Two small charged spheres, A and B, are placed on insulated stands, 0,2 m apart, as shown in the diagram below. They carry charges of -4 x 10^-6 C and +3 x 10^-6 C respectively.

Question

Asked on August 22, 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 · 2026-08-22T14:31:20.396Z

Step 1: Calculate the number of electrons in excess on sphere A. The charge on sphere A is Q_A = -4 × 10^-6 C. The charge of a single electron is e = -1.6 × 10^-19 C. The number of electrons in excess is given by: n = (Q_A)/(e) n = -4 × 10^-6 C-1.6 × 10^-19 C n = 2.5 × 10^13 The number of electrons in excess on sphere A is 2.5 × 10^13. Step 2: Calculate the magnitude of the electrostatic force exerted by sphere A on sphere B. The charges are Q_A = -4 × 10^-6 C and Q_B = +3 × 10^-6 C. The distance between them is r = 0.2 m. Coulomb's constant is k = 9 × 10^9 N·m^2/C^2. The magnitude of the electrostatic force is given by Coulomb's Law: F = (k |Q_A Q_B|)/(r^2) F = (9 × 10^9 N·m^2/C^2) |-4 × 10^-6 C| |+3 × 10^-6 C|(0.2 m)^2 F = (9 × 10^9)(4 × 10^-6)(3 × 10^-6)0.04 N F = 108 × 10^-30.04 N F = 2.7 N The magnitude of the electrostatic force is 2.7 N. Step 3: Describe the term electric field. An electric field is a region around a charged particle or object within which a force would be exerted on other charged particles or objects. It is defined as the electrostatic force per unit positive charge. Step 4: Calculate the magnitude of the net electric field at point M. Point M is 0.1 m to the right of sphere B. The distance from A to M is r_AM = 0.2 m + 0.1 m = 0.3 m. The distance from B to M is r_BM = 0.1 m. The electric field due to sphere A at M (E_A) is: E_A = (k |Q_A|)/(r_AM)^2 = (9 × 10^9 N·m^2/C^2) |-4 × 10^-6 C|(0.3 m)^2 E_A = (36 × 10^3)/(0.09) N/C = 4 × 10^5 N/C Since Q_A is negative, E_A points towards A (to the left). The electric field due to sphere B at M (E_B) is: E_B = (k |Q_B|)/(r_BM)^2 = (9 × 10^9 N·m^2/C^2) |+3 × 10^-6 C|(0.1 m)^2 E_B = (27 × 10^3)/(0.01) N/C = 2.7 × 10^6 N/C Since Q_B is positive, E_B points away from B (to the right). The net electric field at M is the vector sum of E_A and E_B. Since they are in opposite directions, we subtract their magnitudes. Let's take right as positive: E_net = E_B - E_A E_net = (2.7 × 10^6 N/C) - (4 × 10^5 N/C) E_net = (27 × 10^5 N/C) - (4 × 10^5 N/C) E_net = 23 × 10^5 N/C = 2.3 × 10^6 N/C The magnitude of the net electric field at point M is 2.3 × 10^6 N/C. Step 5: Determine if the charge on sphere D is POSITIVE or NEGATIVE. Sphere A has a negative charge (Q_A = -4 × 10^-6 C). Sphere B has a positive charge (Q_B = +3 × 10^-6 C). The force exerted by B on A (F_AB) is attractive, so it points to the right (positive x-direction). The net electrostatic force experienced by sphere A is 7.69 N in the direction shown, which has both a positive x-component and a positive y-component. The force exerted by D on A (F_AD) acts along the y-axis because D is located vertically above A. Since the net force on A has an upward (positive y) component, the force F_AD must be directed upwards. For a negative charge A to experience an upward force from D, D must repel A. This means D must also be a negative charge. Therefore, the charge on sphere D is NEGATIVE. Step 6: Calculate the magnitude of the charge on sphere D. The x-component of the net force on A is solely due to sphere B, as D is on the y-axis relative to A. From Step 2, the magnitude of the force F_AB is 2.7 N. This is the x-component of the net force on A. F_net,x = F_AB = 2.7 N The magnitude of the net force on A is F_net = 7.69 N. We can find the y-component of the net force using the Pythagorean theorem: F_net^2 = F_net,x^2 + F_net,y^2 F_net,y = sqrt(F_net)^2 - F_net,x^2 F_net,y = sqrt((7.69 N))^2 - (2.7 N)^2 F_net,y = sqrt(59.1361 - 7.29) N F_net,y = sqrt(51.8461) N F_net,y ≈ 7.20 N This y-component of the net force is the force exerted by D on A, F_AD. F_AD = 7.20 N The distance between A and D is r_AD = 0.15 m. Using Coulomb's Law for F_AD: F_AD = (k |Q_A Q_D|)/(r_AD)^2 We can rearrange to solve for |Q_D|: |Q_D| = F_AD · r_AD^2k |Q_A| |Q_D| = (7.20 N) (0.15 m)^2(9 × 10^9 N·m^2/C^2) |-4 × 10^-6 C| |Q_D| = (7.20 × 0.0225)/(36 × 10^3) C |Q_D| = (0.162)/(36000) C |Q_D| = 4.5 × 10^-6 C The magnitude of the charge on sphere D is 4.5 × 10^-6 C. Send me the next one 📸

