Question

The electrostatic force between two charges is originally F
. Which expression represents the new electrostatic force between the two charges if one charge doubles and the other one is divided by three?(1 point)
Responses

16F
1 sixth cap f

32F
3 halves cap f

23F
2 thirds cap f

6F

Answer 1

The electrostatic force between two charges is given by Coulomb's Law:

\[
F = k \cdot \frac{q_1 \cdot q_2}{r^2}
\]

where:
- \( F \) is the electrostatic force,
- \( k \) is Coulomb's constant,
- \( q_1 \) and \( q_2 \) are the magnitudes of the two charges, and
- \( r \) is the distance between the charges.

Let's denote the original charges as \( q_1 \) and \( q_2 \). According to the problem:
- One charge doubles: \( q_1' = 2q_1 \)
- The other charge is divided by three: \( q_2' = \frac{q_2}{3} \)

Now we can find the new force \( F' \):

\[
F' = k \cdot \frac{q_1' \cdot q_2'}{r^2} = k \cdot \frac{(2q_1) \cdot \left(\frac{q_2}{3}\right)}{r^2}
\]

Simplifying this expression:

\[
F' = k \cdot \frac{2q_1 \cdot q_2}{3r^2} = \frac{2}{3} \cdot k \cdot \frac{q_1 \cdot q_2}{r^2} = \frac{2}{3} F
\]

Thus, the new electrostatic force \( F' \) is:

\[
F' = \frac{2}{3} F
\]

Therefore, the correct response is:

**2 thirds cap f**