15 Capacitors and Dielectrics

A capacitor is a device designed to store electric charge in a controlled way. Capacitors are found in many everyday technologies, including camera flashes, computers, smartphones, and medical equipment. They are especially useful whenever electrical energy must be stored and then released quickly. In health and biomedical applications, one familiar example is a heart defibrillator, where electrical energy is stored in a capacitor and then delivered as a brief, high-energy pulse to help restore a normal heart rhythm.

Most commercial capacitors consist of two conducting parts placed very close together but separated by an insulating region so that charge cannot flow directly between them. These conducting parts are commonly called plates, even though they are often rolled, folded, or manufactured in other shapes. In idealized examples, the insulating region is simply air, but practical capacitors almost always contain an insulating material between the conductors. This material is called a dielectric, which we will discuss later in this section.

When the terminals of a battery are connected to an initially uncharged capacitor, the battery uses chemical energy to separate charge. Electrons are removed from one conducting plate and deposited onto the other. As a result, equal amounts of opposite charge accumulate on the two plates: one plate carries a charge of [latex]+Q[/latex] while the other carries [latex]-Q[/latex]. Although the capacitor remains electrically neutral overall because the total charge is still zero, it stores separated charge. This separation of charge is what allows the capacitor to store electric potential energy.

The amount of charge [latex]Q[/latex] that a capacitor can store depends on two main factors:

• The applied voltage, which determines how strongly the battery separates charge.
• The physical design of the capacitor, including the size and geometry of the conductors and the material separating them.

To build intuition, consider two identical, parallel conducting plates separated by a small distance. This arrangement is called a parallel plate capacitor. It is one of the most important capacitor models because it clearly illustrates the relationship between voltage, electric field, and stored charge.

In Figure 15.2, the plates carry charges of [latex]+Q[/latex] and [latex]-Q[/latex]. Electric field lines originate on the positive plate and terminate on the negative plate. In diagrams, the number of field lines is only a visual aid, but their density represents the strength of the electric field. Consequently, more stored charge produces a greater density of field lines and therefore a stronger electric field.

For a parallel plate capacitor, the electric field increases as the stored charge increases, so we can write

From the relationship between electric field and voltage for a uniform field,

where [latex]d[/latex] is the plate separation. Since [latex]V\propto E[/latex] and [latex]E\propto Q[/latex], it follows that

This relationship is true for every capacitor: increasing the applied voltage increases the amount of charge stored. However, different capacitors store different amounts of charge at the same voltage because their physical construction differs.
We define the capacitance, denoted by [latex]C[/latex], as a measure of how much charge a capacitor stores for each volt applied across it. The defining relationship is

This is not simply another equation to memorize—it is the definition of capacitance. It shows that the stored charge depends on two independent factors:

• • the applied voltage [latex]V[/latex], and

• • the capacitance [latex]C[/latex], which depends only on the capacitor's geometry and the materials from which it is constructed.

Rearranging the definition gives

The SI unit of capacitance is the farad (F), named after the English scientist Michael Faraday. Since capacitance is defined as charge per unit voltage,

A capacitance of one farad is very large. A 1-F capacitor stores one coulomb of charge with only one volt applied across its terminals. Most capacitors used in electronics have much smaller capacitances, typically ranging from a few picofarads to several millifarads.
Some common metric prefixes used for capacitance are:

• • [latex]1\ \text{mF}=10^{-3}\ \text{F}[/latex] (millifarad)

• • [latex]1\ \mu\text{F}=10^{-6}\ \text{F}[/latex] (microfarad)

• • [latex]1\ \text{nF}=10^{-9}\ \text{F}[/latex] (nanofarad)

• • [latex]1\ \text{pF}=10^{-12}\ \text{F}[/latex] (picofarad)

Different types of capacitors are shown in Figure 15.3. Although larger capacitors often have larger capacitances, physical size alone is not a reliable indicator of capacitance. The amount of charge a capacitor can store depends primarily on its internal design, including the area of the conducting plates, the distance between them, and the dielectric material separating the plates.