Accepted Answer · 2026-08-22T14:31:20.396Z

Step 1: Calculate the number of electrons in excess on sphere A. The charge on sphere A is Q_A = -4 × 10^-6 C. The charge of a single electron is e = -1.6 × 10^-19 C. The number of electrons in excess is given by: n = (Q_A)/(e) n = -4 × 10^-6 C-1.6 × 10^-19 C n = 2.5 × 10^13 The number of electrons in excess on sphere A is 2.5 × 10^13. Step 2: Calculate the magnitude of the electrostatic force exerted by sphere A on sphere B. The charges are Q_A = -4 × 10^-6 C and Q_B = +3 × 10^-6 C. The distance between them is r = 0.2 m. Coulomb's constant is k = 9 × 10^9 N·m^2/C^2. The magnitude of the electrostatic force is given by Coulomb's Law: F = (k |Q_A Q_B|)/(r^2) F = (9 × 10^9 N·m^2/C^2) |-4 × 10^-6 C| |+3 × 10^-6 C|(0.2 m)^2 F = (9 × 10^9)(4 × 10^-6)(3 × 10^-6)0.04 N F = 108 × 10^-30.04 N F = 2.7 N The magnitude of the electrostatic force is 2.7 N. Step 3: Describe the term electric field. An electric field is a region around a charged particle or object within which a force would be exerted on other charged particles or objects. It is defined as the electrostatic force per unit positive charge. Step 4: Calculate the magnitude of the net electric field at point M. Point M is 0.1 m to the right of sphere B. The distance from A to M is r_AM = 0.2 m + 0.1 m = 0.3 m. The distance from B to M is r_BM = 0.1 m. The electric field due to sphere A at M (E_A) is: E_A = (k |Q_A|)/(r_AM)^2 = (9 × 10^9 N·m^2/C^2) |-4 × 10^-6 C|(0.3 m)^2 E_A = (36 × 10^3)/(0.09) N/C = 4 × 10^5 N/C Since Q_A is negative, E_A points towards A (to the left). The electric field due to sphere B at M (E_B) is: E_B = (k |Q_B|)/(r_BM)^2 = (9 × 10^9 N·m^2/C^2) |+3 × 10^-6 C|(0.1 m)^2 E_B = (27 × 10^3)/(0.01) N/C = 2.7 × 10^6 N/C Since Q_B is positive, E_B points away from B (to the right). The net electric field at M is the vector sum of E_A and E_B. Since they are in opposite directions, we subtract their magnitudes. Let's take right as positive: E_net = E_B - E_A E_net = (2.7 × 10^6 N/C) - (4 × 10^5 N/C) E_net = (27 × 10^5 N/C) - (4 × 10^5 N/C) E_net = 23 × 10^5 N/C = 2.3 × 10^6 N/C The magnitude of the net electric field at point M is 2.3 × 10^6 N/C. Step 5: Determine if the charge on sphere D is POSITIVE or NEGATIVE. Sphere A has a negative charge (Q_A = -4 × 10^-6 C). Sphere B has a positive charge (Q_B = +3 × 10^-6 C). The force exerted by B on A (F_AB) is attractive, so it points to the right (positive x-direction). The net electrostatic force experienced by sphere A is 7.69 N in the direction shown, which has both a positive x-component and a positive y-component. The force exerted by D on A (F_AD) acts along the y-axis because D is located vertically above A. Since the net force on A has an upward (positive y) component, the force F_AD must be directed upwards. For a negative charge A to experience an upward force from D, D must repel A. This means D must also be a negative charge. Therefore, the charge on sphere D is NEGATIVE. Step 6: Calculate the magnitude of the charge on sphere D. The x-component of the net force on A is solely due to sphere B, as D is on the y-axis relative to A. From Step 2, the magnitude of the force F_AB is 2.7 N. This is the x-component of the net force on A. F_net,x = F_AB = 2.7 N The magnitude of the net force on A is F_net = 7.69 N. We can find the y-component of the net force using the Pythagorean theorem: F_net^2 = F_net,x^2 + F_net,y^2 F_net,y = sqrt(F_net)^2 - F_net,x^2 F_net,y = sqrt((7.69 N))^2 - (2.7 N)^2 F_net,y = sqrt(59.1361 - 7.29) N F_net,y = sqrt(51.8461) N F_net,y ≈ 7.20 N This y-component of the net force is the force exerted by D on A, F_AD. F_AD = 7.20 N The distance between A and D is r_AD = 0.15 m. Using Coulomb's Law for F_AD: F_AD = (k |Q_A Q_D|)/(r_AD)^2 We can rearrange to solve for |Q_D|: |Q_D| = F_AD · r_AD^2k |Q_A| |Q_D| = (7.20 N) (0.15 m)^2(9 × 10^9 N·m^2/C^2) |-4 × 10^-6 C| |Q_D| = (7.20 × 0.0225)/(36 × 10^3) C |Q_D| = (0.162)/(36000) C |Q_D| = 4.5 × 10^-6 C The magnitude of the charge on sphere D is 4.5 × 10^-6 C. Send me the next one 📸

Two small charged spheres, A and B, are placed on insulated stands, 0,2 m apart, as shown in the diagram below. They carry charges of -4 x 10^-6 C and +3 x 10^-6 C respectively.

Need help with your own homework?

More Physics Questions

Two small charged spheres, A and B, are placed on insulated stands, 0,2 m apart, as shown in the diagram below. They carry charges of -4 x 10^-6 C and +3 x 10^-6 C respectively.

Need help with your own homework?

More Physics Questions