Unit 2 Notes: Capacitors, Dielectric Materials, and Stored Electric Energy

Capacitance

What capacitance is (and what it is not)

A capacitor is any two-conductor system that can store separated electric charge. When you put charge $+Q$ on one conductor and $-Q$ on the other, an electric field forms in the space between them, and a potential difference appears between the conductors.

The central idea is capacitance: how much charge separation you can “hold” per volt of potential difference. The definition is

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

where:

• $C$ is capacitance (units: farads, $\mathrm{F} = \mathrm{C/V}$)
• $Q$ is the magnitude of charge on either conductor (coulombs)
• $\Delta V$ is the potential difference between conductors (volts)

A crucial conceptual point: capacitance is a property of the geometry and the material between the conductors, not of how much charge you happened to place on it. For an ideal capacitor, if you double $Q$, the voltage difference doubles, so the ratio $Q/\Delta V$ stays constant.

What capacitance is not:

• It is not “how much charge the capacitor has.” That’s $Q$.
• It is not “how much energy it has.” That’s related to $Q$ and $\Delta V$ (you’ll derive this later).
• It is not a constant for “two wires in general” unless you specify their shape, separation, and medium.

Why capacitance matters

Capacitance connects electrostatics (fields and potentials) to circuits and energy storage.

• In electrostatics, it lets you convert between charge and potential difference without re-solving the full field every time.
• In circuits, capacitors store and release energy and charge, shaping transient behavior.
• In physical systems (sensors, microphones, touchscreens), changing geometry changes $C$, which is measurable.

How it works: where the voltage comes from

When you separate charge onto two conductors, the field created by that separation does work on additional charges you try to move. That “resistance” to further separation shows up as a growing potential difference. The capacitance tells you how effectively the geometry “spreads out” the field so that a given amount of charge produces a smaller (or larger) voltage.

A useful analogy (with limits): think of capacitance like the “size” of a container for charge per volt. Bigger plates closer together generally mean a larger “container,” i.e., larger $C$.

Notation you’ll see

Combining capacitors (often needed even in “electrostatics” problems)

Even when a problem isn’t explicitly a circuit problem, you may be asked for an equivalent capacitance of a configuration.

Parallel connection (same $\Delta V$ across each capacitor): charges add.

$C_{\text{eq}} = C_1 + C_2 + \cdots$

Series connection (same charge magnitude $Q$ on each capacitor in series): voltages add.

$\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots$

The “why” here is worth internalizing:

• In parallel, each capacitor sees the same voltage, so the total charge drawn from a source is the sum: $Q_{\text{tot}} = (C_1 + C_2)\Delta V$.
• In series, the same charge must accumulate on each internal conductor surface (charge conservation), and the total voltage is the sum of each capacitor’s voltage: $\Delta V_{\text{tot}} = Q(1/C_1 + 1/C_2)$.

Example 1: Finding charge from capacitance

A capacitor has $C = 4.0\ \mu\mathrm{F}$ and is held at $\Delta V = 12\ \mathrm{V}$.

Use the definition:

$Q = C\Delta V$

Compute:

$Q = (4.0\times 10^{-6}\ \mathrm{F})(12\ \mathrm{V}) = 4.8\times 10^{-5}\ \mathrm{C}$

So each plate holds charge magnitude $4.8\times 10^{-5}\ \mathrm{C}$ (one positive, one negative).

Example 2: Equivalent capacitance

Two capacitors $C_1 = 2.0\ \mu\mathrm{F}$ and $C_2 = 6.0\ \mu\mathrm{F}$ are in series.

$\frac{1}{C_{\text{eq}}} = \frac{1}{2.0} + \frac{1}{6.0}$

$\frac{1}{C_{\text{eq}}} = \frac{3}{6} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$

$C_{\text{eq}} = 1.5\ \mu\mathrm{F}$

A quick reasonableness check: series capacitance is always less than the smallest capacitor, which matches $1.5 < 2.0$.

Exam Focus

• Typical question patterns:
  • Given $C$ and $\Delta V$, find $Q$ (or vice versa), often embedded in multi-step electrostatics.
  • Compute equivalent capacitance for simple series/parallel networks.
  • Use conceptual reasoning: “Does increasing plate separation increase or decrease capacitance?”
• Common mistakes:
  • Treating $C$ as if it changes when you change $Q$ (for ideal capacitors, it doesn’t).
  • Mixing up series vs parallel rules (use “same $\Delta V$ in parallel,” “same $Q$ in series”).
  • Forgetting that $Q$ is the magnitude on each plate, not the algebraic sum of charges (which is often zero for an isolated capacitor).

Parallel Plate Capacitors

What a parallel plate capacitor is

A parallel plate capacitor is the simplest (and most testable) capacitor model: two large conducting plates of area $A$ separated by distance $d$, with a uniform medium between them. When $d$ is small compared with the plate dimensions, edge effects are small and the electric field between plates is approximately uniform.

