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

Options

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

Answer

To find the position of the neutral point between two charges, we can follow these steps:

### Step 1: Identify the Charges and Distance
Let the two charges be:
- \( q_1 = 4 \times 10^{-9} \) C (positive charge)
- \( q_2 = -16 \times 10^{-9} \) C (negative charge)

The distance between the two charges is given as 20 cm, which we convert to meters:
\[ d = 20 \, \text{cm} = 0.2 \, \text{m} \]

### Step 2: Define the Position of the Neutral Point
Let the distance from the positive charge \( q_1 \) to the neutral point be \( x \). Consequently, the distance from the negative charge \( q_2 \) to the neutral point will be:
\[ d - x = 0.2 - x \]

### Step 3: Set Up the Equation for Electric Fields
At the neutral point, the electric field due to \( q_1 \) must equal the electric field due to \( q_2 \):
\[ E_1 = E_2 \]

The electric field \( E \) due to a point charge is given by:
\[ E = \frac{k |q|}{r^2} \]
where \( k \) is Coulomb's constant.

Thus, we can write:
\[ \frac{k \cdot |q_1|}{x^2} = \frac{k \cdot |q_2|}{(0.2 - x)^2} \]

### Step 4: Cancel Out the Constant
Since \( k \) appears on both sides of the equation, we can cancel it out:
\[ \frac{4 \times 10^{-9}}{x^2} = \frac{16 \times 10^{-9}}{(0.2 - x)^2} \]

### Step 5: Simplify the Equation
We can simplify this equation:
\[ \frac{4}{x^2} = \frac{16}{(0.2 - x)^2} \]
Cross-multiplying gives us:
\[ 4(0.2 - x)^2 = 16x^2 \]

### Step 6: Expand and Rearrange
Expanding the left side:
\[ 4(0.04 - 0.4x + x^2) = 16x^2 \]
\[ 0.16 - 1.6x + 4x^2 = 16x^2 \]

Rearranging the equation:
\[ 0 = 16x^2 - 4x^2 + 1.6x - 0.16 \]
\[ 0 = 12x^2 + 1.6x - 0.16 \]

### Step 7: Solve the Quadratic Equation
Now we can solve the quadratic equation:
\[ 12x^2 + 1.6x - 0.16 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 12 \), \( b = 1.6 \), and \( c = -0.16 \).

Calculating the discriminant:
\[ b^2 - 4ac = (1.6)^2 - 4 \cdot 12 \cdot (-0.16) \]
\[ = 2.56 + 7.68 = 10.24 \]

Now substituting into the formula:
\[ x = \frac{-1.6 \pm \sqrt{10.24}}{2 \cdot 12} \]
\[ = \frac{-1.6 \pm 3.2}{24} \]

Calculating the two possible values for \( x \):
1. \( x = \frac{1.6}{24} = 0.0667 \) m (approximately)
2. \( x = \frac{-4.8}{24} \) (not valid since distance cannot be negative)

### Step 8: Convert to Centimeters
Convert \( x \) back to centimeters:
\[ x \approx 0.0667 \times 100 = 6.67 \, \text{cm} \]

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

---