3.3 - Capacitors

Introduction to Parallel Plate Capacitors

A capacitor is a device that stores electric charge and energy. The type you’ll see most in AP Physics is the parallel plate capacitor.

The Structure

A parallel plate capacitor is: - Two conducting plates, usually the same size and shape. - Separated by a small distance. - Carrying equal amounts of charge, but with opposite signs. - One plate positive (

+Q

), the other negative (

-Q

The main features: - The electric field between the plates is uniform, meaning parallel, evenly spaced field lines. - Edge effects usually get ignored in AP Physics problems. - The plates carry equal and opposite charges, and that’s what makes the field.

Capacitance

Capacitance is a property of the capacitor that links the charge it stores to the potential difference across its plates.

The Definition

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

where: -

C

= capacitance (F, farads). -

Q

= magnitude of the charge on each plate (C). -

\Delta V

= potential difference between the plates (V).

Units:

1 \text{ farad (F)} = 1 \text{ C/V}

Named for Michael Faraday, who did foundational work in electromagnetism.

What it means: capacitance measures the capacity to store charge at a given voltage. Higher capacitance means you can pack in more charge without raising the voltage.

Capacitance of a Parallel Plate Capacitor

The capacitance depends only on the capacitor’s geometry and the material between the plates:

C = \frac{\kappa \epsilon_0 A}{d}

where: -

\kappa

(kappa) = dielectric constant of whatever’s between the plates (dimensionless). -

\epsilon_0 = 8.85 \times 10^{-12}

C²/(N·m²) = permittivity of free space. -

A

= area of one plate (m²). -

d

= separation between the plates (m).

For air-filled capacitors,

\kappa = 1

, so:

C = \frac{\epsilon_0 A}{d}

What Changes the Capacitance

Reading off

C = \frac{\epsilon_0 A}{d}

Change	Effect on C	Why
Increase the area $A$	$C$ goes up	More room to store charge
Decrease the distance $d$	$C$ goes up	Closer plates interact more strongly
Insert a dielectric ( $\kappa > 1$ )	$C$ goes up	The material boosts charge storage

The key point: capacitance depends ONLY on geometry and materials, NOT on how much charge or voltage you put on it.

Finding Any Capacitance: The Recipe

Every capacitance on the AP exam comes from the same four steps:

Put $+Q$ on one conductor and $-Q$ on the other.

Use Gauss’s law to find $E$ in the gap.

Integrate to get the potential difference, $\Delta V = -\int \vec{E} \cdot d\vec{r}$ (the magnitude is all you need).

Divide: $C = \frac{Q}{\Delta V}$ . The $Q$ always cancels, which is why $C$ depends only on geometry and materials.

Parallel plates:

E = \frac{Q}{\epsilon_0 A}

is uniform (see below), so

\Delta V = Ed = \frac{Qd}{\epsilon_0 A}

and

C = \frac{Q}{\Delta V} = \frac{\epsilon_0 A}{d}

Spherical Capacitors

Two concentric conducting spheres: an inner sphere of radius

a

with charge

+Q

, and an outer shell of radius

b

with charge

-Q

. A Gaussian sphere of radius

r

between them (

a < r < b

) encloses only

+Q

, so the field is the same as a point charge’s:

E(4\pi r^2) = \frac{Q}{\epsilon_0} \quad\Rightarrow\quad E = \frac{Q}{4\pi\epsilon_0 r^2}

Integrate from the outer shell in to the inner sphere to find how much higher the inner sphere’s potential is:

\Delta V = V_a - V_b = -\int_b^a \frac{Q}{4\pi\epsilon_0 r^2}\,dr = \frac{Q}{4\pi\epsilon_0}\left[\frac{1}{r}\right]_b^a = \frac{Q}{4\pi\epsilon_0}\left(\frac{1}{a} - \frac{1}{b}\right) = \frac{Q}{4\pi\epsilon_0}\cdot\frac{b - a}{ab}

C = \frac{Q}{\Delta V} = \frac{4\pi\epsilon_0 ab}{b - a}

Sanity checks: if the gap is tiny (

b - a = d

and

ab \approx a^2

), this becomes

\frac{\epsilon_0 (4\pi a^2)}{d} = \frac{\epsilon_0 A}{d}

, the parallel-plate result. If the outer shell moves out to infinity (

b \to \infty

), it becomes

C = 4\pi\epsilon_0 a

, the capacitance of a single isolated sphere.

