Question

Use these two constants for the question that follows:


e = 1.6 × 10−19 C
k = 8.99 × 109 N m2/C2

Two negative charges are 10−14 m away from each other. Using Coulomb's law, which of the following is the electrical force between these two particles?
2.3 N
−2.3 N
1.4 N
−1.4 N

Answer 1

Coulomb's law states that the electric force \( F \) between two point charges can be calculated using the formula:

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

where:
- \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \) (Coulomb's constant),
- \( q_1 \) and \( q_2 \) are the charges,
- \( r \) is the distance between the charges.

In this case, we have two negative charges, both equal to \( e = 1.6 \times 10^{-19} \, \text{C} \). Since both charges are negative, the force will be repulsive, hence it will be negative.

Let's calculate the force:

1. **Use the values**:
\[
q_1 = -e = -1.6 \times 10^{-19} \, \text{C}
\]
\[
q_2 = -e = -1.6 \times 10^{-19} \, \text{C}
\]
\[
r = 10^{-14} \, \text{m}
\]

2. **Insert values into Coulomb's law**:
\[
F = k \frac{|q_1 q_2|}{r^2} = 8.99 \times 10^9 \frac{(1.6 \times 10^{-19})(1.6 \times 10^{-19})}{(10^{-14})^2}
\]

3. **Calculate**:
\[
F = 8.99 \times 10^9 \frac{(2.56 \times 10^{-38})}{(1 \times 10^{-28})}
\]
\[
F = 8.99 \times 10^9 \times 2.56 \times 10^{-10}
\]
\[
F = 2.3 \, \text{N}
\]

Since both charges are negative, the force is repulsive, which means we express it as a negative value:

\[
F = -2.3 \, \text{N}
\]

Thus, the electrical force between these two particles is **−2.3 N**.