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Conductors and Capacitors
Electricity & Magnetism track · College Board Unit 10. Build the model, derive its consequences, and test the assumptions.
What you’ll learn
- Explain electrostatic conductors and shielding.
- Model parallel-plate capacitors and dielectrics.
- Distinguish fixed-charge and fixed-voltage energy changes.
A conductor reaches electrostatic equilibrium
Mobile charges in a conductor respond to internal electric fields. In electrostatic equilibrium the field inside the conducting material is zero and the whole connected conductor is at one potential. Excess charge resides on surfaces. The field immediately outside a conductor is perpendicular to the surface; a tangential component would move charges.
A cavity containing no charge has zero electrostatic field when enclosed by a conductor in equilibrium. If a charge is inside a cavity, induced inner-surface charge sums to its negative; the conductor’s net charge determines the remaining outer charge. Grounding allows charge exchange with a large reservoir, so grounded and isolated conductors are different constraints.
PAUSE & TRY IT
What is the potential difference between two points in one conductor at equilibrium?
Reveal answer
Zero, although the common potential need not be zero.
Capacitance belongs to the geometry and medium
A capacitor stores separated equal-magnitude opposite charges. Its capacitance is C=, where Q is the magnitude on either plate, not the sum of both plates. For large parallel plates with negligible edge effects, C= in vacuum. Increasing area increases C; increasing separation decreases C.
The approximately uniform gap field is E==. State the idealizations: plate dimensions much larger than separation, nearly uniform field, and negligible fringing. This course’s quantitative capacitor work uses parallel plates rather than derivations for cylindrical or spherical capacitors.
Fringing near edges is neglected. The arrows show the field from the positive plate toward the negative plate.
PAUSE & TRY IT
If both plate area and spacing double, what happens to C?
Reveal answer
It is unchanged because is unchanged.
PAUSE & TRY IT
Why does the capacitor formula use Q rather than 2Q?
Reveal answer
Q denotes charge magnitude on either plate; C relates that separated charge to the potential difference.
Energy is stored in the field
Moving additional charge onto a capacitor takes increasing work because the potential difference grows with charge. Integrating dW=()dq gives U==C(ΔV)=. These expressions are equivalent, but choose the one that keeps the experiment’s fixed quantity explicit.
For a vacuum gap, energy density is u=. Multiplying by volume Ad recovers the capacitor energy. This field description helps explain why separating charged plates requires work when Q is fixed: the field-filled volume increases.
Connected and disconnected are different experiments
If a charged capacitor is disconnected from its battery and leakage is negligible, Q stays constant. Doubling plate spacing halves C, doubles ΔV, and doubles stored energy; external work supplies that energy. The gap field remains in the ideal model.
If the battery remains connected, ΔV stays constant. Doubling spacing halves C and Q, halves E, and halves capacitor energy. Charge and energy flow between the capacitor and battery while an external agent moves the plates. Looking only at capacitor energy is not a full accounting of external work.
Read figure values as text
Isolated capacitor: voltage versus relative plate spacing: 0.5: 0.5; 0.5416666666666666: 0.5416666666666666; 0.5833333333333334: 0.5833333333333334; 0.625: 0.625; 0.6666666666666666: 0.6666666666666666; 0.7083333333333334: 0.7083333333333334; 0.75: 0.75; 0.7916666666666667: 0.7916666666666667; 0.8333333333333333: 0.8333333333333333; 0.875: 0.875; 0.9166666666666667: 0.9166666666666667; 0.9583333333333333: 0.9583333333333333; 1: 1; 1.0416666666666665: 1.0416666666666665; 1.0833333333333335: 1.0833333333333335; 1.125: 1.125; 1.1666666666666665: 1.1666666666666665; 1.2083333333333335: 1.2083333333333335; 1.25: 1.25; 1.2916666666666665: 1.2916666666666665; 1.3333333333333335: 1.3333333333333335; 1.375: 1.375; 1.4166666666666665: 1.4166666666666665; 1.4583333333333335: 1.4583333333333335; 1.5: 1.5; 1.5416666666666667: 1.5416666666666667; 1.5833333333333333: 1.5833333333333333; 1.625: 1.625; 1.6666666666666667: 1.6666666666666667; 1.7083333333333333: 1.7083333333333333; 1.75: 1.75; 1.7916666666666667: 1.7916666666666667; 1.8333333333333333: 1.8333333333333333; 1.875: 1.875; 1.9166666666666667: 1.9166666666666667; 1.9583333333333333: 1.9583333333333333; 2: 2; 2.041666666666667: 2.041666666666667; 2.083333333333333: 2.083333333333333; 2.125: 2.125; 2.166666666666667: 2.166666666666667; 2.208333333333333: 2.208333333333333; 2.25: 2.25; 2.291666666666667: 2.291666666666667; 2.333333333333333: 2.333333333333333; 2.375: 2.375; 2.416666666666667: 2.416666666666667; 2.458333333333333: 2.458333333333333; 2.5: 2.5; 2.5416666666666665: 2.5416666666666665; 2.5833333333333335: 2.5833333333333335; 2.625: 2.625; 2.6666666666666665: 2.6666666666666665; 2.7083333333333335: 2.7083333333333335; 2.75: 2.75; 2.7916666666666665: 2.7916666666666665; 2.8333333333333335: 2.8333333333333335; 2.875: 2.875; 2.9166666666666665: 2.9166666666666665; 2.9583333333333335: 2.9583333333333335; 3: 3
A dielectric polarizes rather than supplying free charge
A dielectric placed in the gap polarizes, reducing the field due to the same free plate charge. If it completely fills the gap, C=. κ is the dielectric constant and is greater than one for the ordinary media considered here.