Example: a spherical capacitor has

a = 5.0

cm and

b = 6.0

cm:

C = \frac{4\pi(8.85 \times 10^{-12})(0.050)(0.060)}{0.060 - 0.050} = 3.3 \times 10^{-11} \text{ F} = 33 \text{ pF}

Cylindrical (Coaxial) Capacitors

Same recipe with a Gaussian cylinder. For an inner cylinder of radius

a

and an outer cylindrical shell of radius

b

, both of length

L

, the field in the gap is

E = \frac{Q}{2\pi\epsilon_0 L r}

, and integrating from

a

b

gives:

C = \frac{2\pi\epsilon_0 L}{\ln(b/a)}

The full derivation (with a dielectric filling the gap) is worked out on the 3.4 Dielectrics page; with air in the gap, just set

\kappa = 1

. These three shapes (parallel plates, concentric spheres, and coaxial cylinders) are the only ones the AP exam asks you to analyze with numbers or equations.

Electric Field in a Capacitor

The field between the plates is uniform, and you can get it two equivalent ways.

Method 1: From the Charge

E = \frac{Q}{\epsilon_0 A}

Method 2: From the Surface Charge Density

E = \frac{\sigma}{\epsilon_0}

where

\sigma = \frac{Q}{A}

is the surface charge density.

Where it comes from (Gauss’s Law):

Apply Gauss’s Law to a flat conducting plate:

EA = \frac{Q_{enclosed}}{\epsilon_0}

Solve for

E

E = \frac{Q}{\epsilon_0 A} = \frac{\sigma}{\epsilon_0}

That’s the pillbox from 3.1: one end of the Gaussian cylinder sits inside the metal plate, where

E = 0

, and the other end sits in the gap. You can also get it from Gauss’s law plus superposition, which also explains why there’s no field outside the plates.

Step 1: one charged sheet. Topic 1.6 (Planar Symmetry) used a pillbox that pokes through a single large sheet to show that the sheet makes a uniform field of

\frac{\sigma}{2\epsilon_0}

on both sides, pointing away from it for positive

\sigma

and toward it for negative

\sigma

Step 2: add the two plates. The positive plate’s field points away from it, and the negative plate’s field points toward it, each with magnitude

\frac{\sigma}{2\epsilon_0}

: - Between the plates: both fields point from

+

-

, so they add:

E = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}

. - Outside the plates (either side): the two fields point opposite ways and cancel:

E = 0

E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}

, directly proportional to the surface charge density on either plate. Double the charge on the plates and you double the field.

Near the edges the field lines bulge outward (called fringing), so the field isn’t uniform there. The uniform-field result works when the plate separation is much smaller than the plates themselves, which is the usual AP assumption.

Relationship to Voltage

The potential difference between the plates connects to the field by:

\Delta V = Ed

which comes from

\Delta V = -\int \vec{E} \cdot d\vec{r}

applied to a uniform field.

Charged Particles Between the Plates

Because the field between the plates is uniform, a charged particle there feels a constant force

\vec{F} = q\vec{E}

, so it has a constant acceleration

\vec{a} = \frac{q\vec{E}}{m}

. That’s the same situation as a ball thrown near Earth’s surface, with

\frac{qE}{m}

playing the role of

g

: - Along the plates (perpendicular to

\vec{E}

), there’s no force, so that velocity component stays constant. - Across the plates (parallel to

\vec{E}

), the particle speeds up or slows down at a constant rate. - The path is a parabola, bending toward the negative plate for a positive charge and toward the positive plate for a negative charge.

For an electron or proton, gravity is tiny next to the electric force, so ignore it. Example Problem 3 below works one out.

Energy Stored in a Capacitor

A charged capacitor holds electric potential energy, which is basically the work it took to pull the charges apart and park them on the plates.

The Energy Formulas

There are three equivalent ways to write it:

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

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

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

Where the other forms come from:

Start with

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

Form 2: substitute

Q = C\Delta V

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

Form 3: substitute

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

U = \frac{1}{2}Q\left(\frac{Q}{C}\right) = \frac{Q^2}{2C}

Which one should you use? - If you know

Q

and

\Delta V

, use

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

. - If you know

C

and

\Delta V

, use

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

. - If you know

Q

and

C

, use

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

Where the ½ Comes From

The stored energy equals the work an external force does to separate the charge, moving it bit by bit from one plate to the other. The catch is that the voltage isn’t constant while you do it. When the capacitor holds charge

q

, the voltage is

\frac{q}{C}

, so moving the next tiny bit

dq

takes work

dW = \frac{q}{C}\,dq

. Add it all up from empty (

q = 0

) to full (

q = Q

U = \int_0^Q \frac{q}{C}\,dq = \frac{1}{C}\left[\frac{q^2}{2}\right]_0^Q = \frac{Q^2}{2C} = \frac{1}{2}Q\Delta V

On a graph of

\Delta V

vs.

q

, that integral is the area under the line. The line is straight with slope

\frac{1}{C}

, so the area is a triangle,

\frac{1}{2}Q\Delta V

. The first bits of charge cross almost no voltage and the last bits cross the full

\Delta V

, so on average each bit crosses half the final voltage. That’s the ½.

