Electric Charge and Electric Force
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AP Physics C: Electricity and Magnetism › Electric Charge and Electric Force
Electric charge is a conserved property of matter measured in coulombs (C); it comes in positive and negative types, and the total charge of an isolated system remains constant. Electric forces act at a distance: like charges repel, unlike charges attract, and the force magnitude depends on charge and separation. In space, two point charges $q_1=+4.0\times10^{-6}\ \text{C}$ and $q_2=+1.0\times10^{-6}\ \text{C}$ are fixed $0.30\ \text{m}$ apart. Coulomb’s Law gives the magnitude: $F=k\frac{|q_1q_2|}{r^2}$. A numerical check: $$F=(8.99\times10^{9})\frac{(4.0\times10^{-6})(1.0\times10^{-6})}{(0.30)^2}$$. If a third charge is later introduced, the principle of superposition states the net force on any charge equals the vector sum of forces from each other charge calculated separately.
How does the force between charges change if the distance is doubled?
It becomes $\tfrac{1}{2}$ as large
It becomes $\tfrac{1}{4}$ as large
It becomes $2$ times as large
It becomes $4$ times as large
Explanation
This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding how electric force varies with distance according to Coulomb's Law. Electric force follows an inverse square relationship with distance, meaning F ∝ 1/r², so doubling the distance causes the force to decrease by a factor of four. In this scenario, two positive charges initially separated by 0.30 m have their separation doubled to 0.60 m. When distance doubles (r → 2r), the force becomes F' = k|q₁q₂|/(2r)² = k|q₁q₂|/4r² = F/4, making it one-fourth as large. Choice B is correct because it accurately applies the inverse square law, recognizing that doubling distance reduces force to 1/4 of its original value. Choice A incorrectly assumes a linear relationship (F ∝ 1/r), while choices C and D wrongly suggest the force increases with distance. To help students: Use graphical representations of 1/r² relationships, practice problems with various distance changes (tripling, halving), and emphasize that this inverse square law appears throughout physics (gravity, light intensity).
Electric charge is a conserved property of matter (C) that can be positive or negative; forces between charges are electric forces that act along the line joining charges. In deep space, three point charges lie on the x-axis: $q_1=+1.0\times10^{-6}\ \text{C}$ at $x=0$, $q_2=+1.0\times10^{-6}\ \text{C}$ at $x=0.20\ \text{m}$, and $q_3=-2.0\times10^{-6}\ \text{C}$ at $x=0.40\ \text{m}$. Coulomb’s Law gives each pairwise force magnitude, e.g. $$F_{21}=k\frac{|q_2q_1|}{(0.20)^2},\quad F_{23}=k\frac{|q_2q_3|}{(0.20)^2}$$ with $k=8.99\times10^{9}$. By superposition, the net force on $q_2$ is the vector sum of forces from $q_1$ and $q_3$.
Determine the net electric force on $q_2$ placed in an electric field.
$0.67\ \text{N}$ to the right
$0.67\ \text{N}$ to the left
$0.22\ \text{N}$ to the left
$0.22\ \text{N}$ to the right
Explanation
This question tests AP Physics C: Electricity and Magnetism skills, specifically applying superposition to find net electric force on a charge due to multiple sources. The principle of superposition requires calculating each pairwise force separately then adding them vectorially. For charge q₂ at x = 0.20 m: force from q₁ is F₂₁ = (8.99×10⁹)(1×10⁻¹²)/(0.04) = 0.225 N rightward (repulsive), and force from q₃ is F₂₃ = (8.99×10⁹)(2×10⁻¹²)/(0.04) = 0.449 N rightward (attractive toward negative q₃). The net force is 0.225 + 0.449 = 0.674 N ≈ 0.67 N to the right. Choice B is correct because both forces on q₂ point rightward: repulsion from q₁ pushes right, and attraction to q₃ pulls right. Choice C incorrectly reverses the direction, while choices A and D have wrong magnitudes. To help students: Always establish a coordinate system and sign convention, draw individual force vectors before adding, and verify that attractive/repulsive nature matches the relative positions and signs of charges.
Electric charge is a conserved property (C) that can be positive or negative; like charges repel and unlike charges attract. Electric force between point charges acts along the line joining them and obeys Coulomb’s Law. Three point charges lie on a line: $q_1=+2.0\times10^{-6}\ \text{C}$ at $x=0$, $q_2=-1.0\times10^{-6}\ \text{C}$ at $x=0.30\ \text{m}$, and $q_3=+2.0\times10^{-6}\ \text{C}$ at $x=0.60\ \text{m}$. For $q_2$, the forces from $q_1$ and $q_3$ have magnitudes $$F_{21}=k\frac{|q_2q_1|}{(0.30)^2},\quad F_{23}=k\frac{|q_2q_3|}{(0.30)^2}$$ and superposition requires adding these vectors to find the net force.
Determine the net electric force on $q_2$ placed in an electric field.