For fixed Q, dielectric insertion increases C and decreases V and U. For fixed V, it increases Q and U; the battery supplies charge and energy. Do not say a dielectric always decreases stored energy without stating which quantity is held fixed.
Read figure values as text
Isolated: fixed Q: 1: 1; 1.0625: 0.9411764705882353; 1.125: 0.8888888888888888; 1.1875: 0.8421052631578947; 1.25: 0.8; 1.3125: 0.7619047619047619; 1.375: 0.7272727272727273; 1.4375: 0.6956521739130435; 1.5: 0.6666666666666666; 1.5625: 0.64; 1.625: 0.6153846153846154; 1.6875: 0.5925925925925926; 1.75: 0.5714285714285714; 1.8125: 0.5517241379310345; 1.875: 0.5333333333333333; 1.9375: 0.5161290322580645; 2: 0.5; 2.0625: 0.48484848484848486; 2.125: 0.47058823529411764; 2.1875: 0.45714285714285713; 2.25: 0.4444444444444444; 2.3125: 0.43243243243243246; 2.375: 0.42105263157894735; 2.4375: 0.41025641025641024; 2.5: 0.4; 2.5625: 0.3902439024390244; 2.625: 0.38095238095238093; 2.6875: 0.37209302325581395; 2.75: 0.36363636363636365; 2.8125: 0.35555555555555557; 2.875: 0.34782608695652173; 2.9375: 0.3404255319148936; 3: 0.3333333333333333; 3.0625: 0.32653061224489793; 3.125: 0.32; 3.1875: 0.3137254901960784; 3.25: 0.3076923076923077; 3.3125: 0.3018867924528302; 3.375: 0.2962962962962963; 3.4375: 0.2909090909090909; 3.5: 0.2857142857142857; 3.5625: 0.2807017543859649; 3.625: 0.27586206896551724; 3.6875: 0.2711864406779661; 3.75: 0.26666666666666666; 3.8125: 0.26229508196721313; 3.875: 0.25806451612903225; 3.9375: 0.25396825396825395; 4: 0.25 • Connected: fixed V: 1: 1; 1.0625: 1.0625; 1.125: 1.125; 1.1875: 1.1875; 1.25: 1.25; 1.3125: 1.3125; 1.375: 1.375; 1.4375: 1.4375; 1.5: 1.5; 1.5625: 1.5625; 1.625: 1.625; 1.6875: 1.6875; 1.75: 1.75; 1.8125: 1.8125; 1.875: 1.875; 1.9375: 1.9375; 2: 2; 2.0625: 2.0625; 2.125: 2.125; 2.1875: 2.1875; 2.25: 2.25; 2.3125: 2.3125; 2.375: 2.375; 2.4375: 2.4375; 2.5: 2.5; 2.5625: 2.5625; 2.625: 2.625; 2.6875: 2.6875; 2.75: 2.75; 2.8125: 2.8125; 2.875: 2.875; 2.9375: 2.9375; 3: 3; 3.0625: 3.0625; 3.125: 3.125; 3.1875: 3.1875; 3.25: 3.25; 3.3125: 3.3125; 3.375: 3.375; 3.4375: 3.4375; 3.5: 3.5; 3.5625: 3.5625; 3.625: 3.625; 3.6875: 3.6875; 3.75: 3.75; 3.8125: 3.8125; 3.875: 3.875; 3.9375: 3.9375; 4: 4
PAUSE & TRY IT
With a battery attached, dielectric insertion holds which quantity fixed?
Reveal answer
The voltage, assuming an ideal battery.
