To find the electric field at the point (3 m, 4 m, 0) due to a charge \( q = -2.0 \, \mu C \) placed at the origin, we can follow these steps:
### Step 1: Identify the position vector
The position vector \( \mathbf{r} \) of the point where we want to find the electric field is given by:
\[
\mathbf{r} = 3 \hat{i} + 4 \hat{j} + 0 \hat{k}
\]
### Step 2: Calculate the magnitude of the position vector
The magnitude of the position vector \( r \) can be calculated using the formula:
\[
r = \sqrt{x^2 + y^2 + z^2}
\]
Substituting the coordinates:
\[
r = \sqrt{3^2 + 4^2 + 0^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, m
\]
### Step 3: Use the formula for the electric field due to a point charge
The electric field \( \mathbf{E} \) due to a point charge is given by:
\[
\mathbf{E} = k \frac{q}{r^2} \hat{r}
\]
where \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) is Coulomb's constant, \( q \) is the charge, and \( \hat{r} \) is the unit vector in the direction of \( \mathbf{r} \).
### Step 4: Calculate the unit vector \( \hat{r} \)
The unit vector \( \hat{r} \) is given by:
\[
\hat{r} = \frac{\mathbf{r}}{r} = \frac{3 \hat{i} + 4 \hat{j}}{5} = 0.6 \hat{i} + 0.8 \hat{j}
\]
### Step 5: Substitute values into the electric field formula
Now, substituting the values into the electric field formula:
\[
\mathbf{E} = k \frac{q}{r^2} \hat{r}
\]
Substituting \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \), \( q = -2.0 \times 10^{-6} \, C \), and \( r = 5 \, m \):
\[
\mathbf{E} = 9 \times 10^9 \frac{-2.0 \times 10^{-6}}{5^2} (0.6 \hat{i} + 0.8 \hat{j})
\]
Calculating \( r^2 \):
\[
r^2 = 25
\]
So,
\[
\mathbf{E} = 9 \times 10^9 \frac{-2.0 \times 10^{-6}}{25} (0.6 \hat{i} + 0.8 \hat{j})
\]
\[
\mathbf{E} = 9 \times 10^9 \times -8 \times 10^{-8} (0.6 \hat{i} + 0.8 \hat{j})
\]
\[
\mathbf{E} = -0.144 \times 10^3 (0.6 \hat{i} + 0.8 \hat{j})
\]
### Step 6: Calculate the components of the electric field
Calculating the components:
\[
\mathbf{E} = -0.144 \times 10^3 (0.6 \hat{i}) - 0.144 \times 10^3 (0.8 \hat{j})
\]
\[
\mathbf{E} = -0.0864 \times 10^3 \hat{i} - 0.1152 \times 10^3 \hat{j}
\]
\[
\mathbf{E} = -86.4 \hat{i} - 115.2 \hat{j} \, \text{N/C}
\]
### Final Result
Thus, the electric field at the point (3 m, 4 m, 0) due to the charge \( q = -2.0 \, \mu C \) is:
\[
\mathbf{E} = -86.4 \hat{i} - 115.2 \hat{j} \, \text{N/C}
\]
