ES.2 Practice Quiz (Lenny)
AP Physics 2 • Unit 10: Electrostatics
Conservation of Charge & Charging Quiz
Topic 10.2 • Formative Assessment
Time: 40 Minutes
Total Score: / 25 pts
Name:
Date:
Period:
Section I: Multiple-Choice Questions (8 Questions • 1 pt each) Choose the single best answer for each question.
A solid neutral copper sphere mounted on an insulating stand is touched by a negatively charged rubber rod. After contact, the sphere is found to have a net negative charge. Which statement correctly describes the microscopic mechanism of this charging process?
(A) Protons transferred from the sphere into the rubber rod.
(B) Electrons transferred from the rubber rod onto the copper sphere.
(C) Equal numbers of protons and electrons were created inside the sphere, but protons decayed.
(D) Positive ions migrated through the crystalline lattice to the interior of the sphere.
A student performs an oil-drop experiment to determine the net charge on several isolated microscopic droplets. Which of the following measured charge values is physically impossible according to the principle of charge quantization?
(A) \( -3.20 \times 10^{-19}\text{ C} \)
(B) \( +4.80 \times 10^{-19}\text{ C} \)
(C) \( -2.40 \times 10^{-19}\text{ C} \)
(D) \( +8.00 \times 10^{-19}\text{ C} \)
A positively charged glass rod is brought near an uncharged plastic sphere hanging from a silk thread. The sphere is attracted toward the rod without making contact. Which of the following best explains why the neutral sphere experiences a net attractive force?
(A) Free electrons in the plastic flow through the bulk of the sphere to the far side.
(B) The molecules of the insulator polarize, causing the closer negative electron clouds to feel a stronger attractive force than the repulsive force on the distant positive nuclei.
(C) The sphere acquires a net negative charge from ionizing radiation in the room air.
(D) Positive charges in the sphere migrate rapidly toward the surface closest to the rod.
Three identical conducting spheres on insulating stands are labeled X, Y, and Z. Sphere X initially carries a charge of \(+6Q\), Sphere Y carries \(-2Q\), and Sphere Z is uncharged (\(0\)). First, Sphere X is touched to Sphere Y and then separated. Next, Sphere Y is touched to Sphere Z and then separated. What are the final charges on each sphere?
(A) \( X = +2Q,\; Y = +1Q,\; Z = +1Q \)
(B) \( X = +2Q,\; Y = +2Q,\; Z = 0 \)
(C) \( X = +4Q,\; Y = +2Q,\; Z = +2Q \)
(D) \( X = +1.33Q,\; Y = +1.33Q,\; Z = +1.33Q \)
AP Physics 2 • Topic 10.2 Assessment Page 1 of 4
Section I: Multiple-Choice Questions (Continued) Topic 10.2: Charging & Conservation
A student intends to give a solid neutral metal sphere on an insulating stand a net positive charge using an available grounding wire and a negatively charged plastic rod. Which procedure successfully accomplishes this goal?
(A) Touch the negative rod to the sphere, then ground the sphere briefly.
(B) Bring the rod near the sphere, remove the rod, and then touch the sphere with the grounding wire.
(C) Connect the grounding wire, touch the negative rod to the wire, and then disconnect the wire.
(D) Bring the rod near the sphere, connect the grounding wire to the sphere, remove the ground wire, and then remove the rod.
An uncharged conducting sphere and an uncharged insulating sphere of identical dimensions are each placed in separate identical external electric fields directed to the right. Which statement correctly compares the internal charge behavior of the two spheres in electrostatic equilibrium?
(A) Free electrons in both spheres drift completely across the diameter until the internal electric field is zero.
(B) Mobile conduction electrons in the conductor shift to the left surface until \(E_{\text{net}} = 0\) inside; in the insulator, electrons are bound and undergo microscopic molecular polarization, resulting in a nonzero internal field.
(C) Neither sphere experiences any redistribution of charge because both objects are electrically neutral.
(D) Positive charges in the conductor move toward the right surface, while no charges shift in the insulator.
Conducting Sphere 1 (radius \(R\)) initially carries a positive charge \(+Q\). Conducting Sphere 2 is larger (radius \(2R\)) and initially uncharged. Both spheres are mounted on insulating stands. Sphere 1 is brought into physical contact with Sphere 2 and then separated far apart. Which statement correctly describes the final charge \(q_2\) on Sphere 2?
(A) \( q_2 = +0.5Q \), because charge divides equally between two contacting conductors regardless of size.
(B) \( q_2 = 0 \), because charge cannot transfer between spheres of differing radii.
(C) \( q_2 > +0.5Q \), because the spheres must reach an equipotential state, requiring the sphere with larger surface area and radius to hold more charge.
