Electric Potential, Voltage, and Capacitance (4C)
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MCAT Chemical and Physical Foundations of Biological Systems › Electric Potential, Voltage, and Capacitance (4C)
An electrophysiology rig models a cell membrane as a capacitor. A membrane patch is approximated as $C=20\ \text{pF}$ and is initially at $V_m=-70\ \text{mV}$. A brief current pulse transfers $+2.0\ \text{pC}$ of charge onto the membrane capacitor (sign defined as increasing the inside potential). Based on $\Delta V=\Delta Q/C$, which outcome is most consistent with the change in electric potential? (Constants: $1\ \text{pC}=10^{-12}\ \text{C}$.)
$V_m$ increases by $0.10\ \text{mV}$, a negligible change because capacitance is small.
$V_m$ does not change because voltage is determined only by resistance, not capacitance.
$V_m$ increases by $0.10\ \text{V}$ (100 mV), moving toward depolarization.
$V_m$ becomes more negative because adding positive charge increases the magnitude of the negative potential.
Explanation
This question tests understanding of voltage changes in biological membrane capacitors when charge is added. The fundamental relationship ΔV = ΔQ/C governs how voltage changes when charge is transferred to a capacitor. With C = 20 pF and ΔQ = +2.0 pC, the voltage change is ΔV = 2.0 pC / 20 pF = 0.10 V = 100 mV. Since positive charge is added (increasing inside potential), the membrane potential increases from -70 mV to -70 + 100 = +30 mV, representing a 100 mV depolarization as stated in choice B. Choice C incorrectly calculates the change as 0.10 mV instead of 0.10 V, a unit conversion error of 1000-fold. To verify membrane potential calculations, always track the sign convention carefully (depolarization means becoming less negative) and convert between V and mV consistently.
A capacitor ($C=1.5\ \mu\text{F}$) is charged by an ideal $10\ \text{V}$ source and then disconnected. The plates are slowly pulled farther apart, reducing capacitance to $0.50\ \mu\text{F}$ while the capacitor remains isolated. Which statement best reflects the principle of capacitance in this system? (Constants: $Q=CV$; assume no charge leakage.)
The voltage decreases because the plates are farther apart, weakening the electric field.
The voltage increases because the same charge is now stored on a smaller capacitance.
The voltage stays the same because the capacitor was originally charged by a $10\ \text{V}$ source.
The voltage becomes zero because changing geometry forces charge to flow off the plates.
Explanation
This question tests understanding of voltage changes when an isolated capacitor's geometry is altered. For an isolated capacitor, charge Q remains constant after disconnection from the source. Initially, Q = CV = (1.5 μF)(10 V) = 15 μC. When capacitance decreases to 0.50 μF due to increased plate separation, the voltage must increase to maintain constant charge: V = Q/C = 15 μC / 0.50 μF = 30 V, making choice B correct. Choice A incorrectly assumes voltage decreases with separation, confusing the effect on electric field strength (E = V/d) with the effect on voltage itself. To analyze isolated capacitor problems, remember that mechanical work done to separate the plates increases the stored electrical energy, which manifests as increased voltage when charge is conserved.
A student compares two capacitor designs for a wearable device. Design X has plate area $A$ and separation $d$; Design Y has plate area $2A$ and separation $2d$, with the same dielectric in both. Each is connected to the same ideal battery (fixed $V$). Which statement is most consistent with the expected stored charge on Y relative to X? (Constants: $C=\varepsilon A/d$; $Q=CV$.)
Y stores more charge because increasing separation increases capacitance by increasing volume.
Y stores the same charge as X because $C_Y=\varepsilon(2A)/(2d)=C_X$ at the same $V$.
Y stores half the charge because doubling separation always halves capacitance regardless of area.
Y stores twice the charge because doubling area always doubles capacitance regardless of separation.
Explanation
This question tests understanding of how geometric scaling affects capacitance and stored charge. The capacitance formula C = εA/d shows that Design Y with area 2A and separation 2d has capacitance C_Y = ε(2A)/(2d) = εA/d = C_X, identical to Design X. Since both capacitors are connected to the same voltage V and Q = CV, they store the same charge, making choice A correct. Choice B incorrectly considers only the area change without accounting for the proportional change in separation, while choice C makes the opposite error. To analyze capacitor scaling problems, always consider all geometric parameters in the capacitance formula and recognize that proportional scaling of all dimensions leaves capacitance unchanged, similar to how electrical resistance scales with length and cross-sectional area.
In a circuit test, a $3.0\ \mu\text{F}$ capacitor is connected in series with a $9.0\ \mu\text{F}$ capacitor across an ideal $12\ \text{V}$ battery. After reaching steady state, which condition would lead to the highest voltage across the $3.0\ \mu\text{F}$ capacitor? (Constants: In series, capacitors carry the same $Q$; $V=Q/C$.)
