AP Physics C Electricity and Magnetism Quiz: Conservation Of Electric Charge

18 questions · exam conditions

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Conservation Of Electric ChargeQuestion 1 of 18

A battery drives a steady current $I=2.0\,\text{A}$ through a single-loop circuit with one resistor. Charge is not created or destroyed; it redistributes in the conducting wires while the surrounding plastic insulation prevents charge leakage. Based on the described process, how does the current in the wire change as it passes through the resistor?

It decreases because charge is used up in the resistor

It increases because the resistor creates charge carriers

It stays

2.0\,\text{A}

everywhere in the loop

It becomes

0\,\text{A}

inside the resistor due to insulation

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AP Physics C Electricity and Magnetism Quiz

AP Physics C Electricity and Magnetism Quiz: Conservation Of Electric Charge

Practice Conservation Of Electric Charge in AP Physics C Electricity and Magnetism with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Conservation Of Electric Charge, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Electricity and Magnetism.

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.

All questions

Question 1

It decreases because charge is used up in the resistor
It increases because the resistor creates charge carriers
It stays $2.0\,\text{A}$ everywhere in the loop (correct answer)
It becomes $0\,\text{A}$ inside the resistor due to insulation

Explanation: This question tests understanding of the conservation of electric charge and current continuity in circuits in AP Physics C. Conservation of charge means that the total electric charge in an isolated system remains constant, which is foundational in analyzing electric phenomena. In this scenario, charge flows continuously through the circuit without accumulation or depletion anywhere, demonstrating that current must be the same at all points in a single loop. Choice C is correct because charge conservation requires that the same amount of charge entering the resistor must exit it, maintaining I = 2.0 A throughout the entire loop. Choice A is incorrect because it wrongly assumes charge is consumed in the resistor, but resistors only convert electrical energy to heat - they don't destroy charge carriers. To help students: Emphasize that current is the flow rate of charge, and in steady state, charge cannot accumulate anywhere in the circuit. Use the analogy of water flow in pipes where the flow rate must be constant in a single pipe regardless of obstacles.

Question 2

A $+3.0\,\text{nC}$ metal sphere touches a neutral identical metal sphere on insulating stands, then they separate. Charge redistributes on conductors until equilibrium, conserving total charge. Based on the described process, what is the resulting charge on each sphere after contact?

$+1.5\,\text{nC}$ each (correct answer)
$+3.0\,\text{nC}$ each
$0\,\text{C}$ each
$-1.5\,\text{nC}$ each

Explanation: This question tests understanding of the conservation of electric charge and charge distribution in conductors in AP Physics C. Conservation of charge means that the total electric charge in an isolated system remains constant, which is foundational in analyzing electric phenomena. In this scenario, when the charged sphere touches the neutral sphere, charge flows between them until they reach the same electric potential, demonstrating how charges redistribute to maintain equilibrium. Choice A is correct because the total charge (+3.0 nC) is conserved and equally distributed between two identical spheres, giving each sphere +1.5 nC. Choice B is incorrect because it wrongly assumes each sphere gets the full original charge, violating conservation by creating charge from nothing. To help students: Emphasize that identical conductors in contact share charge equally, and the total charge before equals the total charge after. Use the analogy of water levels equalizing between connected containers to reinforce the concept of charge equilibrium.

Question 3

A neutral metal sphere on an insulating stand is brought near a negatively charged rod without touching. Electrons in the conductor shift, but none can leave through the insulating stand, so total charge stays zero. Considering the explained phenomenon, which best explains the charge distribution on the sphere while the rod is nearby?

Net negative charge appears because induction creates electrons
Positive near the rod, negative far away; net charge remains zero (correct answer)
Charge stays uniformly zero everywhere because metals cannot polarize
Negative charge collects inside the metal volume, not the surface

Explanation: This question tests understanding of the conservation of electric charge and electrostatic induction in conductors in AP Physics C. Conservation of charge means that the total electric charge in an isolated system remains constant, which is foundational in analyzing electric phenomena. In this scenario, the negative rod induces charge separation in the neutral conductor - electrons are repelled to the far side, leaving positive charges near the rod, demonstrating charge redistribution without charge creation. Choice B is correct because the conductor's free electrons redistribute with positive charge accumulating near the negative rod and negative charge on the far side, while the total charge remains zero. Choice A is incorrect because it wrongly assumes induction creates new electrons, violating charge conservation - charges only redistribute, they aren't created. To help students: Emphasize that induction causes charge separation, not charge creation, and the net charge on an isolated conductor remains unchanged. Use demonstrations with electroscopes to show how charges redistribute without contact.

