Electric Potential, Voltage, and Capacitance (4C)
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MCAT Chemical and Physical Foundations of Biological Systems › Electric Potential, Voltage, and Capacitance (4C)
A lab group reports that a capacitor has a measured value of $2.0\ \text{V}$, and they plan to “increase the capacitance to 5.0 V” by changing plate area. Which statement is most consistent with correct interpretation of electric potential and capacitance?
Capacitance is measured in farads; changing geometry changes $C$, while voltage depends on $Q$ and $C$ via $V=Q/C$.
Capacitance is measured in volts, so increasing plate area should increase it in V.
Capacitance is measured in coulombs; increasing plate area increases charge directly even if the capacitor is isolated.
Capacitance is measured in ohms; increasing plate area increases resistance and therefore voltage.
Explanation
This question tests understanding of the fundamental units and relationships in capacitor systems. The key principle is recognizing that capacitance is measured in farads (F), not volts (V), and represents the ability to store charge per unit voltage: C = Q/V. The students' statement confuses capacitance (a property of the capacitor) with voltage (the potential difference across it). Changing plate area does change capacitance (C ∝ A), but the resulting voltage depends on whether the capacitor is connected to a source or isolated. Choice A incorrectly states capacitance is measured in volts, showing the same unit confusion as the students. To avoid such errors, always verify units: capacitance is in farads, voltage in volts, and charge in coulombs, related by Q = CV.
A neuron membrane can be approximated as a capacitor. In a simplified preparation, the membrane capacitance is $C_m=100\ \text{pF}$ and the membrane potential is held at $V_m=-70\ \text{mV}$ relative to the extracellular fluid. If $C_m$ increases to $200\ \text{pF}$ while the membrane potential is actively clamped at the same $V_m$ (ideal voltage clamp), which statement is most consistent with capacitance behavior? (Use $Q=CV$; sign indicates plate polarity.)
No change occurs because capacitance only affects current through resistors, not charge storage.
The membrane potential becomes less negative because higher capacitance forces voltage toward zero.
The magnitude of stored charge increases because capacitance increases at fixed voltage.
The magnitude of stored charge decreases because capacitance and charge are inversely related.
Explanation
This question tests understanding of capacitance in biological systems under voltage clamp conditions. The principle is that under ideal voltage clamp, the membrane potential V_m is held constant at -70 mV regardless of capacitance changes. When capacitance doubles from 100 pF to 200 pF while voltage is clamped constant, the charge must change according to Q = CV. Since C doubles and V remains at -70 mV, the magnitude of stored charge doubles (from 7 pC to 14 pC). Choice C incorrectly assumes higher capacitance forces voltage toward zero, not recognizing that the voltage clamp actively maintains the potential. To verify biological capacitor problems, identify whether voltage or current is clamped, then apply Q = CV with the appropriate constraint.
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 $0.10\ \text{V}$ (100 mV), moving toward depolarization.
$V_m$ does not change because voltage is determined only by resistance, not capacitance.
$V_m$ becomes more negative because adding positive charge increases the magnitude of the negative potential.
$V_m$ increases by $0.10\ \text{mV}$, a negligible change because capacitance is small.
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.
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 becomes $0\ \text{V}$ because inserting a dielectric neutralizes the plates.
The voltage increases to $18\ \text{V}$ because the capacitance increased.
The voltage remains $6.0\ \text{V}$ because disconnecting the battery fixes the voltage.
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 $2.0,\mu\text{F}$ 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 $V$.)
Stored charge remains constant because the capacitor was already charged.
Voltage increases because decreased capacitance forces $V$ to rise at fixed battery.
Voltage remains $4.0,\text{V}$ while the stored charge decreases.
Voltage decreases because increasing separation always lowers voltage.
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.
In a microfluidic sensor, a capacitor is formed by two electrodes separated by a polymer film. The device is charged to a fixed charge $Q$ and then isolated. If the polymer film is replaced by one with larger relative permittivity $\kappa$ (same thickness and area), which statement best reflects the principle of capacitance in this system? (Constants: $C\propto \kappa$; $V=Q/C$ when isolated.)
Capacitance decreases because permittivity is inversely related to charge storage.
Voltage is unchanged because permittivity affects resistance, not capacitance.
Voltage decreases because capacitance increases while charge is fixed.
Voltage increases because higher permittivity increases the electric field.
Explanation
This question tests understanding of how dielectric permittivity affects capacitance and voltage in an isolated capacitor. For a parallel-plate capacitor, C = κε₀A/d, where κ is the relative permittivity. Increasing κ increases capacitance proportionally. Since the capacitor is charged to fixed charge Q and then isolated, Q remains constant. With Q constant and C increased, the voltage must decrease according to V = Q/C. Choice B incorrectly claims higher permittivity increases electric field, confusing the effect on capacitance with field strength. To analyze dielectric effects, remember that κ appears in the numerator of the capacitance formula, and use the appropriate constraint (Q or V constant) based on whether the capacitor is isolated or connected.
