Capacitors - AP Physics 2

A $5.0\,\mu\text{F}$ capacitor is connected to a $10\,\text{V}$ ideal battery. A dielectric is inserted, increasing the capacitance to $10\,\mu\text{F}$ while still connected. Which statement best describes the charge on the capacitor?
Answer: It increases because $V$ stays $10\,\text{V}$ while $C$ increases. This question tests understanding of Capacitors. When a capacitor remains connected to a battery, the battery maintains a constant voltage across the capacitor. Initially, Q = CV = (5.0 μF)(10 V) = 50 μC. When the dielectric is inserted, capacitance doubles to 10 μF, but voltage remains 10 V because of the battery connection. The new charge is Q = CV = (10 μF)(10 V) = 100 μC, which is an increase. Choice B incorrectly states the battery fixes the charge—batteries fix voltage, not charge, allowing charge to flow as needed. Always remember that connected batteries maintain constant voltage while allowing charge to vary.
A $8.0\,\mu\text{F}$ capacitor is connected to a $6.0\,\text{V}$ battery. Compared to a $4.0\,\mu\text{F}$ capacitor on the same battery, the charge on the $8.0\,\mu\text{F}$ capacitor is
Answer: twice as large because $Q$ is proportional to $C$ at fixed $V$ . 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.
A $4.0\,\mu\text{F}$ capacitor is connected to a $20\,\text{V}$ battery. Which statement best describes the potential difference across the capacitor?
Answer: It is $20\,\text{V}$ because a connected battery fixes the capacitor's voltage This question tests understanding of Capacitors. When a capacitor is connected to a battery, the battery establishes and maintains a specific voltage across the capacitor's plates through the transfer of charge. The voltage across the capacitor equals the battery voltage once equilibrium is reached, regardless of the capacitor's capacitance value. For this 4.0 μF capacitor connected to a 20 V battery, the voltage across the capacitor is simply 20 V. Choice A incorrectly uses V = QC (which should be Q = CV), showing confusion about which quantities are fixed by the battery connection. Always remember that a connected battery determines the voltage across a capacitor, not the capacitance value.
A parallel-plate capacitor is charged and then disconnected from the battery, so it is isolated. The plate separation is reduced by a factor of $2$ without allowing charge to leak. Compared to before, the voltage across the capacitor is
Answer: half as large because the capacitance increases. 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.
Two capacitors, $C_1=2.0\,\mu\text{F}$ and $C_2=4.0\,\mu\text{F}$ , are connected in series across an ideal $12\,\text{V}$ battery and reach equilibrium. Which statement best describes the magnitude of charge on each capacitor?
Answer: They are equal because series capacitors carry the same charge. 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.
A $5.0\,\mu\text{F}$ capacitor is isolated with charge $20\,\mu\text{C}$ . Which statement best describes the voltage across it?
Answer: It is $4.0\,\text{V}$ because $V=Q/C$ . This question tests understanding of Capacitors. For an isolated capacitor with known charge and capacitance, the voltage is determined by rearranging the fundamental relationship Q = CV to get V = Q/C. With Q = 20 μC = 20 × 10⁻⁶ C and C = 5.0 μF = 5.0 × 10⁻⁶ F, the voltage is V = (20 × 10⁻⁶ C)/(5.0 × 10⁻⁶ F) = 4.0 V. Choice A incorrectly multiplies Q and C instead of dividing, showing confusion about the mathematical relationship between these quantities. Always use V = Q/C when finding voltage from known charge and capacitance.
A $4.0\,\mu\text{F}$ capacitor is connected directly across an ideal $9.0\,\text{V}$ battery. A dielectric is inserted fully, increasing the capacitance to $8.0\,\mu\text{F}$ while it remains connected. Compared to before insertion, the charge on the capacitor is
Answer: larger because the voltage stays constant. 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.
A capacitor is connected to a constant $10\,\text{V}$ source. If the plate separation is increased (same area, same dielectric), what happens to the stored charge?