Combine capacitors using charge and voltage constraints
Parallel capacitors share the same potential difference, so their capacitances add. In an ideal series chain with initially neutral isolated interior nodes, charge magnitudes match and potential differences add, giving _eq=i. The smaller series capacitance takes the larger voltage.
After reconnecting charged capacitors, conserve total charge on each isolated conductor group and require a common final potential for connected nodes. Electrostatic energy need not be conserved during redistribution: resistance and electromagnetic radiation can carry energy away even when the connecting wires are treated as ideal for the final state.
Explain electrostatic equilibrium in a conductor
A conductor contains mobile charges. If a nonzero electric field persisted inside its conducting material, charges would continue to move. In electrostatic equilibrium the internal field is zero, the conductor is equipotential, and excess charge resides on surfaces. This statement concerns equilibrium; a current-carrying resistive wire can have an internal electric field.
The field immediately outside an ideal conducting surface is normal to that surface, because a tangential component would drive further surface motion. Surface charge need not be uniformly distributed on an irregular object. Greater curvature can be associated with stronger local fields, but use the geometry supplied rather than assuming every conductor is a sphere.
A cavity and the conducting material around it are not the same region. Charges placed inside a cavity can induce charges on its wall. A Gaussian surface within the metal, where E=0, constrains the net charge it encloses. Apply that constraint with the actual enclosed cavity charge.
PAUSE & TRY IT
Why is a conductor equipotential in electrostatic equilibrium?
Reveal answer
Its internal electric field is zero, so integrating E·dl between points in the conductor gives zero potential difference.
Understand capacitance as a property of geometry and material
Capacitance C= relates the magnitude of separated charge to the potential difference in the linear capacitor model. For ideal parallel plates, C=, neglecting fringing. Larger area permits more charge for the same voltage; greater separation reduces capacitance. C is not simply “how much charge is currently present.”
A dielectric polarizes in response to the field, changing the relation between free charge and field. In the ideal linear model its insertion increases capacitance. The resulting charge and voltage changes depend on whether the capacitor remains connected to a battery or is isolated.
A battery fixes voltage approximately, allowing charge to enter or leave. An isolated capacitor fixes its free charge approximately, allowing voltage to change. Start every geometry-change question by identifying that constraint. Using both constant Q and constant V when C changes is inconsistent.
Combine capacitors and calculate stored energy consistently
Parallel capacitors share the same potential difference, and their charges add, giving an equivalent capacitance equal to the sum. Series capacitors in the standard initially uncharged isolated-node arrangement carry equal charge magnitudes, while potential differences add, giving reciprocal addition.
Stored energy can be written as ½CV2, or ½QV. These forms are equivalent but highlight different fixed quantities. Increasing C at constant Q decreases stored energy, whereas increasing C at constant V increases it. In the battery-connected case, energy exchanged with the battery must be included when discussing the whole process.
Do not decide a force or work direction from capacitor energy alone without stating the external electrical constraint. Mechanical work, field energy and battery exchange can all participate. A consistent system energy account resolves apparent contradictions.
Use boundary and limiting checks on capacitor models
For a simple parallel-plate idealization, the field is approximately uniform away from edges and potential changes linearly across the gap. Fringing matters near edges and when separation is not small compared with plate dimensions. A schematic’s uniform arrows communicate an approximation rather than an exact field everywhere.
If an isolated capacitor’s plate spacing doubles, ideal capacitance halves, voltage doubles and stored field energy doubles. External work is required to separate attracting plates. If the same change occurs at fixed battery voltage, charge decreases instead. These two experiments have different energy exchanges despite identical final plate geometry.
When several dielectric regions are present, identify whether they divide the plate area side by side or divide the gap along the field. The corresponding parallel or series model follows from shared voltage or charge constraints, not simply from the number of materials shown.
Derive parallel-plate capacitance from field and potential
For large parallel plates with equal and opposite charge densities, neglect fringing and use the approximately uniform field E= between them. The potential difference magnitude is Ed. With Q=σA, capacitance C= becomes . The geometry determines C in this ideal model; changing Q alone changes voltage proportionally.
A dielectric changes the relation between free charge and field through polarization. In the simple linear model, inserting a dielectric with relative permittivity κ fully between plates multiplies capacitance by κ. Distinguish bound polarization charge from free charge supplied by a battery.
Check whether the battery remains attached. At fixed voltage, increased capacitance draws additional charge; at fixed charge, the voltage decreases. The stored-energy formulas and C(ΔV) are equivalent, but each makes a different constraint easy to see. Using the wrong fixed quantity reverses an energy prediction.
Analyze conductor cavities and capacitor networks carefully
Inside conducting material at electrostatic equilibrium, the electric field is zero and potential is constant. This does not require the potential to be zero. If a charge lies inside a cavity, induced charge appears on the cavity wall; the cavity itself is not conducting material and can contain a field.
