A charge of `16xx10^(-9) C` is fixed at the origin of coordinates. A second charge of unknow magnitude is at `x=3 m, y=0` and a third charge of `12xx10^(-9) C` is at x=6m, y=0. What is the value of the unknown charge if the resultant field at x=8m, y=0 is 20.25 N/C directed towards positive x-axis ?

Answer

To solve the problem, we need to find the unknown charge \( q \) that results in a net electric field of \( 20.25 \, \text{N/C} \) directed towards the positive x-axis at the point \( (8, 0) \). We will use the principle of superposition for electric fields due to point charges.

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions**:
- Charge \( Q_1 = 16 \times 10^{-9} \, \text{C} \) is at \( (0, 0) \).
- Charge \( Q_2 = q \) (unknown) is at \( (3, 0) \).
- Charge \( Q_3 = 12 \times 10^{-9} \, \text{C} \) is at \( (6, 0) \).

2. **Calculate the Distances**:
- Distance from \( Q_1 \) to point \( (8, 0) \):
\[ r_1 = 8 - 0 = 8 \, \text{m} \]
- Distance from \( Q_2 \) to point \( (8, 0) \):
\[ r_2 = 8 - 3 = 5 \, \text{m} \]
- Distance from \( Q_3 \) to point \( (8, 0) \):
\[ r_3 = 8 - 6 = 2 \, \text{m} \]

3. **Write the Expression for the Electric Field**:
The electric field \( E \) due to a point charge is given by:
\[ E = k \frac{Q}{r^2} \]
where \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

4. **Calculate the Electric Fields**:
- Electric field due to \( Q_1 \):
\[ E_1 = k \frac{16 \times 10^{-9}}{8^2} = 9 \times 10^9 \frac{16 \times 10^{-9}}{64} = 2.25 \, \text{N/C} \, \text{(directed to the right)} \]
- Electric field due to \( Q_2 \):
\[ E_2 = k \frac{q}{5^2} = k \frac{q}{25} \]
- Electric field due to \( Q_3 \):
\[ E_3 = k \frac{12 \times 10^{-9}}{2^2} = 9 \times 10^9 \frac{12 \times 10^{-9}}{4} = 27 \, \text{N/C} \, \text{(directed to the left)} \]

5. **Set Up the Equation for the Resultant Electric Field**:
The resultant electric field \( E_R \) at \( (8, 0) \) is given as \( 20.25 \, \text{N/C} \) directed towards the positive x-axis. Therefore, we can write:
\[ E_1 + E_2 - E_3 = 20.25 \]
Substituting the values:
\[ 2.25 + \frac{9 \times 10^9 q}{25} - 27 = 20.25 \]

6. **Solve for \( q \)**:
Rearranging the equation:
\[ \frac{9 \times 10^9 q}{25} = 20.25 + 27 - 2.25 \]
\[ \frac{9 \times 10^9 q}{25} = 45 \]
\[ 9 \times 10^9 q = 45 \times 25 \]
\[ 9 \times 10^9 q = 1125 \]
\[ q = \frac{1125}{9 \times 10^9} = 125 \times 10^{-9} \, \text{C} \]

7. **Determine the Sign of \( q \)**:
Since the resultant electric field is directed towards the positive x-axis, \( q \) must be negative to counteract the electric field due to \( Q_3 \). Thus, we have:
\[ q = -25 \times 10^{-9} \, \text{C} \]

### Final Answer:
The value of the unknown charge \( q \) is \( -25 \times 10^{-9} \, \text{C} \).