The electric field at (30, 30) cm due to a charge of `-8nC` at the origin in `NC^(-1)` is

Options

A. `-400 (bar(i) + bar(j))`
B. `400 (bar(I) + bar(j))`
C. `-200 sqrt2 (bar(i) + bar(j))`
D. `200 sqrt2 (bar(i) + bar(j))`

Answer

To find the electric field at the point (30 cm, 30 cm) due to a charge of -8 nC located at the origin, we can follow these steps:

### Step 1: Convert the coordinates to meters
The coordinates given are in centimeters. We need to convert them to meters.
- \( x = 30 \, \text{cm} = 0.3 \, \text{m} \)
- \( y = 30 \, \text{cm} = 0.3 \, \text{m} \)

### Step 2: Calculate the position vector
The position vector \( \vec{r} \) from the origin to the point (0.3 m, 0.3 m) is:
\[ \vec{r} = 0.3 \hat{i} + 0.3 \hat{j} \, \text{m} \]

### Step 3: Calculate the magnitude of the position vector
The magnitude of \( \vec{r} \) is calculated using the Pythagorean theorem:
\[ |\vec{r}| = \sqrt{(0.3)^2 + (0.3)^2} = \sqrt{0.09 + 0.09} = \sqrt{0.18} = 0.3\sqrt{2} \, \text{m} \]

### Step 4: Use the formula for electric field
The electric field \( \vec{E} \) due to a point charge \( q \) is given by:
\[ \vec{E} = \frac{1}{4 \pi \epsilon_0} \frac{q}{|\vec{r}|^2} \hat{r} \]
where \( \hat{r} \) is the unit vector in the direction of \( \vec{r} \).

### Step 5: Calculate the unit vector \( \hat{r} \)
The unit vector \( \hat{r} \) is given by:
\[ \hat{r} = \frac{\vec{r}}{|\vec{r}|} = \frac{0.3 \hat{i} + 0.3 \hat{j}}{0.3\sqrt{2}} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \]

### Step 6: Substitute values into the electric field equation
Given:
- \( q = -8 \, \text{nC} = -8 \times 10^{-9} \, \text{C} \)
- \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)

Substituting the values into the electric field equation:
\[ \vec{E} = \frac{1}{4 \pi (8.85 \times 10^{-12})} \frac{-8 \times 10^{-9}}{(0.3\sqrt{2})^2} \left( \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \right) \]

### Step 7: Calculate the magnitude of the electric field
Calculating the denominator:
\[ (0.3\sqrt{2})^2 = 0.18 \, \text{m}^2 \]
Now substituting this into the electric field equation:
\[ \vec{E} = \frac{1}{4 \pi (8.85 \times 10^{-12})} \frac{-8 \times 10^{-9}}{0.18} \left( \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \right) \]

### Step 8: Final calculations
Calculating the constant:
\[ \frac{1}{4 \pi (8.85 \times 10^{-12})} \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \]
Thus:
\[ \vec{E} = 9 \times 10^9 \cdot \frac{-8 \times 10^{-9}}{0.18} \left( \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \right) \]
Calculating:
\[ \vec{E} = \frac{-72 \times 10^0}{0.18} \left( \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \right) = -400 \sqrt{2} \hat{i} - 400 \sqrt{2} \hat{j} \, \text{N/C} \]

### Final Result
The electric field at (30 cm, 30 cm) due to a charge of -8 nC at the origin is:
\[ \vec{E} = -200\sqrt{2} \hat{i} - 200\sqrt{2} \hat{j} \, \text{N/C} \]