Capacitors in parallel share the same potential difference, so their charges add and capacitances add. Capacitors in series carry equal charge magnitudes on ideal initially neutral intermediate conductors; their voltage differences add and reciprocal capacitances add. Trace actual nodes rather than deciding from the visual orientation of symbols.
After finding an equivalent capacitance, return to the individual components if asked for charge, voltage, or energy. Equivalent reduction preserves the terminal relation but does not make each internal capacitor carry the total terminal charge. Check that individual series voltages sum to the source voltage and parallel charges sum to the total.
PAUSE & TRY IT
Is the potential inside a conductor necessarily zero?
Reveal answer
No. It is constant in electrostatic equilibrium. Its value depends on the chosen reference and the conductor’s surroundings.
FROM IDEA TO APPLICATION
Worked examples
An isolated capacitor gains a dielectric
A 6 μF capacitor is charged to 12 V, disconnected, then filled with κ=3 material. Find its final Q, C, V, and energy ratio.
Reveal worked solution
- Initial Q=CV=72 μC and stays fixed.
- Final C=18 μF, hence V==4 V.
- At fixed Q, energy is inversely proportional to C.
Q=72 μC, C=18 μF, V=4 V, U_final/U_initial=.
Two capacitors in series
Capacitors C and 2C are connected in series across voltage V. Find their voltage drops.
Reveal worked solution
- C_eq=, so each charge magnitude is .
- Use Vi=i.
Across C: ; across 2C: .
Compare connected and isolated changes
An ideal capacitor initially has C0, V0 and Q0. Its separation doubles. Compare final Q and V when isolated and when battery-connected.
Reveal worked solution
- The final capacitance is in both ideal geometries.
- Isolated: Q stays Q0, so V==2V0.
- Battery-connected: V stays V0, so Q=()V0=.
The electrical connection determines which quantity stays fixed.
Return from an equivalent capacitor to components
Capacitors of 2 μF and 6 μF are in series across 12 V.
Reveal worked solution
- =+, so Ceq=1.5 μF.
- Q=Ceq V=18 μC on each series capacitor in magnitude.
- V2==9 V and V6==3 V; their sum is 12 V.
The smaller series capacitance has the larger voltage.
Insert a dielectric into an isolated capacitor
A charged isolated capacitor has capacitance C0 and voltage V0. A dielectric triples C.
Reveal worked solution
- Isolation fixes free charge Q=C0V0.
- New voltage is =.
- New stored energy =.
Voltage and stored electrical energy fall to one-third in this model; energy accounting includes mechanical work and any dissipation.
MAKE THE DISTINCTION
Common mistakes, clearer reasoning
The trapDisconnecting a capacitor holds its voltage fixed.
The better explanationIt fixes charge if leakage is negligible; voltage may change with geometry.
The trapElectric field vanishes everywhere inside a hollow conductor even if a charge is in its cavity.
The better explanationIt vanishes in the conducting material, but a charged cavity can contain a field.
RETRIEVE BEFORE YOU REVEAL
Practice checkpoints
Revisit the quick checks from this guide without looking back. Explain why, then reveal the answer.
1. What is the potential difference between two points in one conductor at equilibrium?
Reveal answer
Zero, although the common potential need not be zero.
2. If both plate area and spacing double, what happens to C?
Reveal answer
It is unchanged because is unchanged.
3. Why does the capacitor formula use Q rather than 2Q?
Reveal answer
Q denotes charge magnitude on either plate; C relates that separated charge to the potential difference.
4. With a battery attached, dielectric insertion holds which quantity fixed?
Reveal answer
The voltage, assuming an ideal battery.
5. Why is a conductor equipotential in electrostatic equilibrium?
Reveal answer
Its internal electric field is zero, so integrating E·dl between points in the conductor gives zero potential difference.
6. Is the potential inside a conductor necessarily zero?
Reveal answer
No. It is constant in electrostatic equilibrium. Its value depends on the chosen reference and the conductor’s surroundings.
Key language
- Capacitance
- Separated charge magnitude per potential difference.
- Dielectric
- Insulating material that polarizes in an electric field.
- Grounding
- Electrical connection allowing charge exchange with a large reservoir.
- Fringing
- Nonuniform edge fields neglected in the ideal parallel-plate model.
- Electrostatic equilibrium
- A state with no net macroscopic charge motion in a conductor.
- Field energy density
- Energy stored per volume, ε₀E²/2 in vacuum.
Connect the equations to the physical assumptions. Explain directions, signs, limiting cases, and units before accepting a numerical result.