1 State Coulomb's Law. 7.2 Draw a labelled free-body diagram for sphere P. 7.3.1 Calculate the magnitude of the electrostatic force between sphere P and sphere Q.

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Asked on May 14, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-05-14T03:26:32.286Z

Hey Kim, good to see you again. Question 7 7.1 Coulomb's Law states that the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between their centers. 7.2 A labelled free-body diagram for sphere P: • Weight (F_g or mg): Acting vertically downwards. • Normal force (F_N): Acting vertically upwards from the stand. • Electrostatic force (F_e): Acting horizontally to the left, due to repulsion from sphere Q. 7.3.1 Step 1: Identify the given values and Coulomb's Law formula. The charges are q_P = -30 × 10^-6 C and q_Q = -20 × 10^-6 C. The distance between them is r = 80 cm = 0.80 m. Coulomb's constant is k = 9 × 10^9 N · m^2/C^2. The formula for electrostatic force is F_e = k (|q_P q_Q|)/(r^2). Step 2: Substitute the values into the formula and calculate the force. F_e = (9 × 10^9 N · m^2/C^2) |(-30 × 10^-6 C)(-20 × 10^-6 C)|(0.80 m)^2 F_e = (9 × 10^9) (600 × 10^-12)0.64 F_e = (9 × 10^9) (9.375 × 10^-10) F_e = 8.4375 N The magnitude of the electrostatic force is 8.44 N. 7.3.2 Step 1: Calculate the weight of sphere Q. The mass of sphere Q is m = 90 g = 0.090 kg. The acceleration due to gravity is g = 9.8 m/s^2. F_g = mg = (0.090 kg)(9.8 m/s^2) = 0.882 N Step 2: Use the equilibrium conditions for sphere Q. Sphere Q is in equilibrium, so the net force in both horizontal and vertical directions is zero. From the free-body diagram for Q (not P, as asked in 7.2, but implied for 7.3.2/7.3.3): • Horizontal forces: T = F_e • Vertical forces: T = F_g We have F_e = 8.4375 N and F_g = 0.882 N. Step 3: Calculate the angle . Divide the horizontal equilibrium equation by the vertical equilibrium equation: (T )/(T ) = (F_e)/(F_g) = 8.4375 N0.882 N = 9.5663 = (9.5663) = 84.04^ Step 4: Calculate the tension T. Using the vertical equilibrium equation: T = F_g T = (F_g)/( ) = 0.882 N(84.04^) T = 0.882 N0.1038 T = 8.497 N The tension T in the string is 8.50 N. 7.3.3 The angle was calculated in Step 3 of 7.3.2. = ((F_e)/(F_g)) = (8.4375 N0.882 N) = (9.5663) = 84.04^ The angle is 84.0^. --- Question 8 8.1 The electric field at a point is defined as the electrostatic force experienced per unit positive test charge placed at that point. 8.2 The electric field pattern between spheres A (+5 nC) and B (-4 nC) would show: • Field lines originating from sphere A (positive charge). • Field lines terminating on sphere B (negative charge). • Arrows on the field lines pointing away from A and towards B. • The lines would be denser closer to the charges, especially near A due to its larger magnitude of charge compared to B. • The lines would not cross each other. 8.3 Step 1: Identify the given values and distances. q_A = +5 nC = +5 × 10^-9 C q_B = -4 nC = -4 × 10^-9 C k = 9 × 10^9 N · m^2/C^2 Distance from X to A is r_A = r. Distance from X to B is r_B = r + 3r = 4r. Net electric field at X is E_net = 4128 N · C^-1 West. Step 2: Determine the direction of electric fields due to A and B at point X. • For q_A (positive), the electric field E_A at X points away from A. Since X is to the left of A, E_A points to the left (West). • For q_B (negative), the electric field E_B at X points towards B. Since X is to the left of B, E_B points to the right (East). Step 3: Write the expressions for E_A and E_B. E_A = k (|q_A|)/(r_A^2) = (9 × 10^9) 5 × 10^-9r^2 = (45)/(r^2) (West) E_B = k (|q_B|)/(r_B^2) = (9 × 10^9) 4 × 10^-9(4r)^2 = (9 × 10^9) 4 × 10^-916r^2 = (36)/(16r^2) = (9)/(4r^2) (East) Step 4: Calculate the net electric field at X. Since E_net is West, E_A (West) must be greater than E_B (East). E_net = E_A - E_B 4128 = (45)/(r^2) - (9)/(4r^2) 4128 = (4 × 45 - 9)/(4r^2) 4128 = (180 - 9)/(4r^2) 4128 = (171)/(4r^2) Step 5: Solve for r. 4r^2 = (171)/(4128) 4r^2 = 0.0414244 r^2 = (0.0414244)/(4) r^2 = 0.0103561 r = sqrt(0.0103561) r = 0.10176 m The value of r is 0.102 m. Send me the next one 📸