Example: a 4.0 μF capacitor is charged to 30 V. The charge is

Q = C\Delta V = (4.0 \times 10^{-6})(30) = 1.2 \times 10^{-4}

C, and the stored energy is

U = \frac{1}{2}(1.2 \times 10^{-4})(30) = 1.8 \times 10^{-3}

J, the shaded area in the graph. A battery that charges it does

Q\Delta V = 3.6 \times 10^{-3}

J of work, twice what’s stored. The other half becomes thermal energy in the wires and resistors while the capacitor charges (more on that in RC circuits).

Example Problem 1: Circular Parallel Plate Capacitor

Problem: A capacitor has two circular parallel plates of radius

R

separated by distance

d

. Its capacitance is

C_0

What would the capacitance be if: (a) The radius doubles to

2R

? (b) The distance is cut in half to

d/2

? (c) You do both at once?

Give your answers in terms of

C_0

Solution

Method 1: just substitute

The general capacitance formula:

C = \frac{\epsilon_0 A}{d}

For circular plates,

A = \pi R^2

, so:

C_0 = \frac{\epsilon_0 \pi R^2}{d}

(a) Radius doubled to 2R, distance unchanged

C_a = \frac{\epsilon_0 \pi (2R)^2}{d} = \frac{\epsilon_0 \pi (4R^2)}{d} = 4 \cdot \frac{\epsilon_0 \pi R^2}{d}

C_a = 4C_0

Answer (a): doubling the radius multiplies the capacitance by 4.

(b) Distance halved to d/2, radius unchanged

C_b = \frac{\epsilon_0 \pi R^2}{d/2} = 2 \cdot \frac{\epsilon_0 \pi R^2}{d}

C_b = 2C_0

Answer (b): halving the distance multiplies the capacitance by 2.

(c) Both changes: radius to 2R and distance to d/2

C_c = \frac{\epsilon_0 \pi (2R)^2}{d/2} = \frac{\epsilon_0 \pi (4R^2)}{d/2} = 8 \cdot \frac{\epsilon_0 \pi R^2}{d}

C_c = 8C_0

Answer (c): doing both multiplies the capacitance by 8.

Method 2: the factor-of-one trick

Here’s a faster route. Just figure out what changed and by what factor.

Starting from

C = \epsilon_0 \pi R^2 / d

$\epsilon_0$ is constant, so factor of 1.

$\pi$ is constant, so factor of 1.

$R \to 2R$ , but it’s squared, so factor of $(2)^2 = 4$ .

$d \to d/2$ , and it’s in the denominator, so factor of $\frac{1}{1/2} = 2$ .

Multiply them together:

C_{new} = (1)(1)(4)(2) \cdot C_0 = 8C_0

Same answer, way less work.

Example Problem 2: Energy and Charge Calculations

Problem: A parallel plate capacitor has capacitance

C = 2.0

μF and gets charged up to

\Delta V = 12

Find: (a) The charge on each plate. (b) The energy stored. (c) The electric field, if the plates are

d = 0.50

mm apart.

Solution

(a) Charge on Each Plate

Straight from the definition of capacitance:

Q = C\Delta V

Q = (2.0 \times 10^{-6})(12)

Q = 2.4 \times 10^{-5} \text{ C} = 24 \text{ μC}

Answer (a):

Q = 24

μC on each plate, one positive and one negative.

(b) Energy Stored

We know

C

and

\Delta V

, so use

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

U = \frac{1}{2}(2.0 \times 10^{-6})(12)^2

U = \frac{1}{2}(2.0 \times 10^{-6})(144)

U = 1.44 \times 10^{-4} \text{ J} = 0.144 \text{ mJ}

Answer (b):

U = 1.44 \times 10^{-4}

(c) Electric Field

Use

\Delta V = Ed

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

Convert the distance:

d = 0.50

= 0.50 \times 10^{-3}

E = \frac{12}{0.50 \times 10^{-3}} = \frac{12}{5 \times 10^{-4}}

E = 2.4 \times 10^4 \text{ V/m} = 24 \text{ kV/m}

Answer (c):

E = 2.4 \times 10^4

V/m.

Example Problem 3: An Electron Between the Plates

Problem: An electron enters the gap of a parallel-plate capacitor at

2.0 \times 10^7

m/s, moving parallel to the plates and midway between them. The plates are 4.0 cm long and 1.0 cm apart, with a 45 V potential difference, and the top plate is positive. (a) Find the electron’s acceleration. (b) How far has it been deflected when it leaves the plates? Does it hit a plate? (c) In what direction is it moving as it leaves?

