Two particles A and B having charges ` 8.0 xx 10 ^(-6) C` and `-2.0 xx 10^(-6) C` respectively are held fixed with a. separation of 20cm. Where should a third charged. particle be placed so that it dose not experience a net electric force?.

Options

A. 20 cm
B. 40 cm
C. 60 cm
D. 45 cm

Explanation

To solve the problem, we need to find the position where a third charged particle can be placed such that it does not experience a net electric force due to the two fixed charges A and B. Let's denote the charges as follows:

- Charge A (Q1) = \( 8.0 \times 10^{-6} \, C \) (positive)
- Charge B (Q2) = \( -2.0 \times 10^{-6} \, C \) (negative)
- Separation between A and B = 20 cm = 0.20 m

### Step-by-step Solution:

1. **Understanding the Forces**:
The third charge (let's call it Q) will experience forces due to both charges A and B. The force due to charge A will be repulsive (since A is positive), and the force due to charge B will be attractive (since B is negative).

2. **Positioning the Third Charge**:
The third charge can be placed either between A and B or outside this segment. However, since the charge A is larger than charge B, we expect that the third charge will be placed closer to charge B (the smaller magnitude).

3. **Setting Up the Equations**:
Let’s denote the distance from charge B to the third charge Q as \( x \). Then, the distance from charge A to charge Q will be \( 0.20 + x \).

The forces acting on charge Q due to charges A and B can be expressed as:
- Force due to A (F1):
\[ F_1 = k \frac{|Q_1 \cdot Q|}{(0.20 + x)^2} \]
- Force due to B (F2):
\[ F_2 = k \frac{|Q_2 \cdot Q|}{x^2} \]

4. **Balancing the Forces**:
For the third charge Q to experience no net force, the magnitudes of these forces must be equal:
\[ F_1 = F_2 \]
Therefore, we have:
\[ k \frac{|Q_1 \cdot Q|}{(0.20 + x)^2} = k \frac{|Q_2 \cdot Q|}{x^2} \]
We can cancel \( k \) and \( |Q| \) from both sides (assuming \( Q \neq 0 \)):
\[ \frac{8.0 \times 10^{-6}}{(0.20 + x)^2} = \frac{2.0 \times 10^{-6}}{x^2} \]

5. **Cross-Multiplying**:
Cross-multiplying gives us:
\[ 8.0 \times 10^{-6} \cdot x^2 = 2.0 \times 10^{-6} \cdot (0.20 + x)^2 \]

6. **Simplifying the Equation**:
Dividing both sides by \( 2.0 \times 10^{-6} \):
\[ 4x^2 = (0.20 + x)^2 \]
Expanding the right side:
\[ 4x^2 = 0.04 + 0.40x + x^2 \]
Rearranging gives:
\[ 4x^2 - x^2 - 0.40x - 0.04 = 0 \]
Simplifying:
\[ 3x^2 - 0.40x - 0.04 = 0 \]

7. **Using the Quadratic Formula**:
We can solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -0.40 \), and \( c = -0.04 \):
\[ x = \frac{0.40 \pm \sqrt{(-0.40)^2 - 4 \cdot 3 \cdot (-0.04)}}{2 \cdot 3} \]
\[ = \frac{0.40 \pm \sqrt{0.16 + 0.48}}{6} \]
\[ = \frac{0.40 \pm \sqrt{0.64}}{6} \]
\[ = \frac{0.40 \pm 0.8}{6} \]

This gives two possible solutions:
\[ x = \frac{1.2}{6} = 0.20 \, m \quad \text{(not valid, as it is outside the range)} \]
\[ x = \frac{-0.40}{6} = -0.0667 \, m \quad \text{(valid)} \]

8. **Conclusion**:
The third charge should be placed approximately 0.0667 m (or 6.67 cm) to the left of charge B (outside the segment between A and B).