Question 4

A hollow conducting spherical shell has net charge $+9.0\,\text{nC}$ and no charge inside its cavity. In electrostatic equilibrium, charges redistribute on conductors so the electric field inside the conductor is zero. Considering the explained phenomenon, what is the resulting charge on the inner surface of the shell?

$+9.0\,\text{nC}$
$-9.0\,\text{nC}$
$0\,\text{C}$ (correct answer)
Cannot be determined without the shell radius

Explanation: This question tests understanding of the conservation of electric charge and charge distribution in hollow conductors in AP Physics C. Conservation of charge means that the total electric charge in an isolated system remains constant, which is foundational in analyzing electric phenomena. In this scenario, with no charge inside the cavity, the conductor arranges its charges to ensure zero electric field within the conducting material, demonstrating electrostatic shielding. Choice C is correct because with no charge in the cavity, Gauss's Law applied to a surface within the conductor requires zero net charge on the inner surface, so all +9.0 nC resides on the outer surface. Choice A is incorrect because it wrongly assumes the charge distributes between inner and outer surfaces, but this would create a field inside the conductor, violating electrostatic equilibrium. To help students: Emphasize that charges on conductors arrange to make E = 0 inside the conducting material, and use Gauss surfaces within the conductor to prove charge distribution. Practice with nested conductor problems to reinforce how cavity charges affect surface charge distribution.

Question 5

A charged metal sphere is enclosed by a spherical Gaussian surface of radius $0.10\,\text{m}$ . The surface is then expanded to radius $0.30\,\text{m}$ without crossing any additional charges. Considering the explained phenomenon, how does the electric field at the Gaussian surface change when the radius increases?

It stays the same because flux is constant
It increases proportional to $r^2$
It decreases proportional to $1/r^2$ (correct answer)
It becomes zero because the Gaussian surface is larger

Explanation: This question tests understanding of the conservation of electric charge and the application of Gauss's Law with changing Gaussian surfaces in AP Physics C. Conservation of charge means that the total electric charge in an isolated system remains constant, which is foundational in analyzing electric phenomena. In this scenario, expanding the Gaussian surface doesn't change the enclosed charge, but the surface area increases, demonstrating how electric field varies with distance from a spherical charge distribution. Choice C is correct because for a spherical charge distribution, E ∝ 1/r², so when radius increases from 0.10 m to 0.30 m (factor of 3), the field decreases by a factor of 9. Choice A is incorrect because it wrongly assumes constant flux means constant field, confusing the fact that flux = EA remains constant while both E and A change. To help students: Emphasize that while flux through any closed surface around the same charge is constant, the field strength depends on distance. Use the analogy of light intensity decreasing with distance from a bulb to reinforce the inverse square law.

Question 6

A metal sphere carries net charge $+8.0\,\text{nC}$ . In electrostatic equilibrium, charge resides on the outer surface of the conductor, and the electric field inside the metal is zero. Considering the explained phenomenon, which best explains the charge distribution in the sphere?

Charge is uniformly throughout the metal volume
Charge is only on the outer surface; $E=0$ inside (correct answer)
Charge accumulates at the center; $E$ is maximum inside
Charge stays where placed because metals are insulators

Explanation: This question tests understanding of the conservation of electric charge and electrostatic equilibrium in conductors in AP Physics C. Conservation of charge means that the total electric charge in an isolated system remains constant, which is foundational in analyzing electric phenomena. In this scenario, the metal sphere has reached electrostatic equilibrium where all excess charge resides on the outer surface, demonstrating how charges arrange to minimize energy. Choice B is correct because in electrostatic equilibrium, free charges in a conductor move to the surface, creating zero electric field inside - this is a fundamental property of conductors. Choice A is incorrect because it wrongly assumes charge distributes throughout the volume like in an insulator, which is a common misconception about metallic conductors. To help students: Emphasize that conductors have free electrons that move until E=0 inside, and use Gauss's Law with a surface inside the conductor to prove no charge can exist there. Demonstrate with a hollow conductor and electroscope that charge resides only on the outer surface.

