AP Physics 2 Quiz: Capacitors

What this quiz covers

This quiz focuses on Capacitors, 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 $5.0\,\text{V}$ because the voltage splits between the plates.
2. It is $20\,\text{V}$ because $V=QC$ increases with capacitance.
3. It is $10\,\text{V}$ because it is connected to the battery terminals. (correct answer)
4. It is $0\,\text{V}$ because a capacitor blocks current.

Explanation: This question tests understanding of Capacitors. When a capacitor is connected directly to an ideal battery, the battery maintains a constant voltage across its terminals. The capacitor charges until the voltage across its plates equals the battery voltage, creating an equilibrium where no more charge flows. Therefore, the voltage across the capacitor must be 10 V, matching the battery. Choice D incorrectly states that capacitors block current by having zero voltage—while capacitors do block DC current in steady state, they do so by matching the battery voltage, not by having zero voltage. Always remember that a capacitor connected to a battery will have the same voltage as the battery.

Question 2

1. half as large because the voltage is shared
2. the same because parallel capacitors must have equal charge
3. twice as large because both have the same voltage (correct answer)
4. twice as large because capacitance depends on stored charge

Explanation: This question tests understanding of Capacitors. In a parallel connection, all capacitors experience the same voltage as the battery, which is 10 V for both capacitors. Using Q = CV, the 3.0 μF capacitor stores Q₁ = (3.0 μF)(10 V) = 30 μC, while the 6.0 μF capacitor stores Q₂ = (6.0 μF)(10 V) = 60 μC. Since 60 μC is twice 30 μC, the 6.0 μF capacitor has twice the charge. Choice B incorrectly assumes parallel capacitors must have equal charge, confusing this with the series case. Always apply Q = CV individually to each capacitor in parallel, using the same voltage for all.

Question 3

1. It decreases because the capacitance decreases at fixed voltage (correct answer)
2. It increases because the electric field must increase
3. It remains the same because the battery keeps charge constant
4. It becomes zero because capacitors do not store charge at steady state

Explanation: This question tests understanding of Capacitors. For a parallel-plate capacitor, capacitance is given by C = εA/d, where d is the plate separation. When d doubles, capacitance halves. Since the capacitor remains connected to the battery, the voltage stays constant at 5.0 V. Using Q = CV, if C halves while V remains constant, then Q must also halve, so the charge decreases. Choice C incorrectly assumes the battery keeps charge constant rather than voltage—batteries are constant voltage sources, not constant charge sources. Always identify whether the capacitor stays connected (constant V) or is isolated (constant Q) during changes.

Question 4

1. It becomes $18\,\text{V}$ because capacitance increased
2. It remains $9.0\,\text{V}$ because the capacitor was charged once
3. It becomes $4.5\,\text{V}$ because charge stays constant (correct answer)
4. It becomes $4.5\,\text{V}$ because the charge decreases to keep $C$ constant

Explanation: This question tests understanding of Capacitors. When a capacitor is disconnected from a battery and isolated, the charge Q remains constant because there's no path for charge to flow away. The relationship Q = CV tells us that if Q is constant and C changes, then V must change inversely. Initially, Q = (6.0 μF)(9.0 V) = 54 μC. When capacitance doubles to 12 μF, the new voltage is V = Q/C = 54 μC / 12 μF = 4.5 V. Choice A incorrectly assumes voltage increases with capacitance, missing that charge is conserved when isolated. Always determine whether the capacitor is connected to a battery (constant V) or isolated (constant Q).

Question 5

1. They are equal because series capacitors carry the same charge. (correct answer)
2. The $4.0\,\mu\text{F}$ capacitor has twice the charge.
3. The $2.0\,\mu\text{F}$ capacitor has twice the charge.
4. Both have zero charge because current stops at equilibrium.

Explanation: This question tests understanding of Capacitors. In a series connection, the same charge flows through all components, establishing equal charge magnitudes on each capacitor. This occurs because charge cannot accumulate between capacitors—what flows onto one plate must flow off the connected plate. The voltages differ (inversely proportional to capacitances), but charges are identical. Choice B incorrectly assumes larger capacitance means larger charge in series, confusing series behavior with parallel behavior. Always remember: series means equal charges, parallel means equal voltages.

