Let's solve the problem step by step. --- ## **Step 1: Given Data** - \( q = 7.14 \) fC \( = 7.14 \times 10^{-15} \) C - \( d = 6.00 \) cm \( = .060 \) m - 1 fC (femtocoulomb) \( = 10^{-15} \) C --- ## **Step 2: Analyzing the Configuration** - There are four charges: - \( +q \) above P (at distance \( d \)) - \( +q \) below P (at distance \( d \)) - \( -q \) to the right of P (distance \( d \)), then another \( -q \) a further distance \( d \) to the right (total \( 2d \) from P) - The two positive charges are vertically aligned. - The two negative charges are horizontally aligned (one at \( d \) and one at \( 2d \) to the right of P). --- ## **Step 3: Electric Field Due to Each Charge** The electric field due to a point charge: \[ E = \frac{k|q|}{r^2} \] where \( k = 8.99 \times 10^9 \) N·m²/C². ### **Directions:** - Fields from \( +q \) point away from the charge. - Fields from \( -q \) point toward the charge. Let's assign axes: - \( +y \) is up, \( +x \) is right. --- ### **A. Top \( +q \) at (, +d):** \[ E_{\text{top}} = \frac{kq}{d^2} \quad (\text{downward at P, i.e., in } -\hat{y}) \] ### **B. Bottom \( +q \) at (, -d):** \[ E_{\text{bottom}} = \frac{kq}{d^2} \quad (\text{upward at P, i.e., in } +\hat{y}) \] --- ### **C. Right \( -q \) at (d, ):** \[ E_{\text{right}} = \frac{kq}{d^2} \quad (\text{toward the charge, i.e., +\hat{x}}) \] ### **D. Far Right \( -q \) at (2d, ):** \[ E_{\text{far-right}} = \frac{kq}{(2d)^2} = \frac{kq}{4d^2} \quad (\text{toward the charge, i.e., +\hat{x}}) \] --- ## **Step 4: Net Electric Field Components at P** ### **Y-components:** - \( +q \) at top: \( E_{\text{top}} \) down (\(-y\)) - \( +q \) at bottom: \( E_{\text{bottom}} \) up (\(+y\)) They are equal in magnitude and opposite in direction: \[ E_{y, \text{net}} = \frac{kq}{d^2} - \frac{kq}{d^2} = \] ### **X-components:** - \( -q \) at right: \( E_{\text{right}} \) right (\(+x\)) - \( -q \) at far right: \( E_{\text{far-right}} \) right (\(+x\)) Sum: \[ E_{x, \text{net}} = \frac{kq}{d^2} + \frac{kq}{4d^2} = \frac{5kq}{4d^2} \] --- ## **Step 5: Plug in the Values** \[ E = \frac{5kq}{4d^2} \] Where: - \( k = 8.99 \times 10^9 \) N·m²/C² - \( q = 7.14 \times 10^{-15} \) C - \( d = .060 \) m Calculate \( d^2 \): \[ d^2 = (.060)^2 = .0036 \text{ m}^2 \] Plug in: \[ E = \frac{5 \times 8.99 \times 10^9 \times 7.14 \times 10^{-15}}{4 \times .0036} \] Calculate numerator: \[ 5 \times 8.99 \times 10^9 \times 7.14 \times 10^{-15} = 5 \times 8.99 \times 7.14 \times 10^{-6} = 5 \times 64.1886 \times 10^{-6} = 320.943 \times 10^{-6} = 3.20943 \times 10^{-4} \] Calculate denominator: \[ 4 \times .0036 = .0144 \] Final calculation: \[ E = \frac{3.20943 \times 10^{-4}}{.0144} = .0223 \text{ N/C} \] --- ## **Step 6: Final Answer** ### **Net Electric Field at Point P:** \[ \boxed{.0223 \; \text{N/C} \;\;\text{to the right} \; (+\hat{x})} \] --- **Summary Table:** | Component | Value (N/C) | Direction | |-----------|-------------|-----------| | \(E_x\) | .0223 | right | | \(E_y\) | | — | | **Total** | **.0223** | **right** |