Topic 10.6 Notes – Capacitors

1. What a Parallel-Plate Capacitor Is

A parallel-plate capacitor has:

• Two parallel conducting plates
• A small separation distance $d$
• Equal and opposite charges: one plate has $+Q$, the other $-Q$

When connected to a battery:

• The battery pulls electrons from one plate and pushes them onto the other.
• A potential difference $\Delta V$ forms between the plates.
• An electric field fills the space between them.

The region between plates can be:

• Air (dielectric constant $\kappa \approx 1$)
• Or an insulating material called a dielectric

For AP Physics 2, we:

• Only analyze parallel plates
• Ignore edge effects (assume the field is uniform except very close to edges)

2. Capacitance and What Determines It

What Capacitance Means

Capacitance tells you how much charge a device can store per volt.

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

• Units: farads (F) where $1\,\text{F} = 1\,\text{C/V}$
• Bigger $C$ means more charge stored for the same voltage.
• Capacitance depends only on physical properties, not on how much charge is currently on it.

That last point is huge. Even if $Q$ changes, $C$ does not.

What Determines Capacitance (Parallel Plates)

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

Where:

• $A$ = area of one plate
• $d$ = separation
• $\varepsilon_{0}$ = permittivity of free space
• $\kappa$ = dielectric constant

How each variable affects $C$:

• Larger area $A$ → larger $C$
• Smaller distance $d$ → larger $C$
• Larger $\kappa$ → larger $C$

Proportional reasoning shows up constantly on tests:

• Double $A$ → double $C$
• Double $d$ → half $C$
• Insert dielectric with $\kappa = 4$ → $C$ becomes 4 times larger

Notice none of this mentions voltage or charge. Geometry and material only.

3. Electric Field Between the Plates

Uniform Electric Field

Between large, closely spaced plates, the field is:

• Constant in magnitude
• Constant in direction
• Directed from positive plate to negative plate

Uniform electric field between parallel plates

The evenly spaced arrows represent the constant electric field between the plates. Because the field is uniform, every point between the plates (ignoring edges) experiences the same electric field.

Electric Field Magnitude

Two useful forms:

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

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

From $E = \Delta V/d$:

• Increasing voltage → increases $E$
• Increasing distance (same voltage) → decreases $E$

From the second form:

• Increasing charge → increases $E$
• Inserting a dielectric (larger $\kappa$) → decreases $E$ for the same free charge

Motion of a Charged Particle

If you place a charge between the plates:

$$F = qE$$

$$a = \frac{qE}{m}$$

Since $E$ is constant, force is constant → acceleration is constant.

This is exactly like projectile motion in gravity:

• Electric field ↔ gravitational field
• $qE$ ↔ $mg$

So you use regular kinematics equations. On FRQs, you’ll often need to explain in words that the particle accelerates because the electric force is constant due to the uniform field.

4. Energy Stored in a Capacitor

Charging a capacitor requires work. The battery moves charge against electric repulsion. That work becomes electric potential energy stored in the electric field.

Three equivalent expressions:

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

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

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

Key relationships:

• Energy is proportional to $(\Delta V)^{2}$
• For fixed voltage, larger $C$ means more energy stored
• The $\tfrac{1}{2}$ appears because voltage increases gradually as the capacitor charges

Students often say energy is “stored on the plates.” It’s stored in the electric field between the plates.

5. Dielectrics and Polarization

A dielectric is an insulator placed between the plates.

When inserted:

• Molecules polarize. Their charges shift slightly.
• This creates an induced field opposite the original field.
• The net electric field inside decreases.

Capacitance increases:

$$C_{\text{new}} = \kappa C_{\text{original}}$$

Why? For the same free charge:

• The dielectric reduces the internal field.
• That reduces $\Delta V$.
• Since $C = Q/\Delta V$, a smaller voltage means larger capacitance.

On conceptual questions, they love asking which quantities change when a dielectric is inserted while the battery is still connected versus disconnected. Think carefully about what stays fixed, $Q$ or $\Delta V$.