Apply Coulomb's Law - Physics

Two charges are separated by $r = 0.60\,\text{m}$ . Charge 1 is $q_1 = -7.0\,\mu\text{C}$ and charge 2 is $q_2 = -2.0\,\mu\text{C}$ . Using Coulomb's Law with $k = 9.0 \times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2$ , which statement best describes the force between the charges?
Answer: Repulsive, because the charges have the same sign. 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.
Two identical pith balls are charged and hung so they repel each other. Treat them as point charges with $q_1 = +3.0\,\mu\text{C}$ and $q_2 = +1.0\,\mu\text{C}$ . Their centers are $r = 25\,\text{cm}$ apart. Using $k = 9.0\times10^9\,\text{N}\cdot\text{m}^2/\text{C}^2$ , what is the magnitude of the electric force between them (in N)?
Answer: $0.432\,\text{N}$ 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 = +3.0 μC = 3.0 × 10⁻⁶ C and q₂ = +1.0 μC = 1.0 × 10⁻⁶ C separated by distance r = 25 cm = 0.25 m, substitute into Coulomb's Law: F = k(q₁q₂)/r² = (9.0 × 10⁹ N·m²/C²)(3.0 × 10⁻⁶ C)(1.0 × 10⁻⁶ C)/(0.25 m)² = (9.0 × 10⁹)(3.0 × 10⁻¹²)/(0.0625) = 27.0 × 10⁻³/0.0625 = 0.432 N. Choice B is correct because it properly applies F = k(q₁q₂)/r² with correct values, units, and scientific notation, including the conversion from centimeters to meters. Choice A (4.32 N) has an error in the power of 10 in scientific notation (reports 10 times the correct value), 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.
A student measures the force between two charged spheres and finds the magnitude is $F = 12\,\text{N}$ when they are separated by $r_1 = 0.15\,\text{m}$ . If the charges are unchanged, what will the force be when the separation is increased to $r_2 = 0.30\,\text{m}$ ? (Use Coulomb's law with $F\propto 1/r^2$ .)
Answer: $3\,\text{N}$ 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 = 0.15 m to r₂ = 0.30 m (doubles), the force changes by the ratio F₂/F₁ = (r₁/r₂)² = (0.15/0.30)² = (1/2)² = 1/4, so F₂ = F₁ × (1/4) = 12 N × (1/4) = 3 N. This demonstrates the inverse square relationship: doubling distance doesn't halve the force, it quarters it. Choice D is correct because it correctly applies the inverse square relationship showing that when distance doubles from 0.15 m to 0.30 m, the force becomes 1/4 of its original value: 12 N × (1/4) = 3 N. Choice C (6 N) incorrectly claims the force halves 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.
Two identical metal spheres are charged and placed on insulating stands. Sphere 1 has $q_1 = -5.0\,\mu\text{C}$ and sphere 2 has $q_2 = -1.0\,\mu\text{C}$ . Their centers are separated by $r = 0.40\,\text{m}$ . Which statement correctly describes the electric force between them?
Answer: Repulsive; the spheres push away from each other. 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, 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). 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).
Two identical small spheres are $r = 0.25\,\text{m}$ apart and experience a repulsive electric force of $F = 2.0\,\text{N}$ . If one sphere has charge $q_1 = +4.0\,\mu\text{C}$ , what is the magnitude of the other charge $|q_2|$ ? (Use $k = 9.0 \times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2$ .)
Answer: $3.47\,\mu\text{C}$ 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 q₂ given F = 2.0 N repulsive, q₁ = +4.0 μC, and r = 0.25 m, rearrange Coulomb's Law: |q₂| = F r² / (k |q₁|) = (2.0)(0.0625) / ((9.0 × 10⁹)(4.0 × 10⁻⁶)) = 0.125 / 3.6 × 10⁴ = 3.47 × 10⁻⁶ C = 3.47 μC. Choice B is correct because it properly applies the rearranged formula with correct values, units, and scientific notation. Choice D uses an incorrect value for Coulomb's constant k (like 9.0 × 10¹⁰ instead of 9.0 × 10⁹ N·m²/C²), causing the result to be off by a factor of 10. 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).
Two point charges are separated by $r = 12\,\text{cm}$ . The charges are $q_1 = +9.0\,\text{nC}$ and $q_2 = -4.0\,\text{nC}$ . Using $k = 9.0\times10^9\,\text{N}\cdot\text{m}^2/\text{C}^2$ , what is the magnitude of the electric force between them (in N)?
