Physics Quiz: Apply Coulombs Law

What this quiz covers

This quiz focuses on Apply Coulombs Law, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $0.18\,\text{N}$
2. $1.8\,\text{N}$ (correct answer)
3. $5.4\,\text{N}$
4. $16.2\,\text{N}$

Explanation: 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).

Question 2

1. $0.405\,\text{N}$
2. $4.05\,\text{N}$ (correct answer)
3. $40.5\,\text{N}$
4. $0.081\,\text{N}$

Explanation: 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.

Question 3

1. $8.0\,\text{N}$
2. $0.80\,\text{N}$ (correct answer)
3. $0.080\,\text{N}$
4. $8.0\times10^{-3}\,\text{N}$

Explanation: 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.

Question 4

1. Attractive, because the charges have opposite signs.
2. Repulsive, because the charges have the same sign. (correct answer)
3. Attractive, because the force is proportional to $r^2$.
4. Repulsive, because the force is proportional to $r^2$.

Explanation: 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.

Question 5

1. $F_2 = 2F_1$
2. $F_2 = \tfrac{1}{2}F_1$
3. $F_2 = \tfrac{1}{4}F_1$ (correct answer)
4. $F_2 = \tfrac{1}{8}F_1$

Explanation: 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 to r₂, 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. 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 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). 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.

Question 6

1. $0.80\,\text{N}$ (correct answer)
2. $8.0\,\text{N}$
3. $0.080\,\text{N}$
4. $80\,\text{N}$

Explanation: 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).

Question 7

1. It becomes $2$ times as large.
2. It becomes $\tfrac{1}{2}$ as large.
3. It becomes $\tfrac{1}{4}$ as large. (correct answer)
4. It becomes $\tfrac{1}{8}$ as large.

Explanation: 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).

Question 8

1. $0.87\,\mu\text{C}$
2. $3.47\,\mu\text{C}$ (correct answer)
3. $0.87\,\text{C}$
4. $34.7\,\mu\text{C}$

Explanation: 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).

Question 9

1. $5.18\,\text{N}$ (correct answer)
2. $0.323\,\text{N}$
3. $1.30\,\text{N}$
4. $32.4\,\text{N}$

Explanation: 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).

Question 10

1. $2.88\,\text{N}$
2. $1.44\,\text{N}$
3. $0.36\,\text{N}$
4. $0.18\,\text{N}$ (correct answer)

Explanation: 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.

Question 11

1. It doubles (factor of 2). (correct answer)
2. It quadruples (factor of 4).
3. It is halved (factor of 1/2).
4. It becomes one-fourth (factor of 1/4).

Explanation: 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 one charge doubles from q₁ = +3.0 μC to +6.0 μC while q₂ and r remain unchanged, the force changes by the ratio F₂/F₁ = (q₁,new / q₁,old) = 6.0 / 3.0 = 2, since force is directly proportional to each charge, so the magnitude doubles. Choice A is correct because it correctly applies the proportional relationship showing force doubles when one charge doubles. Choice C incorrectly claims the force is halved, missing the direct proportionality to the product of charges—doubling one charge doubles the force, not halves it. 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.

Question 12

1. $0.090\,\text{m}$
2. $0.284\,\text{m}$ (correct answer)
3. $0.900\,\text{m}$
4. $2.84\,\text{m}$

Explanation: 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).

Question 13

1. $4$
2. $2$
3. $\tfrac{1}{2}$
4. $\tfrac{1}{4}$ (correct answer)

Explanation: 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.

Question 14

1. $2.25\times10^{-5}\,\text{N}$ (correct answer)
2. $2.25\times10^{-3}\,\text{N}$
3. $2.25\times10^{-7}\,\text{N}$
4. $2.25\times10^{-1}\,\text{N}$

Explanation: 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).

Question 15

1. $5.6\,\text{N}$, attractive
2. $5.6\,\text{N}$, repulsive (correct answer)
3. $0.56\,\text{N}$, repulsive
4. $56\,\text{N}$, repulsive

Explanation: 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).

Question 16

1. $0.90\,\text{N}$ (correct answer)
2. $9.0\,\text{N}$
3. $90\,\text{N}$
4. $0.090\,\text{N}$

Explanation: 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 = 8.0 × 10⁻⁶ C and q₂ = -2.0 × 10⁻⁶ C separated by distance r = 0.40 m (converted from 40 cm), substitute into Coulomb's Law: F = k|q₁q₂|/r² = (9.0 × 10⁹ N·m²/C²)(1.6 × 10⁻¹¹)/(0.40)² = (9.0 × 10⁹)(1.6 × 10⁻¹¹)/0.16 = (9.0 × 10⁹)(1.0 × 10⁻¹⁰) = 0.90 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. Choice B forgets to convert centimeters to meters before calculating, using 40 cm = 40 m directly in the formula and getting a result that's off by a factor of 10⁴ (since (40/0.4)² = 10⁴). 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).

Question 17

1. $2.25\,\text{N}$, attractive (correct answer)
2. $2.25\,\text{N}$, repulsive
3. $0.45\,\text{N}$, attractive
4. $45\,\text{N}$, attractive

Explanation: 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 × 10⁻⁶ C and q₂ = -5.0 × 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)(1.0 × 10⁻¹¹)/(0.20)² = (9.0 × 10⁹)(1.0 × 10⁻¹¹)/0.04 = (9.0 × 10⁹)(2.5 × 10⁻¹⁰) = 2.25 N, and since the charges have opposite signs, the force is attractive. Choice A is correct because it accurately identifies the force as attractive based on the signs of the charges and properly applies F = k|q₁q₂|/r² with correct values, units, and scientific notation. Choice B incorrectly identifies the force as repulsive when it should be attractive, 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.

Question 18

1. Attractive, because the charges have opposite signs.
2. Repulsive, because the charges have the same sign. (correct answer)
3. Attractive, because the force is proportional to $r^2$.
4. Repulsive, because the force is proportional to $r^2$.

Explanation: 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.

Question 19

1. It becomes $2$ times as large.
2. It becomes $\tfrac{1}{2}$ as large.
3. It becomes $\tfrac{1}{4}$ as large. (correct answer)
4. It becomes $\tfrac{1}{8}$ as large.

Explanation: 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).

Question 20

1. $0.87\,\mu\text{C}$
2. $3.47\,\mu\text{C}$ (correct answer)
3. $0.87\,\text{C}$
4. $34.7\,\mu\text{C}$

Explanation: 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).