Let's solve each part step by step, **explaining at the end of each step**. --- ## (a) Potential at the Origin ### **Step 1: Formula and Setup** The **electric potential** at a point due to a point charge is: \[ V = \frac{kq}{r} \] Where: - \( k = 8.99 \times 10^9 \) Nm²/C² (Coulomb's constant) - \( q \) = charge - \( r \) = distance from charge to point All four charges are **distance = 1.00 \, \text{m}** from the origin. **Explanation:** The potential from each charge adds algebraically (since potential is a scalar). --- ### **Step 2: Calculation** Sum potentials: \[ V_{\text{total}} = \sum_{i=1}^{4} \frac{kq_i}{1.00} \] Plugging in the values (\(\mu C = 10^{-6} C\)): \[ V_{\text{total}} = k \left(\frac{1.}{1} + \frac{2.}{1} + \frac{4.}{1} + \frac{1.5}{1}\right) \times 10^{-6} = k \cdot 8.5 \times 10^{-6} \] \[ V_{\text{total}} = (8.99 \times 10^9) \times 8.5 \times 10^{-6} \] \[ V_{\text{total}} = 76,415 \, \text{V} \] **Explanation:** Add all the charges (in microcoulombs), multiply by Coulomb's constant, and convert to volts. --- **Final Answer (a):** \[ \boxed{76,415 \, \text{V}} \] --- ## (b) Electric Field at \( q_2 \) ### **Step 1: Formula and Setup** The electric field at \( q_2 \) (positioned at \((,1)\)): \[ \vec{E} = \sum_{i \neq 2} \vec{E}_i = \sum_{i \neq 2} \frac{kq_i}{r_{i2}^2} \hat{r}_{i2} \] Where: - \( r_{i2} \) = distance from \( q_i \) to \( q_2 \) - \( \hat{r}_{i2} \) = unit vector from \( q_i \) to \( q_2 \) The charges and their coordinates: - \( q_1: (1,) \) - \( q_2: (,1) \) - \( q_3: (-1,) \) - \( q_4: (,-1) \) **Explanation:** We'll find the field at \( q_2 \) due to \( q_1, q_3, q_4 \) and add the vectors. --- ### **Step 2: Calculation** **1. From \( q_1 \) at (1,) to \( q_2 \) at (,1):** - Distance: \( r = \sqrt{(1-)^2 + (-1)^2} = \sqrt{2} \) - Direction: from (1,) to (,1) is \( (-1, 1) \), unit: \( (-1/\sqrt{2}, +1/\sqrt{2}) \) \[ E_{q_1 \to q_2} = \frac{k q_1}{(\sqrt{2})^2} = \frac{k \cdot 1. \times 10^{-6}}{2} = 4.495 \times 10^3 \, \text{N/C} \] So, \[ \vec{E}_{q_1 \to q_2} = 4.495 \times 10^3 \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) = (-3,180, +3,180) \, \text{N/C} \] --- **2. From \( q_3 \) at (-1,) to \( q_2 \) at (,1):** - Distance: \( \sqrt{2} \) - Direction: \( (1,1)/\sqrt{2} \) \[ E_{q_3 \to q_2} = \frac{k \cdot 4. \times 10^{-6}}{2} = 1.798 \times 10^4 \, \text{N/C} \] \[ \vec{E}_{q_3 \to q_2} = 1.798 \times 10^4 \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) = (+12,728, +12,728) \, \text{N/C} \] --- **3. From \( q_4 \) at (,-1) to \( q_2 \) at (,1):** - Distance: 2 (vertical) - Direction: \( (,1) \) \[ E_{q_4 \to q_2} = \frac{k \cdot 1.5 \times 10^{-6}}{(2)^2} = \frac{8.99 \times 10^9 \cdot 1.5 \times 10^{-6}}{4} = 3,371 \, \text{N/C} \] \[ \vec{E}_{q_4 \to q_2} = (, 3,371) \] --- **Sum components:** - **x-component:** \( -3,180 + 12,728 + = 9,548 \) - **y-component:** \( 3,180 + 12,728 + 3,371 = 19,279 \) \[ |\vec{E}| = \sqrt{(9,548)^2 + (19,279)^2} \approx 21,478 \, \text{N/C} \] **Explanation:** Add the vector components from each charge, then find the magnitude. --- **Final Answer (b):** \[ \boxed{|\vec{E}_{q_2}| \approx 21,500 \, \text{N/C}} \] --- ## (c) Net Force on \( q_2 \) ### **Step 1: Formula and Setup** The net force on \( q_2 \) is: \[ \vec{F} = q_2 \vec{E}_{q_2} \] **Explanation:** Multiply the value of \( q_2 \) by the net electric field at its position. --- ### **Step 2: Calculation** \[ q_2 = 2. \,\mu C = 2. \times 10^{-6} \, C \] \[ |\vec{F}| = q_2 \cdot |\vec{E}_{q_2}| = (2. \times 10^{-6}) \times 21,478 \approx .04296 \, \text{N} \] **Explanation:** Plug in the calculated electric field and charge to get the force. --- **Final Answer (c):** \[ \boxed{.043 \, \text{N}} \] (Direction is the same as the net electric field at \( q_2 \)).