Let's break down the problem step by step: --- ## **Given Data** - **Charge on left (Q₁):** \( Q_+ = +5. \, \mu \text{C} = 5. \times 10^{-6} \, \text{C} \) - **Charge on right (Q₂):** \( Q_- = -4. \, \mu \text{C} = -4. \times 10^{-6} \, \text{C} \) - **Coordinates:** - \( Q_+ \) at (-8. cm, ) - \( Q_- \) at (8. cm, ) - \( A \) at (, 16. cm) - \( B \) at (, ) --- ## **a. Electric Field at Point A** ### **Step 1: Find Distances** - Distance from \( Q_+ \) to \( A \): \[ r_1 = \sqrt{(8.\,\text{cm})^2 + (16.\,\text{cm})^2} = \sqrt{64 + 256} = \sqrt{320} = 17.89\,\text{cm} = .179\,\text{m} \] - Distance from \( Q_- \) to \( A \): \[ r_2 = \sqrt{(8.\,\text{cm})^2 + (16.\,\text{cm})^2} = .179\,\text{m} \] ### **Step 2: Electric Field Formula** \[ E = \frac{1}{4\pi\epsilon_} \frac{|q|}{r^2} \] where \(\frac{1}{4\pi\epsilon_} = 8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2\) ### **Step 3: Components of the Electric Field** #### **From Q₁ (+)** - Direction: Away from positive charge - x-component: \(-8.\,\text{cm}/r_1\) - y-component: \(16.\,\text{cm}/r_1\) So, \[ E_{1x} = E_1 \frac{-8.}{17.89} \] \[ E_{1y} = E_1 \frac{16.}{17.89} \] where \( E_1 = \frac{8.99 \times 10^9 \times 5. \times 10^{-6}}{(.179)^2} \approx 1.404 \times 10^6 \, \text{N/C} \) So: - \( \frac{-8.}{17.89} = -.447 \) - \( \frac{16.}{17.89} = .894 \) \[ E_{1x} = 1.404 \times 10^6 \times (-.447) = -627,000\,\text{N/C} \] \[ E_{1y} = 1.404 \times 10^6 \times .894 = 1,256,000\,\text{N/C} \] #### **From Q₂ (-)** - Direction: Toward negative charge x-component: \(+8.\,\text{cm}/r_2\) y-component: \(16.\,\text{cm}/r_2\) \[ E_2 = \frac{8.99 \times 10^9 \times 4. \times 10^{-6}}{(.179)^2} \approx 1.123 \times 10^6 \, \text{N/C} \] \[ E_{2x} = 1.123 \times 10^6 \times .447 = 502,000\,\text{N/C} \] \[ E_{2y} = 1.123 \times 10^6 \times .894 = 1,004,000\,\text{N/C} \] --- ### **Step 4: Add the Vectors** \[ E_x = E_{1x} + E_{2x} = -627,000 + 502,000 = -125,000\,\text{N/C} \] \[ E_y = E_{1y} + E_{2y} = 1,256,000 + 1,004,000 = 2,260,000\,\text{N/C} \] \[ \boxed{ \vec{E}_A = (-1.25 \times 10^5 \, \hat{i} + 2.26 \times 10^6 \, \hat{j}) \, \text{N/C} } \] --- ## **b. Charge at B to Make Electric Field at A Zero** Let the charge at B be \( q \). - Distance from B to A = \( 16. \) cm = \( .16 \) m (straight up) - The field at A due to \( q \) at B: \( E_q = \frac{1}{4\pi\epsilon_} \frac{q}{(.16)^2} \) upwards (if \( q > \)), downwards (if \( q < \)) - To cancel \( E_A \), \( E_q \) must have components: - \( E_{qx} = +1.25 \times 10^5 \, \text{N/C} \) - \( E_{qy} = -2.26 \times 10^6 \, \text{N/C} \) But since B is directly below A, it only contributes to the y-component. So, \[ E_{qy} + E_y = \] \[ E_{qy} = - E_y = -2.26 \times 10^6 \, \text{N/C} \] But \( E_{qy} = \frac{1}{4\pi\epsilon_} \frac{q}{(.16)^2} \) So, \[ -2.26 \times 10^6 = \frac{8.99 \times 10^9 \cdot q}{.0256} \] \[ q = \frac{-2.26 \times 10^6 \times .0256}{8.99 \times 10^9} \] \[ q = \frac{-57,856}{8.99 \times 10^9} \] \[ q = -6.44 \times 10^{-6} \, \text{C} = -6.44 \, \mu\text{C} \] --- ## **Final Answers** **a.** \[ \boxed{ \vec{E}_A = (-1.25 \times 10^5 \, \hat{i} + 2.26 \times 10^6 \, \hat{j}) \, \text{N/C} } \] **b.** \[ \boxed{ q = -6.44 \, \mu\text{C} } \] (Must be placed at **B** for the field at **A** to be zero.)