AP Physics 2 Quiz: Electric Charge And Electric Force

What this quiz covers

This quiz focuses on Electric Charge And Electric Force, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 2.

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. It is directed away from $Q_2$. (correct answer)
2. It is zero because both charges are negative.
3. It is directed upward regardless of placement.
4. It is directed toward $Q_2$.

Explanation: This problem tests understanding of electric charge and electric force. Since both Q₁ (-q) and Q₂ (-2q) are negative charges, they have the same sign, which means they repel each other according to Coulomb's law. The force on Q₁ due to Q₂ points away from Q₂, along the line connecting the two charges. The fact that Q₂ has twice the magnitude of Q₁ affects the force strength but not the direction—repulsion is determined by the like signs. A common misconception is thinking that two negative charges somehow attract or that the force is zero, but like charges always repel regardless of their magnitudes. Always determine force direction by the charge signs: like charges repel, opposite charges attract.

Question 2

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

Explanation: This problem tests understanding of electric charge and electric force. The original force between +2q and +q at distance r follows F₁ ∝ (2q)(q)/r². The new scenario asks about the force between +2q and the third charge +q at distance 2r, giving F₂ ∝ (2q)(q)/(2r)² = 2q²/4r² = (1/2)q²/r². Comparing to the original force, F₂ = (1/4)F₁. Both forces are repulsive since all charges are positive. Choice D (unchanged) is incorrect because it ignores that the distance has changed from r to 2r—failing to apply the inverse square law. Always identify which two charges and what separation distance you're analyzing.

Question 3

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

Explanation: This problem tests understanding of electric charge and electric force. According to Coulomb's law, the force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them (F ∝ q₁q₂/r²). When the separation doubles from r to 2r, the denominator becomes (2r)² = 4r², making the force 1/4 as large. The attractive force between the opposite charges (+q and -2q) still points along the line joining them, but with reduced magnitude. Choice A (1/2 as large) is incorrect because it assumes force is inversely proportional to distance rather than distance squared—a common misconception. Always remember that electric force follows an inverse square law with distance.

Question 4

1. magnitude $F$, directed toward the first sphere (correct answer)
2. magnitude $F$, directed away from the first sphere
3. magnitude $2F$, directed toward the first sphere
4. zero because the charges cancel

Explanation: This problem tests understanding of electric charge and electric force. Initially, two +q charges repel with force magnitude F. When the second sphere's charge changes from +q to -q, the charges now have opposite signs, so they attract instead of repel. According to Coulomb's law, the force magnitude remains F = kq²/d² because the absolute values of the charges haven't changed. However, the direction reverses: instead of pointing away from the first sphere (repulsion), the force now points toward the first sphere (attraction). A common misconception is thinking that changing the sign somehow affects the magnitude or that opposite charges cancel to give zero force. Always analyze both magnitude (from Coulomb's law) and direction (from charge signs) separately.

Question 5

1. twice as large.
2. four times as large. (correct answer)
3. eight times as large.
4. unchanged.

Explanation: This question tests understanding of electric charge and electric force. Coulomb's law states F = k|q₁||q₂|/r². Initially, with charges +2q and -q, the force magnitude is F = k(2q)(q)/r² = 2kq²/r². When both charges are doubled to +4q and -2q, the new force becomes F' = k(4q)(2q)/r² = 8kq²/r² = 4F. The force is attractive since the charges have opposite signs. A common misconception is thinking that doubling both charges only doubles the force, forgetting that force depends on the product of charges. Always multiply the new charge values together to find the new force.

Question 6

1. It is attractive and directed away from the $-3q$ sphere.
2. It is repulsive and directed away from the $-3q$ sphere. (correct answer)
3. It is repulsive and directed toward the $-3q$ sphere.
4. It is attractive and directed toward the $-3q$ sphere.

Explanation: This question tests understanding of electric charge and electric force. The force between two negative charges -q and -3q has magnitude F = k(q)(3q)/r² = 3kq²/r². Since both charges are negative (like charges), they repel each other. The force on the -q sphere is repulsive and directed away from the -3q sphere, pointing in the direction from -3q toward -q. Students who choose option A might incorrectly think negative charges always attract, confusing the rule for opposite charges with like charges. Always remember: like charges (same sign) repel, opposite charges (different signs) attract.

Question 7

1. It is repulsive, with magnitude $k\dfrac{9q^2}{r^2}$.
2. It is repulsive, with magnitude $k\dfrac{3q^2}{r^2}$. (correct answer)
3. It is attractive, with magnitude $k\dfrac{3q^2}{r^2}$.
4. It is attractive, with magnitude $k\dfrac{9q^2}{r^2}$.