Question 6

1. smaller because the voltage stays constant.
2. larger because the voltage stays constant. (correct answer)
3. unchanged because capacitance does not affect charge.
4. zero because a capacitor stores current, not charge.

Explanation: This question tests understanding of Capacitors. When a capacitor remains connected to a battery, the voltage across the capacitor is fixed at the battery voltage. The relationship Q = CV shows that charge is directly proportional to capacitance when voltage is constant. Since the capacitance doubles (from 4.0 μF to 8.0 μF) while voltage remains at 9.0 V, the charge must also double. Choice C incorrectly assumes capacitance doesn't affect charge, ignoring the fundamental Q = CV relationship. Always identify whether the battery remains connected (constant voltage) or is disconnected (constant charge) to determine which quantity stays fixed.

Question 7

1. unchanged because isolation prevents voltage change
2. doubled because capacitance increased
3. halved because the charge remains constant (correct answer)
4. halved because the charge decreases to keep capacitance constant

Explanation: This question tests understanding of Capacitors. When a charged capacitor is isolated, its charge Q remains constant because there's no conducting path for charge to leave or enter. Initially, Q = (2.0 μF)(8.0 V) = 16 μC. After inserting the dielectric, the capacitance doubles to 4.0 μF, but Q stays at 16 μC. Using V = Q/C, the new voltage is V = 16 μC / 4.0 μF = 4.0 V, which is half the original voltage. Choice B incorrectly assumes voltage doubles with capacitance, missing that charge is conserved in isolation. Always determine whether changes occur while connected to a battery (constant V) or while isolated (constant Q).

Question 8

1. the same because both are connected to $6.0\,\text{V}$.
2. half as large because larger capacitors store less charge.
3. twice as large because $Q$ is proportional to $C$ at fixed $V$. (correct answer)
4. zero because a capacitor stores current rather than charge.

Explanation: This question tests understanding of Capacitors. When capacitors are connected to the same battery, they experience the same voltage. Using Q = CV, the 4.0 μF capacitor stores Q₁ = (4.0 μF)(6.0 V) = 24 μC, while the 8.0 μF capacitor stores Q₂ = (8.0 μF)(6.0 V) = 48 μC. Since 48 μC is twice 24 μC, the larger capacitor stores twice the charge. Choice B incorrectly assumes larger capacitors store less charge—this reverses the actual relationship where charge is proportional to capacitance at fixed voltage. Always remember that at constant voltage, doubling capacitance doubles the stored charge.

Question 9

1. It stays the same because capacitance depends on geometry and dielectric. (correct answer)
2. It becomes zero because the capacitor stops storing current at higher voltage.
3. It triples because capacitance depends on voltage.
4. It becomes one-third because capacitance depends on charge.

Explanation: This question tests understanding of Capacitors. Capacitance is a property of the capacitor's physical structure, determined by factors like plate area, separation distance, and dielectric material according to C = εA/d. Capacitance does not depend on the applied voltage or stored charge; it remains constant for a given capacitor geometry. When the battery voltage is tripled, the charge will triple according to Q = CV, but the capacitance C itself remains unchanged. Choice A incorrectly assumes capacitance depends on voltage, confusing the cause-and-effect relationship in Q = CV. Always remember that capacitance is determined by geometry and materials, not by the electrical state of the capacitor.

Question 10

1. Both capacitors have the same voltage, $3.0\,\text{V}$. (correct answer)
2. Both capacitors have the same charge because they share a battery.
3. The $4.0\,\mu\text{F}$ capacitor has $8.0\,\text{V}$ across it.
4. The $8.0\,\mu\text{F}$ capacitor stores twice the current of the other.