This model matters because it lets you connect:

• Fields (from Gauss’s law),
• Potential (from the field), and
• Capacitance (from the definition).

How the capacitance formula is built (conceptual derivation)

Assume vacuum (or air approximately) between plates.

1) Surface charge density on plates (magnitude) is

$\sigma = \frac{Q}{A}$

2) The field between two oppositely charged large plates is approximately uniform and given by

$E = \frac{\sigma}{\varepsilon_0}$

3) Potential difference between plates separated by distance $d$ in a uniform field is

$\Delta V = Ed$

Combine these:

$\Delta V = \left(\frac{\sigma}{\varepsilon_0}\right)d = \left(\frac{Q}{A\varepsilon_0}\right)d$

Use $C = Q/\Delta V$:

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

This result is a big deal because it encodes the physical intuition:

• Larger area $A$ gives more “room” for charge, increasing $C$.
• Larger separation $d$ makes it harder to maintain a given $Q$ without raising $\Delta V$, decreasing $C$.

Electric field and potential in a parallel plate capacitor

With no dielectric (vacuum):

$E = \frac{Q}{\varepsilon_0 A}$

and

$\Delta V = \frac{Qd}{\varepsilon_0 A}$

A common way to use these in problems is to treat $E$ as the “bridge” between charge and voltage.

Forces and pressure (why plates attract)

The plates attract because opposite charges pull toward each other. A useful AP Physics C result for the pressure (force per area) on the plates for a vacuum-filled parallel plate capacitor is

$P = \frac{1}{2}\varepsilon_0 E^2$

Then the force magnitude is

$F = PA = \frac{1}{2}\varepsilon_0 E^2 A$

This is essentially the same structure as energy density (you’ll see that later), and it’s a powerful way to connect field energy to mechanical effects.

Example 1: Capacitance from geometry

Plates have area $A = 0.020\ \mathrm{m^2}$ and separation $d = 1.0\ \mathrm{mm} = 1.0\times 10^{-3}\ \mathrm{m}$. Find $C$ (vacuum).

Use

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

with $\varepsilon_0 = 8.85\times 10^{-12}\ \mathrm{F/m}$.

$C = \frac{(8.85\times 10^{-12})(0.020)}{1.0\times 10^{-3}} = 1.77\times 10^{-10}\ \mathrm{F}$

So $C \approx 1.8\times 10^{-10}\ \mathrm{F} = 180\ \mathrm{pF}$.

Example 2: Field and voltage from charge

Using the same capacitor, suppose $Q = 3.6\times 10^{-9}\ \mathrm{C}$ is placed on the plates.

Field:

$E = \frac{Q}{\varepsilon_0 A} = \frac{3.6\times 10^{-9}}{(8.85\times 10^{-12})(0.020)} \approx 2.0\times 10^{4}\ \mathrm{V/m}$

Voltage:

$\Delta V = Ed = (2.0\times 10^{4})(1.0\times 10^{-3}) = 20\ \mathrm{V}$

A consistency check using $Q = C\Delta V$:

$\Delta V = \frac{Q}{C} = \frac{3.6\times 10^{-9}}{1.77\times 10^{-10}} \approx 20\ \mathrm{V}$

Exam Focus

• Typical question patterns:
  • Derive or apply $C = \varepsilon_0 A/d$, often paired with finding $E$ and $\Delta V$.
  • Multi-step problems: given geometry and either $Q$ or $\Delta V$, find the other and then compute energy.
  • Conceptual comparisons: “What happens to $C$ if $d$ doubles?” or “If $A$ halves?”
• Common mistakes:
  • Using $E = \sigma/(2\varepsilon_0)$ between the plates (that’s for a single infinite sheet; two plates give $E = \sigma/\varepsilon_0$ between them).
  • Forgetting the “large plates, small separation” assumption and misapplying the uniform-field model in situations where fringing dominates.
  • Confusing the direction of $E$ with its magnitude; on exams, sign conventions can matter when you compute potentials.

Dielectrics

What a dielectric is

A dielectric is an insulating material placed between capacitor conductors. Common examples are glass, plastic, or ceramic. Dielectrics don’t let free charge flow through them easily, but their molecules can polarize: positive and negative charges within the molecules shift slightly in opposite directions when an external electric field is present.

Why dielectrics matter

Dielectrics are used because they:

• Increase capacitance without changing the capacitor’s physical size.
• Reduce the electric field for a given free charge on the plates, which can help prevent electrical breakdown.
• Provide mechanical support and stability between plates.

In AP Physics C: E&M, dielectrics are also a major test of whether you can reason correctly about what stays constant: $Q$ or $\Delta V$.

