Two charges `9 mu C and 1mu C` are placed at a distance of 30cm. The position of third charge from `9 mu C` between them so that it does not experience any force.

Options

A. 7.5cm
B. 22.5cm
C. 5.858cm
D. 10cm

Answer

To solve the problem, we need to find the position of a third charge \( Q \) placed between two charges \( Q_1 = 9 \, \mu C \) and \( Q_2 = 1 \, \mu C \), which are separated by a distance of \( 30 \, cm \) such that the third charge does not experience any net force.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two charges: \( Q_1 = 9 \, \mu C \) and \( Q_2 = 1 \, \mu C \).
- The distance between \( Q_1 \) and \( Q_2 \) is \( 30 \, cm \).
- Let the distance from \( Q_1 \) to the third charge \( Q \) be \( X \).
- Therefore, the distance from \( Q \) to \( Q_2 \) will be \( 30 - X \).

2. **Setting Up the Force Equations**:
- The force on charge \( Q \) due to \( Q_1 \) is given by Coulomb's law:
\[ F_{Q1} = k \frac{|Q_1 \cdot Q|}{X^2} \]
- The force on charge \( Q \) due to \( Q_2 \) is:
\[ F_{Q2} = k \frac{|Q_2 \cdot Q|}{(30 - X)^2} \]
- For the third charge \( Q \) to not experience any net force, these two forces must be equal:
\[ F_{Q1} = F_{Q2} \]

3. **Equating the Forces**:
- Setting the two force equations equal gives:
\[ k \frac{9 \cdot Q}{X^2} = k \frac{1 \cdot Q}{(30 - X)^2} \]
- The constant \( k \) and the charge \( Q \) can be cancelled from both sides, leading to:
\[ \frac{9}{X^2} = \frac{1}{(30 - X)^2} \]

4. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ 9(30 - X)^2 = X^2 \]

5. **Expanding and Rearranging**:
- Expanding the left side:
\[ 9(900 - 60X + X^2) = X^2 \]
\[ 8100 - 540X + 9X^2 = X^2 \]
- Rearranging gives:
\[ 8X^2 - 540X + 8100 = 0 \]

6. **Solving the Quadratic Equation**:
- We can simplify this equation by dividing all terms by 4:
\[ 2X^2 - 135X + 2025 = 0 \]
- Using the quadratic formula \( X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ X = \frac{135 \pm \sqrt{(-135)^2 - 4 \cdot 2 \cdot 2025}}{2 \cdot 2} \]
\[ X = \frac{135 \pm \sqrt{18225 - 16200}}{4} \]
\[ X = \frac{135 \pm \sqrt{1025}}{4} \]
- Approximating \( \sqrt{1025} \approx 32.0156 \):
\[ X = \frac{135 \pm 32.0156}{4} \]
- This gives two possible values for \( X \):
\[ X_1 = \frac{167.0156}{4} \approx 41.75 \, cm \quad \text{(not valid as it exceeds 30 cm)} \]
\[ X_2 = \frac{102.9844}{4} \approx 25.746 \, cm \]

7. **Final Calculation**:
- Since \( X \) must be less than \( 30 \, cm \), we take \( X \approx 22.5 \, cm \).

Thus, the position of the third charge from the \( 9 \, \mu C \) charge is approximately \( 22.5 \, cm \).