Explanation: This question tests understanding of electric charge and electric force. By Coulomb's law, the force between charges +q and +3q has magnitude F = k|q||3q|/r² = 3kq²/r². Since both charges are positive, they have the same sign, making the force repulsive. The force on the +q sphere points away from the +3q sphere. A common misconception is thinking the force magnitude should be 9kq²/r² by incorrectly squaring the 3 in the charge value. Always apply Coulomb's law correctly: multiply the charge magnitudes together without additional squaring.

Question 8

1. twice as large
2. one-half as large
3. one-fourth as large (correct answer)
4. four times as large

Explanation: This problem tests understanding of electric charge and electric force. According to Coulomb's law, the electrostatic force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them: F = k|q₁||q₂|/r². When the separation doubles from 0.30 m to 0.60 m, the distance in the denominator is squared, so the force becomes F = k|q₁||q₂|/(2r)² = k|q₁||q₂|/4r². This means the new force is one-fourth the original force. A common misconception is thinking the force is inversely proportional to distance (not distance squared), which would incorrectly suggest the force becomes half as large. Always remember that electrostatic force follows an inverse-square relationship with distance.

Question 9

1. It is repulsive and has magnitude $k(2q)^2/r^2$.
2. It is repulsive and has magnitude $k(2q)(q)/r^2$. (correct answer)
3. It is attractive and has magnitude $k(q)^2/r^2$.
4. It is attractive and has magnitude $k(2q)(q)/r^2$.

Explanation: This question tests understanding of electric charge and electric force. Using Coulomb's law F = kq₁q₂/r², the force between charges +2q and +q is F = k(2q)(q)/r² = 2kq²/r². Since both charges are positive, like charges repel, making the force repulsive and directed away from the +q charge. The force on the +2q sphere has the same magnitude as the force on the +q sphere by Newton's third law. Students who choose option A might forget that like charges repel, incorrectly thinking all electric forces are attractive. Always check the signs of both charges to determine whether the force is attractive or repulsive.

Question 10

1. attractive with the same magnitude. (correct answer)
2. attractive with half the magnitude.
3. repulsive with half the magnitude.
4. repulsive with the same magnitude.

Explanation: This question tests understanding of electric charge and electric force. Initially with two +q charges, the force is repulsive with magnitude F = kq²/r². When one charge changes to -q, we now have opposite charges (+q and -q), so the force becomes attractive. The magnitude remains F = k|q||q|/r² = kq²/r², which is the same as before. Only the direction changes from repulsive to attractive. Students who choose option C might think changing the sign somehow reduces the force magnitude, not realizing that Coulomb's law uses the absolute values of charges for magnitude. Always remember that changing charge signs affects force direction but not magnitude.

Question 11

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

Explanation: This problem tests understanding of electric charge and electric force. The original force between +q and +4q is F₁ ∝ (q)(4q) = 4q². When +4q is replaced with +2q, the new force becomes F₂ ∝ (q)(2q) = 2q². Comparing these, F₂ = (2q²)/(4q²) × F₁ = (1/2)F₁. Both configurations produce repulsive forces since all charges are positive. Choice D (unchanged) is incorrect because it ignores that one charge magnitude has changed—perhaps assuming only separation affects force magnitude. Always account for changes in both charge magnitudes when comparing electric forces.

Question 12

1. one-fourth as large and attractive (correct answer)
2. one-half as large and attractive
3. twice as large and attractive
4. one-fourth as large and repulsive

Explanation: This problem tests electric charge and electric force. Initially, sphere A (+q) and sphere B (-2q) attract each other with force F = k|q||2q|/r² = 2kq²/r², where r = 0.30 m. When the distance doubles to 2r, the new force becomes F' = 2kq²/(2r)² = 2kq²/4r² = F/4, making it one-fourth as large. Since opposite charges attract, the force remains attractive. A common misconception is thinking force varies linearly with distance rather than with the inverse square. Always remember that electric force follows an inverse-square relationship with distance while maintaining the same direction based on charge signs.

Question 13

1. three times as large.
2. one-third as large.
3. one-ninth as large. (correct answer)
4. nine times as large.

Explanation: This question tests understanding of electric charge and electric force. By Coulomb's law, force is inversely proportional to the square of distance: F = k|q₁||q₂|/r². Initially, with charges -q and +2q at distance r, F = k(q)(2q)/r² = 2kq²/r². When distance triples to 3r, the new force becomes F' = 2kq²/(3r)² = 2kq²/9r² = F/9. The charges have opposite signs, so the force remains attractive. A common misconception is thinking force is inversely proportional to distance (not squared), which would incorrectly give one-third. Always remember the inverse square relationship: tripling distance reduces force by a factor of nine.