Answer: It decreases because $C$ decreases and $Q=CV$ . This question tests understanding of Capacitors. For a parallel-plate capacitor, capacitance is given by C = εA/d, where ε is the permittivity, A is the plate area, and d is the separation distance. When the plate separation increases, capacitance decreases because C is inversely proportional to d. Since the capacitor remains connected to a constant voltage source, V stays fixed at 10 V while C decreases, causing the stored charge Q = CV to decrease proportionally. Choice C incorrectly assumes the voltage source fixes the charge rather than the voltage, confusing the roles of batteries and isolated capacitors. Always identify whether voltage or charge is held constant based on the circuit configuration.
A $4.0\,\mu\text{F}$ capacitor is connected to a $12\,\text{V}$ ideal battery and fully charged, then the battery is disconnected so the capacitor is isolated. Which statement best describes the capacitor's charge after it is isolated?
Answer: It remains $48\,\mu\text{C}$ because the capacitor is isolated. This question tests understanding of Capacitors. When a capacitor is charged by a battery, charge flows onto the plates until the voltage across the capacitor equals the battery voltage, with Q = CV = (4.0 μF)(12 V) = 48 μC. Once the battery is disconnected, the capacitor becomes isolated, meaning there is no conducting path for charge to flow off the plates. Since charge cannot escape from an isolated capacitor, the charge remains constant at 48 μC. Choice C incorrectly confuses charge with current—capacitors store charge, not current flow. Always remember that for an isolated capacitor, charge is conserved while voltage may change if capacitance changes.
An initially uncharged capacitor is connected to an ideal $6.0\,\text{V}$ battery and reaches electrostatic equilibrium. Without changing the battery, a second identical capacitor is connected in parallel with the first. Which statement best describes the voltage across the original capacitor after the connection?
Answer: It remains $6.0\,\text{V}$ because it is still across the battery. 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.
Two capacitors, $C_1=2.0\,\mu\text{F}$ and $C_2=6.0\,\mu\text{F}$ , are connected in series to a $12\,\text{V}$ battery. Which statement best describes their charges?
Answer: They have equal charge because series capacitors share the same $Q$ . This question tests understanding of Capacitors. When capacitors are connected in series, they must have the same charge Q on their plates because charge conservation requires that any charge leaving one capacitor's plate must accumulate on the adjacent plate of the next capacitor. This creates a chain where all capacitors in series share identical charge regardless of their individual capacitances. The voltages across each capacitor will differ according to V = Q/C, with the smaller capacitor having larger voltage. Choice A incorrectly assumes larger capacitance means more charge in series, confusing series behavior with parallel behavior. Always remember that series capacitors share charge while parallel capacitors share voltage.
A $5.0\,\mu\text{F}$ capacitor is connected to an ideal $10\,\text{V}$ battery. The battery remains connected while the plate separation is increased, reducing the capacitance to $2.5\,\mu\text{F}$ . Compared to before, the charge on the capacitor is
Answer: smaller because the capacitance is smaller at the same voltage. 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.
A $2.5\,\mu\text{F}$ capacitor is connected to a $20\,\text{V}$ ideal battery. Which statement best describes the capacitance if the capacitor becomes more charged?
Answer: It stays $2.5\,\mu\text{F}$ because capacitance depends on geometry and dielectric. This question tests understanding of Capacitors. Capacitance is a property determined by the physical characteristics of the capacitor: plate area, separation distance, and dielectric material between the plates. It does not depend on the amount of charge stored or the applied voltage. Therefore, the capacitance remains 2.5 μF regardless of how much charge accumulates on the plates. Choice A incorrectly suggests capacitance depends on stored charge—this confuses cause and effect, as charge depends on capacitance and voltage, not the reverse. Always remember that capacitance is determined by geometry and materials, not by electrical state.
A capacitor is charged to $12\,\text{V}$ , then disconnected and isolated. If its capacitance is doubled, what happens to its voltage?
