10.6 Capacitors

AP Physics 2 10.6 Capacitors Summary

A parallel-plate capacitor stores equal and opposite charge on two conducting plates, and its capacitance is set by the plate area, the gap between the plates, and the material in between. The two AP Physics 2 capacitor equations you lean on most are $C = \frac{Q}{\Delta V}$ (the definition) and $C = \kappa \varepsilon_0 \frac{A}{d}$ (the physical build). You also need the uniform field between the plates and the stored energy $U_C = \frac{1}{2} Q \Delta V$.

A parallel-plate capacitor stores equal and opposite charge on two plates, and it links the field and potential ideas from earlier in this unit to devices you will meet again in circuits. Questions may ask you to sketch the uniform field between the plates, relate it to potential difference, and connect the math to a verbal or graphical model.

Capacitance Depends on Geometry and Dielectric

• Capacitance is defined by $C = \frac{Q}{\Delta V}$, but its value is fixed by physical build: $C = \kappa \varepsilon_0 \frac{A}{d}$.
• The field between the plates is uniform (except near edges) and equals $E_C = \frac{Q}{\kappa \varepsilon_0 A}$, which is the same as $\frac{\Delta V}{d}$.
• A charged particle between the plates feels constant force, so it moves like a projectile with constant acceleration.
• Stored energy can be written three equivalent ways: $U_C = \frac{1}{2} Q \Delta V = \frac{1}{2} C (\Delta V)^2 = \frac{Q^2}{2C}$.
• Adding a dielectric raises capacitance by a factor of $\kappa$ and sets up an induced field inside the dielectric that opposes the plate field.
• On the AP exam, only parallel-plate capacitors are required, and edge effects are ignored unless a problem says otherwise.

Physical Properties of Parallel-Plate Capacitors

Basic Structure and Function

A parallel-plate capacitor is two conducting plates set parallel to each other with a small gap between them. The plates store equal amounts of charge with opposite signs.

• When connected to a voltage source, one plate gains positive charge while the other gains an equal amount of negative charge.
• An insulating material (dielectric) between the plates keeps charge from flowing straight across the gap.

Capacitance Fundamentals

Capacitance measures how much charge a capacitor stores for a given potential difference. It ties the charge on each plate to the resulting voltage between the plates.

• Capacitance ($C$) is the ratio of charge ($Q$) to voltage ($\Delta V$): $C = \frac{Q}{\Delta V}$
• The unit is the farad (F), where 1 F = 1 coulomb/volt.
• Many real capacitors have values in the microfarad (μF) or picofarad (pF) range.
• Capacitance is a property of the capacitor itself. Putting more charge on the plates does not change $C$; it just raises $\Delta V$ to match.

Physical Factors Affecting Capacitance

The capacitance of a parallel-plate capacitor depends only on its physical build, not on the charge or voltage applied.

It is set by:

• The surface area of the plates ($A$)
• The distance between the plates ($d$)
• The dielectric material between the plates

The relationship is:

$C = \kappa \varepsilon_0 \frac{A}{d}$

Where:

• $\kappa$ is the dielectric constant of the material between plates (dimensionless)
• $\varepsilon_0$ is the electric permittivity of free space ($8.85 \times 10^{-12}$ F/m)
• $A$ is the area of one plate (m²)
• $d$ is the separation distance between plates (m)

This means:

• Doubling the plate area doubles the capacitance.
• Halving the separation distance doubles the capacitance.
• A higher dielectric constant increases capacitance proportionally.

Electric Field Between Capacitor Plates

Uniform Electric Field

The electric field between two charged parallel plates with uniformly distributed charge is constant through most of the space between them.