The $3.0\ \mu\text{F}$ capacitor has the smaller voltage drop because it has the smaller capacitance.
Voltage depends only on the battery, so each capacitor has $12\ \text{V}$.
The $3.0\ \mu\text{F}$ capacitor has the larger voltage drop because it has the smaller capacitance at the same charge.
Each capacitor has $6.0\ \text{V}$ because series elements split voltage equally.
Explanation
This question tests understanding of voltage division in series capacitors. In series, capacitors must carry the same charge Q at steady state due to charge conservation. The equivalent capacitance is 1/C_eq = 1/3.0 + 1/9.0 = 4/9.0, giving C_eq = 2.25 μF. The total charge is Q = C_eq × V_total = 2.25 μF × 12 V = 27 μC. For each capacitor, V = Q/C, so V₁ = 27 μC / 3.0 μF = 9.0 V and V₂ = 27 μC / 9.0 μF = 3.0 V. The 3.0 μF capacitor has the larger voltage drop (9.0 V) because smaller capacitance requires larger voltage to store the same charge, confirming choice A. Choice B reverses this relationship, incorrectly stating smaller capacitance yields smaller voltage. To verify series capacitor problems, check that individual voltages sum to the total (9.0 + 3.0 = 12 V) and remember that voltage divides inversely with capacitance in series.
A materials lab measures the effect of a dielectric on a capacitor while keeping the battery connected (constant $V=5.0\ \text{V}$). Inserting the dielectric increases capacitance from $C_0$ to $2C_0$. Which outcome is most consistent with changes in electric potential energy stored in the capacitor? (Constants: $U=\tfrac{1}{2}CV^2$.)
Stored energy becomes zero because a dielectric prevents charge separation.
Stored energy doubles because $U\propto C$ at fixed $V$.
Stored energy halves because the dielectric reduces the electric field.
Stored energy is unchanged because the battery fixes the voltage.
Explanation
This question tests understanding of energy storage changes when a dielectric is inserted with constant voltage. The energy formula U = ½CV² shows that at fixed voltage, energy is directly proportional to capacitance. When the dielectric doubles capacitance from C₀ to 2C₀ while the battery maintains V = 5.0 V, the stored energy doubles from U₀ = ½C₀V² to U = ½(2C₀)V² = 2U₀, making choice A correct. Choice B incorrectly assumes energy decreases, perhaps confusing this scenario with the isolated capacitor case where inserting a dielectric reduces voltage and energy. To distinguish these cases, remember that with battery connected, the voltage is fixed and extra charge flows to utilize the increased capacitance, while in the isolated case, charge is fixed and voltage must decrease.
A charged capacitor is used to deliver a brief stimulus in a prototype device. The device uses either Capacitor A ($C=1.0\ \mu\text{F}$) or Capacitor B ($C=5.0\ \mu\text{F}$). In both cases, the same charge magnitude $Q=10\ \mu\text{C}$ is placed on the capacitor and the capacitor is isolated. Which statement is most consistent with the resulting voltage across the capacitor? (Constants: $V=Q/C$.)
Voltage depends on resistance in the circuit, so capacitance is irrelevant here.
Both have the same voltage because they carry the same charge.
Capacitor B has the higher voltage because larger capacitance means higher voltage for a given charge.
Capacitor A has the higher voltage because smaller capacitance means higher voltage for a given charge.
Explanation
This question tests understanding of the voltage-charge-capacitance relationship for isolated capacitors. The fundamental equation V = Q/C shows that for a given charge, voltage is inversely proportional to capacitance. With Q = 10 μC constant, Capacitor A yields V_A = 10 μC / 1.0 μF = 10 V while Capacitor B yields V_B = 10 μC / 5.0 μF = 2.0 V. Therefore, the smaller capacitance (Capacitor A) produces the higher voltage, making choice B correct. Choice A reverses this relationship, incorrectly claiming larger capacitance yields higher voltage at constant charge. To verify such problems, remember that capacitance represents the ability to store charge per unit voltage—larger capacitance means more charge storage at the same voltage, or equivalently, less voltage needed to store the same charge.
In a microfluidics experiment, a parallel-plate capacitor ($C=2.0\ \mu\text{F}$) is charged to $V_0=6.0\ \text{V}$ and then disconnected from the battery. A researcher inserts a dielectric slab that increases the capacitance to $6.0\ \mu\text{F}$ while the capacitor remains isolated. Which outcome is most consistent with changes in electric potential across the capacitor? (Constants: $Q=CV$.)
The voltage decreases to $2.0\ \text{V}$ because charge remains constant while capacitance increases.
The voltage remains $6.0\ \text{V}$ because disconnecting the battery fixes the voltage.
The voltage becomes $0\ \text{V}$ because inserting a dielectric neutralizes the plates.
The voltage increases to $18\ \text{V}$ because the capacitance increased.