Question 7

A thin insulating spherical shell has total charge $Q=+4.0\,\text{nC}$ uniformly spread over its surface. Using Gauss's Law with a spherical Gaussian surface of radius $r=0.20\,\text{m}$ centered on the shell, charge is conserved and symmetry makes $E$ constant on the surface. Based on Gauss's Law, what is the electric flux through the Gaussian surface?

$\Phi_E=Q/(4\pi\varepsilon_0)$
$\Phi_E=Q/\varepsilon_0$ (correct answer)
$\Phi_E=4\pi r^2 Q$
$\Phi_E=0$ because the shell is insulating

Explanation: This question tests understanding of the conservation of electric charge and the application of Gauss's Law in AP Physics C. Conservation of charge means that the total electric charge in an isolated system remains constant, which is foundational in analyzing electric phenomena. In this scenario, the charged insulating shell is completely enclosed by the Gaussian surface, demonstrating how Gauss's Law relates enclosed charge to electric flux regardless of the charge distribution details. Choice B is correct because Gauss's Law states that electric flux equals the enclosed charge divided by ε₀, giving ΦE = Q/ε₀ = 4.0×10⁻⁹ C/ε₀. Choice A is incorrect because it wrongly includes an extra factor of 4π, confusing the flux formula with the electric field formula for a point charge. To help students: Emphasize that Gauss's Law depends only on enclosed charge, not on the size of the Gaussian surface or charge distribution. Practice applying Gauss's Law to various symmetric charge distributions to reinforce that flux = Qenc/ε₀ always holds.

Question 8

A conducting spherical shell has net charge $Q=-2.0\,\text{nC}$ . A spherical Gaussian surface of radius $r=0.30\,\text{m}$ is centered on it in air, and symmetry allows Gauss's Law. Considering the described process, if the charge is $-2.0\,\text{nC}$ , what is the resulting electric field magnitude at $0.30\,\text{m}$ ?

$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{|Q|}{r}$
$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{|Q|}{r^2}$ (correct answer)
$E=\dfrac{|Q|}{\varepsilon_0}$
$E=0$ outside because conductors block fields

Explanation: This question tests understanding of the conservation of electric charge and the application of Gauss's Law to spherically symmetric charge distributions in AP Physics C. Conservation of charge means that the total electric charge in an isolated system remains constant, which is foundational in analyzing electric phenomena. In this scenario, the conducting shell's charge distributes uniformly on its surface, creating a spherically symmetric field that can be analyzed using Gauss's Law. Choice B is correct because for a spherical charge distribution, the electric field at distance r is E = |Q|/(4πε₀r²), which for Q = -2.0 nC and r = 0.30 m gives the point charge formula result. Choice A is incorrect because it wrongly uses 1/r instead of 1/r², which would violate Gauss's Law and dimensional analysis for electric field. To help students: Emphasize that spherical conductors create fields identical to point charges at distances outside the conductor. Practice deriving E from Gauss's Law for spherical symmetry to reinforce the 1/r² dependence.

Question 9

A metal sphere has net charge $+2.0\,\mu\text{C}$ ; at $r=0.30\,\text{m}$ , what is $E$ outside, by Gauss's Law?