How polarization changes the field

When you insert a dielectric, the field from the free charges on the plates polarizes the dielectric. That polarization creates bound charges on the dielectric surfaces:

• Bound negative charge appears near the positive plate.
• Bound positive charge appears near the negative plate.

These bound charges produce an electric field that partially cancels the original field inside the dielectric. The net effect is a reduced field magnitude.

A convenient macroscopic parameter is the dielectric constant (also called relative permittivity) $\kappa$, defined by

$\varepsilon = \kappa\varepsilon_0$

where $\varepsilon$ is the permittivity of the material.

For a linear dielectric that completely fills the gap, the field becomes

$E = \frac{\sigma}{\kappa\varepsilon_0}$

and the capacitance becomes

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

So inserting a dielectric multiplies the capacitance by $\kappa$.

Two “boundary condition” scenarios you must distinguish

Many mistakes happen because students implicitly assume the wrong constraint.

Scenario A: Capacitor connected to a battery (fixed voltage)

If the capacitor stays connected to a battery, the battery maintains a constant potential difference $\Delta V$.

• $\Delta V$ stays constant.
• Capacitance increases: $C \to \kappa C$.
• Therefore the free charge increases: $Q = C\Delta V$ increases by a factor of $\kappa$.

Physically: the dielectric reduces the field for the existing charge, which would reduce $\Delta V$, so additional charge flows from the battery onto the plates to bring $\Delta V$ back to the battery’s value.

Scenario B: Isolated capacitor (fixed charge)

If the capacitor is disconnected from any circuit and isolated, the free charge on the plates cannot change.

• $Q$ stays constant.
• Capacitance increases: $C \to \kappa C$.
• Therefore the potential difference decreases: $\Delta V = Q/C$ decreases by a factor of $\kappa$.

Physically: polarization reduces the field, so the same charge separation produces a smaller voltage.

Dielectric slabs and partial filling (idea level)

If a dielectric fills only part of the gap or area, you often model the system as a combination of capacitors (series or parallel), because different regions have different permittivity or separation. In AP problems, the geometry is usually chosen so this reduction is straightforward.

Example 1: Dielectric insertion with battery connected

A parallel plate capacitor has vacuum capacitance $C_0 = 200\ \mathrm{pF}$ and is connected to a $9.0\ \mathrm{V}$ battery. A dielectric with $\kappa = 4.0$ completely fills the space.

New capacitance:

$C = \kappa C_0 = 4.0(200\ \mathrm{pF}) = 800\ \mathrm{pF}$

Since $\Delta V$ stays $9.0\ \mathrm{V}$, the charge becomes

$Q = C\Delta V = (800\times 10^{-12}\ \mathrm{F})(9.0\ \mathrm{V}) = 7.2\times 10^{-9}\ \mathrm{C}$

Compare to before insertion:

$Q_0 = C_0\Delta V = (200\times 10^{-12})(9.0) = 1.8\times 10^{-9}\ \mathrm{C}$

Charge increased by a factor of 4, matching $\kappa$.

Example 2: Dielectric insertion with isolated capacitor

The same capacitor is instead charged to $Q = 2.0\times 10^{-9}\ \mathrm{C}$ and then disconnected from the battery. Insert the same dielectric with $\kappa = 4.0$.

Capacitance increases by 4, so voltage decreases by 4:

$\Delta V = \frac{Q}{C} = \frac{Q}{\kappa C_0} = \frac{1}{\kappa}\frac{Q}{C_0}$

If the initial voltage was

$\Delta V_0 = \frac{Q}{C_0}$

then

$\Delta V = \frac{\Delta V_0}{4.0}$

The key reasoning is the constraint: charge cannot flow onto or off the plates, so $Q$ is fixed.

Exam Focus

• Typical question patterns:
  • “A dielectric is inserted while the capacitor remains connected to a battery” vs “after it is disconnected” and asking for how $Q$, $\Delta V$, $E$, or energy change.
  • Use $C = \kappa\varepsilon_0 A/d$ and relate changes via proportional reasoning.
  • Conceptual questions about polarization and bound charge locations.
• Common mistakes:
  • Forgetting to decide whether $Q$ or $\Delta V$ is held constant before doing any math.
  • Assuming $E = \Delta V/d$ always “stays the same” because $d$ didn’t change; $E$ changes when $\Delta V$ changes.
  • Treating bound charge as if it adds to the free charge; bound charge rearranges internally and modifies the field, but the battery controls free charge only when connected.

Energy Stored in Capacitors

What “stored energy” means here

Separating charge requires work: you must move charges onto a conductor that is already at a higher potential. That work does not disappear; it is stored as electric potential energy associated with the electric field configuration.

For a capacitor, the stored energy can be expressed in several equivalent forms. Knowing when each form is convenient is part of mastery.

