Capacitors - AP Physics 2 | Varsity Tutors

SECTION 1

Historical Context & Motivation

The ability to store electric charge fascinated natural philosophers long before anyone understood electricity at the atomic level. Early experimenters discovered that certain arrangements of conductors could "hold" charge and release it in dramatic sparks, hinting at a deeper relationship between geometry, materials, and the electric field. The capacitor — originally called a "condenser" — evolved from a laboratory curiosity into one of the most essential components in modern electronics, from camera flashes to cardiac defibrillators.

1745 • The Leyden Jar

Pieter van Musschenbroek and Ewald Georg von Kleist independently invented the Leyden jar, a glass jar lined inside and out with metal foil — the first practical capacitor.

1775 • Volta's Electrophorus

Alessandro Volta designed the electrophorus, demonstrating that charge could be stored and transferred repeatedly using flat parallel conductors, foreshadowing the parallel-plate geometry.

1837 • Faraday & Dielectrics

Michael Faraday showed that inserting an insulating material between the plates increases charge storage, introducing the concept of dielectric constant (κ).

1861 • Maxwell's Field Theory

James Clerk Maxwell formalized the idea that capacitors store energy in the electric field between the plates, not on the plates themselves — a cornerstone of electromagnetic theory.

The central question these discoveries address is remarkably practical: How can we store electrical energy without a chemical reaction, and how does the geometry of conductors and the choice of insulator govern that storage? Answering this question leads directly to the physics of capacitance, electric fields, and potential difference — core topics in AP Physics 2.

SECTION 2

Core Principles & Definitions

A capacitor is any system of two conductors carrying equal and opposite charges, separated by an insulating region (which may simply be a vacuum). When a voltage source is connected across the conductors, charge migrates until the potential difference across the plates matches the source voltage. The ratio of stored charge to that voltage defines the device's capacitance, measured in farads (F). Because one farad represents an enormous amount of charge storage, practical capacitors are rated in microfarads (μF), nanofarads (nF), or picofarads (pF).

Capacitance (C)

The ratio Q/V — how much charge a capacitor stores per volt of potential difference. Depends only on geometry and dielectric, not on Q or V.

Dielectric Material

An insulating substance between the plates that becomes polarized in an electric field, reducing the internal field and increasing capacitance by a factor κ.

Electric Field Between Plates

For a parallel-plate capacitor the field is approximately uniform: E = V/d. Energy is stored in this field, not on the plates themselves.

Energy Storage

A charged capacitor stores electrostatic potential energy U = ½CV². This energy can be released quickly, making capacitors essential in pulsed-power applications.

SECTION 3

Visual Explanation — The Parallel-Plate Capacitor

A parallel-plate capacitor with plate area A separated by distance d. The uniform electric field (amber arrows) points from the positive plate to the negative plate. The potential difference ΔV exists across the gap, and the three governing equations are shown at right.

The diagram above captures the essential physics of the ideal parallel-plate capacitor. Equal and opposite charges reside on the inner surfaces of the two plates, and the resulting electric field is nearly uniform throughout the interior (edge effects are neglected in the AP treatment). Notice that the field lines are parallel and equally spaced, confirming that E has the same magnitude and direction everywhere between the plates. The potential drops linearly from the positive plate to the negative plate, so V at any interior point can be found from V = V₀ − Ex, where x is the distance from the positive plate. Because capacitance depends only on geometry and the dielectric, changing the voltage changes Q proportionally but does not alter C itself.

SECTION 4

Mathematical Framework

The quantitative treatment of capacitors rests on a small set of equations that connect charge, voltage, geometry, dielectric properties, and energy. Mastery of these relationships — and knowing when each form is most convenient — is essential for AP Physics 2.

DEFINITION OF CAPACITANCE

C = Q / ΔV

C = capacitance (F), Q = magnitude of charge on one plate (C), ΔV = potential difference across the plates (V). This is the fundamental definition: capacitance is the charge stored per volt.

