A positive charge 50 muC is located in xy plane at a position vector `r_0 = 4hati + 4hatj`. the electric field strength E at a point whose position vector is `vecr = 10hati - 4hatj` is (`vecr_0` and `vecr` are expressed in metre)

Options

A. `(-1.6hati - 3.6hatj) kV/m`
B. `(3.6hati + 1.6hatj) kV/m`
C. `(hati - 3hatj) kV/m`
D. `(2.7hati - 3.6hatj)` kV/m

Answer

To solve the problem of finding the electric field strength \( \mathbf{E} \) at a point due to a positive charge located at a specific position, we can follow these steps:

### Step 1: Identify the Given Information
We have:
- Charge \( q = 50 \, \mu C = 50 \times 10^{-6} \, C \)
- Position vector of the charge \( \mathbf{r_0} = 4 \hat{i} + 4 \hat{j} \) (in meters)
- Position vector of the point where we want to find the electric field \( \mathbf{r} = 10 \hat{i} - 4 \hat{j} \) (in meters)

### Step 2: Calculate the Displacement Vector
The displacement vector \( \mathbf{r'} \) from the charge to the point is given by:
\[ \mathbf{r'} = \mathbf{r} - \mathbf{r_0} \]
Substituting the values:
\[ \mathbf{r'} = (10 \hat{i} - 4 \hat{j}) - (4 \hat{i} + 4 \hat{j}) = (10 - 4) \hat{i} + (-4 - 4) \hat{j} = 6 \hat{i} - 8 \hat{j} \]

### Step 3: Calculate the Magnitude of the Displacement Vector
The magnitude of \( \mathbf{r'} \) is calculated as:
\[ |\mathbf{r'}| = \sqrt{(6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, m \]

### Step 4: Use the Electric Field Formula
The electric field \( \mathbf{E} \) due to a point charge is given by:
\[ \mathbf{E} = \frac{k \cdot q}{|\mathbf{r'}|^2} \hat{r'} \]
where \( k \) is Coulomb's constant \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) and \( \hat{r'} \) is the unit vector in the direction of \( \mathbf{r'} \).

### Step 5: Calculate the Unit Vector
The unit vector \( \hat{r'} \) is given by:
\[ \hat{r'} = \frac{\mathbf{r'}}{|\mathbf{r'}|} = \frac{6 \hat{i} - 8 \hat{j}}{10} = 0.6 \hat{i} - 0.8 \hat{j} \]

### Step 6: Substitute Values into the Electric Field Formula
Now we can substitute the values into the electric field formula:
\[ \mathbf{E} = \frac{(9 \times 10^9) \cdot (50 \times 10^{-6})}{(10)^2} (0.6 \hat{i} - 0.8 \hat{j}) \]
Calculating the scalar part:
\[ \mathbf{E} = \frac{(9 \times 10^9) \cdot (50 \times 10^{-6})}{100} (0.6 \hat{i} - 0.8 \hat{j}) = (4.5 \times 10^5) (0.6 \hat{i} - 0.8 \hat{j}) \]

### Step 7: Final Calculation
Now, calculate the components:
\[ \mathbf{E} = (4.5 \times 10^5 \cdot 0.6) \hat{i} + (4.5 \times 10^5 \cdot -0.8) \hat{j} \]
\[ \mathbf{E} = 2.7 \times 10^5 \hat{i} - 3.6 \times 10^5 \hat{j} \]

### Final Answer
Thus, the electric field strength \( \mathbf{E} \) at the given point is:
\[ \mathbf{E} = 2.7 \times 10^5 \hat{i} - 3.6 \times 10^5 \hat{j} \, \text{N/C} \]

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