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 of the charge and the point of interest
The charge \( q = -2.0 \, \mu C \) is located at the origin (0, 0, 0). The point where we want to find the electric field is (3 m, 4 m, 0).
### Step 2: Calculate the distance \( R \) from the charge to the point
The distance \( R \) can be calculated using the Pythagorean theorem:
\[
R = \sqrt{x^2 + y^2} = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, m
\]
### Step 3: Determine the direction of the electric field
Since the charge is negative, the electric field will point towards the charge. The unit vector \( \hat{n} \) from the point (3, 4) to the origin (0, 0) is given by:
\[
\hat{n} = \frac{(0 - 3) \hat{i} + (0 - 4) \hat{j}}{R} = \frac{-3 \hat{i} - 4 \hat{j}}{5} = -\frac{3}{5} \hat{i} - \frac{4}{5} \hat{j}
\]
### Step 4: Calculate the magnitude of the electric field
The electric field \( E \) due to a point charge is given by the formula:
\[
E = \frac{k |q|}{R^2}
\]
where \( k \) is Coulomb's constant, \( k \approx 8.99 \times 10^9 \, N m^2/C^2 \), and \( |q| = 2.0 \times 10^{-6} \, C \).
Substituting the values:
\[
E = \frac{(8.99 \times 10^9) \times (2.0 \times 10^{-6})}{(5)^2} = \frac{(8.99 \times 10^9) \times (2.0 \times 10^{-6})}{25}
\]
\[
E = \frac{17.98 \times 10^3}{25} = 719.2 \, N/C
\]
### Step 5: Combine the magnitude and direction to find the electric field vector
The electric field vector \( \vec{E} \) is given by:
\[
\vec{E} = E \hat{n} = 719.2 \left(-\frac{3}{5} \hat{i} - \frac{4}{5} \hat{j}\right)
\]
Calculating this gives:
\[
\vec{E} = -431.52 \hat{i} - 575.36 \hat{j} \, N/C
\]
### Final Result
The electric field at the point (3 m, 4 m, 0) is:
\[
\vec{E} = -431.52 \hat{i} - 575.36 \hat{j} \, N/C
\]
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