A charge `q = (-2 x 10^(-9))` is placed at (1 m, 2m, 3m). There is a point `P = (2m, - 3m, 4 m)`
Component of electric field along x-axis at point would be

Options

A. `(1)/(5sqrt3) N//C`
B. `(9)/(5 sqrt3) N//C`
C. `(2)/(9 sqrt3) N//C`
D. `8 sqrt3 N//C`

Answer

To find the component of the electric field along the x-axis at point P due to the charge \( q \) placed at the coordinates (1 m, 2 m, 3 m), we can follow these steps:

### Step 1: Identify the coordinates
- The charge \( q \) is located at \( (1, 2, 3) \).
- The point \( P \) is located at \( (2, -3, 4) \).

### Step 2: Calculate the position vector \( \mathbf{r} \)
The position vector \( \mathbf{r} \) from the charge to point \( P \) is given by:
\[ \mathbf{r} = P - q = (2 - 1) \hat{i} + (-3 - 2) \hat{j} + (4 - 3) \hat{k} = 1 \hat{i} - 5 \hat{j} + 1 \hat{k} \]

### Step 3: Calculate the magnitude of \( \mathbf{r} \)
The magnitude of the position vector \( \mathbf{r} \) is calculated as:
\[ |\mathbf{r}| = \sqrt{(1)^2 + (-5)^2 + (1)^2} = \sqrt{1 + 25 + 1} = \sqrt{27} \]

### Step 4: Calculate the unit vector \( \hat{r} \)
The unit vector \( \hat{r} \) in the direction of \( \mathbf{r} \) is given by:
\[ \hat{r} = \frac{\mathbf{r}}{|\mathbf{r}|} = \frac{1 \hat{i} - 5 \hat{j} + 1 \hat{k}}{\sqrt{27}} = \frac{1}{\sqrt{27}} \hat{i} - \frac{5}{\sqrt{27}} \hat{j} + \frac{1}{\sqrt{27}} \hat{k} \]

### Step 5: Calculate the electric field \( \mathbf{E} \)
The electric field \( \mathbf{E} \) due to the charge \( q \) at point \( P \) is given by:
\[ \mathbf{E} = k \frac{q}{r^2} \hat{r} \]
Where \( k \) is Coulomb's constant (\( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)) and \( r^2 = |\mathbf{r}|^2 = 27 \).

Substituting the values:
\[ \mathbf{E} = 9 \times 10^9 \cdot \frac{-2 \times 10^{-9}}{27} \hat{r} \]
\[ = \frac{-18 \times 10^0}{27} \hat{r} = -\frac{2}{3} \hat{r} \]

### Step 6: Substitute \( \hat{r} \)
Substituting \( \hat{r} \) into the equation:
\[ \mathbf{E} = -\frac{2}{3} \left( \frac{1}{\sqrt{27}} \hat{i} - \frac{5}{\sqrt{27}} \hat{j} + \frac{1}{\sqrt{27}} \hat{k} \right) \]
\[ = -\frac{2}{3\sqrt{27}} \hat{i} + \frac{10}{3\sqrt{27}} \hat{j} - \frac{2}{3\sqrt{27}} \hat{k} \]

### Step 7: Find the x-component of the electric field
The x-component of the electric field \( E_x \) is:
\[ E_x = -\frac{2}{3\sqrt{27}} \]

### Final Result
Thus, the component of the electric field along the x-axis at point \( P \) is:
\[ E_x = -\frac{2}{3\sqrt{27}} \, \text{N/C} \]