10.3 Capacitors

A capacitor stores charge and energy using two conducting surfaces separated by an insulator. The key relationship is $C = \frac{Q}{\Delta V}$, and for a parallel plate capacitor the capacitance depends only on geometry and the material between the plates: $C = \frac{\kappa \varepsilon 0 A}{d}$. Capacitor questions often ask you to connect charge, voltage, field, geometry, dielectric materials, and stored energy.

Why This Matters for the AP Physics C: E&M Exam

Capacitors sit in Unit 10 (Conductors and Capacitors), which carries a meaningful part of the exam. This topic pulls together earlier ideas: you apply Gauss's law and superposition to get the field between the plates, connect potential difference to field, and use energy reasoning from electric potential energy.

On the exam you can expect to:

• Derive symbolic expressions for capacitance, field, and stored energy from first principles.
• Predict how capacitance, charge, field, or energy changes when you adjust area, plate separation, voltage, or the material between the plates.
• Translate between verbal descriptions, equations, diagrams, and graphs. The qualitative/quantitative translation free-response question can use content from any unit, so practicing this kind of reasoning here helps.

Capacitors also set you up for RC circuits in Unit 11, so getting comfortable now pays off later.

Key Takeaways

• Capacitance is defined by $C = \frac{Q}{\Delta V}$ and depends only on the capacitor's geometry and the material between the plates, not on the charge or voltage you apply.
• For a parallel-plate capacitor, $C = \frac{\kappa \varepsilon_0 A}{d}$: larger plate area or smaller separation gives more capacitance.
• The field between the plates is uniform (except near the edges), with magnitude $E = \frac{Q}{\varepsilon_0 A}$, and it points from the positive plate to the negative plate.
• A charged particle between the plates feels a constant force, so its motion looks like projectile motion in a uniform gravitational field.
• Stored energy can be written three equivalent ways: $U_C = \frac{1}{2}Q\Delta V = \frac{1}{2}C(\Delta V)^2 = \frac{Q^2}{2C}$.
• Quantitative work on the exam is limited to parallel-plate, concentric spherical, and coaxial cylindrical capacitors.

Physical Properties of Parallel-Plate Capacitors

Parallel Conducting Surfaces

A parallel-plate capacitor uses two parallel conducting surfaces separated by a distance, so they can store equal amounts of opposite charge. The plates are usually metal and have the same shape and size.

The charges on the two plates attract each other, but the gap between the plates keeps them from combining, which lets the capacitor hold charge. An insulating material (a dielectric) between the plates prevents charge from flowing across the gap.

Capacitance and Charge Storage

Capacitance measures how much charge a capacitor stores for a given potential difference. It connects the magnitude of the charge on each plate to the resulting potential difference from separating those charges.

A capacitor's capacitance depends only on its physical properties: its shape and the dielectric material between the plates. Capacitance is the ratio of the charge stored on either plate to the potential difference between the plates:

$C=\frac{Q}{\Delta V}$

Rearranging gives:

$Q=C\Delta V$

Where:

• $Q$ is the charge stored on each plate (in coulombs)
• $C$ is the capacitance (in farads)
• $\Delta V$ is the potential difference between plates (in volts)

For a parallel-plate capacitor, capacitance is directly proportional to the plate area and inversely proportional to the plate separation distance. The constant of proportionality is the product of the dielectric constant $\kappa$ of the separating material and the electric permittivity of free space $\varepsilon_{0}$:

$C=\frac{\kappa \varepsilon_{0} A}{d}$

Where:

• $C$ represents the capacitance (in farads)
• $\kappa$ is the dielectric constant of the material between the plates (unitless)
• $\varepsilon_{0}$ is the electric permittivity of free space (approximately $8.85 \times 10^{-12}$ $\text{F}/\text{m}$)
• $A$ is the area of one of the plates (in square meters)
• $d$ is the distance between the plates (in meters)

Factors Affecting Capacitance

Several factors set the capacitance of a parallel-plate capacitor:

• Increasing the plate area ($A$) increases capacitance proportionally.
• Decreasing the separation distance ($d$) increases capacitance.
• Using a dielectric material with a higher dielectric constant ($\kappa$) increases capacitance proportionally.

