Let's break the solution into **two steps** as requested. --- ## **Step 1: Calculate the Electric Field Due to Each Charge at Point P** ### **Given Data** - \( q_1 = q_2 = 3e \) - \( q_3 = 2e \) - \( q_4 = -8e \) - Distance \( d = 5.11 \, \mu m = 5.11 \times 10^{-6} \, m \) - Elementary charge \( e = 1.602 \times 10^{-19} \, C \) - Coulomb’s constant \( k = 8.99 \times 10^9 \, N \cdot m^2/C^2 \) ### **Electric Field Due to a Point Charge** \[ E = \frac{k|q|}{r^2} \] All charges are at the same distance \( d \) from \( P \). #### **Electric Fields Magnitude at P (ignore direction for now):** - For \( q_1 \) and \( q_2 \) (both = \( 3e \)): \[ E_{1,2} = \frac{k \cdot 3e}{d^2} \] - For \( q_3 = 2e \): \[ E_3 = \frac{k \cdot 2e}{d^2} \] - For \( q_4 = -8e \) (magnitude \( 8e \)): \[ E_4 = \frac{k \cdot 8e}{d^2} \] #### **Explanation:** All charges are equidistant from \( P \), so the only difference in their field is due to their charge magnitude and sign. --- ## **Step 2: Add the Electric Fields as Vectors** ### **Directions** From the diagram: - \( q_1 \) and \( q_4 \) are along the "upward-right" diagonal. - \( q_2 \) and \( q_3 \) are along the "downward-right" diagonal. - Each diagonal is at 45° to the axes. The electric field from a positive charge points **away** from the charge, and from a negative charge points **toward** the charge. Let's resolve all electric fields into their x and y components. #### **Component Calculation** For each, \( E_x = E \cos(45^\circ) \), \( E_y = E \sin(45^\circ) \). Since \( \cos(45^\circ) = \sin(45^\circ) = \frac{1}{\sqrt{2}} \): - \( q_1, q_4 \): upper diagonal - \( q_2, q_3 \): lower diagonal Let’s assign axes (up is +y, right is +x): | Charge | Value | Direction from P | Field points | x-comp | y-comp | |--------|--------|-------------------|--------------|--------|--------| | q1 | +3e | up-left | down-right | + | - | | q2 | +3e | down-left | up-right | + | + | | q3 | +2e | up-right | down-left | - | - | | q4 | -8e | down-right | *toward* P (up-left) | - | + | Compute components (all at 45°, so multiply by \( 1/\sqrt{2} \)): \[ E_x = \frac{1}{\sqrt{2}} \left[ E_1 + E_2 - E_3 - E_4 \right] \] \[ E_y = \frac{1}{\sqrt{2}} \left[ -E_1 + E_2 - E_3 + E_4 \right] \] Plug values: - \( E_1 = \frac{k 3e}{d^2} \) - \( E_2 = \frac{k 3e}{d^2} \) - \( E_3 = \frac{k 2e}{d^2} \) - \( E_4 = \frac{k 8e}{d^2} \) So, \[ E_x = \frac{1}{\sqrt{2}} \left[ (3 + 3 - 2 - 8)\frac{k e}{d^2} \right] = \frac{1}{\sqrt{2}} \left[ (-4)\frac{k e}{d^2} \right] \] \[ E_y = \frac{1}{\sqrt{2}} \left[ (-3 + 3 - 2 + 8)\frac{k e}{d^2} \right] = \frac{1}{\sqrt{2}} \left[ (6)\frac{k e}{d^2} \right] \] Now, magnitude: \[ E = \sqrt{E_x^2 + E_y^2} \] Plug in the numbers: - \( k = 8.99 \times 10^9 \) - \( e = 1.602 \times 10^{-19} \) - \( d = 5.11 \times 10^{-6} \) Calculate \( \frac{k e}{d^2} \): \[ \frac{k e}{d^2} = \frac{8.99 \times 10^9 \times 1.602 \times 10^{-19}}{(5.11 \times 10^{-6})^2} \] \[ (5.11 \times 10^{-6})^2 = 2.611 \times 10^{-11} \] \[ 8.99 \times 10^9 \times 1.602 \times 10^{-19} = 1.440 \times 10^{-9} \] \[ \frac{1.440 \times 10^{-9}}{2.611 \times 10^{-11}} \approx 55.15 \, N/C \] Now plug into components: \[ E_x = \frac{1}{\sqrt{2}} \times (-4) \times 55.15 = -156.1 \, N/C \] \[ E_y = \frac{1}{\sqrt{2}} \times 6 \times 55.15 = 234.1 \, N/C \] Magnitude: \[ E = \sqrt{(-156.1)^2 + (234.1)^2} = \sqrt{24361 + 54798} = \sqrt{79159} \approx 281.5 \, N/C \] --- ## **Final Answer** **The magnitude of the net electric field at point \( P \) due to the particles is:** \[ \boxed{282 \, \text{N/C}} \] (Rounded to three significant digits.) --- ### **Summary of Steps** 1. **Found the electric field due to each charge at P using Coulomb’s law.** 2. **Resolved the vector components, added them, and calculated the net magnitude.**