Question

Two identical charges have a Coulomb Force between equal to F. If the charges remain the same, but
the Coulomb Force changes to 1/9*F, what can you say happened to the charges? (1 point)
Answers below:
The charges changed polarities

The distance between the charges increased by three times

The distance between the charges increased by nine times

The Coulomb's constant between the charges changed by nine times

Answer 1

The correct answer is: **The distance between the charges increased by three times.**

Explanation: According to Coulomb's Law, the force \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by:

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

where \( k \) is Coulomb's constant.

If the initial force between the charges is \( F \) and is now changed to \( \frac{1}{9}F \), we can set up the relationship as:

\[
\frac{1}{9}F = k \frac{|q_1 q_2|}{r'^2}
\]

where \( r' \) is the new distance. The original force can be expressed as:

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

By setting up the ratio of the two forces:

\[
\frac{1}{9}F = F \left(\frac{r^2}{r'^2}\right)
\]

This simplifies to:

\[
\frac{1}{9} = \frac{r^2}{r'^2}
\]

Cross multiplying gives:

\[
r'^2 = 9 r^2
\]

Taking the square root of both sides:

\[
r' = 3r
\]

This indicates that the distance between the charges increased by three times.