MCAT Chemical and Physical Foundations of Biological Systems Quiz: 4c Electric Potential Voltage Capacitance

What this quiz covers

This quiz focuses on 4c Electric Potential Voltage Capacitance, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

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. Capacitance is measured in volts, so increasing plate area should increase it in V.
2. Capacitance is measured in farads; changing geometry changes $C$, while voltage depends on $Q$ and $C$ via $V=Q/C$. (correct answer)
3. Capacitance is measured in ohms; increasing plate area increases resistance and therefore voltage.
4. Capacitance is measured in coulombs; increasing plate area increases charge directly even if the capacitor is isolated.

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.

Question 2

1. The magnitude of stored charge increases because capacitance increases at fixed voltage. (correct answer)
2. The magnitude of stored charge decreases because capacitance and charge are inversely related.
3. The membrane potential becomes less negative because higher capacitance forces voltage toward zero.
4. No change occurs because capacitance only affects current through resistors, not charge storage.

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.

Question 3

1. $V_m$ becomes more negative because adding positive charge increases the magnitude of the negative potential.
2. $V_m$ increases by $0.10\ \text{V}$ (100 mV), moving toward depolarization. (correct answer)
3. $V_m$ increases by $0.10\ \text{mV}$, a negligible change because capacitance is small.
4. $V_m$ does not change because voltage is determined only by resistance, not capacitance.

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.

Question 4

1. The voltage increases to $18\ \text{V}$ because the capacitance increased.
2. The voltage decreases to $2.0\ \text{V}$ because charge remains constant while capacitance increases. (correct answer)
3. The voltage remains $6.0\ \text{V}$ because disconnecting the battery fixes the voltage.
4. The voltage becomes $0\ \text{V}$ because inserting a dielectric neutralizes the plates.

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.

Question 5

1. Voltage remains $4.0\,\text{V}$ while the stored charge decreases. (correct answer)
2. Voltage decreases because increasing separation always lowers voltage.
3. Voltage increases because decreased capacitance forces $V$ to rise at fixed battery.
4. Stored charge remains constant because the capacitor was already charged.

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.

Question 6

1. Voltage decreases because capacitance increases while charge is fixed. (correct answer)
2. Voltage increases because higher permittivity increases the electric field.
3. Capacitance decreases because permittivity is inversely related to charge storage.
4. Voltage is unchanged because permittivity affects resistance, not capacitance.

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.

Question 7

1. The initial capacitor voltage is set by $V_0$ regardless of capacitance; larger capacitance changes how much charge is stored at $V_0$. (correct answer)
2. A larger capacitance necessarily gives a larger initial voltage because $V=CQ$.
3. A larger capacitance gives a smaller initial voltage because capacitance acts like resistance.
4. 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.

Question 8

1. The smaller capacitor has the larger voltage magnitude across it. (correct answer)
2. The larger capacitor has the larger voltage magnitude across it.
3. Both capacitors have the same voltage because series elements share voltage equally.
4. Each capacitor has $8.0\,\text{V}$ across it because both are connected to the battery.

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.

Question 9

1. The voltage decreases to $6\,\text{V}$ because the charge remains constant while capacitance increases. (correct answer)
2. The voltage increases to $24\,\text{V}$ because the capacitance increases with dielectric insertion.
3. The voltage remains $12\,\text{V}$ because the dielectric does not affect electric potential.
4. The voltage becomes $0\,\text{V}$ because inserting a dielectric neutralizes the plates.

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.

Question 10

1. The magnitude of stored charge decreases over time, lowering voltage. (correct answer)
2. Capacitance increases over time, lowering voltage, even though geometry is unchanged.
3. Resistance increases over time, directly lowering capacitor voltage at fixed charge.
4. Voltage decreases because the electric field reverses direction while charge stays constant.

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.

Question 11

1. Voltage remains fixed by the source, while the stored charge decreases. (correct answer)
2. Voltage increases because lower capacitance causes higher voltage in all capacitors.
3. Voltage decreases because removing dielectric always lowers potential difference to zero.
4. Capacitance increases because charge must flow out to keep voltage constant.

