1. **Problem Statement:** A test charge of +1.0 μC experiences an electric force of $6.0 \times 10^{-6}$ N to the right. a) Find the intensity and direction of the electric field at this point. b) Find the force (magnitude and direction) on a charge of $-7.2 \times 10^{-4}$ C at the same point. 2. **Formula Used:** The electric field $E$ at a point is related to the force $F$ on a test charge $q$ by: $$E = \frac{F}{q}$$ The direction of $E$ is the direction of the force on a positive test charge. 3. **Step-by-step Solution:** **a) Calculate the electric field intensity:** Given: $$F = 6.0 \times 10^{-6} \text{ N (to the right)}$$ $$q = +1.0 \mu C = 1.0 \times 10^{-6} C$$ Substitute into formula: $$E = \frac{6.0 \times 10^{-6}}{1.0 \times 10^{-6}} = 6.0 \text{ N/C}$$ Direction: Since force is to the right on a positive charge, electric field is also to the right. **Answer a):** Electric field intensity is $6.0$ N/C to the right. **b) Calculate the force on the new charge:** Given: $$q_2 = -7.2 \times 10^{-4} C$$ Use formula: $$F_2 = E \times q_2 = 6.0 \times (-7.2 \times 10^{-4}) = -4.32 \times 10^{-3} \text{ N}$$ Magnitude: $$|F_2| = 4.32 \times 10^{-3} \text{ N}$$ Direction: Since $q_2$ is negative, force direction is opposite to electric field direction. Electric field is to the right, so force on $q_2$ is to the left. **Answer b):** Force magnitude is $4.32 \times 10^{-3}$ N to the left. **Final answers:** - a) Electric field intensity = $6.0$ N/C to the right. - b) Force on $-7.2 \times 10^{-4}$ C charge = $4.32 \times 10^{-3}$ N to the left.