## Step 1: Find the Electric Field at the Midpoint Due to Each Charge The distance between the charges is **20. cm**, so the midpoint is **10. cm = .10 m** from each charge. The electric field due to a point charge is given by: \[ E = \frac{k |q|}{r^2} \] - For \( q_1 = +20.\,\mu\text{C} = 20. \times 10^{-6}\,\text{C} \): \[ E_1 = \frac{8.99 \times 10^9 \times 20. \times 10^{-6}}{(.10)^2} = \frac{8.99 \times 10^9 \times 20. \times 10^{-6}}{.01} \] \[ = \frac{8.99 \times 10^9 \times 20. \times 10^{-6}}{.01} = 1.798 \times 10^7\, \text{N/C} \] - For \( q_2 = -8.00\,\mu\text{C} = -8.00 \times 10^{-6}\,\text{C} \): \[ E_2 = \frac{8.99 \times 10^9 \times 8.00 \times 10^{-6}}{(.10)^2} = \frac{8.99 \times 10^9 \times 8.00 \times 10^{-6}}{.01} \] \[ = 7.192 \times 10^6\, \text{N/C} \] **Explanation:** We calculated the electric field magnitude at the midpoint due to each charge using the formula for the electric field of a point charge. --- ## Step 2: Find the Net Electric Field at the Midpoint Both fields point in the same direction (from +20. μC toward -8.00 μC), so **add** the fields: \[ E_{\text{net}} = E_1 + E_2 = 1.798 \times 10^7 + 7.192 \times 10^6 = 2.517 \times 10^7\, \text{N/C} \] **Explanation:** Because the positive charge creates a field pointing away and the negative toward itself, at the midpoint between them both fields point from the positive to the negative charge. We add their magnitudes for the net field. --- ## Final Answer \[ \boxed{2.52 \times 10^7\, \text{N/C}} \] This is the magnitude of the electric field at the midpoint between the charges.