Explanation: This question tests understanding of capacitor behavior when dielectric is removed while connected to a voltage source. When connected to an ideal voltage source, the capacitor voltage remains fixed at the source voltage regardless of capacitance changes. Removing dielectric decreases capacitance (C = κε₀A/d, and κ decreases toward 1). With V constant and C decreased, the stored charge must decrease according to Q = CV. Choice B incorrectly assumes voltage increases with lower capacitance, confusing the connected case with the isolated case. To solve such problems, recognize that voltage sources maintain constant V, forcing charge to adjust when capacitance changes.

Question 12

1. $10\,\text{V}$ across each capacitor because series elements share the same voltage.
2. $5\,\text{V}$ across each capacitor because identical series capacitors split the battery voltage equally. (correct answer)
3. $0\,\text{V}$ across each capacitor because steady state implies no potential difference.
4. Greater than $10\,\text{V}$ across one capacitor because charge accumulates at the midpoint.

Explanation: This question tests understanding of voltage division in series capacitors. For capacitors in series, the same charge Q flows through both, but voltages divide inversely with capacitances. For identical capacitors, the total capacitance is Cₜₒₜₐₗ = C/2, and the charge is Q = CₜₒₜₐₗVₜₒₜₐₗ = (C/2)(10V) = 5CV. Each capacitor has voltage V = Q/C = 5CV/C = 5V. This equal division occurs because identical capacitors in series split the total voltage equally. Choice A incorrectly applies the parallel rule (same voltage) to series capacitors. To solve series capacitor problems, remember that charge is the same through all capacitors, while voltages add to equal the source voltage.

Question 13

1. Voltage remains determined by $Q/C$ because neither charge nor capacitance changes. (correct answer)
2. Voltage must increase because any external electric field adds directly to the capacitor's voltage.
3. Voltage must decrease because external fields always reduce potential differences inside conductors.
4. Capacitance changes because electric field strength determines farads.

Explanation: This question tests understanding of how external fields affect isolated capacitors. An isolated capacitor maintains its stored charge Q and geometric capacitance C = ε₀A/d (determined by plate area, separation, and dielectric). The voltage V = Q/C depends only on these internal parameters. An external uniform field parallel to the internal field does not change Q or C, so the voltage remains unchanged at V₀. The external field may affect the absolute potentials of the plates, but their difference (the capacitor voltage) is unaffected. Choice B incorrectly assumes external fields add to internal voltage. To analyze such problems, focus on what determines capacitor voltage: stored charge and capacitance, both unchanged by external fields.

Question 14

1. Both capacitors have the same voltage, and Capacitor B stores twice the charge of Capacitor A. (correct answer)
2. Capacitor B has twice the voltage because it has twice the capacitance.
3. Capacitor A stores more charge because smaller capacitance concentrates charge.
4. Both store the same charge because parallel elements share charge equally.

Explanation: This question tests understanding of parallel capacitor behavior. In parallel connection, both capacitors experience the same voltage (the battery voltage), while their charges differ according to Q = CV. Capacitor A stores Q_A = CV, while Capacitor B stores Q_B = (2C)V = 2CV = 2Q_A. Thus, Capacitor B stores twice the charge of Capacitor A at the same voltage. Choice B incorrectly assumes larger capacitance means larger voltage, failing to recognize that parallel elements share the same voltage. To analyze parallel capacitors, remember: same voltage across all, charge proportional to individual capacitance.

Question 15

1. Voltage decreases because capacitance increases while charge stays constant. (correct answer)
2. Voltage increases because larger area increases electric field strength.
3. Capacitance decreases because the plates are farther apart in terms of field lines.
4. Voltage is unchanged because only plate separation affects capacitance.

Explanation: This question tests understanding of capacitance changes with plate area in an isolated capacitor. For a parallel-plate capacitor, C = ε₀A/d, so doubling the overlapping area doubles the capacitance. Since the capacitor is charged and then isolated, the charge Q remains constant. With Q constant and C doubled, the voltage must decrease according to V = Q/C, becoming half its original value. Choice B incorrectly assumes larger area increases field strength and voltage, failing to recognize that E = σ/ε₀ = Q/(ε₀A) actually decreases with larger area. To analyze geometric changes in isolated capacitors, identify which parameter changes (A, d, or ε), calculate the new capacitance, then find the new voltage using constant charge.