$0\ \text{N}$
$0.20\ \text{N}$ to the left
$0.40\ \text{N}$ to the left
$0.40\ \text{N}$ to the right
Explanation
This question tests AP Physics C: Electricity and Magnetism skills, specifically applying superposition to find net force on a charge in a symmetric configuration. When multiple charges interact, the net force on any charge is the vector sum of individual forces from all other charges. For q₂ = -1.0×10⁻⁶ C at x = 0.30 m: force from q₁ = +2.0×10⁻⁶ C is F₂₁ = (8.99×10⁹)(2×10⁻¹²)/(0.09) = 0.20 N leftward (attractive), and force from q₃ = +2.0×10⁻⁶ C is F₂₃ = (8.99×10⁹)(2×10⁻¹²)/(0.09) = 0.20 N rightward (attractive). Since q₂ is equidistant from equal positive charges q₁ and q₃, it experiences equal attractive forces in opposite directions, resulting in zero net force. Choice C is correct because the symmetric arrangement creates equal and opposite forces that cancel. Choices A, B, and D incorrectly suggest non-zero net forces, failing to recognize the symmetry. To help students: Identify symmetric configurations before calculating, draw force diagrams to visualize cancellation, and verify that equal charges at equal distances from opposite sides produce zero net force on a charge at the center.
Electric charge (C) is a conserved property that can be positive or negative; like charges repel and unlike charges attract. Electric force is the interaction between charges, directed along the line joining them. Two point charges in space are fixed: $q_1=-5.0\times10^{-6}\ \text{C}$ and $q_2=+2.0\times10^{-6}\ \text{C}$ separated by $r=0.50\ \text{m}$. Coulomb’s Law gives $$F=k\frac{|q_1q_2|}{r^2}=(8.99\times10^{9})\frac{|(-5.0\times10^{-6})(2.0\times10^{-6})|}{(0.50)^2}$$. If additional charges were introduced, superposition would require adding all forces vectorially to find the net force on either charge.
Calculate the force between two charges of $-5.0\ \mu\text{C}$ and $+2.0\ \mu\text{C}$, separated by $0.50\ \text{m}$.
$0.18\ \text{N}$, attractive
$0.72\ \text{N}$, repulsive
$0.36\ \text{N}$, attractive
$0.36\ \text{N}$, repulsive
Explanation
This question tests AP Physics C: Electricity and Magnetism skills, specifically calculating electric force magnitude and determining its nature (attractive/repulsive) using Coulomb's Law. Electric force between charges depends on the product of charge magnitudes and inversely on the square of separation distance. For charges q₁ = -5.0×10⁻⁶ C and q₂ = +2.0×10⁻⁶ C separated by 0.50 m: F = (8.99×10⁹)(10×10⁻¹²)/(0.25) = 0.3596 N ≈ 0.36 N. Since the charges have opposite signs (negative and positive), the force between them is attractive. Choice A is correct because it gives the right magnitude (0.36 N) and correctly identifies the force as attractive between opposite charges. Choice D has the correct magnitude but wrongly states repulsive, while choices B and C have incorrect magnitudes. To help students: Practice identifying force nature from charge signs before calculating, use the absolute value in Coulomb's Law for magnitude then determine direction separately, and reinforce that opposite charges always attract regardless of which is positive or negative.
Electric charge is a conserved property of matter measured in coulombs (C); it can be positive or negative, and interactions between charges produce electric forces. Electric force acts along the line joining charges and follows Coulomb’s Law $F=k\frac{|q_1q_2|}{r^2}$. In space, two point charges $q_1=+6.0\times10^{-6}\ \text{C}$ and $q_2=-1.0\times10^{-6}\ \text{C}$ are separated by $r=0.30\ \text{m}$. Numerical setup: $$F=(8.99\times10^{9})\frac{|(6.0\times10^{-6})(-1.0\times10^{-6})|}{(0.30)^2}$$. If additional charges are introduced, the net force is found by superposition, adding the individual force vectors.
Calculate the force between two charges of $+6.0\ \mu\text{C}$ and $-1.0\ \mu\text{C}$, separated by $0.30\ \text{m}$.
$0.30\ \text{N}$, attractive
$0.60\ \text{N}$, attractive
$0.060\ \text{N}$, attractive
$0.60\ \text{N}$, repulsive
Explanation
This question tests AP Physics C: Electricity and Magnetism skills, specifically calculating electric force between opposite charges using Coulomb's Law. The force magnitude is calculated as F = k|q₁q₂|/r², where the absolute value ensures a positive result for magnitude calculation. For charges q₁ = +6.0×10⁻⁶ C and q₂ = -1.0×10⁻⁶ C separated by 0.30 m: F = (8.99×10⁹)(6.0×10⁻¹²)/(0.09) = 0.599 N ≈ 0.60 N. Since one charge is positive and the other negative (opposite signs), the force between them is attractive. Choice A is correct because it provides the accurate magnitude (0.60 N) and correctly identifies the attractive nature of the force between opposite charges. Choice B has the right magnitude but wrongly states repulsive, while choices C and D have incorrect magnitudes. To help students: Always calculate magnitude first using absolute values, then determine attraction/repulsion from charge signs separately, and practice problems with mixed positive and negative charges to reinforce the concept.