The dielectric constant describes how effectively a material stores electric field energy compared to a vacuum. Air is close to $\kappa \approx 1$, while many solid insulators have higher values. (Dielectrics are covered in detail in Topic 10.4.)

Electric Field Between Plates

When the plate separation is much smaller than the plate dimensions, the electric field between the plates is approximately constant in magnitude and direction, except near the edges where fringing occurs. This uniform field is a key feature of parallel-plate capacitors.

You can find the field magnitude by applying Gauss's law and the principle of superposition to the two charged plates. The derived result is:

$E=\frac{Q}{\varepsilon_{0} A}$

Where:

• $E$ is the electric field magnitude (in $\text{N}/\text{C}$ or $\text{V}/\text{m}$)
• $Q$ is the charge on each plate (in coulombs)
• $\varepsilon_{0}$ is the electric permittivity of free space
• $A$ is the area of one of the plates (in square meters)

The field is directly proportional to the surface charge density on either plate. A charged particle placed between two oppositely charged plates feels a constant electric force $F=qE$, so its acceleration $a=\frac{qE}{m}$ is constant. Its motion is then analogous to projectile motion in a uniform gravitational field: the velocity component parallel to the plates stays constant, while the component along the field changes linearly with time.

Electric Potential Energy Storage

The electric potential energy stored in a capacitor equals the work an external force does to separate the charge onto the plates. That stored energy can be released when the capacitor discharges.

The stored energy $U_{\mathrm{C}}$ is given by:

$U_{\mathrm{C}}=\frac{1}{2} Q \Delta V$

Where:

• $U_{\mathrm{C}}$ is the electric potential energy stored in the capacitor (in joules)
• $Q$ is the charge stored on each plate (in coulombs)
• $\Delta V$ is the electric potential difference between the plates (in volts)

The stored energy depends on both the charge on the plates and the potential difference between them. The factor of $\frac{1}{2}$ appears because the potential difference builds up linearly as charge is added.

Other Forms of the Energy Equation

Using $C = \frac{Q}{\Delta V}$, you can rewrite the stored energy in terms of capacitance and voltage:

$U_{\mathrm{C}}=\frac{1}{2} C (\Delta V)^2$

This shows the stored energy is proportional to the capacitance and to the square of the potential difference, so doubling the voltage quadruples the stored energy.

You can also write the energy in terms of charge and capacitance:

$U_{\mathrm{C}}=\frac{Q^2}{2C}$

This form is handy when you know the charge rather than the voltage. All three energy forms are equivalent, so pick whichever matches your given quantities.

How to Use This on the AP Physics C: E&M Exam

Problem Solving

• Start by writing $C = \frac{Q}{\Delta V}$. It defines capacitance and connects the three core quantities. Solve for whichever one the problem leaves out.
• For geometry questions, use $C = \frac{\kappa \varepsilon_0 A}{d}$ and reason with proportions. If a problem doubles the area, capacitance doubles; if it halves the separation, capacitance doubles again.
• Match your energy formula to your givens: use $\frac{1}{2}Q\Delta V$ with charge and voltage, $\frac{1}{2}C(\Delta V)^2$ with capacitance and voltage, and $\frac{Q^2}{2C}$ with charge and capacitance.
• For a particle moving between the plates, treat it like projectile motion: constant velocity along the plates, constant acceleration $a = \frac{qE}{m}$ across the gap.

