Answer :
Sure, let's solve this step-by-step using Coulomb's Law:
Coulomb's Law formula:
[tex]\[ \vec{F} = k_e \frac{\left|q_1 q_2\right|}{r^2} \][/tex]
Given values:
- Coulomb's constant \( k_e = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)
- Distance between the charges \( r = 0.25 \, \text{m} \)
- We need to assume the charges \( q_1 \) and \( q_2 \). Let's assume:
- \( q_1 = 1 \, \text{C} \)
- \( q_2 = 1 \, \text{C} \)
Now, plug these values into the formula:
[tex]\[ \vec{F} = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \times \frac{\left|1 \, \text{C} \times 1 \, \text{C}\right|}{(0.25 \, \text{m})^2} \][/tex]
First, calculate the denominator:
[tex]\[ (0.25 \, \text{m})^2 = 0.0625 \, \text{m}^2 \][/tex]
Then, calculate the force:
[tex]\[ \vec{F} = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \times \frac{1 \, \text{C}^2}{0.0625 \, \text{m}^2} \][/tex]
Simplify the fraction:
[tex]\[ \frac{1 \, \text{C}^2}{0.0625 \, \text{m}^2} = 16 \, \text{C}^2 / \text{m}^2 \][/tex]
Now multiply:
[tex]\[ \vec{F} = 8.99 \times 10^9 \, \text{N} \cdot 16 = 143.84 \times 10^9 \, \text{N} \][/tex]
So the force between \( q_1 \) and \( q_2 \) is:
[tex]\[ \vec{F} \approx 143840000000 \, \text{N} \][/tex]
Thus, the force \( \vec{F}_2 \) is:
[tex]\[ \vec{F}_2 = -143840000000 \, \text{N} \][/tex]
The negative sign indicates the direction of the force, implying attraction for opposite charges or repulsion for like charges, but since specific charges were not given, we assume the context implies the opposites. Hence, the magnitude of the force is:
[tex]\[ \boxed{143840000000 \, \text{N}} \][/tex]
Coulomb's Law formula:
[tex]\[ \vec{F} = k_e \frac{\left|q_1 q_2\right|}{r^2} \][/tex]
Given values:
- Coulomb's constant \( k_e = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)
- Distance between the charges \( r = 0.25 \, \text{m} \)
- We need to assume the charges \( q_1 \) and \( q_2 \). Let's assume:
- \( q_1 = 1 \, \text{C} \)
- \( q_2 = 1 \, \text{C} \)
Now, plug these values into the formula:
[tex]\[ \vec{F} = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \times \frac{\left|1 \, \text{C} \times 1 \, \text{C}\right|}{(0.25 \, \text{m})^2} \][/tex]
First, calculate the denominator:
[tex]\[ (0.25 \, \text{m})^2 = 0.0625 \, \text{m}^2 \][/tex]
Then, calculate the force:
[tex]\[ \vec{F} = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \times \frac{1 \, \text{C}^2}{0.0625 \, \text{m}^2} \][/tex]
Simplify the fraction:
[tex]\[ \frac{1 \, \text{C}^2}{0.0625 \, \text{m}^2} = 16 \, \text{C}^2 / \text{m}^2 \][/tex]
Now multiply:
[tex]\[ \vec{F} = 8.99 \times 10^9 \, \text{N} \cdot 16 = 143.84 \times 10^9 \, \text{N} \][/tex]
So the force between \( q_1 \) and \( q_2 \) is:
[tex]\[ \vec{F} \approx 143840000000 \, \text{N} \][/tex]
Thus, the force \( \vec{F}_2 \) is:
[tex]\[ \vec{F}_2 = -143840000000 \, \text{N} \][/tex]
The negative sign indicates the direction of the force, implying attraction for opposite charges or repulsion for like charges, but since specific charges were not given, we assume the context implies the opposites. Hence, the magnitude of the force is:
[tex]\[ \boxed{143840000000 \, \text{N}} \][/tex]