PARALLEL-PLATE CAPACITANCE

C = κε₀A / d

κ = dielectric constant (dimensionless, κ ≥ 1), ε₀ = permittivity of free space (8.85 × 10⁻¹² F/m), A = area of one plate (m²), d = separation between plates (m). Capacitance increases with plate area and dielectric constant, and decreases with plate separation.

ENERGY STORED IN A CAPACITOR

U = ½CV² = ½QV = Q² / (2C)

All three forms are equivalent; choose whichever contains the two known quantities. U has units of joules (J). The energy resides in the electric field between the plates.

ELECTRIC FIELD BETWEEN PLATES

E = ΔV / d = Q / (κε₀A)

The field is uniform and directed from the positive plate to the negative plate. Inserting a dielectric reduces E by the factor 1/κ for a fixed charge Q on the plates.

ℹ

AP Exam Tip

SECTION 5

Capacitors in Series and Parallel

Circuits rarely contain a single capacitor. When multiple capacitors are wired together, the combination can be reduced to a single equivalent capacitance using rules that mirror — but are inverted relative to — the combination rules for resistors. The two fundamental arrangements are parallel (same voltage, charges add) and series (same charge, voltages add).

Left: capacitors in parallel share the same voltage; their capacitances add directly. Right: capacitors in series carry the same charge; their reciprocals add. Note the analogy inversion relative to resistor combination rules.

Property	Parallel	Series
Shared quantity	Voltage (ΔV)	Charge (Q)
Additive quantity	Charge: Q_total = Σ $Q_i$	Voltage: ΔV_total = ΣΔ $V_i$
Equivalent capacitance	C_eq = C₁ + C₂ + ...	1/C_eq = 1/C₁ + 1/C₂ + ...
C_eq vs. individual C	Always larger than the largest	Always smaller than the smallest
Resistor analogy	Opposite of parallel resistors	Opposite of series resistors

Comparison of parallel and series capacitor combinations

SECTION 6

Worked Example — Dielectric Insertion

A parallel-plate capacitor with plate area A = 0.020 m² and separation d = 1.0 mm is connected to a 12 V battery. After charging, the battery is disconnected and a dielectric with κ = 4.0 is inserted between the plates. Find: (a) the original capacitance, (b) the charge on the plates, (c) the new capacitance, (d) the new voltage, and (e) the energy stored before and after the dielectric is inserted.

Worked example

1
Step 1 — Original Capacitance

Use C = ε₀A/d with no dielectric (κ = 1). C = (8.85 × 10⁻¹² F/m)(0.020 m²) / (1.0 × 10⁻³ m).

C₀ = 1.77 × 10⁻¹⁰ F ≈ 177 pF
2
Step 2 — Charge on the Plates

While still connected to the battery: Q = C₀ΔV = (1.77 × 10⁻¹⁰ F)(12 V).

Q = 2.12 × 10⁻⁹ C ≈ 2.12 nC
3
Step 3 — New Capacitance with Dielectric

Inserting the dielectric multiplies capacitance by κ: C_new = κC₀ = 4.0 × 177 pF.

C_new = 708 pF
4
Step 4 — New Voltage

The battery is disconnected, so Q is fixed. ΔV_new = Q / C_new = 2.12 × 10⁻⁹ C / 7.08 × 10⁻¹⁰ F.

ΔV_new = 3.0 V (reduced by factor of κ = 4)
5
Step 5 — Energy Comparison

Before: U₀ = ½C₀V₀² = ½(1.77 × 10⁻¹⁰)(12)² = 1.27 × 10⁻⁸ J. After: U_new = ½C_new V_new² = ½(7.08 × 10⁻¹⁰)(3.0)² = 3.19 × 10⁻⁹ J. Alternatively, U_new = Q²/(2C_new) gives the same answer.

