1. **Problem statement:** Calculate the electric field strength at a point 1.00 cm to the left of the middle charge (+1.50 μC) in a system of three charges: +6.00 μC at 3.00 cm left of middle, +1.50 μC in the middle, and -2.00 μC at 2.00 cm right of middle. 2. **Formula:** The electric field due to a point charge is given by: $$E = \frac{k|q|}{r^2}$$ where $k = 8.99 \times 10^9$ N·m²/C², $q$ is the charge, and $r$ is the distance from the charge to the point. 3. **Important rules:** - Electric field direction is away from positive charges and toward negative charges. - The net electric field is the vector sum of fields from all charges. 4. **Calculate distances from each charge to the point:** - Distance from +6.00 μC charge: $3.00 \text{ cm} + 1.00 \text{ cm} = 4.00 \text{ cm} = 0.04 \text{ m}$ - Distance from +1.50 μC charge: $1.00 \text{ cm} = 0.01 \text{ m}$ - Distance from -2.00 μC charge: $2.00 \text{ cm} + 1.00 \text{ cm} = 3.00 \text{ cm} = 0.03 \text{ m}$ 5. **Calculate electric field magnitudes:** - From +6.00 μC: $$E_1 = \frac{8.99 \times 10^9 \times 6.00 \times 10^{-6}}{(0.04)^2} = \frac{8.99 \times 10^9 \times 6.00 \times 10^{-6}}{0.0016} = 3.37 \times 10^7 \text{ N/C}$$ - From +1.50 μC: $$E_2 = \frac{8.99 \times 10^9 \times 1.50 \times 10^{-6}}{(0.01)^2} = \frac{8.99 \times 10^9 \times 1.50 \times 10^{-6}}{0.0001} = 1.35 \times 10^8 \text{ N/C}$$ - From -2.00 μC: $$E_3 = \frac{8.99 \times 10^9 \times 2.00 \times 10^{-6}}{(0.03)^2} = \frac{8.99 \times 10^9 \times 2.00 \times 10^{-6}}{0.0009} = 1.998 \times 10^7 \text{ N/C}$$ 6. **Determine directions of each field at the point:** - $E_1$: from +6.00 μC, field points away from positive charge, so to the right. - $E_2$: from +1.50 μC, field points away from positive charge, so to the right. - $E_3$: from -2.00 μC, field points toward negative charge, so to the right. All fields point to the right, so net field magnitude is sum: $$E_{net} = E_1 + E_2 + E_3 = 3.37 \times 10^7 + 1.35 \times 10^8 + 1.998 \times 10^7 = 1.8878 \times 10^8 \text{ N/C}$$ 7. **Answer for (a):** The magnitude of the electric field at the point is approximately $$1.89 \times 10^8 \text{ N/C}$$ 8. **For (b), force on charge $q = -2.26 \times 10^{-6} C$ placed at this point:** $$F = |q| E_{net} = 2.26 \times 10^{-6} \times 1.8878 \times 10^8 = 426.7 \text{ N}$$ 9. **Direction of force:** Since the charge is negative and the field points right, force is opposite to field direction, so to the left. **Final answers:** - (a) Electric field magnitude: $1.89 \times 10^8$ N/C - (b) Force magnitude: 426.7 N - Force direction: left