Two charges `4muC` and `1muC` are seperated by 16m. Where do you place a third charge so that it doesn't experience any force.

Answer

To solve the problem of where to place a third charge so that it doesn't experience any force due to the two given charges (4 µC and 1 µC) separated by 16 m, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Charges and Distance**:
- Let \( Q_1 = 4 \, \mu C = 4 \times 10^{-6} \, C \)
- Let \( Q_2 = 1 \, \mu C = 1 \times 10^{-6} \, C \)
- The distance between \( Q_1 \) and \( Q_2 \) is \( d = 16 \, m \).

2. **Define the Position of the Third Charge**:
- Let the position of the third charge \( Q \) be at a distance \( x \) from \( Q_1 \) (the 4 µC charge).
- The distance from \( Q_2 \) (the 1 µC charge) will then be \( 16 - x \).

3. **Set Up the Force Equations**:
- The force on charge \( Q \) due to \( Q_1 \) is given by:
\[ F_{Q_1} = k \frac{|Q_1 Q|}{x^2} \]
- The force on charge \( Q \) due to \( Q_2 \) is given by:
\[ F_{Q_2} = k \frac{|Q_2 Q|}{(16 - x)^2} \]
- For the charge \( Q \) to experience no net force, these two forces must be equal:
\[ F_{Q_1} = F_{Q_2} \]

4. **Equate the Forces**:
\[ k \frac{|Q_1 Q|}{x^2} = k \frac{|Q_2 Q|}{(16 - x)^2} \]
- Since \( k \) and \( Q \) are common on both sides, they can be canceled out:
\[ \frac{4 \times 10^{-6}}{x^2} = \frac{1 \times 10^{-6}}{(16 - x)^2} \]

5. **Simplify the Equation**:
\[ \frac{4}{x^2} = \frac{1}{(16 - x)^2} \]
- Cross-multiplying gives:
\[ 4(16 - x)^2 = x^2 \]

6. **Expand and Rearrange**:
\[ 4(256 - 32x + x^2) = x^2 \]
\[ 1024 - 128x + 4x^2 = x^2 \]
\[ 3x^2 - 128x + 1024 = 0 \]

7. **Use the Quadratic Formula**:
- The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, \( a = 3 \), \( b = -128 \), and \( c = 1024 \):
\[ x = \frac{128 \pm \sqrt{(-128)^2 - 4 \cdot 3 \cdot 1024}}{2 \cdot 3} \]
\[ x = \frac{128 \pm \sqrt{16384 - 12288}}{6} \]
\[ x = \frac{128 \pm \sqrt{4096}}{6} \]
\[ x = \frac{128 \pm 64}{6} \]

8. **Calculate the Two Possible Values of \( x \)**:
- First value:
\[ x = \frac{192}{6} = 32 \, m \]
- Second value:
\[ x = \frac{64}{6} = \frac{32}{3} \, m \approx 10.67 \, m \]

9. **Determine Valid Positions**:
- Since \( x = 32 \, m \) is outside the range of the two charges (16 m apart), the valid position for the third charge is:
\[ x = \frac{32}{3} \, m \text{ from } Q_1 \]

### Final Answer:
The third charge should be placed at a distance of \( \frac{32}{3} \, m \) (approximately 10.67 m) from the charge \( Q_1 \) (4 µC) towards \( Q_2 \) (1 µC) to experience no net force.