Energy decreases by a factor of κ = 4. The "lost" energy went into the work done pulling the dielectric into the gap (the field does positive work on the dielectric).

SECTION 7

Dielectric Effects — Battery Connected vs. Disconnected

One of the most tested conceptual areas on the AP Physics 2 exam involves reasoning about what happens to various capacitor quantities when a dielectric is inserted. The answer depends critically on whether the capacitor remains connected to a voltage source. The table below summarizes both scenarios for a dielectric of constant κ > 1.

Quantity	Battery Connected (ΔV fixed)	Battery Disconnected (Q fixed)
Capacitance C	Increases by κ	Increases by κ
Charge Q	Increases by κ	Unchanged
Voltage ΔV	Unchanged	Decreases by κ
Electric field E	Unchanged (ΔV/d same)	Decreases by κ
Energy U	Increases by κ	Decreases by κ

Effect of inserting a dielectric slab (κ > 1)

κ=3 gives C 10→30 nF; connected V=100 V, Q 1→3 µC, U 50→150 µJ; isolated Q=1 µC, V=33.3 V, U=16.7 µJ. For complete insertion of a dielectric with κ = 3, capacitance triples. The connected capacitor holds voltage fixed; the isolated capacitor holds charge fixed.

SECTION 8

Connection to RC Circuits and Beyond

In AP Physics 2, capacitors appear not only as static energy-storage devices but also in RC circuits, where a resistor controls the rate at which the capacitor charges or discharges. The time constant τ = RC sets the characteristic timescale: after one time constant the capacitor has charged to about 63% of its final voltage. Although the exponential charging equation itself is treated qualitatively on the AP exam, understanding how C and R jointly govern circuit behavior is essential for experimental design and qualitative FRQs.

Concept	Static Capacitor (This Lesson)	RC Circuit (Advanced)
Focus	Energy storage, field, and geometry	Time-dependent charging/discharging
Key equation	C = Q/ΔV ; U = ½CV²	V(t) = V₀(1 − $e^{−t/RC}$ )
Math level	Algebra	Exponential functions (qualitative on AP 2)
AP exam relevance	MCQ and FRQ every year	Experimental design FRQs, graph interpretation

Static capacitor concepts vs. RC circuit extensions

Beyond the AP course, capacitors connect to the concept of energy density in electromagnetic fields (u = ½ε₀E²), AC circuit impedance, and the displacement current that Maxwell added to Ampère's law. Each of these extensions builds directly on the static-capacitor foundations developed here, so a solid grasp of this material will pay dividends throughout electromagnetism.

100 µF at 300 V stores 4.5 J; separate charging and flash discharge paths. A capacitor stores energy while charging and releases energy into a flash tube on discharge; some energy may be lost.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A parallel-plate capacitor is fully charged and then disconnected from the battery. The plates are then slowly pulled apart to double the separation d. Which of the following correctly describes the changes to capacitance C, charge Q, and voltage ΔV?

A. C halves, Q unchanged, ΔV doubles
B. C halves, Q halves, ΔV unchanged
C. C doubles, Q unchanged, ΔV halves
D. C halves, Q doubles, ΔV quadruples

Show answer (A)

A is correct. Because the battery is disconnected, Q is fixed (charge has nowhere to go). Since C = ε₀A/d, doubling d halves C. Then ΔV = Q/C doubles because Q is constant and C halved. B is wrong because Q cannot change on an isolated capacitor. C is wrong because increasing d decreases (not increases) C. D is wrong because Q remains unchanged — conservation of charge requires this on an isolated conductor.

Question 2: BASIC CALCULATION

A 4.7 μF capacitor is connected to a 9.0 V battery and fully charged. How much energy is stored in the capacitor?