Explanation: This question tests understanding of Capacitors. When capacitors are connected in parallel, they share the same voltage across their terminals because their positive plates connect to the same point and their negative plates connect to the same point. Both capacitors experience the full battery voltage of 3.0 V. However, they store different amounts of charge according to Q = CV: the 4.0 μF capacitor stores Q₁ = (4.0 μF)(3.0 V) = 12 μC, while the 8.0 μF capacitor stores Q₂ = (8.0 μF)(3.0 V) = 24 μC. Choice B incorrectly assumes parallel capacitors share the same charge, confusing parallel with series behavior. Always remember that parallel capacitors share voltage while series capacitors share charge.

Question 11

1. It stays the same because the voltage source fixes the charge.
2. It becomes zero because increasing area reduces stored current.
3. It halves because the larger area spreads charge more thinly.
4. It doubles because $C$ doubles and $Q=CV$. (correct answer)

Explanation: This question tests understanding of Capacitors. For a parallel-plate capacitor, capacitance is proportional to plate area: C = εA/d. When the plate area doubles while separation remains constant, the capacitance doubles from 10 μF to 20 μF. Since the capacitor remains connected to the 5.0 V source, the voltage stays constant while the increased capacitance allows more charge to be stored according to Q = CV. The new charge is Q_new = (20 μF)(5.0 V) = 100 μC, which is double the original charge of 50 μC. Choice A incorrectly assumes charge decreases with larger area, misunderstanding how increased capacitance affects charge storage. Always use Q = CV and determine which quantities change based on the circuit configuration.

Question 12

1. It increases because the capacitance increases
2. It decreases because the capacitance increases (correct answer)
3. It remains constant because isolation fixes the voltage
4. It becomes zero because the capacitor stores current instead of charge

Explanation: This question tests understanding of Capacitors. For a parallel-plate capacitor, capacitance is C = εA/d, where A is the plate area. When area increases, capacitance increases proportionally. Since the capacitor is isolated, charge Q remains constant. Using V = Q/C, if C increases while Q stays constant, then V must decrease inversely. This makes physical sense: spreading the same charge over a larger area reduces the electric field and thus the voltage. Choice C incorrectly claims isolation fixes voltage rather than charge—isolation prevents charge flow, not voltage changes. Always use V = Q/C for isolated capacitors, remembering that Q is constant during isolation.

Question 13

1. unchanged because voltage is fixed for an isolated capacitor
2. half as large because capacitance increases when extra charge is added
3. twice as large because $V=Q/C$ and capacitance is unchanged (correct answer)
4. zero because adding charge cancels the stored current in the capacitor

Explanation: This question tests understanding of Capacitors. For an isolated capacitor, the capacitance C remains constant because it depends only on the physical geometry and dielectric material, not on the charge or voltage. When charge is added to an isolated capacitor, we use V = Q/C to find the new voltage. If the initial charge Q gives voltage V = Q/C, then doubling the charge to 2Q gives V_new = (2Q)/C = 2(Q/C) = 2V, which is twice the original voltage. Choice B incorrectly claims that capacitance increases when charge is added, misunderstanding that capacitance is a fixed property of the capacitor's construction. Always remember that for isolated capacitors, C is constant and V = Q/C shows voltage is proportional to charge.

Question 14

1. It is half as large because the voltage is the same. (correct answer)
2. It is twice as large because the capacitance is smaller.
3. It is the same because the battery sets the current.
4. It is zero because capacitors only store voltage.

Explanation: This question tests understanding of Capacitors. When a capacitor is connected to a battery, the voltage across it equals the battery voltage. Using Q = CV, the original charge is Q₁ = 6.0 μF × 12 V = 72 μC. The new capacitor has Q₂ = 3.0 μF × 12 V = 36 μC, which is half the original charge. Choice B incorrectly inverts the relationship, thinking smaller capacitance means larger charge at the same voltage. Always apply Q = CV directly: at constant voltage, charge is proportional to capacitance.

Question 15

1. It doubles because the dielectric increases capacitance.
2. It halves because the charge remains constant. (correct answer)
3. It stays at $V$ because the dielectric does not affect voltage.
4. It becomes zero because the battery was removed.