Solution

(a) The field is uniform:

E = \frac{\Delta V}{d} = \frac{45}{0.010} = 4.5 \times 10^3

V/m, pointing down (from the positive top plate to the negative bottom plate). The electron is negative, so the force on it points up, toward the positive plate. Gravity’s 10 m/s² is nothing next to this, so ignore it:

a = \frac{eE}{m_e} = \frac{(1.60 \times 10^{-19})(4.5 \times 10^3)}{9.11 \times 10^{-31}} = 7.9 \times 10^{14} \text{ m/s}^2

(b) Treat it like a horizontally launched projectile. Along the plates there’s no force, so the time between the plates is

t = \frac{L}{v_0} = \frac{0.040}{2.0 \times 10^7} = 2.0 \times 10^{-9}

s. Across the plates it starts with zero velocity and has constant acceleration:

y = \frac{1}{2}at^2 = \frac{1}{2}(7.90 \times 10^{14})(2.0 \times 10^{-9})^2 = 1.6 \times 10^{-3} \text{ m}

Answer (b): it’s deflected 1.6 mm upward. It started 5.0 mm from each plate, so it misses the top plate and leaves the gap.

(c) The velocity component across the plates is

v_y = at = (7.90 \times 10^{14})(2.0 \times 10^{-9}) = 1.6 \times 10^6

m/s, so:

\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{1.58 \times 10^6}{2.0 \times 10^7}\right) = 4.5^\circ \text{ above the original direction}

Once the electron leaves the plates, where the field is zero, it keeps going in a straight line at that angle.

Proportionality Relationships

Knowing how the variables lean on each other lets you solve problems fast.

For Capacitance: $C = \frac{\epsilon_0 A}{d}$

C \propto A \quad \text{(capacitance scales with area)}

C \propto \frac{1}{d} \quad \text{(capacitance goes inversely with distance)}

For Energy: Three Forms

From

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

U \propto C \quad \text{(at constant voltage)}

U \propto (\Delta V)^2 \quad \text{(at constant capacitance)}

From

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

U \propto Q^2 \quad \text{(at constant capacitance)}

U \propto \frac{1}{C} \quad \text{(at constant charge)}

Important: whether energy goes up or down when

C

changes depends entirely on what’s being held constant, charge or voltage.

Summary of Key Formulas

Quantity	Formula	Units
Capacitance (definition)	$C = \frac{Q}{\Delta V}$	F (farads)
Capacitance (geometry)	$C = \frac{\epsilon_0 A}{d}$	F
Electric field	$E = \frac{Q}{\epsilon_0 A} = \frac{\sigma}{\epsilon_0}$	V/m or N/C
Voltage-field relation	$\Delta V = Ed$	V
Energy (form 1)	$U = \frac{1}{2}Q\Delta V$	J
Energy (form 2)	$U = \frac{1}{2}C(\Delta V)^2$	J
Energy (form 3)	$U = \frac{Q^2}{2C}$	J

Common Mistakes

Confusing capacitance with charge or voltage. Capacitance $C$ is a fixed property baked into the geometry. Changing $Q$ or $\Delta V$ doesn’t change it, unless you physically rebuild the capacitor.

Forgetting to square the radius when the area changes. Double the radius and the area goes up by a factor of 4, not 2, because $A = \pi R^2$ . Always square the geometric dimensions that show up in area formulas.

Getting the energy proportionality backwards. Whether $U$ rises or falls when $C$ changes depends on what’s held fixed. Constant voltage means $U \propto C$ . Constant charge means $U \propto 1/C$ . Know your constraint.

Reaching for the wrong energy formula. Pick the one that matches what you actually know. Don’t force $U = \frac{1}{2}Q\Delta V$ when you only have $C$ and $\Delta V$ . Use $U = \frac{1}{2}C(\Delta V)^2$ instead.

Unit conversion slip-ups. Always get everything into SI: cm to m, mm to m, μF to F, kV to V. Mixed units are the single most common source of numerical errors in capacitor problems.

Back Next

3.3 - Capacitors

Introduction to Parallel Plate Capacitors

The Structure

Capacitance

The Definition

Capacitance of a Parallel Plate Capacitor

What Changes the Capacitance

Finding Any Capacitance: The Recipe

Spherical Capacitors

Cylindrical (Coaxial) Capacitors

Electric Field in a Capacitor

Method 1: From the Charge

Method 2: From the Surface Charge Density

Relationship to Voltage

Charged Particles Between the Plates

Energy Stored in a Capacitor

The Energy Formulas

Where the ½ Comes From

Example Problem 1: Circular Parallel Plate Capacitor

Solution

Example Problem 2: Energy and Charge Calculations

Solution

Example Problem 3: An Electron Between the Plates

Solution

Proportionality Relationships

For Capacitance: C=ϵ0AdC = \frac{\epsilon_0 A}{d}C=dϵ0​A​

For Energy: Three Forms

Summary of Key Formulas

Common Mistakes

For Capacitance: $C = \frac{\epsilon_0 A}{d}$