Answer: $2.25\times10^{-5}\,\text{N}$ 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 = 9.0 × 10⁻⁹ C and q₂ = -4.0 × 10⁻⁹ C separated by distance r = 0.12 m (converted from 12 cm), substitute into Coulomb's Law: F = k|q₁q₂|/r² = (9.0 × 10⁹ N·m²/C²)(3.6 × 10⁻¹⁷)/(0.12)² = (9.0 × 10⁹)(3.6 × 10⁻¹⁷)/0.0144 = (9.0 × 10⁹)(2.5 × 10⁻¹⁵) = 2.25 × 10⁻⁵ N. Choice A is correct because it properly applies F = k|q₁q₂|/r² with correct values, units, and scientific notation, including converting centimeters to meters and nanocoulombs to coulombs. Choice C forgets to convert nanocoulombs to coulombs before calculating, using nC directly in the formula and getting a result that's off by a factor of 10⁶ (since $10^{-9}$ vs 1). 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).
A charged tape strip has charge $q_1 = +4.0\,\mu\text{C}$ and a second strip has charge $q_2 = -1.5\,\mu\text{C}$ . They are $r = 0.20\,\text{m}$ apart. Using $k = 9.0\times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2$ , what is the magnitude of the electric force between the strips?
Answer: $1.35\,\text{N}$ 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 = -1.5 μC = -1.5 × 10⁻⁶ C separated by distance r = 0.20 m, substitute into Coulomb's Law: F = k|q₁q₂|/r² = (9.0 × 10⁹ N·m²/C²)|（4.0 × 10⁻⁶ C)(-1.5 × 10⁻⁶ C)|/(0.20 m)² = (9.0 × 10⁹)(6.0 × 10⁻¹²)/(0.04) = (54.0 × 10⁻³)/(0.04) = 1.35 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 0.135 instead of 1.35), 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). 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.
A charge $q_2 = +4.0\,\mu\text{C}$ is placed $r = 0.50\,\text{m}$ to the right of a charge $q_1 = +1.0\,\mu\text{C}$ . Using Coulomb's Law ( $k = 9.0\times10^9\,\text{N}\cdot\text{m}^2/\text{C}^2$ ), what is the magnitude and direction of the force on $q_1$ due to $q_2$ ?
Answer: $0.144\,\text{N}$ to the left 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 on q₁ = +1.0 μC = 1.0 × 10⁻⁶ C due to q₂ = +4.0 μC = 4.0 × 10⁻⁶ C separated by distance r = 0.50 m, substitute into Coulomb's Law: F = k(q₁q₂)/r² = (9.0 × 10⁹ N·m²/C²)(1.0 × 10⁻⁶ C)(4.0 × 10⁻⁶ C)/(0.50 m)² = (9.0 × 10⁹)(4.0 × 10⁻¹²)/(0.25) = 36.0 × 10⁻³/0.25 = 0.144 N. Since both charges are positive, they repel each other, so q₁ experiences a force to the left (away from q₂). Choice A is correct because it properly applies F = k(q₁q₂)/r² with correct values, units, and scientific notation, and correctly identifies the force direction as to the left since same-sign charges repel. Choice B incorrectly identifies the force as to the right when it should be to the left, confusing the rule: same sign charges repel (push apart), so q₁ is pushed away from q₂ (to the left). 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.
Two point charges have $q_1 = +9.0\,\mu\text{C}$ and $q_2 = -1.0\,\mu\text{C}$ . Using Coulomb's law with $k = 9.0\times10^9\,\text{N}\cdot\text{m}^2/\text{C}^2$ , what separation distance $r$ (in meters) would produce an electric force magnitude of $F = 1.0\,\text{N}$ between them?
Answer: $0.284\,\text{m}$ 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 separation r for charges q₁ = +9.0 μC and q₂ = -1.0 μC to produce F = 1.0 N, rearrange Coulomb's Law: r² = k|q₁q₂|/F, so r = √[ (9.0 × 10⁹)(9.0 × 10⁻⁶)(1.0 × 10⁻⁶) / 1.0 ] = √[ (9.0 × 10⁹)(9.0 × 10⁻¹²) ] = √(8.1 × 10⁻²) = 0.284 m. Choice B is correct because it properly applies the rearranged F = k|q₁q₂|/r² with correct values, units, and scientific notation to solve for r. Choice A uses r in the denominator instead of r², calculating something like r = k|q₁q₂|/F, which misses the inverse square relationship and makes the distance too small. 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).
Two small charged spheres can be treated as point charges. Sphere A has charge $q_1 = +4.0\,\mu\text{C}$ and sphere B has charge $q_2 = +2.0\,\mu\text{C}$ . Their centers are separated by $r = 0.30\,\text{m}$ . Using Coulomb's law with $k = 9.0\times10^9\,\text{N}\cdot\text{m}^2/\text{C}^2$ , what is the magnitude of the electric force between them (in N)?
