Capacitors
Review capacitor charge, voltage, and energy before combining capacitors and interpreting charging.
Before this connection: Potential difference and stored energy.
AP Physics 2 Topic 10.6 · College Board CED printed p. 52, 53 / PDF p. 59, 60 · See this topic in the course outline
A capacitor separates charge and stores energy. Its geometry sets capacitance; its connection to the rest of the system determines what stays fixed when you change it.
For another explanation, read Capacitors. Use the corrected parallel-plate geometry discussion to compare C, Q, and voltage.
For another explanation, read Stored Energy and Dielectrics. Use the corrected energy and dielectric discussion with an explicit fixed-Q or fixed-voltage constraint.
Before you start
- Potential difference, uniform electric fields, and work
- Constant-acceleration kinematics
Charge separation and the geometry that sets C
A parallel-plate capacitor has two separated conducting surfaces with equal and opposite plate charges +Q and −Q in the ideal model. Here Q means the positive magnitude on either plate, not the net charge of the complete capacitor. Its capacitance is , in farads (F = C/V). In a linear capacitor, increasing voltage increases charge proportionally; that does not change C.
For plate overlap area A, separation d, and a uniform dielectric filling the gap, . The physical shape, spacing, and material set capacitance. Doubling area doubles C; doubling separation halves C. Keep overlap area distinct from the combined area of both plates. We analyze parallel plates and ignore edge effects unless specified, with d much smaller than plate dimensions.
A uniform field gives projectile-style motion
Far from the edges, uniformly charged parallel plates create a field of constant direction and magnitude, pointing from the positive plate to the negative one. In a uniformly filled dielectric, . An increase in d does not by itself tell you what happens to E: at fixed Q and A it stays the same, whereas at fixed voltage it decreases.
A charge q between these plates experiences constant force qE⃗ and acceleration . Choose one axis along E⃗ and the other perpendicular. The parallel coordinate has constant acceleration; the perpendicular coordinate has constant velocity if no other force acts. For horizontal entry at speed v₀ into a vertical field, , so the path is parabolic. A negative charge bends opposite the field. Ignore gravity only when the electric acceleration dominates or the problem says to.
Stored energy is the work of separating charge
Stored electric energy is the work required to separate charge in an ideal quasistatic assembly. Imagine an external agent raising the capacitor voltage gradually as Q grows, with negligible kinetic energy and dissipation. Early in the assembly the capacitor voltage is small; each later unit of separated charge requires more work. For a linear capacitor, . This energy belongs to the separated-charge configuration and its field.
A graph of the capacitor’s voltage against Q is a straight line from the origin with slope 1/C. Its triangular area, ½QΔV, is the increase in stored field energy and the minimum quasistatic external assembly work. It is not automatically all the energy supplied by a battery.
A constant-voltage battery delivering Q at voltage ΔV supplies , while an initially empty capacitor stores only half that energy. In ordinary resistor-mediated charging, the other half becomes thermal energy. A graph of battery voltage against delivered charge is horizontal, whereas the capacitor-voltage graph rises. The two areas answer different energy questions.
Dielectric insertion: say what is held fixed
A dielectric polarizes: bound negative charge shifts toward the positive plate and bound positive charge toward the negative plate. The induced field inside the dielectric opposes the field from the plate free charges. Fully filling the gap raises C by the factor κ. For the same free Q it lowers E and voltage. If a battery holds voltage fixed, extra free charge flows onto the plates, so the net field ΔV/d remains unchanged for fixed d.
| Quantity | Isolated capacitor (fixed Q) | Ideal battery connected (fixed ΔV) |
|---|---|---|
| C | κ times | κ times |
| Q | Unchanged | κ times |
| ΔV and E | 1/κ times | Unchanged |
| Stored U | 1/κ times | κ times |
For the isolated case the decrease in stored energy can become mechanical work as the dielectric is pulled in, or heat if motion is damped. In the battery case, energy also crosses the boundary from the battery. Increasing stored energy does not mean all battery work becomes stored energy; mechanical work can occur too. Always state the system and connection before using an energy formula.
One capacitor, two dielectric outcomes
An air-gap capacitor has A = 0.020 m² and d = 1.0 mm and is charged to 12 V. Find C, Q, E, and U. Then fully insert κ = 3 either with the battery disconnected or connected.
Take κ ≈ 1 for air and neglect edges. C = ε₀A/d = 1.77 × 10⁻¹⁰ F. Q = CΔV = 2.12 × 10⁻⁹ C.
Why this step? The geometry fixes C; the charging voltage then fixes Q.
