This question tests understanding of Coulomb's Law, which describes the electric force between charged objects. Coulomb's Law states that the electric force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them: F = k(q₁q₂)/r², where k = 9.0 × 10⁹ N·m²/C² is Coulomb's constant, q₁ and q₂ are the charges in Coulombs, r is the distance between them in meters, and F is the force in Newtons—the force is attractive if the charges have opposite signs (one positive, one negative) and repulsive if they have the same sign (both positive or both negative). To find the electric force between charges q₁ = +8.0 μC and q₂ = -2.0 μC separated by distance r = 40 cm, first convert units: 8.0 μC = 8.0 × 10⁻⁶ C, -2.0 μC = -2.0 × 10⁻⁶ C, 40 cm = 0.40 m; substitute into Coulomb's Law for magnitude: |F| = (9.0 × 10⁹ N·m²/C²)(8.0 × 10⁻⁶ C)(2.0 × 10⁻⁶ C)/(0.40 m)² = (9.0 × 10⁹)(1.6 × 10⁻¹¹)/0.16 = 1.44 × 10⁻¹ / 0.16 = 0.90 N. Choice A is correct because it properly applies F = k|q₁q₂|/r² with correct values, units, and scientific notation. Choice B forgets to convert centimeters to meters before calculating, using 40 m directly in the formula and getting a result that's off by a factor of 10⁴ in r². When applying Coulomb's Law: (1) convert all charges to Coulombs (1 μC = 10⁻⁶ C, 1 nC = 10⁻⁹ C) and distances to meters (1 cm = 0.01 m), (2) substitute carefully into F = k(q₁q₂)/r² keeping track of scientific notation, (3) determine direction from charge signs (opposite signs → attractive, same signs → repulsive), and (4) verify your result makes sense (typical classroom charges in μC at cm distances give forces in mN to N range, while atomic charges at atomic distances give enormous forces). The inverse square law means electric force decreases rapidly with distance: double the separation and force drops to 1/4, triple it and force drops to 1/9—this is why charged objects interact strongly when close but the force becomes negligible at larger distances (unlike gravity which, though also following inverse square law, acts over astronomical distances because masses are so much larger and always attractive).

This question tests understanding of Coulomb's Law, which describes the electric force between charged objects. Coulomb's Law states that the electric force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them: F = k(q₁q₂)/r², where k = 9.0 × 10⁹ N·m²/C² is Coulomb's constant, q₁ and q₂ are the charges in Coulombs, r is the distance between them in meters, and F is the force in Newtons—the force is attractive if the charges have opposite signs (one positive, one negative) and repulsive if they have the same sign (both positive or both negative). The charges in this scenario have opposite signs: the rod has q₁ = +6.0 μC (positive) and the sphere has q₂ = -3.0 μC (negative), which means the electric force is attractive—attractive forces pull the charges together (each charge experiences force toward the other), while repulsive forces push them apart, but in both cases the magnitude is given by |F| = k|q₁||q₂|/r². Choice A is correct because it accurately identifies the force as attractive based on the signs of the charges—when one charge is positive and the other is negative, they attract each other. Choice B incorrectly identifies the force as repulsive, confusing the rule: opposite sign charges attract (pull together), same sign charges repel (push apart). When applying Coulomb's Law: (1) convert all charges to Coulombs (1 μC = 10⁻⁶ C, 1 nC = 10⁻⁹ C) and distances to meters (1 cm = 0.01 m), (2) substitute carefully into F = k(q₁q₂)/r² keeping track of scientific notation, (3) determine direction from charge signs (opposite signs → attractive, same signs → repulsive), and (4) verify your result makes sense (typical classroom charges in μC at cm distances give forces in mN to N range, while atomic charges at atomic distances give enormous forces). Key relationships: (a) force proportional to each charge (double q₁ → double F), (b) force proportional to product of charges (double both → quadruple F), (c) force inversely proportional to r² (double r → F becomes 1/4), and (d) by Newton's Third Law, the force on q₁ due to q₂ equals the force on q₂ due to q₁ in magnitude but opposite in direction—both charges feel the same magnitude force whether attractive or repulsive.