$E=0\,\text{N/C}$ , because charge is on the surface
$E=\dfrac{kQ}{r^2}=2.0\times10^5\,\text{N/C}$ (correct answer)
$E=\dfrac{Q}{\varepsilon_0 r}=7.5\times10^5\,\text{N/C}$
$E=\dfrac{kQ}{r}=6.0\times10^4\,\text{N/C}$

Explanation: This question tests understanding of the conservation of electric charge and applying Gauss's Law to find electric fields outside charged conductors in AP Physics C. Conservation of charge ensures all excess charge resides on the conductor's surface, creating an electric field outside identical to a point charge. In this scenario, the metal sphere's charge distribution creates a radially symmetric field that can be analyzed using a spherical Gaussian surface. Choice B is correct because it correctly applies the principle that outside a spherical conductor, E = kQ/r² = (9×10⁹)(2×10⁻⁶)/(0.30)² = 2.0×10⁵ N/C, treating all charge as if concentrated at the center. Choice C is incorrect because it uses an incorrect formula (missing 4π factor), which is a common error when students mix up different forms of field equations. To help students: Emphasize that conductors with spherical symmetry produce fields identical to point charges when viewed from outside. Practice using both Coulomb's law and Gauss's Law approaches to verify they give the same result for spherical charge distributions.

Question 10

A charged insulating rod is brought near a metal sphere on an insulating stand; based on induction, how does the sphere's net charge change?

It becomes oppositely charged because electrons jump across the air gap
It stays neutral overall, but charges redistribute on its surface (correct answer)
It gains the same sign of charge as the rod without any grounding
Its net charge increases because polarization adds charge to the system

Explanation: This question tests understanding of the conservation of electric charge and electrostatic induction without grounding in AP Physics C. Conservation of charge means that in an isolated system (sphere on insulating stand), the total charge cannot change without physical contact or grounding. In this scenario, the charged rod induces charge separation on the metal sphere through electrostatic forces, but no charge transfer occurs across the air gap. Choice B is correct because it correctly applies the principle that induction only redistributes existing charges - the sphere remains neutral overall but develops opposite charges on different regions of its surface. Choice A is incorrect because it wrongly assumes electrons can jump across air gaps under normal conditions, which is a common error when students confuse induction with conduction. To help students: Emphasize the distinction between induction (charge redistribution) and conduction (charge transfer through contact). Practice identifying when systems are isolated (no grounding) versus when charge can flow, reinforcing that air is an excellent insulator.

Question 11

A neutral conductor is placed in a uniform external electric field; which statement best describes its charge distribution at equilibrium?

Excess charge accumulates uniformly throughout the conductor's volume
Positive and negative charges separate on opposite surfaces; net charge remains zero (correct answer)
Charges remain fixed because conductors behave like insulators in fields
New charge is created on the conductor to cancel the external field

Explanation: This question tests understanding of the conservation of electric charge and charge redistribution in conductors placed in external fields in AP Physics C. Conservation of charge means that the total electric charge in an isolated neutral conductor remains zero, but charges can redistribute internally. In this scenario, the external field causes free electrons to move opposite to the field direction until equilibrium is reached, demonstrating how charges rearrange while maintaining zero net charge. Choice B is correct because it correctly describes that positive and negative charges separate to opposite surfaces (polarization) while the conductor remains neutral overall, creating an internal field that cancels the external field. Choice A is incorrect because it wrongly assumes charge distributes uniformly throughout the volume, which is a common error when students forget that charges in conductors move to surfaces. To help students: Emphasize that conductors in electrostatic equilibrium have zero internal field, achieved by surface charge redistribution. Use diagrams showing induced surface charges creating fields that exactly cancel external fields inside the conductor.

Question 12

A $+1.0\,\mu\text{C}$ charge is placed at center of an insulating spherical shell; considering Gauss's Law, what is flux through the shell?

$\Phi_E=0$ , because insulators block electric fields
$\Phi_E=\dfrac{Q}{\varepsilon_0}=1.13\times10^5\,\text{N}\cdot\text{m}^2/\text{C}$ (correct answer)
$\Phi_E=\dfrac{kQ}{r^2}$ , because flux equals field magnitude
$\Phi_E$ increases with shell thickness because more material encloses charge

Explanation: This question tests understanding of the conservation of electric charge and the application of Gauss's Law to insulating materials in AP Physics C. Conservation of charge ensures the charge remains fixed at the center, while Gauss's Law determines the flux through any enclosing surface. In this scenario, the insulating shell doesn't affect the flux calculation since Gauss's Law depends only on enclosed charge, not on the material of the Gaussian surface. Choice B is correct because it correctly applies Gauss's Law: Φ_E = Q_enclosed/ε₀ = (1.0×10⁻⁶)/(8.85×10⁻¹²) = 1.13×10⁵ N·m²/C, independent of whether the shell is conducting or insulating. Choice A is incorrect because it wrongly assumes insulators block electric fields or affect flux calculations, which is a common error when students confuse material properties with field propagation. To help students: Emphasize that Gauss's Law applies universally - the flux depends only on enclosed charge, not on intervening materials. Practice with both conducting and insulating shells to reinforce that material type doesn't affect flux through surfaces.

