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 find the electric field strength \( \vec{E} \) at a point given a positive charge, we can follow these steps:

### Step 1: Identify the given values
- Charge \( Q = 50 \, \mu C = 50 \times 10^{-6} \, C \)
- Position vector of the charge \( \vec{r_0} = 4 \hat{i} + 4 \hat{j} \) (in meters)
- Position vector where we need to find the electric field \( \vec{r} = 10 \hat{i} - 4 \hat{j} \) (in meters)

### Step 2: Calculate the displacement vector \( \vec{D} \)
The displacement vector \( \vec{D} \) from the charge to the point of interest is calculated as:
\[ \vec{D} = \vec{r} - \vec{r_0} \]
Substituting the values:
\[ \vec{D} = (10 \hat{i} - 4 \hat{j}) - (4 \hat{i} + 4 \hat{j}) = (10 - 4) \hat{i} + (-4 - 4) \hat{j} = 6 \hat{i} - 8 \hat{j} \]

### Step 3: Calculate the magnitude of \( \vec{D} \)
The magnitude of \( \vec{D} \) is given by:
\[ |\vec{D}| = \sqrt{(6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, m \]

### Step 4: Calculate the unit vector \( \hat{D} \)
The unit vector \( \hat{D} \) in the direction of \( \vec{D} \) is:
\[ \hat{D} = \frac{\vec{D}}{|\vec{D}|} = \frac{6 \hat{i} - 8 \hat{j}}{10} = 0.6 \hat{i} - 0.8 \hat{j} \]

### Step 5: Use the formula for electric field \( \vec{E} \)
The electric field \( \vec{E} \) due to a point charge is given by:
\[ \vec{E} = k \frac{Q}{|\vec{D}|^2} \hat{D} \]
Where \( k \) (Coulomb's constant) is \( 9 \times 10^9 \, N m^2/C^2 \).

### Step 6: Substitute the values into the formula
Substituting the known values:
\[ \vec{E} = 9 \times 10^9 \frac{50 \times 10^{-6}}{10^2} (0.6 \hat{i} - 0.8 \hat{j}) \]
Calculating:
\[ \vec{E} = 9 \times 10^9 \frac{50 \times 10^{-6}}{100} (0.6 \hat{i} - 0.8 \hat{j}) \]
\[ \vec{E} = 9 \times 10^9 \times 0.5 \times 10^{-6} (0.6 \hat{i} - 0.8 \hat{j}) \]
\[ \vec{E} = 4.5 \times 10^3 (0.6 \hat{i} - 0.8 \hat{j}) \]
\[ \vec{E} = (4.5 \times 0.6 \times 10^3) \hat{i} - (4.5 \times 0.8 \times 10^3) \hat{j} \]
\[ \vec{E} = 2700 \hat{i} - 3600 \hat{j} \, V/m \]

### Step 7: Convert to kV/m
To convert to kV/m, we divide by 1000:
\[ \vec{E} = 2.7 \hat{i} - 3.6 \hat{j} \, kV/m \]

### Final Result
The electric field strength at the point is:
\[ \vec{E} = 2.7 \hat{i} - 3.6 \hat{j} \, kV/m \]