(D) \( q_2 < +0.5Q \), because excess charge repels more strongly toward the smaller sphere where curvature is sharper.
In a nuclear physics reaction chamber, an alpha particle (\(q_\alpha = +2e\)) collides with a neutral target atom. Immediately after the collision, four total particles emerge: two free electrons (\(-1e\) each), one singly ionized proton-bearing fragment (\(+1e\)), and an unknown recoil nucleus \(X\). Assuming the system is completely isolated, what is the net electric charge of nucleus \(X\)?
(A) \( +1e \)
(B) \( +2e \)
(C) \( +3e \)
(D) \( +4e \)
AP Physics 2 • Topic 10.2 Assessment Page 2 of 4
Section II: Free-Response
Question 1: Process of Charging & Charge Distribution (8 Points)
Suggested Time: 15 min • 8 pts
Two identical uncharged conducting spheres, Sphere A and Sphere B, rest on insulating stands in physical contact with one another. A glass rod carrying a large positive net charge \(+Q_{\text{rod}}\) is brought near the left side of Sphere A without touching it, as shown in the diagram.
(a) [2 pts] Draw Charge Distribution: Rod held near Sphere A (spheres still touching)
On the diagram below, sketch the distribution of excess positive (\(+\)) and negative (\(-\)) charges on Sphere A and Sphere B while the positive rod is held in place. Clearly show where charge accumulates.
Rod (\(+Q\))
Sphere A
Sphere B
(b) [3 pts] Explanation of Microscopic Mechanism:
While the rod is held in place, Sphere B is moved several centimeters to the right so it no longer touches Sphere A. The rod is then removed far away. State the final sign of the net charge on Sphere A and Sphere B, and explain the microscopic movement of fundamental charged particles that produced this result.
(c) [3 pts] Grounding Analysis:
Suppose the experiment is repeated from the beginning with both spheres neutral and touching. While the positive rod is held near Sphere A, a student briefly connects a grounding wire from the right edge of Sphere B to ground, and then removes the wire. Finally, the rod is removed. What is the net charge on the two-sphere system? Justify your claim by referencing the path of electron flow.
AP Physics 2 • Topic 10.2 Assessment Page 3 of 4
Section II: Free-Response
Question 2: Charge Quantization & Conservation (9 Points)
Suggested Time: 15 min • 9 pts
A student team replicates Robert Millikan's oil-drop experiment to investigate the fundamental unit of charge. The team suspends charged oil droplets between two parallel plates and records the net charge for six droplets in the table below.
| Droplet |
1 |
2 |
3 |
4 |
5 |
6 |
| Charge \(q\;(\times 10^{-19}\text{ C})\) |
\(-4.80\) |
\(+1.60\) |
\(-6.40\) |
\(-2.40\) |
\(-3.20\) |
\(+8.00\) |
(a) [3 pts] Identifying Impossible Fractional Charge:
One of the droplet measurements contains an experimental error that physically contradicts the principle of charge quantization. Identify which droplet contains this error and justify your selection using quantitative reasoning based on the elementary charge \(e\).
(b) [3 pts] Number of Fundamental Charge Carriers:
For Droplet 3 (\(q = -6.40 \times 10^{-19}\text{ C}\)), determine whether the droplet possesses an excess or deficit of electrons, and calculate the exact integer number \(N\) of excess or deficit electrons residing on the droplet.
(c) [3 pts] Macroscopic Charge Sharing & Carrier Count:
Two identical macroscopic metal spheres on insulating stands initially carry charges \(Q_1 = -16.0\text{ nC}\) and \(Q_2 = +6.4\text{ nC}\). The two spheres are brought into contact and then separated. Calculate the final charge on each sphere and the total number of electrons transferred between the spheres during contact, identifying the direction of electron flow.
AP Physics 2 • Topic 10.2 Assessment Page 4 of 4
ES.2 Practice Quiz Key
AP Physics 2 • Scoring Guide & Key
Charge Transfer Quiz • Answer Key
Topic 10.2: Conservation of Charge & the Process of Charging
Teacher Master
Total Points: 25 pts
Section I: Multiple-Choice Detailed Explanations 1 point each (8 points total)
Question 1: Microscopic Mechanism of Conduction Correct: (B)
Explanation: In solid metallic conductors (copper), positive atomic nuclei are bound rigidly in crystal lattice sites. Only valence conduction electrons are mobile. When touched by a negatively charged rod, excess electrons on the rod transfer onto the copper sphere due to mutual electrostatic repulsion.
Distractor Rationale: (A) & (D) are common misconceptions; protons never migrate in solids. (C) violates baryon and charge conservation.
