To solve the problem, we need to find the point on the line joining two like charges where the net force experienced by a test charge placed at that point is zero. We have two charges: \( Q_1 = 1 \times 10^{-9} \, \text{C} \) and \( Q_2 = 9 \times 10^{-9} \, \text{C} \), separated by a distance of \( r = 1 \, \text{m} \).
### Step-by-step Solution:
1. **Identify the Charges and Distance**:
- Let \( Q_1 = 1 \times 10^{-9} \, \text{C} \) (located at point A).
- Let \( Q_2 = 9 \times 10^{-9} \, \text{C} \) (located at point B).
- The distance between the charges \( r = 1 \, \text{m} \).
2. **Assume a Point C**:
- Let point C be at a distance \( x \) from charge \( Q_1 \) (point A).
- Therefore, the distance from charge \( Q_2 \) (point B) to point C will be \( r - x = 1 - x \).
3. **Apply Coulomb’s Law**:
- The force \( F_{AC} \) exerted by \( Q_1 \) on a test charge \( q \) at point C:
\[
F_{AC} = k \frac{Q_1 q}{x^2}
\]
- The force \( F_{BC} \) exerted by \( Q_2 \) on the same test charge \( q \) at point C:
\[
F_{BC} = k \frac{Q_2 q}{(1 - x)^2}
\]
4. **Set the Forces Equal**:
- For the net force to be zero at point C, the magnitudes of these forces must be equal:
\[
F_{AC} = F_{BC}
\]
- Thus, we have:
\[
k \frac{Q_1 q}{x^2} = k \frac{Q_2 q}{(1 - x)^2}
\]
- We can cancel \( k \) and \( q \) from both sides:
\[
\frac{Q_1}{x^2} = \frac{Q_2}{(1 - x)^2}
\]
5. **Substitute the Values**:
- Substituting \( Q_1 = 1 \times 10^{-9} \) and \( Q_2 = 9 \times 10^{-9} \):
\[
\frac{1 \times 10^{-9}}{x^2} = \frac{9 \times 10^{-9}}{(1 - x)^2}
\]
6. **Cross Multiply**:
- Cross multiplying gives:
\[
1 \times 10^{-9} (1 - x)^2 = 9 \times 10^{-9} x^2
\]
- Dividing both sides by \( 10^{-9} \):
\[
(1 - x)^2 = 9x^2
\]
7. **Expand and Rearrange**:
- Expanding the left side:
\[
1 - 2x + x^2 = 9x^2
\]
- Rearranging gives:
\[
1 - 2x + x^2 - 9x^2 = 0
\]
\[
-8x^2 - 2x + 1 = 0
\]
- Multiplying through by -1:
\[
8x^2 + 2x - 1 = 0
\]
8. **Use the Quadratic Formula**:
- Applying the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 8 \), \( b = 2 \), \( c = -1 \):
\[
x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 8 \cdot (-1)}}{2 \cdot 8}
\]
\[
x = \frac{-2 \pm \sqrt{4 + 32}}{16}
\]
\[
x = \frac{-2 \pm \sqrt{36}}{16}
\]
\[
x = \frac{-2 \pm 6}{16}
\]
9. **Calculate Possible Values for x**:
- This gives two possible solutions:
\[
x = \frac{4}{16} = 0.25 \, \text{m} \quad \text{(valid, between A and B)}
\]
\[
x = \frac{-8}{16} = -0.5 \, \text{m} \quad \text{(not valid, outside the range)}
\]
10. **Conclusion**:
- The point where the force experienced by a charge is zero is at a distance of \( 0.25 \, \text{m} \) from charge \( Q_1 \).
### Final Answer:
The point on the line joining the charges where the force experienced by a charge placed at that point is zero is **0.25 m from charge \( Q_1 \)**.