Answer: It halves because $V=Q/C$ and $Q$ stays constant. This question tests understanding of Capacitors. When a capacitor is disconnected from a battery and isolated, the charge Q on its plates remains constant because there's no conducting path for charge to leave or enter. The relationship V = Q/C shows that voltage is inversely proportional to capacitance when charge is fixed. If capacitance doubles while charge stays constant, the voltage must halve to maintain the equality: V_new = Q/(2C) = (1/2)(Q/C) = V_old/2. Choice C incorrectly assumes isolated capacitors maintain constant voltage, which only occurs when connected to a voltage source. Always determine whether the capacitor is connected to a battery or isolated before analyzing changes.
A $10\,\mu\text{F}$ capacitor is connected to a $3.0\,\text{V}$ battery and reaches steady state. Which statement best describes the voltage across the capacitor?
Answer: $3.0\,\text{V}$ because it is in parallel with the battery This question tests understanding of Capacitors. When a capacitor reaches steady state connected to a battery, the voltage across the capacitor equals the battery voltage. This occurs because charge flows until the potential difference across the capacitor matches the battery's EMF, at which point current stops flowing. Therefore, the voltage across the capacitor is 3.0 V, matching the battery. Choice C incorrectly claims the voltage is zero, confusing the fact that current is zero at steady state with voltage being zero—these are different quantities. Always remember that at steady state, a capacitor's voltage equals the battery voltage, while current through it is zero.
A $3.0\,\mu\text{F}$ capacitor is connected across a $9.0\,\text{V}$ battery. Which conclusion is correct about the charge on one plate?
Answer: It is $27\,\mu\text{C}$ because $Q=CV$ for a capacitor connected to a voltage source This question tests understanding of Capacitors. The fundamental relationship for capacitors is Q = CV, where Q is the charge, C is the capacitance, and V is the voltage across the capacitor. When a capacitor is connected to a battery, the battery maintains a constant voltage across the capacitor's plates, causing charge to accumulate until Q = CV is satisfied. For this 3.0 μF capacitor connected to a 9.0 V battery, the charge is Q = (3.0 × 10⁻⁶ F)(9.0 V) = 27 × 10⁻⁶ C = 27 μC. Choice C shows the misconception that charge equals voltage numerically, ignoring the role of capacitance in determining how much charge accumulates. Always use Q = CV to find the charge on a capacitor when you know both capacitance and voltage.
A $3.0\,\mu\text{F}$ capacitor is connected to a $12\,\text{V}$ battery. The capacitor is replaced by a $6.0\,\mu\text{F}$ capacitor on the same battery. Compared to before, the voltage across the capacitor is
Answer: unchanged because it remains connected to the $12\,\text{V}$ battery. This question tests understanding of Capacitors. When a capacitor is connected to a battery, the battery maintains a constant voltage across its terminals regardless of what capacitor is connected. Whether using a 3.0 μF or 6.0 μF capacitor, the voltage across it will be 12 V to match the battery. The larger capacitor will store more charge (Q = CV), but the voltage remains unchanged at 12 V. Choice A incorrectly assumes larger capacitance reduces voltage—this would only occur for an isolated capacitor with fixed charge. Always determine whether the capacitor is connected to a battery or isolated.
A capacitor is isolated after being charged to a fixed charge $Q$ . If the plate area is increased while isolation is maintained, which statement best describes the capacitor's voltage?
Answer: It decreases because the capacitance increases 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.
Two capacitors, $3.0\,\mu\text{F}$ and $6.0\,\mu\text{F}$ , are connected in parallel across a $10\,\text{V}$ battery. Compared to the $3.0\,\mu\text{F}$ capacitor, the charge on the $6.0\,\mu\text{F}$ capacitor is
Answer: twice as large because both have the same voltage 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.
A $10\,\mu\text{F}$ capacitor is connected to a $5.0\,\text{V}$ source. If the plate area is doubled (same separation), what happens to the capacitor's charge?
Answer: It doubles because $C$ doubles and $Q=CV$ . 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.

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