Find out electridc filed intensity at point `A(1,0,2)` due to a point charge `-20muC` situated at pont `B(0,sqrt(2),1)`:

Options

A. `-22.5xx10^(3)(hati-sqrt(2)hatj+hatk)`
B. `8.5xx10^(3)(hati+sqrt(2)hatj+hatk)`
C. `22.5xx10^(3)(hati+sqrt(2)hatj-hatk)`
D. `8.5xx10^(3)(hati-sqrt(2)hatj+hatk)`

Explanation

To find the electric field intensity at point A(1, 0, 2) due to a point charge of -20 μC situated at point B(0, √2, 1), we can follow these steps:

### Step 1: Identify the coordinates of points A and B
- Point A has coordinates \( A(1, 0, 2) \).
- Point B has coordinates \( B(0, \sqrt{2}, 1) \).

### Step 2: Determine the position vectors of points A and B
- The position vector of point A, \( \vec{OA} \), is given by:
\[ \vec{OA} = 1 \hat{i} + 0 \hat{j} + 2 \hat{k} = \hat{i} + 2 \hat{k} \]
- The position vector of point B, \( \vec{OB} \), is given by:
\[ \vec{OB} = 0 \hat{i} + \sqrt{2} \hat{j} + 1 \hat{k} = \sqrt{2} \hat{j} + \hat{k} \]

### Step 3: Calculate the vector \( \vec{r} \) from B to A
- The vector \( \vec{r} \) from point B to point A is calculated as:
\[ \vec{r} = \vec{OA} - \vec{OB} = (\hat{i} + 2 \hat{k}) - (0 \hat{i} + \sqrt{2} \hat{j} + \hat{k}) = \hat{i} - \sqrt{2} \hat{j} + (2 - 1) \hat{k} \]
\[ \vec{r} = \hat{i} - \sqrt{2} \hat{j} + \hat{k} \]

### Step 4: Calculate the magnitude of \( \vec{r} \)
- The magnitude \( r \) of the vector \( \vec{r} \) is given by:
\[ r = |\vec{r}| = \sqrt{(1)^2 + (-\sqrt{2})^2 + (1)^2} = \sqrt{1 + 2 + 1} = \sqrt{4} = 2 \]

### Step 5: Calculate the electric field intensity \( \vec{E} \)
- The electric field intensity \( \vec{E} \) due to a point charge is given by:
\[ \vec{E} = k \frac{Q}{r^2} \hat{r} \]
where \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) and \( Q = -20 \times 10^{-6} \, \text{C} \).
- The unit vector \( \hat{r} \) in the direction of \( \vec{r} \) is:
\[ \hat{r} = \frac{\vec{r}}{|\vec{r}|} = \frac{\hat{i} - \sqrt{2} \hat{j} + \hat{k}}{2} \]

### Step 6: Substitute values into the electric field equation
- The electric field intensity becomes:
\[ \vec{E} = k \frac{Q}{r^2} \hat{r} = 9 \times 10^9 \frac{-20 \times 10^{-6}}{(2)^2} \left(\frac{\hat{i} - \sqrt{2} \hat{j} + \hat{k}}{2}\right) \]
\[ = 9 \times 10^9 \frac{-20 \times 10^{-6}}{4} \left(\frac{\hat{i} - \sqrt{2} \hat{j} + \hat{k}}{2}\right) \]
\[ = 9 \times 10^9 \frac{-20 \times 10^{-6}}{8} (\hat{i} - \sqrt{2} \hat{j} + \hat{k}) \]
\[ = -22.5 \times 10^3 (\hat{i} - \sqrt{2} \hat{j} + \hat{k}) \]

### Step 7: Write the final expression for the electric field
- Thus, the electric field intensity at point A is:
\[ \vec{E} = -22.5 \times 10^3 \hat{i} + 22.5 \times 10^3 \sqrt{2} \hat{j} - 22.5 \times 10^3 \hat{k} \]

### Final Answer
The electric field intensity at point A due to the charge at point B is:
\[ \vec{E} = -22.5 \times 10^3 \hat{i} + 22.5 \times 10^3 \sqrt{2} \hat{j} - 22.5 \times 10^3 \hat{k} \]