Electric charge (C) can be positive or negative; electric force is the interaction between charges and follows Coulomb’s Law $F=k\dfrac{|q_1q_2|}{r^2}$. Two like point charges $q_1=+2.0,\mu\text{C}$ and $q_2=+2.0,\mu\text{C}$ are $0.30,\text{m}$ apart on a line. A third point charge $q_3=+1.0,\mu\text{C}$ is placed exactly midway between them. By superposition, the net force on $q_3$ equals the vector sum of the forces from $q_1$ and $q_2$. Explain the effect of introducing $q_3$ between the like charges by determining the net force on $q_3$.
Nonzero, toward $q_2$ because $q_2$ repels more
$0,\text{N}$ because $q_3$ has smaller charge
$0,\text{N}$ because forces cancel
Nonzero, toward $q_1$ because $q_1$ repels more
Explanation
This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding electric force superposition with multiple charges. When multiple charges interact, the net force on any charge is the vector sum of all individual forces acting on it, following the principle of superposition. In this scenario, two identical positive charges q₁ = q₂ = +2.0 μC are 0.30 m apart, and a third positive charge q₃ = +1.0 μC is placed exactly midway between them. Choice A is correct because q₃ experiences equal magnitude forces from both q₁ and q₂ (same charges at same distance), but these forces point in opposite directions along the line, causing them to cancel completely. Choices B and C are incorrect because they assume one charge repels more than the other, which is impossible given the symmetry of the setup. To help students: Use symmetry arguments to identify when forces cancel, draw clear vector diagrams showing force directions, and practice problems with charges arranged in symmetric configurations.
Electric charge (C) comes in positive and negative; electric force is the push/pull between charges and depends on separation by Coulomb’s Law $F\propto 1/r^2$. Two point charges $q_1=+1.0,\mu\text{C}$ and $q_2=+3.0,\mu\text{C}$ are initially separated by $r=0.20,\text{m}$, producing force magnitude $F_0$ on each. If the distance is doubled to $0.40,\text{m}$ while charges remain fixed, superposition would still apply if other charges existed, but here only the distance changes. How does the force magnitude change?
Becomes $\tfrac{1}{4}F_0$
Becomes $\tfrac{1}{2}F_0$
Becomes $4F_0$
Becomes $2F_0$
Explanation
This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding how electric force varies with distance according to the inverse square law. Coulomb's Law states that electric force is inversely proportional to the square of the distance between charges, meaning F ∝ 1/r². In this scenario, two positive charges initially separated by 0.20 m have their distance doubled to 0.40 m, and we need to determine how the force changes. Choice D is correct because when distance doubles, the force becomes 1/(2)² = 1/4 of its original value, so F = F₀/4. Choice B is incorrect because it suggests force increases with distance, which violates the inverse square law. To help students: Use proportional reasoning to avoid unnecessary calculations, practice problems where only one variable changes, and reinforce that doubling distance always reduces force to one-fourth.
Electric charge is measured in coulombs (C) and can be positive or negative; electric force between point charges follows Coulomb’s Law and acts along the line between them. Three charges are on a line: $q_1=+4.0,\mu\text{C}$ at $x=0$, $q_2=-4.0,\mu\text{C}$ at $x=0.30,\text{m}$, and $q_3=+1.0,\mu\text{C}$ at $x=0.60,\text{m}$. By superposition, the net force on $q_2$ is $\vec F_{21}+\vec F_{23}$. Calculate the net force on $q_2$ (take +x right).
$+2.0,\text{N}$
$+1.2,\text{N}$
$0,\text{N}$
$-1.2,\text{N}$
Explanation
This question tests AP Physics C: Electricity and Magnetism skills, specifically applying superposition to find net force on a middle charge from two others. Vector addition of forces requires careful attention to both magnitude and direction, especially when dealing with attractive and repulsive forces simultaneously. In this scenario, q₂ = -4.0 μC at x = 0.30 m experiences forces from q₁ = +4.0 μC at x = 0 and q₃ = +1.0 μC at x = 0.60 m. Choice B is correct because F₂₁ = k(4×10⁻⁶)(4×10⁻⁶)/(0.30)² = 1.6 N leftward (opposite charges attract) and F₂₃ = k(4×10⁻⁶)(1×10⁻⁶)/(0.30)² = 0.40 N rightward (opposite charges attract), giving net force = -1.6 + 0.40 = -1.2 N (leftward). Choice A is incorrect because it has the wrong sign, indicating rightward motion when the net force is actually leftward. To help students: Establish clear sign conventions before calculating, draw force vectors to visualize the problem, and always verify that the final answer makes physical sense given the charge configuration.