 

 

Parallel Plate Capacitor

The parallel plate capacitor shown in Figure 15.4 has two identical conducting plates, each with surface area [latex]A[/latex], separated by a distance [latex]d[/latex]. In this ideal model, the space between the plates is air or vacuum. When a voltage [latex]V[/latex] is applied, charge separates so that one plate stores [latex]+Q[/latex] and the other stores [latex]-Q[/latex].
We can understand how capacitance depends on geometry without carrying out a detailed mathematical derivation. If the conducting plates have a larger surface area, the same amount of charge is distributed over a larger region. This reduces the repulsive forces between like charges on each plate, making it possible to store more charge for the same applied voltage. Therefore, capacitance increases with plate area [latex]A[/latex].
If the plates are closer together, the opposite charges on facing plates attract more strongly. This also makes it easier to store charge at a given voltage. Therefore, capacitance increases as the plate separation [latex]d[/latex] decreases.

A complete analysis using Gauss's law and the relationship between electric field and potential difference shows that the capacitance of an ideal parallel plate capacitor, with air or vacuum between the plates, is

[latex]C=\epsilon_0\frac{A}{d}[/latex]

 

 

Capacitance of a Parallel Plate Capacitor

 

[latex]C=\epsilon_0\frac{A}{d}[/latex]

For an ideal parallel plate capacitor with air or vacuum between the plates, capacitance is directly proportional to plate area and inversely proportional to plate separation.

Here [latex]\epsilon_0[/latex] is the permittivity of free space:

[latex]\epsilon_0=8.85\times10^{-12}\ \text{F/m}[/latex]

The small numerical value of [latex]\epsilon_0[/latex] is one reason the farad is such a large unit. To obtain a large capacitance using parallel plates, one generally needs a very large plate area, an extremely small separation distance, or both.
This formula is most accurate when the plates are large compared with their separation and when edge effects, or field spreading near the edges, can be neglected. In real devices, the same general trend remains true: larger plate area and smaller plate separation increase capacitance.

 

Example 15.1: Capacitance and Charge Stored in a Parallel Plate Capacitor

(a) What is the capacitance of a parallel plate capacitor with metal plates, each of area [latex]1.00\ \text{m}^2[/latex], separated by 1.00 mm? (b) What charge is stored in this capacitor if a voltage of [latex]3.00\times10^{3}\ \text{V}[/latex] is applied to it?

Strategy

Use the parallel plate formula to find the capacitance [latex]C[/latex]. Then use the definition of capacitance, [latex]Q=CV[/latex], to find the stored charge.

Solution for (a)

Convert the plate separation to meters:

[latex]d=1.00\ \text{mm}=1.00\times10^{-3}\ \text{m}[/latex]

Now compute the capacitance:

[latex]C=\epsilon_0\frac{A}{d}=(8.85\times10^{-12}\ \text{F/m})\frac{1.00\ \text{m}^2}{1.00\times10^{-3}\ \text{m}}[/latex]

 

[latex]C=8.85\times10^{-9}\ \text{F}=8.85\ \text{nF}[/latex]

 

Discussion for (a)

Even with plates as large as [latex]1.00\ \text{m}^2[/latex], the capacitance is only a few nanofarads because the farad is such a large unit. To create larger capacitances, engineers use designs with very large effective plate area, very thin separations, and dielectric materials between the conducting plates.

Solution for (b)

Use the definition of capacitance:

[latex]Q=CV[/latex]

 

[latex]Q=(8.85\times10^{-9}\ \text{F})(3.00\times10^{3}\ \text{V})[/latex]

 

[latex]Q=2.66\times10^{-5}\ \text{C}=26.6\ \mu\text{C}[/latex]

 

Discussion for (b)

This stored charge is only slightly larger than typical static electricity charges. Increasing the voltage further would eventually cause breakdown of the air between the plates because [latex]E=V/d[/latex]. Once breakdown occurs, charge can leak through the air as a spark or electrical discharge.

