Electric Charge and Electric Force - AP Physics 2

Two point charges are separated by distance $d$ : $Q_1=+q$ and $Q_2=-3q$ . Which statement best describes the direction of the force on $Q_1$ due to $Q_2$ ?
Answer: It is directed toward $Q_2$ . This problem tests understanding of electric charge and electric force. Since Q₁ is positive (+q) and Q₂ is negative (-3q), the charges have opposite signs, which means they attract each other according to Coulomb's law. The force on Q₁ due to Q₂ points toward Q₂, along the line connecting the two charges. The magnitude of charge Q₂ being three times larger than Q₁ affects the force magnitude but not the direction—attraction is determined solely by the opposite signs. A common misconception is thinking unequal charges somehow affect the direction or that the force has no direction, but Coulomb forces always act along the line joining the charges. Always determine force direction by considering whether charges attract (opposite signs) or repel (same signs).
Two point charges $+q$ and $-q$ are separated by distance $r$ . If the distance becomes $2r$ , which statement best describes the force magnitude on either charge?
Answer: It becomes one-fourth as large. This question tests understanding of electric charge and electric force. According to Coulomb's law, the force between two charges is inversely proportional to the square of the distance between them: F = kq₁q₂/r². When the distance doubles from r to 2r, the denominator becomes (2r)² = 4r², making the force F/4 or one-fourth as large. The force remains attractive since opposite charges (+q and -q) attract each other. Students who choose option B might incorrectly think force is inversely proportional to distance (not distance squared), leading them to believe doubling distance halves the force. Always remember that electric force follows an inverse-square relationship with distance.
Two point charges $Q_1=-q$ and $Q_2=-2q$ are separated by distance $d$ . Which statement best describes the direction of the force on $Q_1$ due to $Q_2$ ?
Answer: It is directed away from $Q_2$ . 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.
Charges $+q$ and $-2q$ are separated by distance $r$ . Compared to the force magnitude when the charges are $+q$ and $-q$ at $r$ , the force is
Answer: twice as large. This question tests understanding of electric charge and electric force. For charges +q and -q at distance r, the force magnitude is F₁ = kq²/r². For charges +q and -2q at the same distance r, the force magnitude is F₂ = k(q)(2q)/r² = 2kq²/r². Comparing these, F₂ = 2F₁, so the force is twice as large. The force remains attractive in both cases since opposite charges attract. Students who choose option B might think changing one charge doesn't affect the force, forgetting that force depends on the product of both charges. Always calculate the product of the charges when comparing electric forces.
Two charges $+2q$ and $-q$ are separated by $0.20\ \text{m}$ . If the separation becomes $0.10\ \text{m}$ , the force magnitude is
Answer: four times as large This problem tests electric charge and electric force. Initially, charges +2q and -q attract with force F = k(2q)(q)/(0.20)² = 2kq²/0.04. When separation becomes 0.10 m (halved), the new force is F' = 2kq²/(0.10)² = 2kq²/0.01 = 4F, making it four times as large. The force remains attractive since the charges have opposite signs. A common mistake is thinking halving distance doubles the force (answer C), forgetting the inverse-square relationship. Always square the distance ratio: (1/2)² = 1/4, so force increases by factor of 4.
A charge $+q$ and a charge $-q$ are separated by distance $r$ . If the distance is reduced to $\tfrac{1}{2}r$ , compared to before, the force magnitude becomes
Answer: four times as large. This question tests understanding of electric charge and electric force. According to Coulomb's law, electric force is inversely proportional to the square of the distance: F = kq₁q₂/r². When distance is reduced from r to r/2, the new force becomes F' = kq²/(r/2)² = kq²/(r²/4) = 4kq²/r² = 4F. The charges +q and -q have opposite signs, so the force remains attractive. A common misconception is thinking that halving the distance doubles the force, forgetting the inverse square relationship. Always square the distance ratio when calculating how force changes with distance.
Two identical small spheres are separated by $r$ . One has charge $+q$ and the other $+3q$ . Which statement best describes the force on the $+q$ sphere due to the $+3q$ sphere?
Answer: It is repulsive, with magnitude $k\dfrac{3q^2}{r^2}$ . 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.
Two point charges $+q$ and $-q$ are separated by distance $r$ . Compared to the force magnitude on either charge, what happens if both charges are tripled to $+3q$ and $-3q$ while $r$ stays the same?
Answer: It becomes $9$ times as large. 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.
Two point charges $+q$ and $-2q$ are separated by distance $r$ . If the magnitude of each charge is halved and $r$ stays the same, the force magnitude becomes
Answer: one-fourth as large. This question tests understanding of electric charge and electric force. Initially, with charges +q and -2q, the force magnitude is F = k(q)(2q)/r² = 2kq²/r². When both charge magnitudes are halved to +q/2 and -q, the new force becomes F' = k(q/2)(q)/r² = kq²/2r² = F/4. The force remains attractive since the charges keep opposite signs. A common misconception is thinking that halving both charges halves the force, forgetting that force depends on the product of charges. Always multiply the new charge values together and compare to the original product.
Charge $+2q$ and charge $+q$ are separated by distance $r$ . A third charge $+q$ is placed so it is a distance $2r$ from $+2q$ . Compared to the original force magnitude between $+2q$ and $+q$ at $r$ , what is the new force magnitude?
Answer: It becomes $\tfrac{1}{4}$ as large. 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.
