A point charge `q = - 8.0 nC` is located at the origin. Find the electric field vector at the point `x = 1.2 m, y = -1.6 m. `

Explanation

To find the electric field vector at the point \( (x = 1.2 \, \text{m}, y = -1.6 \, \text{m}) \) due to a point charge \( q = -8.0 \, \text{nC} \) located at the origin, we can follow these steps:

### Step 1: Identify the coordinates of the charge and the point of interest.
- The point charge \( q \) is located at the origin \( (0, 0) \).
- The point where we want to find the electric field is \( (1.2, -1.6) \).

### Step 2: Calculate the distance \( R \) from the charge to the point.
- The distance \( R \) can be calculated using the distance formula:
\[ R = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
where \( (x_1, y_1) = (0, 0) \) and \( (x_2, y_2) = (1.2, -1.6) \).

\[ R = \sqrt{(1.2 - 0)^2 + (-1.6 - 0)^2} = \sqrt{(1.2)^2 + (-1.6)^2} = \sqrt{1.44 + 2.56} = \sqrt{4} = 2 \, \text{m} \]

### Step 3: Determine the unit vector in the direction from the charge to the point.
- The unit vector \( \hat{n} \) in the direction from the charge to the point is given by:
\[ \hat{n} = \frac{(x, y)}{R} = \frac{(1.2, -1.6)}{2} = (0.6, -0.8) \]

### Step 4: Calculate the electric field magnitude \( E \).
- The electric field \( E \) due to a point charge is given by:
\[ E = \frac{k |q|}{R^2} \]
where \( k \) (Coulomb's constant) is approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \) and \( |q| = 8.0 \, \text{nC} = 8.0 \times 10^{-9} \, \text{C} \).

\[ E = \frac{(8.99 \times 10^9) \times (8.0 \times 10^{-9})}{(2)^2} = \frac{(8.99 \times 10^9) \times (8.0 \times 10^{-9})}{4} \]
\[ E = \frac{71.92}{4} = 17.98 \, \text{N/C} \]

### Step 5: Determine the direction of the electric field.
- Since the charge is negative, the electric field points towards the charge. Thus, the electric field vector \( \vec{E} \) is:
\[ \vec{E} = -E \hat{n} = -17.98 (0.6, -0.8) = (-10.788, 14.384) \, \text{N/C} \]

### Final Result:
The electric field vector at the point \( (1.2, -1.6) \) is:
\[ \vec{E} = (-10.79, 14.38) \, \text{N/C} \]