \)
- \( d = 7 \, \text{\AA} = 7 \times 10^{-10} \, \text{m} \)
[tex]\[
F_{c, \text{Set 2}} = \frac{(8.987551787 \times 10^9) \left(45 \times 15\right)}{(7 \times 10^{-10})^2}
\][/tex]
From the numerical calculation, we get:
[tex]\[
F_{c, \text{Set 2}} \approx 123808111356.60242 \, \text{N}
\][/tex]
### Comparison:
- \( F_{c, \text{Set 1}} \approx 82538740904.40161 \, \text{N} \)
- \( F_{c, \text{Set 2}} \approx 123808111356.60242 \, \text{N} \)
Since \( F_{c, \text{Set 2}} > F_{c, \text{Set 1}} \), Set 2 has more Coulombic force.
Therefore, the correct answer is:
Set 2
