Let's solve each part step by step, with explanations after each step: --- ### **a. Direction of the Electric Field at Point P (.6 m from the origin)** #### **Step 1:** The electric field due to a positive charge points **away** from the charge. - Since Q (the charge at the origin) is positive, the electric field at point P (which is on the x-axis at x = .6 m) will point **away from the origin**, i.e., in the **positive x-direction**. **Explanation:** Positive charges create electric fields that radiate outward, so at any point on the positive x-axis, the field points to the right (positive x-direction). --- ### **b. Magnitude of the Electric Field at Point P** #### **Step 2:** Use Coulomb's law for the electric field: \[ E = \frac{kQ}{r^2} \] Where: - \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \) - \( Q = +6 \, \mu\text{C} = +6 \times 10^{-6} \, \text{C} \) - \( r = .6 \, \text{m} \) Plug in the values: \[ E = \frac{8.99 \times 10^9 \times 6 \times 10^{-6}}{(.6)^2} \] \[ E = \frac{8.99 \times 10^9 \times 6 \times 10^{-6}}{.36} \] \[ E = \frac{53.94 \times 10^3}{.36} \] \[ E = 149,833 \, \text{N/C} \] **Explanation:** The electric field decreases with the square of the distance from the source charge. --- ### **Final Answers** a. **Direction:** Positive x-direction (away from the origin). b. **Magnitude:** \( \approx 1.5 \times 10^5 \) N/C (rounded to 2 significant figures). Let me know if you'd like to continue with parts c, d, and e!

# Electric Field and Force Calculations Let's solve all parts of the problem step by step. --- ## **a. Direction of the Electric Field at Point P (0.6 m from the origin)** ### **Step 1:** The electric field created by a positive charge points **away** from the charge. - Since the charge \( Q = +6 \, \mu\text{C} \) is at the origin, the electric field at point P (0.6 m away) will point **in the positive x-direction**. **Explanation:** Electric fields from positive charges radiate outward. Therefore, at point P, located on the positive x-axis, the field direction is to the right. --- ## **b. Magnitude of the Electric Field at Point P** ### **Step 2:** Use Coulomb's law to calculate the electric field: \[ E = \frac{kQ}{r^2} \] Where: - \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \) - \( Q = +6 \, \mu\text{C} = +6 \times 10^{-6} \, \text{C} \) - \( r = 0.6 \, \text{m} \) Calculating: \[ E = \frac{8.99 \times 10^9 \times 6 \times 10^{-6}}{(0.6)^2} \] \[ E = \frac{53.94 \times 10^3}{0.36} \] \[ E \approx 149,833 \, \text{N/C} \] **Explanation:** The electric field's magnitude is inversely proportional to the square of the distance from the charge. As you move away from the charge, the field strength diminishes. --- ## **c. Direction of the Electric Force on Charge \( q \)** ### **Step 3:** The force on a charge in an electric field is given by: \[ F = qE \] Where \( q = +1 \, \text{pC} = 1 \times 10^{-12} \, \text{C} \). Since \( q \) is also positive, the direction of the electric force will be the same as the direction of the electric field. **Direction:** The force on charge \( q \) will also point **in the positive x-direction**. **Explanation:** A positive test charge experiences a force in the direction of the electric field created by another positive charge. --- ## **d. Magnitude of the Electric Force on Charge \( q \)** ### **Step 4:** Calculate the electric force using: \[ F = qE \] Substituting the known values: \[ F = (1 \times 10^{-12} \, \text{C})(149,833 \, \text{N/C}) \] \[ F \approx 149.83 \, \text{nN} \] **Explanation:** The force magnitude is directly proportional to the charge and the electric field strength. A small test charge in a strong electric field experiences a discernible force. --- ## **e. Acceleration of Charge \( q \)** ### **Step 5:** Use Newton’s second law: \[ F = ma \] Where: - \( F \approx 149.83 \, \text{nN} = 149.83 \times 10^{-9} \, \text{N} \) - \( m = 2.5 \, \text{g} = 2.5 \times 10^{-3} \, \text{kg} \) Rearranging for acceleration \( a \): \[ a = \frac{F}{m} = \frac{149.83 \times 10^{-9}}{2.5 \times 10^{-3}} \] \[ a \approx 59.93 \times 10^{-6} \, \text{m/s}^2 \] \[ a \approx 59.93 \, \mu\text{m/s}^2 \] **Explanation:** The acceleration is calculated from the force experienced by the charge and its mass. A small mass results in a higher acceleration for the same force. --- ## **Final Answers** - **a. Direction of Electric Field:** Positive x-direction (away from the origin). - **b. Magnitude of Electric Field:** \( \approx 149,833 \, \text{N/C} \). - **c. Direction of Electric Force on Charge \( q \):** Positive x-direction. - **d. Magnitude of Electric Force on Charge \( q \):** \( \approx 149.83 \, \text{nN} \). - **e. Acceleration of Charge \( q \):** \( \approx 59.93 \, \mu\text{m/s}^2 \).