Two like charges of magnitude `1xx10^(-9) ` coulomb and `9xx10^(-9)` coulomb are separated by a distance of 1 meter. The point on the line joining the charges , where the force experienced by a charge placed at that point is zero , is

Options

A. 0.25 m from the charge `1xx10^(-9)C`
B. 0.75 m from the charge `9xx10^(-9)C`
C. 0.25 m from the charge `1xx10^(-9)C, 0.75` m from charge `9xx10^(-9)C`
D. at all points on the line joining the charges

Answer

To solve the problem, we need to find the point on the line joining two like charges where the net electric force experienced by a test charge placed at that point is zero.

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions**:
- Let \( Q_1 = 1 \times 10^{-9} \, \text{C} \) (Charge 1)
- Let \( Q_2 = 9 \times 10^{-9} \, \text{C} \) (Charge 2)
- The distance between the two charges is \( d = 1 \, \text{m} \).

2. **Define the Position of the Test Charge**:
- Let \( x \) be the distance from \( Q_1 \) to the point where the test charge \( q \) is placed.
- Therefore, the distance from \( Q_2 \) to the test charge will be \( (1 - x) \).

3. **Set Up the Equation for Electric Fields**:
- The electric field \( E_1 \) due to charge \( Q_1 \) at the point is given by:
\[ E_1 = \frac{k \cdot Q_1}{x^2} \]
- The electric field \( E_2 \) due to charge \( Q_2 \) at the point is given by:
\[ E_2 = \frac{k \cdot Q_2}{(1 - x)^2} \]

4. **Equate the Electric Fields**:
- For the net force to be zero, the magnitudes of the electric fields must be equal:
\[ E_1 = E_2 \]
- Substituting the expressions for \( E_1 \) and \( E_2 \):
\[ \frac{k \cdot Q_1}{x^2} = \frac{k \cdot Q_2}{(1 - x)^2} \]
- The constant \( k \) cancels out:
\[ \frac{Q_1}{x^2} = \frac{Q_2}{(1 - x)^2} \]

5. **Substituting the Values of Charges**:
- Substituting \( Q_1 = 1 \times 10^{-9} \) and \( Q_2 = 9 \times 10^{-9} \):
\[ \frac{1 \times 10^{-9}}{x^2} = \frac{9 \times 10^{-9}}{(1 - x)^2} \]
- Simplifying gives:
\[ \frac{1}{x^2} = \frac{9}{(1 - x)^2} \]

6. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ (1 - x)^2 = 9x^2 \]

7. **Expanding and Rearranging**:
- Expanding the left side:
\[ 1 - 2x + x^2 = 9x^2 \]
- Rearranging gives:
\[ 1 - 2x - 8x^2 = 0 \]
- This is a quadratic equation:
\[ 8x^2 + 2x - 1 = 0 \]

8. **Using the Quadratic Formula**:
- The quadratic formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 8, b = 2, c = -1 \).
- Discriminant \( D = b^2 - 4ac = 2^2 - 4 \cdot 8 \cdot (-1) = 4 + 32 = 36 \).
- Thus:
\[ x = \frac{-2 \pm 6}{16} \]
- This gives two solutions:
\[ x = \frac{4}{16} = \frac{1}{4} \quad \text{and} \quad x = \frac{-8}{16} = -\frac{1}{2} \]

9. **Interpreting the Results**:
- The positive solution \( x = \frac{1}{4} \, \text{m} \) is valid and indicates the position from \( Q_1 \).
- The negative solution \( x = -\frac{1}{2} \) is not physically meaningful in this context.

### Final Answer:
The point on the line joining the charges where the force experienced by a charge placed at that point is zero is \( 0.25 \, \text{m} \) from the charge \( Q_1 \) (1 x \( 10^{-9} \, \text{C} \)).