Capacitors
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AP Physics 2 › Capacitors
A $2.0,\mu\text{F}$ capacitor is connected to a $10,\text{V}$ ideal battery. Which statement best describes the voltage across the capacitor?
It is $10,\text{V}$ because it is connected to the battery terminals.
It is $5.0,\text{V}$ because the voltage splits between the plates.
It is $20,\text{V}$ because $V=QC$ increases with capacitance.
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.
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
twice as large because capacitance depends on stored charge
the same because parallel capacitors must have equal charge
half as large because the voltage is shared
twice as large because both have the same voltage
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.
A capacitor is connected to an ideal $5.0,\text{V}$ battery and reaches steady state. If the plate separation is doubled while still connected, which statement best describes the charge on the capacitor?
It increases because the electric field must increase
It remains the same because the battery keeps charge constant
It decreases because the capacitance decreases at fixed voltage
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.
A $6.0,\mu\text{F}$ capacitor is charged to $9.0,\text{V}$ by a battery, then disconnected and isolated. If the capacitance is increased to $12,\mu\text{F}$, which conclusion is correct about the final voltage?
It becomes $4.5,\text{V}$ because the charge decreases to keep $C$ constant
It remains $9.0,\text{V}$ because the capacitor was charged once
It becomes $18,\text{V}$ because capacitance increased
It becomes $4.5,\text{V}$ because charge stays 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).
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
twice as large because $Q$ is proportional to $C$ at fixed $V$.
zero because a capacitor stores current rather than charge.
half as large because larger capacitors store less charge.
the same because both are connected to $6.0,\text{V}$.
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.
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?
They are equal because series capacitors carry the same charge.
The $2.0,\mu\text{F}$ capacitor has twice the charge.
Both have zero charge because current stops at equilibrium.
The $4.0,\mu\text{F}$ capacitor has twice the charge.
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.
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?
$0,\text{V}$ because no current flows through a capacitor at steady state
It depends on the stored charge because voltage is always constant for a capacitor
$3.0,\text{V}$ because it is in parallel with the battery
$30,\text{V}$ because voltage increases with capacitance
Explanation
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 $4.0,\mu\text{F}$ 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
larger because the voltage stays constant.
smaller because the voltage stays constant.
unchanged because capacitance does not affect charge.
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.
A capacitor has charge $Q$ while connected to a battery. If the battery voltage is tripled and the capacitor is unchanged, what happens to $C$?
It triples because capacitance depends on voltage.
It stays the same because capacitance depends on geometry and dielectric.
It becomes zero because the capacitor stops storing current at higher voltage.
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.
Two capacitors, $C_1=4.0,\mu\text{F}$ and $C_2=8.0,\mu\text{F}$, are connected in parallel to a $3.0,\text{V}$ battery. Which conclusion is correct?
Both capacitors have the same voltage, $3.0,\text{V}$.
The $8.0,\mu\text{F}$ capacitor stores twice the current of the other.
Both capacitors have the same charge because they share a battery.
The $4.0,\mu\text{F}$ capacitor has $8.0,\text{V}$ across it.
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.