Question 13

A conducting spherical shell has net charge $+5.0\,\mu\text{C}$ ; considering Gauss's Law, what is $E$ inside the metal?

$E=\dfrac{kQ}{r^2}$ , because enclosed charge is nonzero
$E=0\,\text{N/C}$ , because charges reside on the surface at equilibrium (correct answer)
$E$ is constant and nonzero throughout the metal
$E$ depends on shell thickness, not on charge

Explanation: This question tests understanding of the conservation of electric charge and the application of Gauss's Law inside conductors in AP Physics C. Conservation of charge combined with electrostatic equilibrium requires that all excess charge on a conductor resides on its outer surface, leaving the interior field-free. In this scenario, the conducting shell's charge distributes entirely on its surface, creating zero field in the metal itself regardless of the charge amount. Choice B is correct because it correctly applies the principle that E = 0 inside any conductor at equilibrium, as free charges would move if any field existed, contradicting the equilibrium condition. Choice A is incorrect because it wrongly assumes Gauss's Law gives a non-zero field inside the conductor, which is a common error when students forget that charges reside only on surfaces. To help students: Emphasize that conductors in electrostatic equilibrium have E = 0 everywhere inside the material itself. Use Gauss's Law with surfaces inside the conductor to show that since no charge is enclosed, the field must be zero.

Question 14

A neutral metal sphere is inside a spherical Gaussian surface; a charged rod outside approaches. Based on Gauss's Law, how does flux change?

Flux increases because the external rod increases the field everywhere
Flux stays $0$ because the enclosed net charge remains $0$ (correct answer)
Flux becomes negative because field lines enter the surface
Flux depends on the Gaussian surface radius, so it must change

Explanation: This question tests understanding of the conservation of electric charge and Gauss's Law for systems with external influences in AP Physics C. Conservation of charge means the neutral sphere remains neutral (zero net charge) regardless of external fields, which is key to applying Gauss's Law correctly. In this scenario, while the external rod induces charge separation on the sphere's surface, the net charge enclosed by the Gaussian surface remains zero. Choice B is correct because it correctly applies Gauss's Law: flux depends only on enclosed charge (Φ = Q_enc/ε₀ = 0), not on external charges or induced polarization, so flux remains zero. Choice A is incorrect because it wrongly assumes external charges affect flux through a surface not enclosing them, which is a common error when students confuse field strength with flux. To help students: Emphasize that Gauss's Law considers only charges inside the Gaussian surface - external charges create fields but don't contribute to flux. Practice with various charge configurations to reinforce the distinction between local field changes and net flux.

Question 15

A $-2.0\,\mu\text{C}$ metal sphere touches an identical $+8.0\,\mu\text{C}$ sphere; after separation, what is each sphere's charge?

$+3.0\,\mu\text{C}$ each, by charge conservation and symmetry (correct answer)
$+8.0\,\mu\text{C}$ and $-2.0\,\mu\text{C}$ , because charges stay with originals
$+5.0\,\mu\text{C}$ each, because only positive charge moves
$+6.0\,\mu\text{C}$ each, because contact creates extra charge

Explanation: This question tests understanding of the conservation of electric charge when identical conductors make contact in AP Physics C. Conservation of charge means that the total electric charge before contact equals the total after, which is fundamental when analyzing charge redistribution. In this scenario, two identical spheres with different charges touch, allowing charge to flow until they reach the same potential, then separate with equal charges. Choice A is correct because it correctly applies charge conservation: total initial charge is -2.0 + 8.0 = +6.0 μC, which divides equally between identical spheres giving +3.0 μC each, ensuring both conservation and symmetry. Choice B is incorrect because it wrongly assumes charges remain with their original spheres, which is a common error when students don't recognize that contact allows complete charge redistribution. To help students: Emphasize that identical conductors must have equal charge after separation due to symmetry. Practice with various initial charge combinations to reinforce that total charge is conserved while individual charges change.