Answer: $0.80\,\text{N}$ 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 and q₂ = +2.0 μC separated by distance r = 0.30 m, first convert units: 4.0 μC = 4.0 × 10⁻⁶ C, 2.0 μC = 2.0 × 10⁻⁶ C; substitute into Coulomb's Law: F = (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)/0.090 = 7.2 × 10⁻² / 0.090 = 0.80 N. Choice A is correct because it properly applies F = k|q₁q₂|/r² with correct values, units, and scientific notation. Choice B uses an incorrect value for Coulomb's constant k (like 9.0 × 10¹⁰ instead of 9.0 × 10⁹ N·m²/C²), causing the result to be off by a factor of 10. 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).
Two small conducting spheres can be treated as point charges. Sphere A has charge $q_1 = +3.0\,\mu\text{C}$ and sphere B has charge $q_2 = +6.0\,\mu\text{C}$ . Their centers are separated by $r = 0.30\,\text{m}$ . Using Coulomb's Law with $k = 9.0\times10^9\,\text{N}\cdot\text{m}^2/\text{C}^2$ , what is the magnitude of the electric force between the spheres (in N)?
Answer: $1.8\,\text{N}$ 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).
Two identical metal spheres carry charges $q_1 = +4.0\,\mu\text{C}$ and $q_2 = +4.0\,\mu\text{C}$ . When their centers are $r_1 = 0.50\,\text{m}$ apart, the force is $F_1$ . They are moved so their centers are $r_2 = 1.0\,\text{m}$ apart, producing force $F_2$ . What is the ratio $F_2/F_1$ ?
Answer: $\tfrac{1}{4}$ 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 = 0.50 m to r₂ = 1.0 m, the force changes by the ratio F₂/F₁ = (r₁/r₂)²—for example, if distance doubles from r to 2r, the ratio is (r/2r)² = (1/2)² = 1/4, so the force becomes one-quarter of its original value. Choice D is correct because it correctly applies the inverse square relationship showing force becomes 1/4 when distance doubles. Choice C 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). 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.
Two small spheres have charges $q_1 = +7.0\,\text{nC}$ and $q_2 = -5.0\,\text{nC}$ and are separated by $r = 2.0\,\text{cm}$ . Using $k = 9.0\times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2$ and converting units correctly, what is the magnitude of the electric force between them?
Answer: $7.9\times 10^{-4}\,\text{N}$ 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 = +7.0 nC = 7.0 × 10⁻⁹ C and q₂ = -5.0 nC = -5.0 × 10⁻⁹ C separated by distance r = 2.0 cm = 0.02 m, substitute into Coulomb's Law: F = k|q₁q₂|/r² = (9.0 × 10⁹ N·m²/C²)|（7.0 × 10⁻⁹ C)(-5.0 × 10⁻⁹ C)|/(0.02 m)² = (9.0 × 10⁹)(35.0 × 10⁻¹⁸)/(0.0004) = (315.0 × 10⁻⁹)/(4 × 10⁻⁴) = 7.875 × 10⁻⁴ N ≈ 7.9 × 10⁻⁴ N. Choice A is correct because it properly applies F = k|q₁q₂|/r² with correct values, units, and scientific notation, including proper conversion of nanocoulombs to coulombs and centimeters to meters. Choice B has an error in the power of 10 in scientific notation (reports 10⁻⁶ instead of 10⁻⁴), 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). 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.
Two small charged spheres can be treated as point charges. Sphere A has charge $q_1 = +2.0\,\mu\text{C}$ and sphere B has charge $q_2 = +6.0\,\mu\text{C}$ . Their centers are separated by $r = 0.30\,\text{m}$ . Using Coulomb's Law with $k = 9.0\times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2$ , what is the magnitude of the electric force between them (in N)?
Answer: $1.2\,\text{N}$ 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 = 2.0 × 10⁻⁶ C and q₂ = +6.0 μC = 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)(2.0 × 10⁻⁶ C)(6.0 × 10⁻⁶ C)/(0.30 m)² = (9.0 × 10⁹)(12.0 × 10⁻¹²)/(0.09) = (108.0 × 10⁻³)/(0.09) = 1.2 N. Choice A is correct because it properly applies F = k(q₁q₂)/r² with correct values, units, and scientific notation. Choice D uses r in the denominator instead of r², calculating F = kq₁q₂/r, which misses the inverse square relationship and makes the force too large by a factor of r = 0.30. 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.
Two isolated spheres are separated by $r = 0.50\,\text{m}$ . Sphere 1 has $q_1 = -3.0\,\mu\text{C}$ and sphere 2 has $q_2 = -2.0\,\mu\text{C}$ . Which statement correctly describes the electric force between them?
Answer: Repulsive; the spheres push away from each other. 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.
Two point charges $q_1 = +3.0\,\mu\text{C}$ and $q_2 = +6.0\,\mu\text{C}$ are separated by $r_1 = 0.20\,\text{m}$ . The charges are then moved so the separation becomes $r_2 = 0.40\,\text{m}$ . By what factor does the electric force change? (Use the inverse-square relationship.)