E = 12/(1.0 × 10⁻³) = 1.20 × 10⁴ N/C. U = ½C(12)² = 1.27 × 10⁻⁸ J.
Why this step? The field is uniform in the central gap, and stored energy uses the final voltage squared.
After insertion C′ = 3C. Disconnected: Q′ = Q, ΔV′ = 4 V, E′ = 4.00 × 10³ N/C, U′ = 4.25 × 10⁻⁹ J. Connected: ΔV′ = 12 V, Q′ = 6.37 × 10⁻⁹ C, E′ = 1.20 × 10⁴ N/C, U′ = 3.82 × 10⁻⁸ J.
Why this step? Different constraints select Q²/(2C) or CΔV²/2; polarization is present in both cases.
Answer and check
The dielectric triples capacitance in both cases. Stored energy decreases by three at fixed Q and increases by three at fixed voltage.
A sideways beam in a vertical field
A particle with q = +2.0 × 10⁻⁹ C and m = 4.0 × 10⁻⁶ kg enters a central plate gap horizontally at 2.0 m/s. The field is 6.0 × 10³ N/C downward over a horizontal length 0.080 m. Find its downward deflection at the far end. Neglect gravity and assume it stays in the gap.
Choose +y downward. ay = qE/m = 3.0 m/s²; ax = 0.
Why this step? A uniform electric field supplies constant vertical acceleration.
t = L/vx = 0.040 s, because the horizontal velocity stays 2.0 m/s.
Why this step? No horizontal force changes the transit speed.
y = ½ayt² = 0.0024 m = 2.4 mm.
Why this step? The initial vertical velocity is zero; use constant-acceleration kinematics.
Answer and check
The deflection is 2.4 mm downward. The assumed gap must be large enough for that path to avoid hitting a plate.
Uniform applies to the central gap
Between broad oppositely charged parallel plates, with uniform charge in the ideal model and separation much smaller than their lateral size, contributions from the plates combine to give approximately the same field magnitude and direction throughout the central gap. The field points from the positive plate to the negative plate. A map there uses parallel arrows of equal length on a common scale; moving nearer one plate within that central region does not by itself make the field stronger.
The finite plates end. Near those edges, symmetry changes and the field fringes: direction and magnitude vary, so central E = ΔV/d and constant-acceleration motion are not global rules. A qualitative fringing sketch indicates direction only unless it supplies a magnitude scale. For a particle wholly inside the central domain, qE⃗/m is constant. With transverse initial velocity, its parabolic trajectory shares the constant-acceleration structure of a projectile in uniform gravity; the charge sign determines its bending direction. However, entering an edge region or striking a plate ends that calculation’s validity.
One field at three central locations
Broad horizontal plates have +12 V on the upper plate and 0 V on the lower plate, separated by 3.0 mm. Their lateral size is 0.20 m. Points A, B and C lie deep in the central region at different heights. Air is approximated as vacuum; charges are uniform in the plate model.
- The plate separation is much smaller than its lateral size: 0.0030/0.20 = 0.015. The central uniform-field approximation is reasonable far from edges.
- E = 12/0.0030 = 4.0 × 10³ N/C downward. A, B and C receive equal-length downward arrows, although they are at different distances from the plates.
- A +1.0 nC nonperturbing charge at any of these points feels 4.0 μN downward; a −1.0 nC charge feels 4.0 μN upward.
- An observation point beside the plate end does not satisfy the central-domain assumption. Its field may tilt and have different magnitude; do not assign 4.0 × 10³ N/C downward automatically.
Answer and check
Central E is 4.0 × 10³ N/C downward at all three points; the stated equal-direction/equal-magnitude map does not describe the edges.
Make the model respond
Focused Physics 2 model
Capacitor: connected or isolated
Predict first. Before adding dielectric, decide whether the battery fixes voltage or isolation fixes charge.
C = κε₀A/d; Q = CV; E = V/d; U = ½CV² = Q²/(2C)
On a narrow screen, scroll within a figure to read its labels. The table and explanation provide the same physics in text.
The battery maintains 12 V and supplies or removes charge as C changes. Q = 1.063 × 10^−9 C and U = 6.375 × 10^−9 J.