This question tests understanding of Coulomb's Law, which describes the electric force between charged objects. Coulomb's Law states that the electric force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them: F = k(q₁q₂)/r², where k = 9.0 × 10⁹ N·m²/C² is Coulomb's constant, q₁ and q₂ are the charges in Coulombs, r is the distance between them in meters, and F is the force in Newtons—the force is attractive if the charges have opposite signs (one positive, one negative) and repulsive if they have the same sign (both positive or both negative). To find the magnitude of the electric force between charges q₁ = 3.0 × 10⁻⁶ C and q₂ = 6.0 × 10⁻⁶ C separated by distance r = 0.30 m, substitute into Coulomb's Law: F = k|q₁q₂|/r² = (9.0 × 10⁹ N·m²/C²)(1.8 × 10⁻¹¹)/(0.30)² = (9.0 × 10⁹)(1.8 × 10⁻¹¹)/0.09 = (9.0 × 10⁹)(2.0 × 10⁻¹⁰) = 1.8 N. Choice B is correct because it properly applies F = k|q₁q₂|/r² with correct values, units, and scientific notation. Choice A has an error in the power of 10 in scientific notation (reports 0.18 instead of 1.8), likely from incorrectly adding or subtracting exponents when multiplying or dividing powers of 10 during the calculation. When applying Coulomb's Law: (1) convert all charges to Coulombs (1 μC = 10⁻⁶ C, 1 nC = 10⁻⁹ C) and distances to meters (1 cm = 0.01 m), (2) substitute carefully into F = k(q₁q₂)/r² keeping track of scientific notation, (3) determine direction from charge signs (opposite signs → attractive, same signs → repulsive), and (4) verify your result makes sense (typical classroom charges in μC at cm distances give forces in mN to N range, while atomic charges at atomic distances give enormous forces). The inverse square law means electric force decreases rapidly with distance: double the separation and force drops to 1/4, triple it and force drops to 1/9—this is why charged objects interact strongly when close but the force becomes negligible at larger distances (unlike gravity which, though also following inverse square law, acts over astronomical distances because masses are so much larger and always attractive).

This question tests understanding of Coulomb's Law, which describes the electric force between charged objects. Coulomb's Law states that the electric force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them: F = k(q₁q₂)/r², where k = 9.0 × 10⁹ N·m²/C² is Coulomb's constant, q₁ and q₂ are the charges in Coulombs, r is the distance between them in meters, and F is the force in Newtons—the force is attractive if the charges have opposite signs (one positive, one negative) and repulsive if they have the same sign (both positive or both negative). To find the electric force between charges q₁ = +6.0 μC and q₂ = -3.0 μC separated by distance r = 0.20 m, first convert units: 6.0 μC = 6.0 × 10⁻⁶ C, -3.0 μC = -3.0 × 10⁻⁶ C; substitute into Coulomb's Law for magnitude: |F| = (9.0 × 10⁹ N·m²/C²)(6.0 × 10⁻⁶ C)(3.0 × 10⁻⁶ C)/(0.20 m)² = (9.0 × 10⁹)(1.8 × 10⁻¹¹)/0.040 = 1.62 × 10⁻¹ / 0.040 = 4.05 N. Choice B is correct because it properly applies F = k|q₁q₂|/r² with correct values, units, and scientific notation. Choice A forgets to convert microcoulombs to coulombs before calculating, using 6.0 and 3.0 directly in the formula and getting a result that's off by a factor of 10⁻¹² in the product but adjusted incorrectly. When applying Coulomb's Law: (1) convert all charges to Coulombs (1 μC = 10⁻⁶ C, 1 nC = 10⁻⁹ C) and distances to meters (1 cm = 0.01 m), (2) substitute carefully into F = k(q₁q₂)/r² keeping track of scientific notation, (3) determine direction from charge signs (opposite signs → attractive, same signs → repulsive), and (4) verify your result makes sense (typical classroom charges in μC at cm distances give forces in mN to N range, while atomic charges at atomic distances give enormous forces). Key relationships: (a) force proportional to each charge (double q₁ → double F), (b) force proportional to product of charges (double both → quadruple F), (c) force inversely proportional to r² (double r → F becomes 1/4), and (d) by Newton's Third Law, the force on q₁ due to q₂ equals the force on q₂ due to q₁ in magnitude but opposite in direction—both charges feel the same magnitude force whether attractive or repulsive.

