A positive point charge `50mu C` is located in the plane xy at a point with radius vector `vecr_(0)=2hat(i)+3hat(j)`. The electric field vector `vec(E )` at a point with radius vector `vec(r )=8 hat(i)-5 hat(j)`, where `r_(0)` and r are expressed in meter, is

Options

A. `(1.4 hat(i)-2.6 hat(j))kNC^(-1)`
B. `(1.4 hat(i)+2.6 hat(j))kNC^(-1)`
C. `(2.7hat(i)-3.6 hat(j))kNC^(-1)`
D. `(2.7 hat(i)+3.6 hat(j))kNC^(-1)`

Explanation

To find the electric field vector \( \vec{E} \) at a given point due to a point charge, we can use the formula for the electric field created by a point charge:

\[ \vec{E} = k \frac{q}{r^3} \vec{r} \]

where:
- \( k \) is Coulomb's constant, approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \),
- \( q \) is the charge,
- \( \vec{r} \) is the position vector from the charge to the point where the electric field is being calculated,
- \( r \) is the magnitude of \( \vec{r} \).

### Step 1: Identify the given values
- Charge \( q = 50 \, \mu C = 50 \times 10^{-6} \, C \)
- Position of the charge \( \vec{r_0} = 2\hat{i} + 3\hat{j} \) (in meters)
- Point where we want to find the electric field \( \vec{r} = 8\hat{i} - 5\hat{j} \) (in meters)

### Step 2: Calculate the vector \( \vec{r} - \vec{r_0} \)
We need to find the vector from the charge to the point where we want to calculate the electric field:

\[ \vec{r} - \vec{r_0} = (8\hat{i} - 5\hat{j}) - (2\hat{i} + 3\hat{j}) \]

Calculating this gives:

\[ \vec{r} - \vec{r_0} = (8 - 2)\hat{i} + (-5 - 3)\hat{j} = 6\hat{i} - 8\hat{j} \]

### Step 3: Calculate the magnitude of \( \vec{r} - \vec{r_0} \)
Now, we find the magnitude \( r \):

\[ r = |\vec{r} - \vec{r_0}| = \sqrt{(6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{m} \]

### Step 4: Calculate the electric field vector \( \vec{E} \)
Now we can substitute into the electric field formula:

\[ \vec{E} = k \frac{q}{r^3} (\vec{r} - \vec{r_0}) \]

Substituting the known values:

\[ \vec{E} = 9 \times 10^9 \frac{50 \times 10^{-6}}{10^3} (6\hat{i} - 8\hat{j}) \]

Calculating \( \frac{50 \times 10^{-6}}{10^3} = 50 \times 10^{-9} \)

Now substituting this back:

\[ \vec{E} = 9 \times 10^9 \times 50 \times 10^{-9} (6\hat{i} - 8\hat{j}) \]
\[ \vec{E} = 450 (6\hat{i} - 8\hat{j}) \, \text{kN/C} \]

### Step 5: Simplify the electric field vector
Now we can distribute \( 450 \):

\[ \vec{E} = 2700\hat{i} - 3600\hat{j} \, \text{kN/C} \]

### Step 6: Convert to kilo Newtons per Coulomb
Since the answer needs to be in kN/C, we divide by 1000:

\[ \vec{E} = 2.7\hat{i} - 3.6\hat{j} \, \text{kN/C} \]

### Final Answer
The electric field vector \( \vec{E} \) at the point is:

\[ \vec{E} = 2.7\hat{i} - 3.6\hat{j} \, \text{kN/C} \]