 

Cell Membranes as Capacitors

Another biologically important example of electric potential involves the cell's plasma membrane. A membrane separates a cell from its surroundings and selectively controls ion movement. Many cells maintain a membrane potential of about [latex]-70\ \text{mV}[/latex], largely due to an excess of negative charge inside the cell and a different distribution of ions outside the cell.
When a nerve cell is stimulated, ion channels open and ions move across the membrane. This produces a rapid change in membrane potential that travels along the neuron as an electrical signal.
The plasma membrane is extremely thin, typically about 7 to 10 nm thick. That means even a small voltage difference corresponds to a very large electric field across the membrane. Using [latex]E=V/d[/latex] with a typical membrane thickness of [latex]8\ \text{nm}[/latex],

[latex]E=\frac{V}{d}=\frac{-70\times10^{-3}\ \text{V}}{8\times10^{-9}\ \text{m}}=-9\times10^{6}\ \text{V/m}[/latex]

This electric field is comparable to, and even larger than, the breakdown field of air. Cells can sustain such strong fields because the membrane is a specialized dielectric material and the separation distance is extraordinarily small.
This example illustrates why capacitors provide an excellent model for biological membranes. The conducting fluids inside and outside the cell act like the capacitor plates, while the thin lipid membrane acts as the dielectric. Many electrical phenomena in physiology—including nerve impulses, muscle contraction, and the electrical activity of the heart—depend on this capacitor-like behavior of cell membranes.

Notice that the dielectric constant of air is very close to 1. As a result, an air-filled capacitor behaves almost like a vacuum-filled capacitor in terms of capacitance. The major difference is that air can undergo electrical breakdown if the electric field becomes too strong.
For a parallel plate capacitor, the relationship between electric field and applied voltage is

If the electric field exceeds the dielectric strength of the material, the material begins to conduct electricity. In air, this occurs because molecules become ionized, producing free electrons and ions that allow charge to move through the gap. The maximum electric field a material can withstand before this occurs is called its dielectric strength. Values for several common materials are listed in Table 15.1.
Dielectric strength is extremely important because it limits the maximum voltage that can safely be applied across a capacitor. Even if a capacitor has a large capacitance, increasing the voltage indefinitely is impossible because the dielectric will eventually break down and the capacitor will discharge.
For example, consider the capacitor from Example 15.1, where the plate separation is [latex]d=1.00\ \text{mm}[/latex]. Using the dielectric strength of air, the maximum safe voltage is

If the air is replaced by Teflon, whose dielectric strength is approximately [latex]60\times10^{6}\ \text{V/m}[/latex], the same capacitor could withstand approximately [latex]60\,000\ \text{V}[/latex] before breakdown. Thus, the same physical capacitor can safely operate at a much higher voltage simply by changing the dielectric material.
The dielectric provides two important advantages:

1. 1. It increases the capacitance by a factor of [latex]\kappa[/latex]. For Teflon, [latex]\kappa\approx2.1[/latex], so the capacitance becomes about 2.1 times larger than it is with air.

1. 1. Its greater dielectric strength allows a much larger voltage to be applied before electrical breakdown occurs. Since the maximum stored charge is given by [latex]Q_{\max}=CV_{\max}[/latex], increasing both the capacitance and the maximum operating voltage greatly increases the amount of charge—and therefore energy—that the capacitor can store.

Using the capacitance from Example 15.1, the maximum charge stored by the Teflon-filled capacitor is

This is approximately 42 times more charge than the air-filled capacitor could store at its breakdown voltage. This example illustrates why nearly all practical capacitors use dielectric materials: they not only increase the capacitance, but also allow the capacitor to operate safely at much higher voltages, dramatically increasing its energy-storage capability.

Microscopically, how does a dielectric increase capacitance? The key idea is polarization. When an electric field is applied across a dielectric, the positive and negative charges within its atoms or molecules shift slightly in opposite directions. Although the material remains electrically neutral overall, this tiny separation of charge creates bound charges on the surfaces of the dielectric.
These bound charges modify the electric field inside the capacitor and allow more free charge to accumulate on the conducting plates for the same applied voltage.
A simple model is illustrated in Figure 15.5. The capacitor plates carry free charges [latex]+Q[/latex] and [latex]-Q[/latex]. The molecules of the dielectric become polarized so that the side of each molecule nearest the positive plate becomes slightly negative, while the side nearest the negative plate becomes slightly positive.
The resulting layer of bound charge has the opposite sign to the nearby conducting plate. These bound charges attract additional free charge from the battery onto the plates. As a result, the capacitor stores more charge while the applied voltage remains unchanged. Since capacitance is defined as [latex]C=Q/V[/latex], the capacitance increases.