This question tests understanding of Coulomb's Law, which describes the electric force between charged objects. Coulomb's Law states that the electric force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them: F = k(q₁q₂)/r², where k = 9.0 × 10⁹ N·m²/C² is Coulomb's constant, q₁ and q₂ are the charges in Coulombs, r is the distance between them in meters, and F is the force in Newtons—the force is attractive if the charges have opposite signs (one positive, one negative) and repulsive if they have the same sign (both positive or both negative). To find the electric force between charges q₁ = +2.0 μC and q₂ = +5.0 μC separated by distance r = 0.30 m, first convert to SI units: 2.0 × 10⁻⁶ C and 5.0 × 10⁻⁶ C, then substitute into Coulomb's Law: F = (9.0 × 10⁹)(2.0 × 10⁻⁶)(5.0 × 10⁻⁶)/(0.30)² = (9.0 × 10⁹)(1.0 × 10⁻¹¹)/0.09 = 9.0 × 10⁻² / 0.09 = 1.0 N. Choice A is correct because it properly applies F = k|q₁q₂|/r² with correct values, units, and scientific notation. Choice D has an error in the power of 10 in scientific notation (reports 1.0 × 10⁻⁵ instead of 1.0), likely from incorrectly adding or subtracting exponents when multiplying or dividing powers of 10 during the calculation. When applying Coulomb's Law: (1) convert all charges to Coulombs (1 μC = 10⁻⁶ C, 1 nC = 10⁻⁹ C) and distances to meters (1 cm = 0.01 m), (2) substitute carefully into F = k(q₁q₂)/r² keeping track of scientific notation, (3) determine direction from charge signs (opposite signs → attractive, same signs → repulsive), and (4) verify your result makes sense (typical classroom charges in μC at cm distances give forces in mN to N range, while atomic charges at atomic distances give enormous forces). The inverse square law means electric force decreases rapidly with distance: double the separation and force drops to 1/4, triple it and force drops to 1/9—this is why charged objects interact strongly when close but the force becomes negligible at larger distances (unlike gravity which, though also following inverse square law, acts over astronomical distances because masses are so much larger and always attractive).

This question tests understanding of Coulomb's Law, which describes the electric force between charged objects. Coulomb's Law states that the electric force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them: F = k(q₁q₂)/r², where k = 9.0 × 10⁹ N·m²/C² is Coulomb's constant, q₁ and q₂ are the charges in Coulombs, r is the distance between them in meters, and F is the force in Newtons—the force is attractive if the charges have opposite signs (one positive, one negative) and repulsive if they have the same sign (both positive or both negative). The charges in this scenario have same signs: both negative (q₁ = -3.0 μC and q₂ = -2.0 μC), which means the electric force is repulsive—repulsive forces push the charges apart (each charge experiences force away from the other), while attractive forces pull them together (each charge experiences force toward the other), but in both cases the magnitude is given by |F| = k|q₁||q₂|/r². Choice B is correct because it accurately identifies the force as repulsive based on the signs of the charges. Choice C incorrectly claims there is no force because both charges are negative, missing the fundamental principle that like charges (both positive or both negative) repel each other—the force doesn't cancel, it's repulsive with magnitude F = k|q₁||q₂|/r². When applying Coulomb's Law: (1) convert all charges to Coulombs (1 μC = 10⁻⁶ C, 1 nC = 10⁻⁹ C) and distances to meters (1 cm = 0.01 m), (2) substitute carefully into F = k(q₁q₂)/r² keeping track of scientific notation, (3) determine direction from charge signs (opposite signs → attractive, same signs → repulsive), and (4) verify your result makes sense (typical classroom charges in μC at cm distances give forces in mN to N range, while atomic charges at atomic distances give enormous forces). Key relationships: (a) force proportional to each charge (double q₁ → double F), (b) force proportional to product of charges (double both → quadruple F), (c) force inversely proportional to r² (double r → F becomes 1/4), and (d) by Newton's Third Law, the force on q₁ due to q₂ equals the force on q₂ due to q₁ in magnitude but opposite in direction—both charges feel the same magnitude force whether attractive or repulsive.

