Question

which set has more coulombic force? remember : ( f_{c}=\frac{k(q_{1}q_{2})}{d^{2}} ) set 2 set 1 the sets have an equal amount of coulombic force.

Explanation:

Step1: Calculate the product of charges for Set 1

For Set 1, \(q_1 = 40\space C\) and \(q_2=65\space C\). The product \(q_1q_2=40\times65 = 2600\space C^{2}\)

Step2: Calculate the product of charges for Set 2

For Set 2, \(q_1 = 70\space C\) and \(q_2 = 65\space C\). The product \(q_1q_2=70\times65=4550\space C^{2}\)

Step3: Analyze the Coulomb's force formula

Since \(F_c=\frac{k(q_1q_2)}{d^{2}}\), and \(k\) (Coulomb's constant) and \(d\) (distance, \(d = 2\space \mathring{A}\) for both sets) are the same for both sets. The force is directly proportional to the product \(q_1q_2\)

Answer:

Set 2