Two small charged spheres have charges $+q$ and $-q$ separated by $d$ . If both charges are tripled while the distance remains $d$ , the force magnitude becomes
Answer: nine times as large This problem tests understanding of electric charge and electric force. Initially, the force between charges +q and -q separated by distance d is F = kq²/d². When both charges are tripled to +3q and -3q while keeping distance d constant, the new force becomes F = k(3q)(3q)/d² = 9kq²/d². Comparing the new force to the original: (9kq²/d²)/(kq²/d²) = 9, so the new force is nine times as large. A common misconception is thinking that tripling both charges only triples the force, forgetting that force depends on the product of charges, so both factors of 3 multiply together. Always remember that when both charges change by the same factor, the force changes by the square of that factor.
Two point charges $+q$ and $-q$ are separated by distance $r$ . Compared to the force magnitude at $r$ , what is the force magnitude if the separation is increased to $3r$ ?
Answer: It becomes $\tfrac{1}{9}$ as large. This problem tests understanding of electric charge and electric force. According to Coulomb's law, force is inversely proportional to the square of separation: F ∝ 1/r². When distance increases from r to 3r, the new force becomes F_new ∝ 1/(3r)² = 1/9r². This makes the force 1/9 as large as the original. The attractive force between opposite charges (+q and -q) maintains its direction along the connecting line. Choice A (1/3 as large) is incorrect because it assumes inverse proportionality to distance rather than distance squared—a fundamental misconception about Coulomb's law. Always remember that tripling the distance reduces electric force by a factor of nine.
Charges $Q_A=+3q$ and $Q_B=-q$ are separated by distance $r$ . Compared to $+q$ and $-q$ at the same $r$ , the force magnitude is
Answer: three times as large. This question tests understanding of electric charge and electric force. Coulomb's law shows that force magnitude is proportional to the product of the charge magnitudes: F = k|q₁||q₂|/r². For charges +3q and -q, the force magnitude is F = k(3q)(q)/r² = 3kq²/r². Comparing this to the reference case of +q and -q, which gives F_ref = kq²/r², we see the new force is 3 times larger. A common misconception is thinking the force depends on the sum or difference of charges rather than their product. Always multiply the charge magnitudes together when calculating electric force.
Two small spheres carry charges $+2q$ and $+q$ and are separated by distance $r$ . Which statement best describes the force on the $+2q$ sphere?
Answer: It is repulsive and has magnitude $k(2q)(q)/r^2$ . 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.
A charge $+Q$ is fixed, and a charge $+q$ is placed a distance $d$ to its right. Which statement best describes the direction of the force on $+q$ due to $+Q$ ?
Answer: To the right, away from $+Q$ . This problem tests understanding of electric charge and electric force. Like charges repel each other, so the positive charge +q experiences a repulsive force away from the positive charge +Q. Since +Q is to the left of +q, the force on +q points to the right, away from +Q. The force acts along the line joining the two charges, following Newton's third law. Choice D (zero force) is incorrect because it assumes like charges don't interact—a misconception that confuses the net force on a charge with the individual forces between charges. Always determine force direction by considering whether charges attract (opposite signs) or repel (same signs).
Two point charges $+2q$ and $-q$ are separated by $d$ . If the distance is reduced to $\tfrac{1}{2}d$ with charges unchanged, the force magnitude becomes
Answer: four times as large This problem tests understanding of electric charge and electric force. Initially, the force between charges +2q and -q separated by distance d is F = k(2q)(q)/d² = 2kq²/d². When the distance is reduced to d/2, the new force becomes F = k(2q)(q)/(d/2)² = 2kq²/(d²/4) = 8kq²/d². Comparing the new force to the original: (8kq²/d²)/(2kq²/d²) = 4, so the new force is four times as large. A common misconception is thinking that halving the distance doubles the force, forgetting that force depends on 1/r², so halving r quadruples the force. Always square the distance ratio when calculating how force changes with distance.
Two small charged objects have charges $-q$ and $-4q$ separated by distance $r$ . Which statement best describes the force on the $-q$ charge due to the $-4q$ charge?
Answer: It is repulsive, with magnitude $k\dfrac{4q^2}{r^2}$ . This question tests understanding of electric charge and electric force. By Coulomb's law, the force between two charges depends on both the product of charges and their signs. Since both charges are negative (-q and -4q), they have the same sign, making the force repulsive. The magnitude is F = k|q₁||q₂|/r² = k|-q||-4q|/r² = 4kq²/r². A common misconception is thinking that two negative charges attract each other, similar to how a positive and negative charge attract. Always remember that like charges (same sign) repel, while opposite charges attract.
Charges $+Q$ and $+Q$ are separated by distance $d$ . A third charge $+2Q$ replaces one of them at the same separation. Compared to the original force magnitude, the new force magnitude is
Answer: twice as large 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.
Charges $Q_1=-q$ and $Q_2=+2q$ are separated by distance $r$ . If the distance is tripled to $3r$ , compared to before, the force magnitude becomes
Answer: one-ninth as large. 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.
Two small spheres are $0.30\,\text{m}$ apart: sphere 1 has charge $+q$ and sphere 2 has charge $-q$ . Without changing the charges, the separation is doubled to $0.60\,\text{m}$ . Compared to the original force magnitude, the new electrostatic force magnitude is
Answer: one-fourth as large 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.

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