Question

Question 21
Using the equation for Coulomb's Law, what happens to the force between two objects when you double the distance between them?
F= k q1q2/r^2
\large F=k\frac{q_1q_2}{r^2}

a
it decreases by 1/4
b
it doubles
c
it decreases by half
d
it quadruples

Answer 1

According to Coulomb's Law, the force \( F \) between two charged objects is given by the equation:

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

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

If you double the distance \( r \), the new distance becomes \( 2r \). Substituting this into the equation gives:

\[
F' = k \frac{q_1 q_2}{(2r)^2} = k \frac{q_1 q_2}{4r^2}
\]

This shows that the new force \( F' \) is:

\[
F' = \frac{1}{4} \left( k \frac{q_1 q_2}{r^2} \right) = \frac{1}{4} F
\]

Therefore, when you double the distance between the two objects, the force decreases to one-fourth of its original value.

The correct answer is:

**a) it decreases by 1/4**.