Question 16

1. Capacitance must have increased during the transient. (correct answer)
2. Capacitance must have decreased during the transient.
3. Resistance must have increased because $V=IR$.
4. Charge must have decreased because voltage decreased.

Explanation: This question tests understanding of the V = Q/C relationship for isolated capacitors. When voltage decreases while charge magnitude remains constant, the only way to satisfy V = Q/C is if capacitance C increases. This could occur through various mechanisms in real devices (temperature effects, mechanical changes, etc.). Since Q is constant and V decreases, C must increase proportionally to maintain the equality. Choice B incorrectly suggests capacitance decreased, which would increase voltage at constant charge. To analyze such observations, apply V = Q/C algebraically: if Q is constant and V decreases, then C must increase.

Question 17

1. Voltage becomes $0$ because charge redistributes until both plates reach the same potential. (correct answer)
2. Voltage increases because charge is forced onto the plates by the conductor.
3. Voltage remains unchanged because capacitance is a geometric property.
4. Capacitance becomes infinite because the plates are shorted.

Explanation: This question tests understanding of charge neutralization in capacitors. When a conductor connects both plates, it provides a path for charge to flow from the positive plate to the negative plate. Since the plates initially have equal and opposite charges (+Q and -Q), the charges neutralize each other completely, leaving both plates uncharged. With Q = 0, the voltage V = Q/C = 0V, regardless of capacitance. Choice C incorrectly assumes voltage remains unchanged, ignoring that the conductor allows charge redistribution. To analyze such scenarios, recognize that conductors connecting capacitor plates allow charge equalization, resulting in zero net charge and zero voltage.

Question 18

1. Capacitor X, because smaller capacitance gives larger voltage at fixed charge. (correct answer)
2. Capacitor Y, because larger capacitance always produces larger voltage.
3. Both have the same voltage because they store the same charge.
4. Neither develops a voltage unless a resistor is in series to set the potential difference.

Explanation: This question tests understanding of the relationship between voltage, charge, and capacitance for isolated capacitors. The key principle is V = Q/C, which shows that voltage is inversely proportional to capacitance when charge is fixed. For capacitor X: V_X = 4.0 μC / 1.0 μF = 4.0 V, while for capacitor Y: V_Y = 4.0 μC / 4.0 μF = 1.0 V. Therefore, capacitor X with smaller capacitance has the higher voltage. Choice B incorrectly assumes larger capacitance always produces larger voltage, confusing the relationship when Q is fixed versus when V is fixed. To check similar problems, remember that for fixed charge, smaller capacitance means larger voltage (like a smaller container holding the same amount creates higher pressure).

Question 19

1. The voltage doubles because higher capacitance produces higher potential difference.
2. The voltage halves because the dielectric reduces the electric field.
3. The charge stored doubles because voltage is fixed and capacitance increases. (correct answer)
4. The charge stored is unchanged because the battery fixes charge, not voltage.

Explanation: This question tests understanding of capacitor behavior when connected to a battery while dielectric properties change. The key principle is that a connected battery maintains constant voltage regardless of capacitance changes. When capacitance doubles due to the higher dielectric constant material, and voltage remains fixed at V = 3.0 V, the charge must change according to Q = CV. Since C doubles and V is constant, Q also doubles. Choice D incorrectly claims the battery fixes charge rather than voltage, misunderstanding the fundamental property of voltage sources. To verify problems with connected capacitors, remember that batteries fix voltage, so any change in capacitance directly changes the stored charge proportionally.

Question 20

1. The voltage increases because bringing plates closer increases electric potential difference at fixed charge.
2. The voltage decreases because capacitance increases while charge is fixed. (correct answer)
3. The charge decreases because capacitance increases, forcing charge to leave the plates.
4. The voltage is unchanged because only the battery can change the potential difference.

Explanation: This question tests understanding of voltage changes in an isolated capacitor when geometry changes. The key principle is that for an isolated capacitor, charge Q is fixed while capacitance changes with geometry as C ∝ 1/d. When plate separation decreases from d to d/2, capacitance doubles. Since V = Q/C and Q is constant while C doubles, voltage must halve. Choice A incorrectly assumes voltage increases when plates come closer, confusing the effect on electric field (which does increase) with the effect on voltage (which decreases due to increased capacitance). To verify isolated capacitor problems, remember that Q is conserved, so increasing C always decreases V proportionally.