Two charges `4 xx 10^(-9)C and -16 xx 10^(-9)C` are separated by a distance 20cm in air. The position of the neutral point from the small charge is

Options

A. 40/3 cm
B. 20/3 cm
C. 40cm
D. 10/3 cm

Answer

To find the position of the neutral point between two charges, we can use the concept of electric fields. The neutral point is where the electric fields due to both charges cancel each other out.

### Step-by-Step Solution:

1. **Identify the Charges and Distance**:
- Let \( q_1 = 4 \times 10^{-9} \, \text{C} \) (positive charge).
- Let \( q_2 = -16 \times 10^{-9} \, \text{C} \) (negative charge).
- The distance between the two charges is \( d = 20 \, \text{cm} = 0.2 \, \text{m} \).

2. **Determine the Position of the Neutral Point**:
- Let \( x \) be the distance from the positive charge \( q_1 \) to the neutral point.
- The distance from the negative charge \( q_2 \) to the neutral point will then be \( d - x = 0.2 - x \).

3. **Set Up the Electric Field Equations**:
- The electric field due to \( q_1 \) at the neutral point is given by:
\[ E_1 = \frac{k \cdot |q_1|}{x^2} \]
- The electric field due to \( q_2 \) at the neutral point is given by:
\[ E_2 = \frac{k \cdot |q_2|}{(0.2 - x)^2} \]
- At the neutral point, \( E_1 = E_2 \):
\[ \frac{k \cdot |q_1|}{x^2} = \frac{k \cdot |q_2|}{(0.2 - x)^2} \]

4. **Cancel the Constant \( k \)**:
- Since \( k \) appears on both sides, we can cancel it out:
\[ \frac{|q_1|}{x^2} = \frac{|q_2|}{(0.2 - x)^2} \]

5. **Substitute the Values of the Charges**:
- Substitute \( |q_1| = 4 \times 10^{-9} \) and \( |q_2| = 16 \times 10^{-9} \):
\[ \frac{4 \times 10^{-9}}{x^2} = \frac{16 \times 10^{-9}}{(0.2 - x)^2} \]

6. **Cross-Multiply**:
- Cross-multiplying gives:
\[ 4 \times 10^{-9} \cdot (0.2 - x)^2 = 16 \times 10^{-9} \cdot x^2 \]

7. **Simplify the Equation**:
- Dividing both sides by \( 4 \times 10^{-9} \):
\[ (0.2 - x)^2 = 4x^2 \]

8. **Expand and Rearrange**:
- Expanding the left side:
\[ 0.04 - 0.4x + x^2 = 4x^2 \]
- Rearranging gives:
\[ 0 = 3x^2 + 0.4x - 0.04 \]

9. **Use the Quadratic Formula**:
- Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 3 \), \( b = 0.4 \), and \( c = -0.04 \).
- Calculate the discriminant:
\[ b^2 - 4ac = (0.4)^2 - 4 \cdot 3 \cdot (-0.04) = 0.16 + 0.48 = 0.64 \]
- Now, apply the quadratic formula:
\[ x = \frac{-0.4 \pm \sqrt{0.64}}{2 \cdot 3} = \frac{-0.4 \pm 0.8}{6} \]
- This gives two possible solutions:
\[ x_1 = \frac{0.4}{6} = \frac{2}{30} = \frac{1}{15} \, \text{m} \, (\text{valid}) \]
\[ x_2 = \frac{-1.2}{6} \, (\text{not valid as distance cannot be negative}) \]

10. **Convert to Centimeters**:
- Convert \( x \) to centimeters:
\[ x = \frac{1}{15} \times 100 \approx 6.67 \, \text{cm} \]

### Final Answer:
The position of the neutral point from the small charge \( q_1 \) is approximately **6.67 cm**.