Two point charges of `+16 muC and -9 mu C` are placed 8 cm apart in air. Determine the position of the point at which the resultant electric field is zero.

Explanation

To determine the position where the resultant electric field is zero due to two point charges of +16 µC and -9 µC placed 8 cm apart, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions:**
- Let the charge +16 µC be at point A and the charge -9 µC be at point B.
- The distance between the charges A and B is 8 cm.

2. **Determine Possible Regions for Zero Electric Field:**
- The electric field due to a positive charge points away from the charge, while the electric field due to a negative charge points towards the charge.
- Therefore, the electric field can potentially be zero in the following regions:
- To the left of charge A (not likely as the field due to +16 µC will dominate).
- Between the two charges (not likely as the fields will not cancel).
- To the right of charge B (this is a potential region).

3. **Set Up the Equation for Electric Fields:**
- Let the point where the electric field is zero be at a distance \( x \) from charge B (the -9 µC charge).
- The distance from charge A (the +16 µC charge) to this point will then be \( 8 + x \) cm.
- The electric field due to charge A at this point is given by:
\[ E_1 = \frac{k \cdot 16 \times 10^{-6}}{(8 + x)^2} \]
- The electric field due to charge B at this point is given by:
\[ E_2 = \frac{k \cdot 9 \times 10^{-6}}{x^2} \]

4. **Set the Electric Fields Equal:**
- For the electric field to be zero, the magnitudes of \( E_1 \) and \( E_2 \) must be equal:
\[ \frac{k \cdot 16 \times 10^{-6}}{(8 + x)^2} = \frac{k \cdot 9 \times 10^{-6}}{x^2} \]
- The \( k \) and \( 10^{-6} \) terms cancel out:
\[ \frac{16}{(8 + x)^2} = \frac{9}{x^2} \]

5. **Cross-Multiply to Solve for x:**
- Cross-multiplying gives:
\[ 16x^2 = 9(8 + x)^2 \]
- Expanding the right side:
\[ 16x^2 = 9(64 + 16x + x^2) \]
\[ 16x^2 = 576 + 144x + 9x^2 \]
- Rearranging the equation:
\[ 16x^2 - 9x^2 - 144x - 576 = 0 \]
\[ 7x^2 - 144x - 576 = 0 \]

6. **Use the Quadratic Formula to Solve for x:**
- The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, \( a = 7 \), \( b = -144 \), and \( c = -576 \):
\[ x = \frac{144 \pm \sqrt{(-144)^2 - 4 \cdot 7 \cdot (-576)}}{2 \cdot 7} \]
\[ x = \frac{144 \pm \sqrt{20736 + 16128}}{14} \]
\[ x = \frac{144 \pm \sqrt{36864}}{14} \]
\[ x = \frac{144 \pm 192}{14} \]

7. **Calculate the Possible Values for x:**
- The two possible solutions are:
\[ x = \frac{336}{14} = 24 \text{ cm} \quad \text{(valid)} \]
\[ x = \frac{-48}{14} \quad \text{(not valid, as distance cannot be negative)} \]

8. **Conclusion:**
- The position where the electric field is zero is 24 cm to the right of the -9 µC charge.

### Final Answer:
The resultant electric field is zero at a point located 24 cm to the right of the -9 µC charge.