Free Response

• Show the derivation, not just the answer. For the field between the plates, state that you are applying Gauss's law to one plate and superposing the fields, then arrive at $E = \frac{Q}{\varepsilon_0 A}$.
• Be explicit about what stays fixed. If a capacitor stays connected to a battery, $\Delta V$ is held constant; if it is isolated after charging, $Q$ is held constant. Naming the fixed quantity controls which formulas you use.
• For translation-style questions, make a clear claim with reasoning first, then derive the supporting equation, then connect the two. Practicing this verbal-to-math flow on capacitor scenarios builds the skill the exam rewards.

Common Trap

• Do not treat capacitance as something that changes when you change the charge or voltage. For a fixed capacitor, $C$ is set by geometry and material; changing $Q$ just changes $\Delta V$ in step.
• Watch your units. Convert millimeters to meters and microfarads to farads before plugging in, or your answer will be off by powers of ten.

Practice Problem 1: Capacitance Calculation

Solution: Use the parallel-plate capacitance equation: $C = \frac{\kappa \varepsilon_0 A}{d}$

Given:

• Area of plates, A = 0.02 m²
• Separation distance, d = 0.5 mm = 0.0005 m
• Dielectric constant, κ = 2.5
• Permittivity of free space, ε₀ = 8.85 × 10⁻¹² F/m

Substituting these values: $C = \frac{(2.5)(8.85 \times 10^{-12} \text{ F/m})(0.02 \text{ m}^2)}{0.0005 \text{ m}}$ $C = 8.85 \times 10^{-10} \text{ F}$ $C = 0.885 \text{ nF}$

Practice Problem 2: Energy Storage

Solution: Use the energy equation that matches the givens (capacitance and voltage): $U_C = \frac{1}{2}C(\Delta V)^2$

Given:

• Capacitance, C = 4.7 μF = 4.7 × 10⁻⁶ F
• Potential difference, ΔV = 12 V

Substituting these values: $U_C = \frac{1}{2} \times 4.7 \times 10^{-6} \times (12)^2$ $U_C = \frac{1}{2} \times 4.7 \times 10^{-6} \times 144$ $U_C = 3.384 \times 10^{-4}$ J $U_C = 338.4$ μJ

Practice Problem 3: Electric Field Calculation

Solution: Use the field equation for parallel plates: $E = \frac{Q}{\varepsilon_0 A}$

Given:

• Charge on each plate, Q = 5 × 10⁻⁸ C
• Area of plates, A = 0.01 m²
• Permittivity of free space, ε₀ = 8.85 × 10⁻¹² F/m

Substituting these values: $E = \frac{5 \times 10^{-8}}{8.85 \times 10^{-12} \times 0.01}$ $E = \frac{5 \times 10^{-8}}{8.85 \times 10^{-14}}$ $E = 5.65 \times 10^5$ N/C (or V/m)

Common Misconceptions

• "Adding more charge increases capacitance." It does not. Capacitance is fixed by the plate area, separation, and dielectric. Adding charge raises $\Delta V$ so that $C = \frac{Q}{\Delta V}$ stays the same.
• "The field between the plates depends on how far apart the plates are." The magnitude $E = \frac{Q}{\varepsilon_0 A}$ depends on charge and area, not separation. Moving the plates apart changes $\Delta V = Ed$, not $E$ itself (for fixed charge).
• "The factor of one-half in the energy equation is just a constant." It comes from the potential difference building up linearly as charge is added, so the average voltage during charging is half the final value.
• "Energy and charge scale the same way with voltage." Charge scales linearly with voltage ($Q = C\Delta V$), but energy scales with the square ($U_C = \frac{1}{2}C(\Delta V)^2$), so doubling voltage quadruples energy.
• "A battery-connected capacitor and an isolated capacitor behave the same when you change the geometry." They do not. A connected capacitor holds $\Delta V$ fixed while $Q$ changes; an isolated one holds $Q$ fixed while $\Delta V$ changes.

Related AP Physics C: E&M Guides

• 10.1 Electrostatics with Conductors
• 10.2 Redistribution of Charge Between Conductors
• 10.4 Dielectrics