The field in a parallel-plate capacitor:

• Has constant magnitude at all points between the plates (away from the edges)
• Points uniformly from the positive plate to the negative plate
• Forms parallel, evenly spaced field lines perpendicular to the plates 📏

Electric Field Strength

When the plate separation is much smaller than the plate dimensions, the field strength is:

$E_C = \frac{Q}{\kappa \varepsilon_0 A}$

You can also write it in terms of voltage and gap:

$E_C = \frac{\Delta V}{d}$

The field strength:

• Is measured in volts per meter (V/m)
• Increases with more charge on the plates
• Decreases when a dielectric material is inserted while the charge $Q$ stays fixed (an isolated capacitor); if a battery keeps $\Delta V$ fixed, $E_C = \frac{\Delta V}{d}$ is unchanged
• Stays the same across the gap (except near the edges)

Motion of Charged Particles Between Plates

A charged particle placed between the plates feels a constant force, so it has constant acceleration. This is why its motion shares characteristics with the projectile motion of a mass near Earth's surface.

• The electric force on the particle is $F_e = qE_C$, where $q$ is the charge of the particle.
• That force gives constant acceleration $a = \frac{qE_C}{m}$, where $m$ is the particle's mass.
• You can analyze the motion with the same constant-acceleration equations used for projectiles:
  • $v = v_0 + at$
  • $x = x_0 + v_0t + \frac{1}{2}at^2$
  • $v^2 = v_0^2 + 2a(x - x_0)$

Energy Storage in Capacitors

Work and Energy in Capacitors

The energy stored in a capacitor equals the work done by an external force to separate the charges on the plates.

As charge builds up on the plates:

• The potential difference rises.
• Each added bit of charge takes more work to move.
• Energy is stored in the electric field between the plates.
• That stored energy can be released when the capacitor discharges. ⚡️

Calculating Stored Energy

The stored energy can be written in several equivalent forms:

$U_C = \frac{1}{2}Q\Delta V = \frac{1}{2}C(\Delta V)^2 = \frac{Q^2}{2C}$

These show that:

• Energy increases with the square of the voltage.
• Energy increases linearly with capacitance (for a fixed voltage).
• The factor of $\frac{1}{2}$ appears because the potential difference builds gradually as charge accumulates, so the average voltage during charging is half the final value.

Dielectrics in Capacitors

Effects of Dielectric Materials

Adding a dielectric between the plates changes the capacitor's behavior.

When a dielectric is inserted:

• The capacitance increases by a factor of $\kappa$.
• The molecules in the dielectric become polarized in the external field.
• This polarization creates an induced field inside the dielectric that points opposite to the field between the plates.

Polarization in Dielectrics

• Because the induced field partially cancels the field between the plates, the same charge produces a smaller potential difference, so capacitance increases, consistent with $C = Q/\Delta V$.
• If the capacitor stays connected to a battery, the voltage is fixed, so it stores more charge.
• If the capacitor is isolated (disconnected), the charge stays the same and the potential difference drops.

Common dielectric materials and their approximate constants (examples, for context):

• Air ($\kappa \approx 1.0$)
• Paper ($\kappa \approx 3.5$)
• Glass ($\kappa \approx 5\text{-}10$)
• Ceramic (can range widely)
• Water ($\kappa \approx 80$) 🧱

Setting Up and Solving Capacitor Problems

Free Response: When a problem changes a capacitor's geometry or dielectric, decide first whether $Q$ or $\Delta V$ is held constant. If it stays wired to a battery, $\Delta V$ is fixed. If it is disconnected, $Q$ is fixed. From there, use $C = \kappa \varepsilon_0 \frac{A}{d}$ to find the new capacitance, then $C = \frac{Q}{\Delta V}$ to track what changes.

When asked to explain why capacitance changes, point to geometry and dielectric, not to charge or voltage. Capacitance is a property of the build.

Problem Solving:

• For a charged particle between plates, find $E_C = \frac{\Delta V}{d}$, then force $F = qE_C$, then acceleration $a = \frac{F}{m}$, then use kinematics. This is the same flow as projectile motion.
• For energy questions, pick the energy form that matches your known quantities. If you know $C$ and $\Delta V$, use $\frac{1}{2}C(\Delta V)^2$. If you know $Q$ and $C$, use $\frac{Q^2}{2C}$.
• Watch units: convert mm and cm to meters and μF and pF to farads before plugging in.

Common Trap: Inserting a dielectric does not always do the "same" thing. The result depends on whether the battery is still connected. Read the setup carefully before deciding what stays constant.