Explanation
This question tests understanding of electric potential changes when a dielectric is inserted into an isolated capacitor. The fundamental principle is that for an isolated capacitor (disconnected from the battery), the charge Q remains constant while capacitance changes. In this system, the initial charge is Q = CV = (2.0 μF)(6.0 V) = 12 μC, and this charge remains fixed after disconnection. When the dielectric increases capacitance to 6.0 μF, the new voltage becomes V = Q/C = 12 μC / 6.0 μF = 2.0 V, making choice B correct. Choice A incorrectly assumes voltage increases with capacitance, failing to recognize that charge is conserved in an isolated system. To verify this type of problem, always check whether the capacitor is connected (V constant) or isolated (Q constant), then apply the appropriate constraint to find the new electrical quantity.
A lab group studies a capacitor ($C=4.0\ \mu\text{F}$) initially uncharged. It is connected to an ideal $12\ \text{V}$ battery and reaches steady state, then the battery is removed and the capacitor is immediately connected across a different ideal $6.0\ \text{V}$ battery with opposite polarity (terminals swapped). Which condition would lead to the highest magnitude of voltage across the capacitor immediately after the swap? (Assume ideal wires; constants: $Q=CV$.)
The capacitor voltage is $18\ \text{V}$ immediately after the swap because the two battery voltages add across it instantly.
The capacitor voltage is $6\ \text{V}$ immediately after the swap because it is now connected to a $6\ \text{V}$ battery.
The capacitor voltage is $12\ \text{V}$ immediately after the swap because capacitor voltage cannot change instantaneously.
The capacitor voltage is $0\ \text{V}$ immediately after the swap because opposite polarity cancels the charge instantly.
Explanation
This question tests understanding of capacitor voltage behavior during rapid switching between sources. The key principle is that capacitor voltage cannot change instantaneously because doing so would require infinite current (I = C dV/dt). Initially, the capacitor reaches steady state at 12 V with the first battery, storing Q = (4.0 μF)(12 V) = 48 μC. When immediately connected to the 6.0 V battery with opposite polarity, the capacitor voltage remains at 12 V for an instant before beginning to discharge and recharge, making choice A correct. Choice B incorrectly assumes the voltage jumps instantly to match the new source, violating the continuity requirement for capacitor voltage. To analyze such problems, remember that capacitors oppose instantaneous voltage changes just as inductors oppose instantaneous current changes, and the initial condition persists at t = 0+ immediately after any switch.
A parallel-plate capacitor is used as a tunable element in a sensor. It is connected to an ideal battery that maintains a constant potential difference of $9.0\ \text{V}$. The plate separation is increased while plate area and dielectric remain unchanged. Which outcome is most consistent with the capacitor’s behavior during this change? (Constants: $C=\varepsilon A/d$; $Q=CV$.)
Capacitance increases and the stored charge increases.
Capacitance decreases and the stored charge increases.
Capacitance is unchanged because the battery fixes the voltage.
Capacitance decreases and the stored charge decreases.
Explanation
This question tests understanding of how capacitance changes affect a capacitor connected to a constant voltage source. The capacitance formula C = εA/d shows that increasing plate separation d decreases capacitance since they are inversely proportional. With the battery maintaining constant voltage V = 9.0 V, the stored charge must adjust according to Q = CV. As capacitance decreases due to increased separation, the charge Q must also decrease to maintain the fixed voltage, making choice A correct. Choice B incorrectly reverses the relationship between separation and capacitance, while choice D fails to recognize that capacitance depends on geometry even with fixed voltage. To analyze variable capacitor problems, first determine whether voltage or charge is held constant, then apply the appropriate capacitance formula to predict how the other quantity changes.
A $2.0,\mu\text{F}$ capacitor is charged to $4.0,\text{V}$ and remains connected to the battery. The plate separation is increased, decreasing the capacitance to $1.0,\mu\text{F}$. Which outcome is most consistent with changes in electric potential across the capacitor? (Assume ideal battery maintains constant $V$.)
Voltage increases because decreased capacitance forces $V$ to rise at fixed battery.
Voltage decreases because increasing separation always lowers voltage.
Stored charge remains constant because the capacitor was already charged.
Voltage remains $4.0,\text{V}$ while the stored charge decreases.
Explanation
This question tests understanding of capacitor behavior when geometry changes while connected to a battery. When the capacitor remains connected to the battery, the voltage is fixed at 4.0V regardless of capacitance changes. Increasing plate separation decreases capacitance from 2.0 μF to 1.0 μF. Since V is constant and C decreases, the stored charge must decrease according to Q = CV: from Q₁ = (2.0 μF)(4.0V) = 8.0 μC to Q₂ = (1.0 μF)(4.0V) = 4.0 μC. Choice D incorrectly assumes charge remains constant when connected to a battery. To solve such problems, identify the constraint (V constant when connected to battery) and calculate how other quantities change.