Explanation: This question tests understanding of Capacitors. Initially, the capacitor has charge Q = CV and is then isolated, making Q constant. When a dielectric doubles the capacitance, we have a new situation where Q = C'V', with C' = 2C. Since charge is conserved (Q = Q'), we get CV = 2CV', which gives V' = V/2. Choice C incorrectly assumes dielectrics don't affect voltage, missing that voltage must adjust to conserve charge when capacitance changes. Always track which quantity (Q or V) remains constant based on whether the capacitor is isolated or connected.

Question 16

1. twice as large because the capacitance decreases.
2. half as large because the capacitance increases. (correct answer)
3. unchanged because the separation does not affect voltage.
4. zero because isolated capacitors cannot maintain voltage.

Explanation: This question tests understanding of Capacitors. When a capacitor is isolated, its charge Q remains constant. For a parallel-plate capacitor, C = ε₀A/d, so halving the separation doubles the capacitance. Since Q = CV and Q is constant while C doubles, the voltage V must be halved to maintain the same charge. Choice A incorrectly predicts voltage doubles, reversing the relationship between capacitance and voltage for constant charge. Always use Q = CV with the constraint that matches the physical situation: constant V when connected to a battery, constant Q when isolated.

Question 17

1. It doubles because the capacitance increases. (correct answer)
2. It halves because the electric field decreases.
3. It stays the same because the battery fixes the charge.
4. It becomes zero because increasing area reduces current storage.

Explanation: This question tests understanding of Capacitors. For a parallel-plate capacitor, capacitance is C = ε₀A/d, where A is plate area and d is separation. When connected to a battery, voltage V remains constant. Since Q = CV and doubling the area doubles the capacitance while V stays fixed, the charge must double. Choice C incorrectly claims the battery fixes the charge rather than the voltage, confusing the roles of connected versus isolated capacitors. Always remember that a connected battery maintains constant voltage, allowing charge to adjust with capacitance changes.

Question 18

1. $10\,\text{V}$ because each capacitor keeps its original voltage.
2. $5.0\,\text{V}$ because charge redistributes over twice the capacitance. (correct answer)
3. $20\,\text{V}$ because the charges add while capacitance stays the same.
4. $0\,\text{V}$ because capacitors cannot share charge without a battery.

Explanation: This question tests understanding of Capacitors. When an isolated charged capacitor is connected to an identical uncharged capacitor, charge redistributes until both reach the same voltage. Initially, Q₁ = C×10V = 20 μC on the first capacitor. When connected in parallel, total charge (20 μC) is conserved but spreads across total capacitance (2.0 + 2.0 = 4.0 μF). Using Q = CV, we get V = Q/C = 20 μC / 4.0 μF = 5.0 V. Choice A incorrectly assumes each capacitor maintains its original voltage, ignoring charge redistribution. Always apply charge conservation when analyzing isolated capacitor systems.

Question 19

1. It decreases because the total capacitance increases.
2. It increases because the charge splits between capacitors.
3. It remains $6.0\,\text{V}$ because it is still across the battery. (correct answer)
4. It becomes $0\,\text{V}$ because parallel capacitors cancel voltage.

Explanation: This question tests understanding of Capacitors. When capacitors are connected in parallel to a battery, they all share the same voltage as the battery. The fundamental principle is that parallel components have identical voltages across them. Adding a second capacitor in parallel doesn't change the voltage across the first capacitor—both capacitors will have 6.0 V across them. Choice A incorrectly suggests voltage decreases, confusing the effect on total capacitance with individual voltages. Always remember that parallel connection means equal voltages, while series connection means equal charges.

Question 20

1. larger because the plate separation is larger.
2. smaller because the capacitance is smaller at the same voltage. (correct answer)
3. unchanged because the battery fixes the charge.
4. zero because increasing separation stops charge storage.

Explanation: This question tests understanding of Capacitors. With the battery connected, voltage remains constant at 10 V. Initially, Q₁ = 5.0 μF × 10 V = 50 μC. After increasing plate separation, capacitance decreases to 2.5 μF, giving Q₂ = 2.5 μF × 10 V = 25 μC, which is half the original charge. Choice A incorrectly relates charge to plate separation rather than capacitance, missing that Q = CV directly. Always use the fundamental relationship Q = CV with the appropriate constraint: here, V is constant because the battery remains connected.