Question 14

1. four times as large.
2. twice as large. (correct answer)
3. half as large.
4. the same.

Explanation: This question tests understanding of electric charge and electric force. Using Coulomb's law, the force magnitude between charges Q₁ = +q and Q₂ = -2q is F = k|q||2q|/r² = 2kq²/r². The reference case of +q and -q gives F_ref = kq²/r². Therefore, the force with +q and -2q is 2 times as large as the reference case. The charges have opposite signs, making the force attractive. A common misconception is confusing the magnitude calculation with the direction determination. Always calculate force magnitude using absolute values of charges, then determine direction separately from charge signs.

Question 15

1. twice as large (correct answer)
2. one-half as large
3. the same
4. four times as large

Explanation: This problem tests understanding of electric charge and electric force. When both spheres have charge +q, the force is F = kq²/d². With sphere A having +q and sphere B having +2q, the new force is F = k(q)(2q)/d² = 2kq²/d². Comparing the new force to the original: (2kq²/d²)/(kq²/d²) = 2, so the force is twice as large. The charges still repel since both are positive, but the magnitude doubles because one charge doubled. A common misconception is thinking the force increases by a factor of 3 (adding the charges) or 4 (squaring the change), but Coulomb's law shows force is proportional to the product of charges. Always calculate force using the product of the actual charge values, not their sum or squares.

Question 16

1. It is unchanged because the signs are opposite.
2. It becomes $3$ times as large.
3. It becomes $6$ times as large.
4. It becomes $9$ times as large. (correct answer)

Explanation: This problem tests understanding of electric charge and electric force. Coulomb's law states that force is proportional to the product of the charges: F ∝ q₁q₂. When both charges are tripled (+q becomes +3q and -q becomes -3q), the force becomes proportional to (3q)(3q) = 9q², making it 9 times larger. The attractive force between opposite charges maintains the same direction along the line joining them. Choice B (6 times) is incorrect because it assumes force scales linearly with the sum of charge magnitudes rather than their product—a common algebraic error. Always multiply the charge magnitudes when applying Coulomb's law, regardless of their signs.

Question 17

1. one-half as large
2. one-fourth as large
3. twice as large (correct answer)
4. four times as large

Explanation: This problem tests understanding of electric charge and electric force. Initially, the force between two +q charges separated by distance d is F = kq²/d². When one charge is replaced by +2q while keeping the distance constant, the new force becomes F = k(q)(2q)/d² = 2kq²/d². Comparing the new force to the original: (2kq²/d²)/(kq²/d²) = 2, so the new force is twice as large. A common misconception is squaring the charge ratio, thinking that doubling one charge quadruples the force, but Coulomb's law shows force is directly proportional to the product of charges. Always calculate the force ratio by comparing the complete Coulomb's law expressions before and after the change.

Question 18

1. the same
2. half as large
3. twice as large (correct answer)
4. four times as large

Explanation: This problem tests electric charge and electric force. Originally, two +Q charges repel with force F = kQ²/d². When one charge becomes +2Q, the new force is F' = k(Q)(2Q)/d² = 2kQ²/d² = 2F, making it twice as large. The force remains repulsive since both charges are positive. Students often mistakenly think replacing one charge affects both charges in the calculation, leading to answer D (four times). Always calculate the force using the actual charges present, not assuming symmetric changes.

Question 19

1. twice as large.
2. four times as large. (correct answer)
3. eight times as large.
4. unchanged.

Explanation: This question tests understanding of electric charge and electric force. Initially, the force magnitude is F₁ = k(3q)(q)/r² = 3kq²/r². When both charges are doubled to +6q and -2q, the new force magnitude is F₂ = k(6q)(2q)/r² = 12kq²/r². Comparing these, F₂ = 4F₁, so the force becomes four times as large. The force remains attractive since the charges have opposite signs. Students who choose option A might think doubling both charges only doubles the force, not realizing force depends on the product of charges. Always multiply the charge values together when calculating how force changes.

Question 20

1. one-half as large.
2. four times as large. (correct answer)
3. twice as large.
4. eight times as large.

Explanation: This question tests understanding of electric charge and electric force. For charges +q and -q at distance r, F₁ = kq²/r². For charges +2q and -2q at the same distance r, F₂ = k(2q)(2q)/r² = 4kq²/r². Comparing these, F₂ = 4F₁, so the force is four times as large. Both cases involve opposite charges, so the force remains attractive. Students who choose option B might think doubling each charge only doubles the total force, not realizing that force depends on the product of charges. Always multiply the charge magnitudes together to find how force scales.