Let’s solve the problem step-by-step: ## Given: - \( q_1 = 1.500 \, \mu C = 1.5 \times 10^{-6} \, C \) - \( q_2 = 4.000 \, \mu C = 4.0 \times 10^{-6} \, C \) - \( q_3 = -4.000 \, \mu C = -4.0 \times 10^{-6} \, C \) - Distances from the diagram: - \( r_{12} = 4.000 \, \text{cm} = 0.040 \, \text{m} \) - \( r_{13} = 5.000 \, \text{cm} = 0.050 \, \text{m} \) - \( r_{23} = 4.000 \, \text{cm} = 0.040 \, \text{m} \) - Coordinates: - \( q_1 \) at (0, 0) - \( q_2 \) at (0.040, 0) - \( q_3 \) at (0.040, 0.030) ### a. **Find the Electric Field at Point P (at the location of \( q_3 \))** #### **Step 1: Calculate the field at P due to \( q_1 \) and \( q_2 \)** - Ignore the field due to \( q_3 \) at its own location. #### **Field due to \( q_1 \) at P** - Distance \( r_{13} = 0.050 \) m - Direction: from \( q_1 \) to \( q_3 \): \( \vec{r}_{13} = (0.040, 0.030) \) m - Magnitude: \[ E_{13} = \frac{k |q_1|}{r_{13}^2} = \frac{8.99 \times 10^9 \times 1.5 \times 10^{-6}}{(0.05)^2} = \frac{8.99 \times 10^9 \times 1.5 \times 10^{-6}}{0.0025} = \frac{13.485 \times 10^3}{0.0025} = 5.394 \times 10^6 \, \text{N/C} \] - Direction: \[ \hat{r}_{13} = \frac{(0.04,0.03)}{0.05} = (0.8, 0.6) \] So, \[ \vec{E}_{13} = 5.394 \times 10^6 \, \text{N/C} \cdot (0.8, 0.6) = (4.315 \times 10^6, 3.236 \times 10^6) \, \text{N/C} \] #### **Field due to \( q_2 \) at P** - Distance \( r_{23} = 0.030 \) m (vertical) - \( q_2 \) at (0.040, 0), \( q_3 \) at (0.040, 0.030) So, displacement vector: (0, 0.030) - Magnitude: \[ E_{23} = \frac{k |q_2|}{r_{23}^2} = \frac{8.99 \times 10^9 \times 4.0 \times 10^{-6}}{(0.03)^2} = \frac{35.96 \times 10^{3}}{0.0009} = 39.956 \times 10^6 \, \text{N/C} \] - Direction: Positive y-direction \[ \vec{E}_{23} = (0, 39.956 \times 10^6) \, \text{N/C} \] #### **Total Field at P** \[ \vec{E}_{\text{total}} = \vec{E}_{13} + \vec{E}_{23} = (4.315 \times 10^6, 3.236 \times 10^6 + 39.956 \times 10^6) = (4.315 \times 10^6, 43.192 \times 10^6) \, \text{N/C} \] **Magnitude:** \[ |\vec{E}_{\text{total}}| = \sqrt{(4.315 \times 10^6)^2 + (43.192 \times 10^6)^2} \approx \sqrt{18.635 \times 10^{12} + 1,867.56 \times 10^{12}} \approx \sqrt{1,886.195 \times 10^{12}} \approx 43.43 \times 10^6 \, \text{N/C} \] **Direction:** \[ \theta = \tan^{-1}\left(\frac{43.192}{4.315}\right) \approx \tan^{-1}(10.01) \approx 84.3^\circ \text{ above the x-axis} \] --- ### b. **Electric Potential at Point P** Potential is scalar, so just add: \[ V_P = V_{1P} + V_{2P} \] \[ V_{1P} = \frac{k q_1}{r_{13}} = \frac{8.99 \times 10^9 \times 1.5 \times 10^{-6}}{0.05} = \frac{13.485 \times 10^{3}}{0.05} = 269,700 \, \text{V} \] \[ V_{2P} = \frac{k q_2}{r_{23}} = \frac{8.99 \times 10^9 \times 4.0 \times 10^{-6}}{0.03} = \frac{35.96 \times 10^{3}}{0.03} = 1,198,667 \, \text{V} \] \[ V_P = 269,700 + 1,198,667 = 1,468,367 \, \text{V} \] --- ### c. **Total Potential Energy of the System** \[ U = k \left( \frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}} \right) \] Plug in: - \( q_1 = 1.5 \times 10^{-6} \) - \( q_2 = 4.0 \times 10^{-6} \) - \( q_3 = -4.0 \times 10^{-6} \) - \( r_{12} = 0.04 \) m, \( r_{13} = 0.05 \) m, \( r_{23} = 0.04 \) m \[ U = 8.99 \times 10^9 \left( \frac{1.5 \times 10^{-6} \times 4.0 \times 10^{-6}}{0.04} + \frac{1.5 \times 10^{-6} \times -4.0 \times 10^{-6}}{0.05} + \frac{4.0 \times 10^{-6} \times -4.0 \times 10^{-6}}{0.04} \right) \] Calculate each term: - \( \frac{1.5 \times 10^{-6} \times 4.0 \times 10^{-6}}{0.04} = \frac{6.0 \times 10^{-12}}{0.04} = 1.5 \times 10^{-10} \) - \( \frac{1.5 \times 10^{-6} \times -4.0 \times 10^{-6}}{0.05} = \frac{-6.0 \times 10^{-12}}{0.05} = -1.2 \times 10^{-10} \) - \( \frac{4.0 \times 10^{-6} \times -4.0 \times 10^{-6}}{0.04} = \frac{-16.0 \times 10^{-12}}{0.04} = -4.0 \times 10^{-10} \) Sum: \[ 1.5 \times 10^{-10} - 1.2 \times 10^{-10} - 4.0 \times 10^{-10} = -3.7 \times 10^{-10} \] Multiply by \( 8.99 \times 10^9 \): \[ U = 8.99 \times 10^9 \times -3.7 \times 10^{-10} = -3.33 \, \text{J} \] --- ## **Final Answers:** **a. Electric Field at Point P:** \[ |\vec{E}_P| \approx 43.4 \times 10^6 \, \text{N/C} \,\, \text{at} \,\, 84.3^\circ \text{ above the x-axis} \] **b. Electric Potential at P:** \[ V_P \approx 1,468,367 \, \text{V} \] **c. Total Potential Energy:** \[ U \approx -3.33 \, \text{J} \] --- **If you need more detail on any step or want the vector breakdown for part (a), let me know!**

Electric field is a vector quantity representing the force per unit charge experienced by a test charge placed at a point due to other charges. Electric potential is a scalar quantity indicating the work done per unit charge in bringing a test charge from infinity to that point. The total electrostatic potential energy of a system is the sum of potential energies due to all pairs of interacting charges, reflecting the system's overall stored electrical energy.