Deriving the energy formula (why there is a factor of 1/2)

Imagine charging a capacitor from $0$ up to a final charge $Q$. When the capacitor has an intermediate charge $q$, its voltage is

$V(q) = \frac{q}{C}$

To add an additional tiny amount of charge $dq$, the incremental work is

$dU = V(q)dq = \frac{q}{C}dq$

Integrate from $q = 0$ to $q = Q$:

$U = \int_0^Q \frac{q}{C}dq = \frac{1}{C}\left(\frac{Q^2}{2}\right) = \frac{Q^2}{2C}$

Using $Q = C\Delta V$ gives the other common forms:

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

and

$U = \frac{1}{2}Q\Delta V$

All three are equivalent; you choose based on what you know.

Where the energy “lives”: energy density in the electric field

In field theory, energy is stored in the electric field throughout space (or throughout the dielectric) rather than “inside the plates.” The energy density (energy per volume) in a region with field magnitude $E$ is

$u = \frac{1}{2}\varepsilon E^2$

For a vacuum-filled parallel plate capacitor, $\varepsilon = \varepsilon_0$, and the volume between plates is $Ad$, so

$U = u(Ad) = \frac{1}{2}\varepsilon_0 E^2 Ad$

This is consistent with $U = \frac{1}{2}C(\Delta V)^2$ because $C = \varepsilon_0 A/d$ and $E = \Delta V/d$.

How dielectrics affect energy (connected vs isolated again)

Energy changes are a favorite conceptual test because the answer flips depending on the constraint.

Battery connected (constant $\Delta V$)

Use

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

If a dielectric is inserted, $C$ increases by $\kappa$ while $\Delta V$ stays fixed, so

$U$ increases by a factor of $\kappa$.

This can feel surprising because some students expect the dielectric to “reduce energy” by reducing field. The resolution is that the battery supplies additional charge and does additional work; the system’s stored energy can increase.

Isolated capacitor (constant $Q$)

Use

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

If a dielectric is inserted, $C$ increases by $\kappa$ while $Q$ stays fixed, so

$U$ decreases by a factor of $\kappa$.

Here the field magnitude decreases and there is no battery to pump in extra charge, so energy stored in the field drops.

Example 1: Compute stored energy from $C$ and $\Delta V$

A capacitor has $C = 5.0\ \mu\mathrm{F}$ and is charged to $\Delta V = 10\ \mathrm{V}$.

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

$U = \frac{1}{2}(5.0\times 10^{-6})(10^2) = 2.5\times 10^{-4}\ \mathrm{J}$

Example 2: Energy change when inserting a dielectric

A capacitor with capacitance $C_0$ is charged to voltage $\Delta V$ and then disconnected (isolated). A dielectric with $\kappa = 3.0$ is inserted fully.

Initially:

$U_0 = \frac{1}{2}C_0(\Delta V)^2$

But after disconnection, it’s more direct to use $Q$-fixed reasoning. The initial charge is

$Q = C_0\Delta V$

After insertion, $C = \kappa C_0$, so

$U = \frac{Q^2}{2C} = \frac{(C_0\Delta V)^2}{2\kappa C_0} = \frac{C_0(\Delta V)^2}{2\kappa}$

So

$U = \frac{U_0}{\kappa}$

With $\kappa = 3.0$, the energy drops to one-third its original value.

A common deeper question: why a dielectric can be “pulled in”

If a dielectric slab is partially inserted, it is often pulled further into the capacitor. One way to understand this: the system tends to move toward lower potential energy.

• For an isolated capacitor (fixed $Q$), inserting dielectric increases $C$ and lowers $U = Q^2/(2C)$, so the system can reduce energy by pulling the dielectric in.
• For a battery-connected capacitor (fixed $\Delta V$), the capacitor’s stored energy increases, but the battery-field system can still lower its total energy depending on work done by the battery. In AP-style problems, you’ll usually be told which energy accounting to use or asked qualitatively about direction of force.

Exam Focus

• Typical question patterns:
  • Compute $U$ using whichever form matches given information: $U = Q^2/(2C)$, $U = (1/2)C(\Delta V)^2$, or $U = (1/2)Q\Delta V$.
  • Compare energies before/after changing $A$, $d$, or inserting a dielectric under “battery connected” vs “isolated” conditions.
  • Use energy density $u = (1/2)\varepsilon E^2$ for field-based reasoning, sometimes to connect to force/pressure.
• Common mistakes:
  • Using $U = (1/2)C(\Delta V)^2$ in a situation where $\Delta V$ is not constant (isolated capacitor after geometry/dielectric changes).
  • Dropping the factor $1/2$ (it comes from integrating the increasing voltage as the capacitor charges).
  • Confusing “energy stored in the capacitor” with “work done by the battery” during a dielectric insertion; they are not always the same because the battery can supply or absorb energy.