A capacitor is used to deliver a brief stimulus pulse in a circuit. The capacitor is precharged to $V_0$ and then connected across a load. The student claims that increasing capacitance will increase the initial voltage delivered to the load for the same precharge. Which statement is most consistent with electric potential in this scenario? (Assume the precharge voltage $V_0$ is fixed by the charging source.)
The initial capacitor voltage is set by $V_0$ regardless of capacitance; larger capacitance changes how much charge is stored at $V_0$.
A larger capacitance gives a smaller initial voltage because capacitance acts like resistance.
A larger capacitance necessarily gives a larger initial voltage because $V=CQ$.
Capacitance has units of volts, so increasing it directly increases voltage.
Explanation
This question tests understanding of initial conditions in capacitor discharge circuits. When a capacitor is precharged to V₀ by a charging source, its initial voltage is V₀ regardless of its capacitance value. What capacitance does affect is how much charge Q = CV₀ is stored and how long the voltage can be maintained during discharge. A larger capacitance stores more charge at the same V₀, providing more energy and longer discharge time, but the initial voltage delivered to the load is still V₀. Choice B incorrectly assumes larger capacitance increases initial voltage, confusing energy storage with voltage. To analyze precharge scenarios, remember: the charging source sets the voltage; capacitance determines stored charge at that voltage.
A $1.0,\mu\text{F}$ capacitor and a $3.0,\mu\text{F}$ capacitor are connected in series across a $8.0,\text{V}$ battery. After steady state, which statement is most consistent with voltage distribution in series capacitors? (No calculations required.)
The larger capacitor has the larger voltage magnitude across it.
Each capacitor has $8.0,\text{V}$ across it because both are connected to the battery.
Both capacitors have the same voltage because series elements share voltage equally.
The smaller capacitor has the larger voltage magnitude across it.
Explanation
This question tests understanding of voltage division in series capacitors. In series, capacitors share the same charge Q, but voltages divide inversely with capacitances: V₁/V₂ = C₂/C₁. For C₁ = 1.0 μF and C₂ = 3.0 μF, we have V₁/V₂ = 3.0/1.0 = 3, meaning the smaller capacitor has three times the voltage of the larger one. Since V₁ + V₂ = 8.0V and V₁ = 3V₂, we get V₁ = 6.0V and V₂ = 2.0V. The smaller capacitor (1.0 μF) has the larger voltage magnitude. Choice C incorrectly assumes equal voltage division regardless of capacitance values. To solve series problems, remember: inverse voltage division - smaller C gets larger V.
In a benchtop setup, a $4.0,\mu\text{F}$ capacitor is charged by a $12,\text{V}$ battery, then disconnected from the battery (isolated) while remaining connected to its plates. A dielectric slab is inserted fully between the plates, increasing the capacitance to $8.0,\mu\text{F}$ without allowing charge to leak. Which outcome is most consistent with changes in electric potential across the capacitor? (Constants: $Q=CV$.)
The voltage becomes $0,\text{V}$ because inserting a dielectric neutralizes the plates.
The voltage decreases to $6,\text{V}$ because the charge remains constant while capacitance increases.
The voltage remains $12,\text{V}$ because the dielectric does not affect electric potential.
The voltage increases to $24,\text{V}$ because the capacitance increases with dielectric insertion.
Explanation
This question tests understanding of voltage changes when a dielectric is inserted into an isolated capacitor. The fundamental principle is that for an isolated capacitor, charge Q remains constant while capacitance changes according to C = κC₀, where κ is the dielectric constant. Since the capacitance doubles from 4.0 μF to 8.0 μF, we can determine κ = 2. With Q constant and C doubled, the voltage must change according to V = Q/C, so the new voltage is V₂ = Q/(2C₀) = V₀/2 = 12V/2 = 6V. Choice B incorrectly assumes voltage increases with capacitance, failing to recognize the inverse relationship when charge is fixed. To verify this type of problem, always check whether the capacitor is isolated (Q constant) or connected to a battery (V constant), then apply the appropriate constraint to V = Q/C.
A charged isolated capacitor is observed to have its voltage drift downward over time due to small leakage through the dielectric. Over that interval, the geometry is unchanged. Which statement is most consistent with the change in electric potential? (Assume capacitance remains constant.)
Capacitance increases over time, lowering voltage, even though geometry is unchanged.
Resistance increases over time, directly lowering capacitor voltage at fixed charge.
Voltage decreases because the electric field reverses direction while charge stays constant.
The magnitude of stored charge decreases over time, lowering voltage.
Explanation
This question tests understanding of voltage drift in leaky capacitors. For an isolated capacitor with small leakage, charge slowly escapes through the imperfect dielectric, decreasing the stored charge magnitude |Q| over time. Since the geometry is unchanged, capacitance C remains constant. With Q decreasing and C constant, the voltage must decrease according to V = Q/C. This explains the observed voltage drift. Choice B incorrectly suggests capacitance changes despite unchanged geometry. To analyze leakage effects, recognize that charge loss in isolated capacitors directly reduces voltage when capacitance is constant.