# Step-by-Step Solution **Given:** - Two point charges on the x-axis: - \( -q = -3.20 \times 10^{-6} \) C at \( x = -4.50 \) m - \( q = 3.20 \times 10^{-6} \) C at \( x = 4.50 \) m - Point \( P \) at coordinates \( (, 2.00) \) m - Find the x- and y-components of the net electric field at \( P \). --- ## 1. **Find the Position Vectors (from each charge to P)** Let’s denote: - Charge 1 at \( (-4.50, ) \) m - Charge 2 at \( (4.50, ) \) m - Point \( P \) at \( (, 2.00) \) m ### Distance from each charge to \( P \): \[ r = \sqrt{(x_P - x_q)^2 + (y_P - y_q)^2} \] For both charges: \[ r_1 = r_2 = \sqrt{( - 4.50)^2 + (2.00 - )^2} = \sqrt{20.25 + 4} = \sqrt{24.25} \approx 4.925 \text{ m} \] --- ## 2. **Calculate the Electric Field from Each Charge** The electric field from a point charge: \[ \vec{E} = \frac{kq}{r^2} \hat{r} \] - \( k = 8.99 \times 10^9 \) N·m²/C² ### **Find Direction Unit Vectors** For charge 1 (\( -q \)): \[ \vec{r}_1 = ( - (-4.50), 2.00 - ) = (4.5, 2.) \] \[ \hat{r}_1 = \frac{(4.5, 2.)}{4.925} \] For charge 2 (\( +q \)): \[ \vec{r}_2 = ( - 4.50, 2.00 - ) = (-4.5, 2.) \] \[ \hat{r}_2 = \frac{(-4.5, 2.)}{4.925} \] --- ## 3. **Calculate Electric Fields** ### **Magnitude:** \[ E = \frac{k|q|}{r^2} = \frac{8.99 \times 10^9 \times 3.20 \times 10^{-6}}{(4.925)^2} \] \[ (4.925)^2 \approx 24.25 \] \[ E = \frac{8.99 \times 10^9 \times 3.20 \times 10^{-6}}{24.25} = \frac{28.768 \times 10^{3}}{24.25} \approx 1,187 \text{ N/C} \] --- ## 4. **Component Form for Each Field** ### **Unit vectors:** \[ \hat{r}_1 = \left( \frac{4.5}{4.925}, \frac{2.}{4.925} \right) \approx (.914, .406) \] \[ \hat{r}_2 = \left( \frac{-4.5}{4.925}, \frac{2.}{4.925} \right) \approx (-.914, .406) \] ### **Electric field due to Charge 1 (\( -q \)), points *toward* the charge:** \[ \vec{E}_1 = \frac{k|q|}{r^2}(-\hat{r}_1) \] \[ \vec{E}_1 = 1,187 \times (-.914, -.406) = (-1,085, -482) \text{ N/C} \] ### **Electric field due to Charge 2 (\( +q \)), points *away* from the charge:** \[ \vec{E}_2 = \frac{k|q|}{r^2}(\hat{r}_2) \] \[ \vec{E}_2 = 1,187 \times (-.914, .406) = (-1,085, 482) \text{ N/C} \] --- ## 5. **Sum the Components** \[ E_x = E_{1x} + E_{2x} = (-1,085) + (-1,085) = -2,170 \text{ N/C} \] \[ E_y = E_{1y} + E_{2y} = (-482) + (482) = \text{ N/C} \] --- ## 6. **Final Answers** ### **(a) x-component of net field:** \[ \boxed{-2,170\ \text{N/C}} \] ### **(b) y-component of net field:** \[ \boxed{\ \text{N/C}} \] --- ## **Summary** - **(a) x-component:** \(-2,170\) N/C - **(b) y-component:** \(\) N/C --- **Diagram Description:** Two charges are on the x-axis, one negative at \(x=-4.5\) m and one positive at \(x=4.5\) m. The field is calculated at the point \(y=2.\) m, \(x=\). The fields from both charges at this point add in the negative x-direction, and their y-components cancel.