A point charge of `100mu C` is placed at `3 hat i+4 hat j m`. Find the electric field intensity due to this charges at a point located at `9 hat i+12 hat j m`.

Options

A. `8000Vm^(-1)`
B. `9000Vm^(-1)`
C. `2250Vm^(-1)`
D. `4500Vm^(-1)`

Explanation

To solve the problem of finding the electric field intensity due to a point charge at a specified location, we will follow these steps:

### Step 1: Identify the Given Information
- Point charge \( Q = 100 \, \mu C = 100 \times 10^{-6} \, C \)
- Position of the charge \( \vec{r_i} = 3 \hat{i} + 4 \hat{j} \, m \)
- Position where we want to find the electric field \( \vec{r_f} = 9 \hat{i} + 12 \hat{j} \, m \)

### Step 2: Calculate the Displacement Vector
The displacement vector \( \vec{r} \) from the charge to the point where we want to find the electric field is given by:
\[ \vec{r} = \vec{r_f} - \vec{r_i} \]
Substituting the values:
\[ \vec{r} = (9 \hat{i} + 12 \hat{j}) - (3 \hat{i} + 4 \hat{j}) = (9 - 3) \hat{i} + (12 - 4) \hat{j} = 6 \hat{i} + 8 \hat{j} \]

### Step 3: Calculate the Magnitude of the Displacement Vector
The magnitude of the displacement vector \( r \) is calculated using the Pythagorean theorem:
\[ r = \sqrt{(6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, m \]

### Step 4: Calculate the Electric Field Intensity
The electric field intensity \( E \) due to a point charge is given by the formula:
\[ E = \frac{k \cdot Q}{r^2} \]
Where \( k \) (Coulomb's constant) is approximately \( 9 \times 10^9 \, N \cdot m^2/C^2 \).

Substituting the values:
\[ E = \frac{9 \times 10^9 \cdot (100 \times 10^{-6})}{(10)^2} \]
Calculating the denominator:
\[ (10)^2 = 100 \]
Now substituting back:
\[ E = \frac{9 \times 10^9 \cdot 100 \times 10^{-6}}{100} = 9 \times 10^9 \cdot 10^{-6} = 9 \times 10^3 \, N/C \]

### Step 5: Final Result
The electric field intensity at the point located at \( 9 \hat{i} + 12 \hat{j} \, m \) due to the point charge is:
\[ E = 9000 \, N/C \, \text{or} \, 9000 \, V/m \]