A. 1.9 × 10⁻⁴ J
B. 4.2 × 10⁻⁵ J
C. 3.8 × 10⁻⁴ J
D. 2.1 × 10⁻² J

Show answer (A)

A is correct. U = ½CV² = ½(4.7 × 10⁻⁶ F)(9.0 V)² = ½(4.7 × 10⁻⁶)(81) = 1.90 × 10⁻⁴ J. B is wrong — it appears to use U = ½CV (missing the square on V). C is wrong — it uses U = CV² (missing the factor of ½). D is wrong by roughly two orders of magnitude, likely from a unit conversion error with μF.

Question 3: INTERMEDIATE

Three capacitors — C₁ = 2.0 μF, C₂ = 3.0 μF, and C₃ = 6.0 μF — are connected in series across a 12 V battery. What is the voltage across C₁?

A. 2.0 V
B. 4.0 V
C. 6.0 V
D. 12 V

Show answer (C)

C is correct. First find C_eq: 1/C_eq = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1, so C_eq = 1.0 μF. The charge on each capacitor in series is Q = C_eq × ΔV_total = (1.0 μF)(12 V) = 12 μC. The voltage across C₁ is ΔV₁ = Q/C₁ = 12 μC / 2.0 μF = 6.0 V. A (2.0 V) incorrectly assumes equal voltage sharing. B (4.0 V) divides the total voltage by 3. D (12 V) incorrectly assigns the total voltage to one capacitor — that would only be true if C₁ were alone.

Question 4: APPLIED

A student designs a parallel-plate capacitor for an experiment. She uses two square aluminum plates (side length 0.30 m) separated by a 0.50 mm sheet of Mylar (κ = 3.1). The capacitor is charged to 50 V. (a) Calculate the capacitance. (b) Calculate the energy stored. (c) Determine the magnitude of the electric field in the Mylar.

Show answer

Rubric: (1 pt) A = (0.30 m)² = 0.090 m². (1 pt) C = κε₀A/d = (3.1)(8.85 × 10⁻¹² F/m)(0.090 m²)/(5.0 × 10⁻⁴ m) = 4.93 × 10⁻⁹ F ≈ 4.9 nF. (1 pt) U = ½CV² = ½(4.93 × 10⁻⁹)(50)² = 6.2 × 10⁻⁶ J ≈ 6.2 μJ. (1 pt) E = ΔV/d = 50 V / (5.0 × 10⁻⁴ m) = 1.0 × 10⁵ V/m. The dielectric fills the entire gap, so the macroscopic field E = ΔV/d is also the field within the Mylar.

Question 5: CRITICAL THINKING

A group of students wants to determine the dielectric constant κ of an unknown plastic sheet experimentally. They have access to a parallel-plate capacitor with removable plates, a set of spacers of various thicknesses, a DC power supply, a voltmeter, and a charge sensor (capable of measuring the charge on the capacitor after it is disconnected from the power supply). (a) Describe a procedure the students could follow to determine κ. (b) State what measurements should be taken and how they should be analyzed (including any graph). (c) Identify one source of systematic error and explain its effect on the calculated value of κ. (d) Explain how the students could modify the experiment to reduce this error.

Show answer

Rubric (4 pts): (1 pt) Procedure: Connect capacitor to power supply at a known voltage V₀. Disconnect the power supply (so Q is fixed). Insert the plastic sheet fully between the plates and record the new voltage

V_d

using the voltmeter. (1 pt) Analysis: With Q constant, κ = V₀ /

V_d

. For higher precision, repeat for several values of V₀ and plot V₀ (y) vs.

V_d

(x); the slope equals κ. (1 pt) Systematic error: Fringe (edge) fields extend beyond the geometric plate area, making the effective capacitance larger than C = κε₀A/d. If the dielectric sheet is smaller than the plates, not all field lines pass through it, causing κ to be underestimated. (1 pt) Reduction: Ensure the dielectric sheet extends well beyond the plate edges. Alternatively, minimize d relative to plate dimensions so fringe fields contribute a negligible fraction of total capacitance.