Question 2: Impossibility of Fractional Elementary Charge Correct: (C)
Explanation: Charge is fundamentally quantized in integer multiples of elementary charge: \(q = \pm N e\), where \(e = 1.60 \times 10^{-19}\text{ C}\). For droplet (C), \(\frac{-2.40 \times 10^{-19}\text{ C}}{-1.60 \times 10^{-19}\text{ C}} = 1.50\). Because isolated fractional elementary charges do not exist, holding 1.5 electrons is physically impossible.
Distractor Rationale: (A) has \(N = 2\), (B) has \(N = 3\) deficit electrons, and (D) has \(N = 5\) deficit electrons; all represent valid whole integers.
Question 3: Induced Polarization in Insulators Correct: (B)
Explanation: In insulators, electrons cannot flow freely through the bulk. However, the external field from the positive rod induces atomic dipole polarization: negative electron clouds displace closer to the rod, while positive nuclei shift farther away. Because electric force scales as \(1/r^2\), attraction on closer negative charges exceeds repulsion on distant positive charges, resulting in net attraction.
Distractor Rationale: (A) wrongly claims free bulk electron flow in plastic. (C) is unrelated. (D) assumes mobile positive charges.
Question 4: Sequential Conduction & Charge Sharing Correct: (A)
Explanation: When identical conductors touch, charge divides equally: First, X (\(+6Q\)) touches Y (\(-2Q\)): \(Q_{\text{tot}} = +4Q \implies q_X = q_Y = +2Q\). Next, Y (\(+2Q\)) touches Z (\(0\)): \(Q_{\text{tot}} = +2Q \implies q_Y = q_Z = +1Q\). Final distribution: \(q_X = +2Q,\; q_Y = +1Q,\; q_Z = +1Q\). Total charge is \(+4Q\), demonstrating conservation.
Distractor Rationale: (B) neglects contact between Y and Z. (C) fails to divide charge. (D) assumes simultaneous three-way contact.
AP Physics 2 • Scoring Guide Page 1 of 4
Section I: Multiple-Choice Detailed Explanations (Continued) Topic 10.2 • Questions 5 - 8
Question 5: Charging by Induction with Grounding Correct: (D)
Explanation: Charging by induction sequence: (1) Bring negative rod near sphere → repels free electrons to far side; (2) Ground the sphere → repelled electrons escape down into Earth; (3) Disconnect ground → locks in the electron deficit while the rod still polarizes; (4) Remove rod → positive net charge distributes across sphere.
Distractor Rationale: In (B), removing the rod before breaking the ground allows electrons from Earth to return, leaving the sphere neutral. In (A), conduction transfers negative charge.
Question 6: Conductor vs. Insulator Internal Field & Equilibrium Correct: (B)
Explanation: In a conductor, mobile conduction electrons drift macroscopically opposite to \(\vec{E}\) until the internal electric field created by separated surface charges exactly cancels the external field (\(E_{\text{net}} = 0\)). In an insulator, electrons cannot leave atoms; they only shift within molecules (dielectric polarization), reducing but not eliminating the internal field (\(E_{\text{net}} > 0\)).
Distractor Rationale: (A) incorrectly claims free electron flow occurs in insulators. (C) ignores induced polarization. (D) attributes mobility to positive nuclei.
Question 7: Charge Sharing Between Unequal Conductors Correct: (C)
Explanation: Touching conductors must reach identical electric potentials: \(V_1 = V_2\). Since \(V = \frac{k q}{r}\), we have \(\frac{k q_1}{R} = \frac{k q_2}{2R} \implies q_2 = 2 q_1\). By charge conservation, \(q_1 + q_2 = +Q \implies 3q_1 = +Q \implies q_1 = \frac{1}{3}Q,\; q_2 = \frac{2}{3}Q \approx +0.67Q\). The larger sphere holds greater than half the charge (\(q_2 > +0.5Q\)).
Distractor Rationale: (A) is a common misconception assuming equal 50/50 division; equal division occurs only when spheres have equal radii.
Question 8: Nuclear / Subatomic Charge Conservation Correct: (C)
Explanation: In an isolated system, net electric charge is conserved: \(\sum Q_{\text{initial}} = \sum Q_{\text{final}}\). Initial charge: \(q_\alpha + q_{\text{atom}} = +2e + 0 = +2e\). Final charge: \(2(-1e) + (+1e) + q_X = -1e + q_X\). Equating charges: \(+2e = -1e + q_X \implies q_X = +3e\).
Distractor Rationale: (A) neglects the negative sign of the electrons. (B) assumes fragment \(X\) alone takes initial charge. (D) adds electron charges as positive.
AP Physics 2 • Scoring Guide Page 2 of 4
Section II: Free-Response Scoring Guide
Question 1: Process of Charging & Charge Distribution (8 Points)
8 Points Total
(a) Charge Distribution Diagram [2 Points] Skills: 1.A, 10.2.A.1
+++
−−− A .