Answer: It becomes $\tfrac{1}{4}$ as large 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 = (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 D is correct because it correctly applies the inverse square relationship showing force becomes 1/4 when distance doubles. Choice C incorrectly claims the force halves 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).
Two point charges are separated by $r = 0.50\,\text{m}$ . The magnitude of the electric force between them is $F = 0.72\,\text{N}$ . If the distance between the charges is doubled to $1.00\,\text{m}$ (charges unchanged), what is the new force magnitude?
Answer: $0.18\,\text{N}$ 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 = 0.50 m to r₂ = 1.00 m, the force changes by the ratio F₂/F₁ = (r₁/r₂)²—for example, if distance doubles from r to 2r, the ratio is (r/2r)² = (1/2)² = 1/4, so the force becomes one-quarter of its original value of 0.72 N, giving F₂ = 0.72 / 4 = 0.18 N; this demonstrates the inverse square relationship: doubling distance doesn't halve the force, it quarters it. Choice D is correct because it correctly applies the inverse square relationship showing force becomes 1/4 when distance doubles. Choice B incorrectly claims the force is halved 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). 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.
Two point charges are fixed in space at a separation of $r = 0.25\,\text{m}$ . The charges are $q_1 = +3.0\,\mu\text{C}$ and $q_2 = +12\,\mu\text{C}$ . Using $k = 9.0\times10^9\,\text{N}\cdot\text{m}^2/\text{C}^2$ , what is the magnitude of the electric force between them (in N)?
Answer: $5.18\,\text{N}$ 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: $F = k q_1 q_2 / r^2$ , where $k = 9.0 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2$ is Coulomb's constant, $q_1$ and $q_2$ 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 $q_1 = 3.0 \times 10^{-6} \, \text{C}$ and $q_2 = 12 \times 10^{-6} \, \text{C}$ separated by distance $r = 0.25 \, \text{m}$ , substitute into Coulomb's Law: $F = k |q_1 q_2| / r^2 = (9.0 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2)(3.6 \times 10^{-11}) / (0.25)^2 = (9.0 \times 10^9)(3.6 \times 10^{-11}) / 0.0625 = (9.0 \times 10^9)(5.76 \times 10^{-10}) = 5.18 \, \text{N}$ . Choice A is correct because it properly applies $F = k |q_1 q_2| / r^2$ with correct values, units, and scientific notation. Choice B has an error in the power of 10 in scientific notation (reports 0.323 instead of 5.18), 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 ( $1 \mu\text{C} = 10^{-6} \, \text{C}$ , $1 \, \text{nC} = 10^{-9} \, \text{C}$ ) and distances to meters ( $1 \, \text{cm} = 0.01 \, \text{m}$ ), (2) substitute carefully into $F = k(q_1 q_2) / r^2$ 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).
Two small spheres are separated by $r = 0.15\,\text{m}$ . Sphere 1 has $q_1 = -7.0\,\mu\text{C}$ and sphere 2 has $q_2 = -2.0\,\mu\text{C}$ . Using $k = 9.0\times10^9\,\text{N}\cdot\text{m}^2/\text{C}^2$ , what is the electric force between them (give magnitude and whether it is attractive or repulsive)?
Answer: $5.6\,\text{N}$ , 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 = -7.0 × 10⁻⁶ C and q₂ = -2.0 × 10⁻⁶ C separated by distance r = 0.15 m, substitute into Coulomb's Law: F = k|q₁q₂|/r² = (9.0 × 10⁹ N·m²/C²)(1.4 × 10⁻¹¹)/(0.15)² = (9.0 × 10⁹)(1.4 × 10⁻¹¹)/0.0225 = (9.0 × 10⁹)(6.222 × 10⁻¹⁰) = 5.6 N, and since the charges have the same signs, the force is repulsive. Choice B is correct because it accurately identifies the force as repulsive based on the signs of the charges and properly applies F = k|q₁q₂|/r² with correct values, units, and scientific notation. 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). 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).
Two point charges are separated by a distance $r$ . If $q_1$ is doubled while $q_2$ and $r$ stay the same, by what factor does the magnitude of the electric force change? (Use Coulomb's Law $F = k|q_1 q_2|/r^2$ .)
Answer: It becomes $2$ times as large. 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). Since F = k(q₁q₂)/r² and only q₁ is doubled while q₂ and r remain constant, the new force F_new = k(2q₁)(q₂)/r² = 2k(q₁q₂)/r² = 2F_original. This demonstrates that force is directly proportional to each charge: doubling one charge doubles the force. Choice B is correct because it properly applies F = k(q₁q₂)/r² recognizing that force is proportional to q₁. Choice A incorrectly claims the force becomes half as large when q₁ is doubled, getting the relationship backwards—force increases proportionally with charge, not inversely. 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.

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