This question tests understanding of Coulomb's Law, which describes the electric force between charged objects. Coulomb's Law states that the electric force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them: F = k(q₁q₂)/r², where k = 9.0 × 10⁹ N·m²/C² is Coulomb's constant, q₁ and q₂ are the charges in Coulombs, r is the distance between them in meters, and F is the force in Newtons—the force is attractive if the charges have opposite signs (one positive, one negative) and repulsive if they have the same sign (both positive or both negative). To find the electric force between charges q₁ = +4.0 μC = 4.0 × 10⁻⁶ C and q₂ = +2.0 μC = 2.0 × 10⁻⁶ C separated by distance r = 0.30 m, substitute into Coulomb's Law: F = k(q₁q₂)/r² = (9.0 × 10⁹ N·m²/C²)(4.0 × 10⁻⁶ C)(2.0 × 10⁻⁶ C)/(0.30 m)² = (9.0 × 10⁹)(8.0 × 10⁻¹²)/(0.09) = 72.0 × 10⁻³/0.09 = 0.80 N. Choice B is correct because it properly applies F = k(q₁q₂)/r² with correct values, units, and scientific notation. Choice A (8.0 N) has an error in the power of 10 in scientific notation (reports 10⁰ instead of 10⁻¹), likely from incorrectly adding or subtracting exponents when multiplying or dividing powers of 10 during the calculation. When applying Coulomb's Law: (1) convert all charges to Coulombs (1 μC = 10⁻⁶ C, 1 nC = 10⁻⁹ C) and distances to meters (1 cm = 0.01 m), (2) substitute carefully into F = k(q₁q₂)/r² keeping track of scientific notation, (3) determine direction from charge signs (opposite signs → attractive, same signs → repulsive), and (4) verify your result makes sense (typical classroom charges in μC at cm distances give forces in mN to N range, while atomic charges at atomic distances give enormous forces). The inverse square law means electric force decreases rapidly with distance: double the separation and force drops to 1/4, triple it and force drops to 1/9—this is why charged objects interact strongly when close but the force becomes negligible at larger distances.

This question tests understanding of Coulomb's Law, which describes the electric force between charged objects. Coulomb's Law states that the electric force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them: F = k(q₁q₂)/r², where k = 9.0 × 10⁹ N·m²/C² is Coulomb's constant, q₁ and q₂ are the charges in Coulombs, r is the distance between them in meters, and F is the force in Newtons—the force is attractive if the charges have opposite signs (one positive, one negative) and repulsive if they have the same sign (both positive or both negative). The charges in this scenario have the same signs (both negative), which means the electric force is repulsive—repulsive forces push the charges apart (each charge experiences force away from the other), while attractive forces pull them together (each charge experiences force toward the other), but in both cases the magnitude is given by |F| = k|q₁||q₂|/r². Choice B is correct because it accurately identifies the force as repulsive based on the signs of the charges. Choice A incorrectly identifies the force as attractive when it should be repulsive, confusing the rule: opposite sign charges attract (pull together), same sign charges repel (push apart). When applying Coulomb's Law: (1) convert all charges to Coulombs (1 μC = 10⁻⁶ C, 1 nC = 10⁻⁹ C) and distances to meters (1 cm = 0.01 m), (2) substitute carefully into F = k(q₁q₂)/r² keeping track of scientific notation, (3) determine direction from charge signs (opposite signs → attractive, same signs → repulsive), and (4) verify your result makes sense (typical classroom charges in μC at cm distances give forces in mN to N range, while atomic charges at atomic distances give enormous forces). Key relationships: (a) force proportional to each charge (double q₁ → double F), (b) force proportional to product of charges (double both → quadruple F), (c) force inversely proportional to r² (double r → F becomes 1/4), and (d) by Newton's Third Law, the force on q₁ due to q₂ equals the force on q₂ due to q₁ in magnitude but opposite in direction—both charges feel the same magnitude force whether attractive or repulsive.

