A total of 9 `mu`C charge in distributed between two point objects. When there objects are placed at a distance of 0.1 m, they apply 18 N of force of repulsion on each other. What is the ratio of magnitude of charges on both the point object?

Options

A. 15/12
B. 15/16
C. `10//12`
D. none

Explanation

To solve the problem, we need to find the ratio of the magnitudes of charges \( Q_1 \) and \( Q_2 \) distributed between two point objects, given that the total charge is \( 9 \, \mu C \) and the force of repulsion between them is \( 18 \, N \) at a distance of \( 0.1 \, m \).

### Step-by-Step Solution:

1. **Identify Given Values:**
- Total charge: \( Q_1 + Q_2 = 9 \, \mu C = 9 \times 10^{-6} \, C \)
- Distance between charges: \( r = 0.1 \, m \)
- Force of repulsion: \( F = 18 \, N \)

2. **Apply Coulomb's Law:**
Coulomb's law states that the force \( F \) between two point charges is given by:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r^2} \]
Where \( \epsilon_0 \) (the permittivity of free space) is approximately \( 8.85 \times 10^{-12} \, C^2/(N \cdot m^2) \).

3. **Substitute Known Values:**
Substitute \( F = 18 \, N \), \( r = 0.1 \, m \), and \( \frac{1}{4 \pi \epsilon_0} \approx 9 \times 10^9 \, N \cdot m^2/C^2 \):
\[ 18 = 9 \times 10^9 \frac{Q_1 Q_2}{(0.1)^2} \]

4. **Simplify the Equation:**
\[ 18 = 9 \times 10^9 \frac{Q_1 Q_2}{0.01} \]
\[ 18 = 9 \times 10^{11} Q_1 Q_2 \]
\[ Q_1 Q_2 = \frac{18}{9 \times 10^{11}} = 2 \times 10^{-11} \, C^2 \]

5. **Set Up the System of Equations:**
Now we have two equations:
- \( Q_1 + Q_2 = 9 \times 10^{-6} \) (1)
- \( Q_1 Q_2 = 2 \times 10^{-11} \) (2)

6. **Express \( Q_2 \) in terms of \( Q_1 \):**
From equation (1):
\[ Q_2 = 9 \times 10^{-6} - Q_1 \]

7. **Substitute \( Q_2 \) into Equation (2):**
\[ Q_1 (9 \times 10^{-6} - Q_1) = 2 \times 10^{-11} \]
\[ 9 \times 10^{-6} Q_1 - Q_1^2 = 2 \times 10^{-11} \]
Rearranging gives:
\[ Q_1^2 - 9 \times 10^{-6} Q_1 + 2 \times 10^{-11} = 0 \]

8. **Solve the Quadratic Equation:**
Using the quadratic formula \( Q_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = -9 \times 10^{-6} \), \( c = 2 \times 10^{-11} \):
\[ Q_1 = \frac{9 \times 10^{-6} \pm \sqrt{(9 \times 10^{-6})^2 - 4 \cdot 1 \cdot 2 \times 10^{-11}}}{2 \cdot 1} \]
\[ Q_1 = \frac{9 \times 10^{-6} \pm \sqrt{81 \times 10^{-12} - 8 \times 10^{-11}}}{2} \]
\[ Q_1 = \frac{9 \times 10^{-6} \pm \sqrt{1 \times 10^{-12}}}{2} \]
\[ Q_1 = \frac{9 \times 10^{-6} \pm 1 \times 10^{-6}}{2} \]

9. **Calculate \( Q_1 \) and \( Q_2 \):**
- \( Q_1 = \frac{10 \times 10^{-6}}{2} = 5 \times 10^{-6} \, C \)
- \( Q_2 = 9 \times 10^{-6} - 5 \times 10^{-6} = 4 \times 10^{-6} \, C \)

10. **Find the Ratio:**
\[ \text{Ratio} = \frac{Q_1}{Q_2} = \frac{5 \times 10^{-6}}{4 \times 10^{-6}} = \frac{5}{4} \]

### Final Answer:
The ratio of the magnitudes of the charges on both point objects is \( \frac{5}{4} \).