Certainly! Let's solve the problem step-by-step using Coulomb's Law:
Coulomb's Law provides the magnitude of the electrostatic force [tex]\( F \)[/tex] between two point charges. The formula is given by:
[tex]\[
F = k_e \frac{|q_1 q_3|}{r^2}
\][/tex]
where:
- [tex]\( k_e \)[/tex] is Coulomb's constant, which has the value [tex]\( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)[/tex].
- [tex]\( q_1 \)[/tex] and [tex]\( q_3 \)[/tex] are the magnitudes of the point charges.
- [tex]\( r \)[/tex] is the distance between the two charges.
Here are the given values:
- [tex]\( k_e = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)[/tex]
- [tex]\( q_1 = 1 \, \text{C} \)[/tex] (Coulombs)
- [tex]\( q_3 = 1 \, \text{C} \)[/tex] (Coulombs)
- [tex]\( r = 0.55 \, \text{m} \)[/tex] (meters)
Using these values in Coulomb's Law formula, we get:
[tex]\[
F = 8.99 \times 10^9 \frac{|1 \cdot 1|}{0.55^2}
\][/tex]
First, calculate the square of the distance [tex]\( r \)[/tex]:
[tex]\[
r^2 = (0.55)^2 = 0.3025 \, \text{m}^2
\][/tex]
Next, substitute this value along with the charges and Coulomb's constant into the formula:
[tex]\[
F = 8.99 \times 10^9 \frac{1}{0.3025}
\][/tex]
Now, compute the division inside the formula:
[tex]\[
\frac{1}{0.3025} \approx 3.3058
\][/tex]
Then, multiply this result by Coulomb's constant:
[tex]\[
F = 8.99 \times 10^9 \times 3.3058 \approx 29719008264.462807 \, \text{N}
\][/tex]
Thus, the electrostatic force [tex]\( \vec{F}_3 \)[/tex] between the charges [tex]\( q_1 \)[/tex] and [tex]\( q_3 \)[/tex] is approximately:
[tex]\[
\boxed{29719008264.462807 \, \text{N}}
\][/tex]