Read the model data and comparisons
| Quantity | Value | Unit | Meaning or limit |
|---|---|---|---|
| Area | 0.01 | m² | |
| Separation | 0.001 | m | |
| Relative permittivity | 1 | ||
| Capacitance | 8.854 × 10^−11 | F | |
| Charge magnitude on either plate | 1.063 × 10^−9 | C | |
| Voltage across plates | 12 | V | |
| Field magnitude between plates | 12,000 | V/m | |
| Stored field energy | 6.375 × 10^−9 | J | |
| Isolated reference charge Q₀ | 1.063 × 10^−9 | C | Prepared at 12 V, A = 100 cm², d = 1 mm, κ = 1; then disconnected |
Increasing area or permittivity raises capacitance; increasing separation lowers it. With a connected ideal battery, fixed voltage means charge and stored energy increase with C. With the isolated reference charge held fixed, voltage and stored energy decrease as C increases. The battery exchanges energy in the connected case, so changes in capacitor energy alone are not the work done by an external agent.
Assumptions and limits. Ideal parallel plates; edge fringing neglected. Linear, uniform dielectric fully fills the gap without changing d. Battery condition means a 12 V ideal source remains attached. Isolated condition means the charge prepared at the explicitly stated reference geometry is retained; no leakage. Energy is the capacitor field energy, not total battery-plus-capacitor energy.
Try it from memory
Write your answer and reason before opening the worked answer. Your self-assessment is a study decision; it is not an automatically graded score.
Question 1
Double plate separation at fixed area. Compare C, Q, E, and U for a battery-connected capacitor.
Worked answer and reason
C halves; Q halves; E halves; U halves. The ideal battery keeps ΔV fixed.
Apply C ∝ 1/d, Q = CΔV, E = ΔV/d, and U = ½CΔV² under one consistent constraint.
Question 2
A neutral dielectric is inserted into an isolated capacitor. Why can the field weaken without changing the plate free charge?
Worked answer and reason
Polarization creates bound surface charges whose field opposes the free-charge field. Q stays fixed, C rises, and voltage and stored energy decrease.
Bound-charge redistribution inside a neutral dielectric does not require a transfer of plate charge.
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Catch the tempting mistake
- A capacitor stores a net charge Q.
- The ideal plates carry +Q and −Q; Q labels a plate-charge magnitude and the complete capacitor has zero net charge.
- Adding a dielectric always decreases capacitor energy.
- It decreases energy at fixed charge but increases it at fixed voltage; the battery participates in the latter energy transfer.
Put it to work
Give the physical reason as well as the calculation. The worked solutions are on a separate page.
- Problem 11 · geometry, energy, and graph area
- Problem 12 · constraint comparison and projectile motion
- Problem 16 · capacitance experiment and controlled comparison
Explain and apply the model
Try these problems before opening the solutions. Show the physical model, your calculation or sketch, and the reason your conclusion follows.
Is the plate field uniform everywhere?
Two broad horizontal conducting plates have uniform equal-and-opposite charge in the idealized model. The upper plate is at 12 V and the lower at 0 V. Separation d = 0.020 m and lateral size is 1.0 m. Points A, B and C lie far from all edges: A is near the upper plate, B is midway, and C near the lower plate. Point D is beside a finite plate end. Use the central approximation, air κ ≈ 1, and no other forces.
(a) Calculate the central field. Draw arrows at A, B and C on one magnitude scale and state whether both magnitude and direction are constant. Explain the plate-sign direction and why A is not assigned a larger field just because it is nearer the upper plate.
(b) Sketch the qualitative field near an end of the plates and explain why the constant central field cannot be assigned automatically at D. State the geometric and location conditions needed for your central approximation, and label any edge sketch direction-only if its arrow lengths have no common magnitude scale.
(c) A particle with q = −2.0 nC and m = 4.0 × 10⁻⁶ kg enters the central gap at the midpoint with horizontal speed 0.50 m/s and zero vertical speed. Use x right and y upward. Find acceleration and its deflection over 0.080 m of central travel. Check for plate collision, and state why this calculation does not determine its path through an entry or exit fringe.
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For the teacher
- Require a constraint statement before factor-of-change predictions.
- Ask students to connect the triangular area of the V versus Q graph to work.
- Check that projectile calculations satisfy the stated gap and negligible-edge assumptions.
- Distinguish the area under capacitor voltage versus charge (stored energy/minimum quasistatic assembly work) from constant-voltage battery energy QV. In ordinary RC charging the remainder is resistor heating.
Source objectives and scope
- 10.6.A — Describe the physical properties of a parallel-plate capacitor.
While other shapes are also able to separate charges, only the analysis and descriptions of parallel-plate capacitors are required for AP Physics 2. Edge effects will be ignored unless explicitly stated otherwise.
Suggested skills are teaching guidance. The course applies all science practices across its objectives.
Science references for this teaching
- College Board AP Physics 2 Course and Exam Description, Unit 10 (PDF pages 50–61) — Official topic order, learning objectives, all essential-knowledge clauses, algebra-based scope boundaries; equations visually inspected in local source pages 50,54,56–61.