Practice Problem 1: Capacitance Calculation

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

Given:

• Area of plates, $A = 0.02$ m²
• Separation distance, $d = 0.5$ mm = $5 \times 10^{-4}$ m
• Dielectric constant of air, $\kappa = 1.0$
• Electric permittivity of free space, $\varepsilon_0 = 8.85 \times 10^{-12}$ F/m

Substituting: $C = 1.0 \times 8.85 \times 10^{-12} \text{ F/m} \times \frac{0.02 \text{ m}^2}{5 \times 10^{-4} \text{ m}}$ $C = 8.85 \times 10^{-12} \text{ F/m} \times 40 \text{ m}$ $C = 3.54 \times 10^{-10} \text{ F} = 354 \text{ pF}$

Practice Problem 2: Electric Field and Charged Particle Motion

Solution: First find the field strength: $E = \frac{\Delta V}{d} = \frac{300 \text{ V}}{0.02 \text{ m}} = 15{,}000 \text{ V/m}$

The force on the proton: $F = qE = 1.6 \times 10^{-19} \text{ C} \times 15{,}000 \text{ V/m} = 2.4 \times 10^{-15} \text{ N}$

The acceleration: $a = \frac{F}{m} = \frac{2.4 \times 10^{-15} \text{ N}}{1.67 \times 10^{-27} \text{ kg}} = 1.44 \times 10^{12} \text{ m/s}^2$

Since acceleration is constant, use: $v^2 = v_0^2 + 2a(x - x_0)$

With $v_0 = 0$ and $(x - x_0) = 0.02$ m: $v^2 = 0 + 2 \times 1.44 \times 10^{12} \text{ m/s}^2 \times 0.02 \text{ m}$ $v^2 = 5.76 \times 10^{10} \text{ m}^2/\text{s}^2$ $v = 2.4 \times 10^5 \text{ m/s}$

Practice Problem 3: Energy Storage in a Capacitor

Solution: Use the energy equation: $U_C = \frac{1}{2}C(\Delta V)^2$

Given:

• Capacitance, $C = 4.7$ μF = $4.7 \times 10^{-6}$ F
• Potential difference, $\Delta V = 12$ V

Substituting: $U_C = \frac{1}{2} \times 4.7 \times 10^{-6} \text{ F} \times (12 \text{ V})^2$ $U_C = \frac{1}{2} \times 4.7 \times 10^{-6} \text{ F} \times 144 \text{ V}^2$ $U_C = 3.38 \times 10^{-4} \text{ J} = 0.338 \text{ mJ}$

This energy is stored in the electric field between the capacitor plates.

Common Misconceptions

• Misconception: Capacitance depends on charge or voltage.
  Actually: Capacitance does not depend on charge or voltage. Adding charge raises $\Delta V$ so that $C = \frac{Q}{\Delta V}$ stays the same. Only geometry and dielectric set $C$.
• Misconception: A dielectric always lowers the voltage.
  Actually: A dielectric does not always lower the voltage. The voltage drops only if the capacitor is isolated. If it stays on a battery, the voltage is fixed and the stored charge goes up instead.
• Misconception: The field between the plates is stronger in the middle and weaker at the edges.
  Actually: The field between the plates is not stronger in the middle and weaker at the edges in the way some students assume. It is uniform across the gap, and edge effects (fringe fields) are ignored unless a problem states otherwise.
• Misconception: The $\frac{1}{2}$ in the energy formula is optional.
  Actually: The $\frac{1}{2}$ in the energy formula is not optional. It comes from the voltage building up gradually as charge is added, so the work is based on the average voltage, not the final voltage.
• Misconception: The two formulas describe different electric fields.
  Actually: $E_C = \frac{\Delta V}{d}$ and $E_C = \frac{Q}{\kappa \varepsilon_0 A}$ are the same field, just written with different known quantities. Pick whichever matches what the problem gives you.

Related AP Physics 2 Guides

• 10.1 Electric Charge and Electric Force
• 10.2 The Process of Charging
• 10.3 Electric Fields
• 10.5 Electric Potential
• 10.7 Conservation of Electric Energy
• 10.4 Electric Potential Energy