Question 16

A $+6.0\,\mu\text{C}$ rod touches a neutral metal sphere on an insulating stand; after separation, what is the sphere's charge?

$+0\,\text{C}$ , because charge returns to the rod
$+3.0\,\mu\text{C}$ , assuming equal sharing by identical conductors (correct answer)
$+6.0\,\mu\text{C}$ , because all charge transfers to the sphere
$+12\,\mu\text{C}$ , because contact creates additional charge

Explanation: This question tests understanding of the conservation of electric charge and charge distribution when conductors make contact in AP Physics C. Conservation of charge means that the total electric charge in an isolated system remains constant, which is foundational in analyzing electric phenomena. In this scenario, when a charged rod touches a neutral conductor, charge redistributes between them until they reach the same electric potential. Choice B is correct because it correctly applies the principle that identical conductors will share charge equally, resulting in each having half the original charge (+3.0 μC each), ensuring charge is conserved and correctly distributed. Choice C is incorrect because it wrongly assumes all charge transfers to the sphere, which is a common error when students overlook that both objects are conductors that must reach equilibrium. To help students: Emphasize that identical conductors share charge equally when in contact, and use the principle that total charge before equals total charge after. Practice with scenarios involving different conductor sizes to reinforce that charge sharing depends on capacitance ratios.

Question 17

Based on induction, a $-4.0\,\mu\text{C}$ rod nears a grounded metal sphere; after removing ground, what is sphere's net charge?

$-4.0\,\mu\text{C}$ , because contact transfers charge
$0\,\text{C}$ , because polarization cannot leave net charge
$+4.0\,\mu\text{C}$ , because electrons flowed to ground (correct answer)
$-8.0\,\mu\text{C}$ , because induction doubles charge

Explanation: This question tests understanding of the conservation of electric charge and electrostatic induction in grounded conductors in AP Physics C. Conservation of charge means that the total electric charge in an isolated system remains constant, but grounding allows charge to flow to/from Earth. In this scenario, the negative rod induces positive charges on the near side of the sphere by repelling electrons through the ground connection, demonstrating how charges rearrange to maintain equilibrium. Choice C is correct because it correctly applies the principle that electrons flow to ground when repelled by the negative rod, leaving the sphere with a net positive charge of +4.0 μC after the ground is removed. Choice B is incorrect because it wrongly assumes polarization alone cannot leave net charge, which is a common error when students overlook the role of grounding in allowing charge transfer. To help students: Emphasize the sequence of events in grounding - first induction occurs, then charge flows through ground, finally removing ground traps the remaining charge. Use diagrams showing electron flow direction to reinforce that grounding provides a path for charge movement.

Question 18

Considering a charged conducting sphere, based on Gauss's Law, what is the electric flux through any enclosing spherical surface?

$\Phi_E = Q/\varepsilon_0$ , independent of radius (correct answer)
$\Phi_E = kQ/r^2$ , because flux equals field
$\Phi_E = 0$ , because conductors block electric fields
$\Phi_E$ increases with surface area $4\pi r^2$

Explanation: This question tests understanding of the conservation of electric charge and the application of Gauss's Law to conducting spheres in AP Physics C. Conservation of charge combined with Gauss's Law states that electric flux through any closed surface depends only on the enclosed charge, not on the surface size or shape. In this scenario, a charged conducting sphere has all its charge on the surface, but any Gaussian surface enclosing it will have the same flux. Choice A is correct because it correctly applies Gauss's Law: Φ_E = Q_enclosed/ε₀, which is independent of the Gaussian surface radius as long as it encloses all the charge. Choice D is incorrect because it wrongly assumes flux increases with surface area, which is a common error when students confuse flux (scalar) with field strength (which does decrease with distance). To help students: Emphasize that Gauss's Law relates flux to enclosed charge only - the size of the Gaussian surface doesn't matter. Practice calculating flux for various surface sizes to reinforce that while E-field decreases with r², the product E·A remains constant.