Accepted Answer · 2026-05-14T03:26:32.286Z

Hey Kim, good to see you again. Question 7 7.1 Coulomb's Law states that the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between their centers. 7.2 A labelled free-body diagram for sphere P: • Weight (F_g or mg): Acting vertically downwards. • Normal force (F_N): Acting vertically upwards from the stand. • Electrostatic force (F_e): Acting horizontally to the left, due to repulsion from sphere Q. 7.3.1 Step 1: Identify the given values and Coulomb's Law formula. The charges are q_P = -30 × 10^-6 C and q_Q = -20 × 10^-6 C. The distance between them is r = 80 cm = 0.80 m. Coulomb's constant is k = 9 × 10^9 N · m^2/C^2. The formula for electrostatic force is F_e = k (|q_P q_Q|)/(r^2). Step 2: Substitute the values into the formula and calculate the force. F_e = (9 × 10^9 N · m^2/C^2) |(-30 × 10^-6 C)(-20 × 10^-6 C)|(0.80 m)^2 F_e = (9 × 10^9) (600 × 10^-12)0.64 F_e = (9 × 10^9) (9.375 × 10^-10) F_e = 8.4375 N The magnitude of the electrostatic force is 8.44 N. 7.3.2 Step 1: Calculate the weight of sphere Q. The mass of sphere Q is m = 90 g = 0.090 kg. The acceleration due to gravity is g = 9.8 m/s^2. F_g = mg = (0.090 kg)(9.8 m/s^2) = 0.882 N Step 2: Use the equilibrium conditions for sphere Q. Sphere Q is in equilibrium, so the net force in both horizontal and vertical directions is zero. From the free-body diagram for Q (not P, as asked in 7.2, but implied for 7.3.2/7.3.3): • Horizontal forces: T = F_e • Vertical forces: T = F_g We have F_e = 8.4375 N and F_g = 0.882 N. Step 3: Calculate the angle . Divide the horizontal equilibrium equation by the vertical equilibrium equation: (T )/(T ) = (F_e)/(F_g) = 8.4375 N0.882 N = 9.5663 = (9.5663) = 84.04^ Step 4: Calculate the tension T. Using the vertical equilibrium equation: T = F_g T = (F_g)/( ) = 0.882 N(84.04^) T = 0.882 N0.1038 T = 8.497 N The tension T in the string is 8.50 N. 7.3.3 The angle was calculated in Step 3 of 7.3.2. = ((F_e)/(F_g)) = (8.4375 N0.882 N) = (9.5663) = 84.04^ The angle is 84.0^. --- Question 8 8.1 The electric field at a point is defined as the electrostatic force experienced per unit positive test charge placed at that point. 8.2 The electric field pattern between spheres A (+5 nC) and B (-4 nC) would show: • Field lines originating from sphere A (positive charge). • Field lines terminating on sphere B (negative charge). • Arrows on the field lines pointing away from A and towards B. • The lines would be denser closer to the charges, especially near A due to its larger magnitude of charge compared to B. • The lines would not cross each other. 8.3 Step 1: Identify the given values and distances. q_A = +5 nC = +5 × 10^-9 C q_B = -4 nC = -4 × 10^-9 C k = 9 × 10^9 N · m^2/C^2 Distance from X to A is r_A = r. Distance from X to B is r_B = r + 3r = 4r. Net electric field at X is E_net = 4128 N · C^-1 West. Step 2: Determine the direction of electric fields due to A and B at point X. • For q_A (positive), the electric field E_A at X points away from A. Since X is to the left of A, E_A points to the left (West). • For q_B (negative), the electric field E_B at X points towards B. Since X is to the left of B, E_B points to the right (East). Step 3: Write the expressions for E_A and E_B. E_A = k (|q_A|)/(r_A^2) = (9 × 10^9) 5 × 10^-9r^2 = (45)/(r^2) (West) E_B = k (|q_B|)/(r_B^2) = (9 × 10^9) 4 × 10^-9(4r)^2 = (9 × 10^9) 4 × 10^-916r^2 = (36)/(16r^2) = (9)/(4r^2) (East) Step 4: Calculate the net electric field at X. Since E_net is West, E_A (West) must be greater than E_B (East). E_net = E_A - E_B 4128 = (45)/(r^2) - (9)/(4r^2) 4128 = (4 × 45 - 9)/(4r^2) 4128 = (180 - 9)/(4r^2) 4128 = (171)/(4r^2) Step 5: Solve for r. 4r^2 = (171)/(4128) 4r^2 = 0.0414244 r^2 = (0.0414244)/(4) r^2 = 0.0103561 r = sqrt(0.0103561) r = 0.10176 m The value of r is 0.102 m. Send me the next one 📸

1 State Coulomb's Law. 7.2 Draw a labelled free-body diagram for sphere P. 7.3.1 Calculate the magnitude of the electrostatic force between sphere P and sphere Q.

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1 State Coulomb's Law. 7.2 Draw a labelled free-body diagram for sphere P. 7.3.1 Calculate the magnitude of the electrostatic force between sphere P and sphere Q.

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