This question tests understanding of Coulomb's Law, which describes the electric force between charged objects. Coulomb's Law states that the electric force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them: F = k(q₁q₂)/r², where k = 9.0 × 10⁹ N·m²/C² is Coulomb's constant, q₁ and q₂ are the charges in Coulombs, r is the distance between them in meters, and F is the force in Newtons—the force is attractive if the charges have opposite signs (one positive, one negative) and repulsive if they have the same sign (both positive or both negative). When the distance between two charges changes from r₁ = 0.20 m to r₂ = 0.40 m, the force changes by the ratio F₂/F₁ = (r₁/r₂)² = (0.20/0.40)² = (0.5)² = 0.25, so the force becomes one-quarter of its original value. Choice C is correct because it correctly applies the inverse square relationship showing force becomes 1/4 when distance doubles. Choice B incorrectly claims the force becomes 1/2 when distance doubles, missing the inverse square law—when distance doubles, force becomes 1/4 (not 1/2), and when distance triples, force becomes 1/9 (not 1/3). When applying Coulomb's Law: (1) convert all charges to Coulombs (1 μC = 10⁻⁶ C, 1 nC = 10⁻⁹ C) and distances to meters (1 cm = 0.01 m), (2) substitute carefully into F = k(q₁q₂)/r² keeping track of scientific notation, (3) determine direction from charge signs (opposite signs → attractive, same signs → repulsive), and (4) verify your result makes sense (typical classroom charges in μC at cm distances give forces in mN to N range, while atomic charges at atomic distances give enormous forces). The inverse square law means electric force decreases rapidly with distance: double the separation and force drops to 1/4, triple it and force drops to 1/9—this is why charged objects interact strongly when close but the force becomes negligible at larger distances (unlike gravity which, though also following inverse square law, acts over astronomical distances because masses are so much larger and always attractive).

This question tests understanding of Coulomb's Law, which describes the electric force between charged objects. Coulomb's Law states that the electric force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them: F = k(q₁q₂)/r², where k = 9.0 × 10⁹ N·m²/C² is Coulomb's constant, q₁ and q₂ are the charges in Coulombs, r is the distance between them in meters, and F is the force in Newtons—the force is attractive if the charges have opposite signs (one positive, one negative) and repulsive if they have the same sign (both positive or both negative). When the distance between two charges changes from r₁ = 0.40 m to r₂ = 0.80 m (doubles), the force changes by the ratio F₂/F₁ = (r₁/r₂)² = (0.40/0.80)² = (1/2)² = 1/4, so the force becomes one-quarter of its original value. This demonstrates the inverse square relationship: doubling distance doesn't halve the force, it quarters it. Choice C is correct because it correctly applies the inverse square relationship showing force becomes 1/4 when distance doubles. Choice B incorrectly claims the force becomes 1/2 as large when distance doubles, missing the inverse square law—when distance doubles, force becomes 1/4 (not 1/2), and when distance triples, force becomes 1/9 (not 1/3). When applying Coulomb's Law: (1) convert all charges to Coulombs (1 μC = 10⁻⁶ C, 1 nC = 10⁻⁹ C) and distances to meters (1 cm = 0.01 m), (2) substitute carefully into F = k(q₁q₂)/r² keeping track of scientific notation, (3) determine direction from charge signs (opposite signs → attractive, same signs → repulsive), and (4) verify your result makes sense (typical classroom charges in μC at cm distances give forces in mN to N range, while atomic charges at atomic distances give enormous forces). The inverse square law means electric force decreases rapidly with distance: double the separation and force drops to 1/4, triple it and force drops to 1/9—this is why charged objects interact strongly when close but the force becomes negligible at larger distances (unlike gravity which, though also following inverse square law, acts over astronomical distances because masses are so much larger and always attractive).