Let’s break this problem down step by step: --- ## **Problem Restatement** Two charged particles are on the y-axis: - \( q_1 = 3.20 \times 10^{-19} \) C at \( y = +4.50 \) cm - \( q_2 = -3.20 \times 10^{-19} \) C at \( y = -4.50 \) cm Find the **x and y components** of the **electric field** at point \( P \) at \( x = 2.00 \) cm on the x-axis. --- ## **Step 1: Positions and Distances** - **Point \( P \):** \( (2.00, ) \) cm - **\( q_1 \):** \( (, +4.50) \) cm - **\( q_2 \):** \( (, -4.50) \) cm Distance from \( P \) to each charge (by Pythagoras): \[ r = \sqrt{(2.00)^2 + (4.50)^2} = \sqrt{4 + 20.25} = \sqrt{24.25} \approx 4.925 \text{ cm} = .04925 \text{ m} \] --- ## **Step 2: Electric Field Due to a Point Charge** \[ \vec{E} = \frac{k q}{r^2} \hat{r} \] where \( k = 8.99 \times 10^9 \) N·m²/C². --- ## **Step 3: Components of the Electric Field** ### **For \( q_1 \) at (, +4.50) cm:** - \( \Delta x = 2.00 \) cm, \( \Delta y = -4.50 \) cm (from \( q_1 \) to \( P \)) - The vector from \( q_1 \) to \( P \) is \( (2.00, -4.50) \) cm Unit vector: \[ \hat{r}_1 = \frac{(2.00, -4.50)}{4.925} \] \[ \hat{r}_1 = (.406, -.914) \] Magnitude: \[ E_1 = \frac{k |q_1|}{r^2} = \frac{8.99 \times 10^9 \times 3.20 \times 10^{-19}}{(.04925)^2} \] \[ E_1 = \frac{8.99 \times 10^9 \times 3.20 \times 10^{-19}}{.002425} \approx 1.187 \times 10^{-6} \text{ N/C} \] Components: \[ E_{1x} = E_1 \cdot .406, \quad E_{1y} = E_1 \cdot (-.914) \] \[ E_{1x} = 1.187 \times 10^{-6} \times .406 \approx 4.82 \times 10^{-7} \text{ N/C} \] \[ E_{1y} = 1.187 \times 10^{-6} \times (-.914) \approx -1.085 \times 10^{-6} \text{ N/C} \] --- ### **For \( q_2 \) at (, -4.50) cm:** - \( \Delta x = 2.00 \) cm, \( \Delta y = +4.50 \) cm (from \( q_2 \) to \( P \)) - The vector from \( q_2 \) to \( P \) is \( (2.00, 4.50) \) cm Unit vector: \[ \hat{r}_2 = \frac{(2.00, 4.50)}{4.925} = (.406, .914) \] But \( q_2 \) is negative, so the electric field points **toward** the charge (opposite to \( \hat{r}_2 \)): \[ \vec{E}_2 = -E_2 \hat{r}_2 \] So, use \( E_2 \) (same magnitude as \( E_1 \)), but components are negative: \[ E_{2x} = -E_2 \cdot .406 = -4.82 \times 10^{-7} \text{ N/C} \] \[ E_{2y} = -E_2 \cdot .914 = -1.085 \times 10^{-6} \text{ N/C} \] --- ## **Step 4: Add the Components** **Total \( x \)-component:** \[ E_x = E_{1x} + E_{2x} = 4.82 \times 10^{-7} + (-4.82 \times 10^{-7}) = \] **Total \( y \)-component:** \[ E_y = E_{1y} + E_{2y} = (-1.085 \times 10^{-6}) + (-1.085 \times 10^{-6}) = -2.17 \times 10^{-6} \text{ N/C} \] --- ## **Step 5: Final Answers** ### **(a) \( x \)-component:** \[ \boxed{ \ \text{N/C}} \] ### **(b) \( y \)-component:** \[ \boxed{-2.17 \times 10^{-6} \ \text{N/C}} \] --- ### **Summary Table** | Component | Value (N/C) | |-----------|----------------------| | \( E_x \) | \( \) | | \( E_y \) | \( -2.17 \times 10^{-6} \) | --- Let me know if you need a diagram or further clarification!