- OpenStax University Physics Volume 2, 8.5 Molecular Model of a Dielectric — Independent verification that bound polarization leaves a dielectric net neutral and its induced field opposes the free-charge field. All lesson prose and problems are original.
- Corrected original reading: Charge and Matter — Use the corrected charge and Coulomb-law explanations to compare signed charges and pair forces.
- Corrected original reading: Conductors and Insulators — Use the corrected charge-mobility discussion to explain contact, polarization, and grounding with a system boundary.
- Corrected original reading: The Electric Field — Use the corrected point-source and vector-superposition discussion to practice fields and forces. General quantitative fields within insulators are outside the AP core.
- Corrected original reading: The Electric Potential and Potential Energy — Use the corrected algebraic point-charge energy discussion to connect configuration changes with external and electric work.
- Corrected original reading: Calculating Potentials — Use the corrected point-charge scalar sums to practice potential.
- Corrected original reading: Equipotentials, Field lines, and Gradients — Use the corrected contour maps and algebraic field estimates here; calculus-gradient material is an optional extension beyond the AP core.
- Corrected original reading: Capacitors — Use the corrected parallel-plate geometry discussion to compare C, Q, and voltage.
- Corrected original reading: Stored Energy and Dielectrics — Use the corrected energy and dielectric discussion with an explicit fixed-Q or fixed-voltage constraint.
Glossary and deeper reading · 13 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
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Electric charge
Electric charge is a conserved property that determines electric interactions. It can be positive or negative and is measured in coulombs; ordinary net charges arise from an imbalance of charged particles. For example, transferring electrons between initially neutral objects leaves one negatively charged and the other positively charged.
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Capacitance
Capacitance is the plate-charge magnitude divided by the potential difference for an ideal capacitor. It is measured in farads and depends on geometry and the material between the conductors. For example, increasing the overlap area of ideal parallel plates increases their capacitance when separation and dielectric stay fixed.
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Potential difference
Potential difference is the change in electric potential between two locations. It represents electric potential-energy change per unit charge and is measured in volts. For example, the potential difference across a resistor describes the energy transferred per coulomb passing through it.
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Electric field
An electric field assigns the electric force per unit positive test charge to each location. The field is produced by the sources and is a vector, while the force on a negative charge points opposite the field. For example, the electric field between ideal parallel capacitor plates can accelerate an electron opposite the field arrows.
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Acceleration
Acceleration is the vector rate of change of velocity with time. It can change an object’s speed, direction, or both, and it is measured in meters per second squared. For example, a satellite moving at constant speed in a circular orbit accelerates toward the center.
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Kinematics
Kinematics describes motion using position, velocity, acceleration, and time without first analyzing its forces. The familiar constant-acceleration equations apply only over intervals in which that acceleration is constant. For example, a ball falling near Earth with negligible air resistance can be modeled with constant vertical acceleration.
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Magnitude
Magnitude is the nonnegative size of a vector quantity. It does not specify direction, and it differs from a signed component. For example, a westward velocity can have a negative x component while its speed, the velocity magnitude, stays positive.
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Force
A force describes an interaction that can change an object’s momentum. It is a vector measured in newtons, so both size and direction matter when forces are added. For example, a rope can pull a cart while the floor exerts friction in another direction.
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Mechanical work
Mechanical work is energy transferred by a force through displacement of its point of application. For a constant force on a particle, it depends on the force component along the particle’s displacement, and it can be positive, negative, or zero. For example, a magnetic force perpendicular to a particle’s displacement does zero work on that particle.
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Kinetic energy
Kinetic energy is energy associated with motion relative to a chosen reference frame. For a nonrelativistic translating object it is one half of its mass times speed squared. For example, doubling a cart’s speed quadruples its translational kinetic energy when its mass stays fixed.
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Heat
Heat is energy transferred between systems because of a temperature difference. It describes a transfer rather than a substance or energy store contained in an object. For example, when a warmer mug transfers energy to a cooler hand, the mug loses internal energy and the hand gains it.
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Velocity
Velocity describes how position changes with time and includes direction. Instantaneous velocity describes one moment, while average velocity is displacement divided by elapsed time. For example, a car turning a corner can change velocity while maintaining the same speed.
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Projectile motion
Projectile motion describes an object after launch when gravity is the only significant force in the chosen model. Near Earth, horizontal and vertical motion can be analyzed separately using the same time interval and approximately constant downward acceleration. For example, a ball rolling off a table continues horizontally while its downward velocity increases.