Let's solve the problem step by step: **Given:** - Charges: \( q_1 = -4.0\, \mu C \) (at bottom left) \( q_2 = +6.0\, \mu C \) (at top) \( q_3 = +7.0\, \mu C \) (at bottom right) - Side length of triangle: \( a = 0.60\, m \) **Task:** Calculate the **net electrostatic force** on the \( q_3 \) charge. --- ### **Step 1: Find the force between \( q_3 \) and \( q_1 \) (\( F_{31} \))** Use Coulomb's Law: \[ F = k_e \frac{|q_1 q_2|}{r^2} \] Where \( k_e = 8.99 \times 10^9 \) N·m²/C². - \( |q_3| = 7.0 \times 10^{-6} \) C - \( |q_1| = 4.0 \times 10^{-6} \) C - Distance \( r = a = 0.60\, m \) \[ F_{31} = 8.99 \times 10^9 \cdot \frac{(7.0 \times 10^{-6})(4.0 \times 10^{-6})}{(0.60)^2} \] \[ = 8.99 \times 10^9 \cdot \frac{28 \times 10^{-12}}{0.36} \] \[ = 8.99 \times 10^9 \cdot 77.78 \times 10^{-12} \] \[ = 699 \times 10^{-3} \] \[ = 0.699\, \text{N} \] **Direction**: \( q_1 \) is negative, \( q_3 \) is positive, so force is attractive (toward \( q_1 \)). --- ### **Step 2: Find the force between \( q_3 \) and \( q_2 \) (\( F_{32} \))** - \( |q_2| = 6.0 \times 10^{-6} \) C - Distance between \( q_3 \) and \( q_2 \): This is the side of the equilateral triangle, so \( r = a = 0.60\, m \) \[ F_{32} = 8.99 \times 10^9 \cdot \frac{(7.0 \times 10^{-6})(6.0 \times 10^{-6})}{(0.60)^2} \] \[ = 8.99 \times 10^9 \cdot \frac{42 \times 10^{-12}}{0.36} \] \[ = 8.99 \times 10^9 \cdot 116.67 \times 10^{-12} \] \[ = 1049 \times 10^{-3} \] \[ = 1.05\, \text{N} \] **Direction**: Both are positive, so force is repulsive (away from \( q_2 \)). --- ### **Step 3: Find the angle between the forces** Since it's an equilateral triangle: - \( F_{31} \) is along the base (horizontal, left). - \( F_{32} \) is at 60° above the base (since the triangle is equilateral). --- ### **Step 4: Resolve the forces into components** #### \( F_{31} \): - \( x \): \( -0.699\, \text{N} \) (to the left) - \( y \): \( 0 \) #### \( F_{32} \): - \( x \): \( 1.05 \cos(60^\circ) = 1.05 \times 0.5 = 0.525 \, \text{N} \) (to the left) - \( y \): \( 1.05 \sin(60^\circ) = 1.05 \times 0.866 = 0.91 \, \text{N} \) (up) But since it is repulsive, it is directed **away** from \( q_2 \), so the direction from \( q_3 \) to \( q_2 \) is up and to the left. So both x-components are to the left (negative), y-component is up (positive). --- ### **Step 5: Sum the components** - **Total x-component:** \[ F_x = -0.699 + (-0.525) = -1.224\, \text{N} \] - **Total y-component:** \[ F_y = 0 + 0.91 = 0.91\, \text{N} \] --- ### **Step 6: Find the magnitude of the net force** \[ F_{net} = \sqrt{F_x^2 + F_y^2} = \sqrt{(-1.224)^2 + (0.91)^2} = \sqrt{1.498 + 0.828} = \sqrt{2.326} \] \[ F_{net} \approx 1.53 \, \text{N} \] --- ### **Step 7: Find the direction (angle above the horizontal)** \[ \tan \theta = \frac{F_y}{|F_x|} = \frac{0.91}{1.224} = 0.744 \] \[ \theta = \tan^{-1}(0.744) \approx 36.6^\circ \] --- ## **Final Answer:** **The magnitude of the net force on the \( 7.0\, \mu C \) charge is approximately \( \boxed{1.53 \, \text{N}} \), directed \( \boxed{36.6^\circ} \) above the negative x-axis (to the left and up).**