. B +++
• Point 1: For showing negative charges (\(-\)) accumulated along the left surface of Sphere A (closest to the positive rod).
• Point 2: For showing equal number of positive charges (\(+\)) along the right surface of Sphere B, with the interior and contact junction remaining net neutral.
(b) Microscopic Movement & Separation [3 Points] Skills: 3.B, 10.2.A.2
Exemplar Response: Sphere A acquires a net negative charge; Sphere B acquires an equal-magnitude net positive charge. Because the spheres were touching, free conduction electrons throughout both spheres were attracted toward the positively charged rod, crossing from Sphere B into Sphere A. Pulling Sphere B away traps these extra electrons on A and leaves a deficit on B. Positive atomic nuclei (protons) remain stationary.
• Point 1: Identifying that Sphere A is net negative and Sphere B is net positive.
• Point 2: Explaining that electrons migrated from Sphere B to Sphere A due to attraction by the positive rod.
• Point 3: Explicitly stating that positive charges (protons/nuclei) do NOT move.
(c) Grounding Sequence Analysis [3 Points] Skills: 3.C, 10.2.A.3
Exemplar Response: The two-sphere system acquires a net negative charge. When grounded, the positive rod attracts mobile electrons from the Earth into the spheres, neutralizing positive charges on Sphere B and adding excess electrons to the system. Disconnecting ground traps these incoming electrons before the rod is removed.
• Point 1: Stating that the final net charge on the two-sphere system is negative.
• Point 2: Explaining that electrons flow from ground into the spheres (attracted by the positive rod).
• Point 3: Explaining that disconnecting the ground before removing the rod traps the excess electrons.
AP Physics 2 • Scoring Guide Page 3 of 4
Section II: Free-Response Scoring Guide
Question 2: Charge Quantization & Conservation (9 Points)
9 Points Total
(a) Identifying Impossible Fractional Charge [3 Points] Skills: 2.C, 10.1.A.1
Exemplar Response: Droplet 4 (\(q = -2.40 \times 10^{-19}\text{ C}\)) contains the error. Charge is quantized by \(q = n e\), where \(n \in \mathbb{Z}\). For Droplet 4, \(n = \frac{-2.40 \times 10^{-19}\text{ C}}{-1.60 \times 10^{-19}\text{ C}} = 1.50\). Because elementary charge \(e\) is indivisible in isolated systems, half-integer multiples are physically impossible.
• Point 1:Correctly identifying Droplet 4.
• Point 2:Quantitative calculation showing the non-integer ratio \(q/e = 1.50\).
• Point 3:Explicitly citing the physical law of charge quantization (indivisibility of \(e\)).
(b) Electron Count on Droplet 3 [3 Points] Skills: 2.B, 10.2.A.2
Exemplar Response: Since net charge is negative (\(q = -6.40 \times 10^{-19}\text{ C}\)), the droplet has an excess of electrons. Calculation: \(N = \frac{|q|}{e} = \frac{6.40 \times 10^{-19}\text{ C}}{1.60 \times 10^{-19}\text{ C}} = 4\). The droplet carries exactly 4 excess electrons.
• Point 1:Correctly identifying an excess of electrons due to negative charge.
• Point 2:Applying \(N = |q|/e\) with correct values substituted.
• Point 3:Calculating the exact integer value of 4 electrons.
(c) Macroscopic Charge Sharing & Carrier Count [3 Points] Skills: 2.B, 2.C, 10.2.A.1
Exemplar Response: Total charge: \(Q_{\text{tot}} = -16.0\text{ nC} + 6.4\text{ nC} = -9.6\text{ nC}\). Identical spheres divide charge equally: \(Q_{1,\text{f}} = Q_{2,\text{f}} = \frac{-9.6\text{ nC}}{2} = \mathbf{-4.8\text{ nC}}\). Charge transferred to Sphere 2: \(\Delta Q_2 = -4.8\text{ nC} - (+6.4\text{ nC}) = -11.2\text{ nC}\). Electrons transferred: \(N_e = \frac{11.2 \times 10^{-9}\text{ C}}{1.60 \times 10^{-19}\text{ C}} = \mathbf{7.0 \times 10^{10}\text{ electrons}}\), moving from Sphere 1 to Sphere 2.
• Point 1:Correct final charge calculation of \(\mathbf{-4.8\text{ nC}}\) on each sphere.
• Point 2:Calculating transferred electrons: \(\mathbf{7.0 \times 10^{10}\text{ electrons}}\).
• Point 3:Specifying flow direction: from Sphere 1 into Sphere 2.
AP Physics 2 • Scoring Guide Page 4 of 4