Another way to understand the same effect is to focus on the electric field. With a dielectric present, some electric field lines terminate on the bound charges within the dielectric instead of extending directly from one plate to the other. Consequently, the net electric field between the plates is smaller than it would be in vacuum, even when the free charge on the plates is unchanged.
Since the voltage across the capacitor is related to the electric field by

[latex]V=Ed[/latex]

a smaller electric field produces a smaller voltage across the same plate separation. Because capacitance is defined as [latex]C=Q/V[/latex], reducing the voltage while keeping the stored charge constant increases the capacitance.
The dielectric constant can also be expressed in terms of the electric field:

[latex]\kappa=\frac{E_0}{E}[/latex]

where [latex]E_0[/latex] is the electric field that would exist with vacuum (or approximately air) between the plates, and [latex]E[/latex] is the electric field after the dielectric has been inserted.
It is important to remember that the dielectric does not create additional charge. Instead, it allows the battery to move more charge onto the capacitor plates while maintaining the same applied voltage. This is the fundamental reason why inserting a dielectric increases a capacitor's capacitance.

 

Things Great and Small

The Submicroscopic Origin of Polarization. Polarization is the separation of positive and negative charge within an atom or molecule. A useful introductory model pictures an atom as a small positive nucleus surrounded by negatively charged electrons. Although this "planetary" picture is not completely accurate, it successfully explains many electrostatic phenomena.
When an external electric field is applied, the electron cloud shifts slightly relative to the nucleus. This displacement creates an induced dipole: one side of the atom becomes slightly negative while the opposite side becomes slightly positive. Although the atom remains electrically neutral overall, this small separation of charge allows it to interact with nearby electric fields and charged objects.

 

A more complete description, introduced later in the chapter on atomic physics, treats electrons not as particles moving in fixed orbits but as quantum-mechanical probability clouds. Nevertheless, the essential idea remains the same: an external electric field distorts the average electron distribution. The atom remains neutral overall, but the induced separation of charge allows it to interact with nearby electric fields.
Some molecules are naturally polarized even in the absence of an external electric field. These are called polar molecules. Water is the most important polar molecule in biology. A water molecule consists of one oxygen atom bonded to two hydrogen atoms in a bent ("boomerang") shape. Because oxygen attracts electrons more strongly than hydrogen, the electron density is concentrated near the oxygen atom. As a result, the oxygen end of the molecule is slightly negative, while the hydrogen ends are slightly positive.
Because water molecules already possess a permanent separation of charge, they align readily in electric fields and contribute strongly to polarization effects. This is why water has a very large dielectric constant ([latex]\kappa\approx80[/latex]), as shown in Table 15.1.
For students in the health sciences, water's polarity has important consequences. It helps screen the electric fields produced by charged biomolecules, influences the motion of ions in cells and tissues, and plays a central role in nerve conduction, muscle contraction, and many other biological processes. Water molecules in humid air also facilitate ionization, which is one reason humid air undergoes electrical breakdown more readily than dry air.

 

Interactive Exploration: Capacitor Lab

Capacitors store electrical energy by separating positive and negative charges on two conducting plates. In this simulation, you will investigate how the capacitance of a capacitor depends on the plate area, the distance between the plates, the applied voltage, and the dielectric material placed between them. You will also observe how these factors affect the electric field, stored charge, and stored energy.
Experiment by changing one variable at a time and observe how the capacitor responds. Compare your observations with the relationships developed in this chapter, including the definitions of capacitance, the equation for a parallel plate capacitor, and the effect of dielectric materials.

 

 

Guided Exploration

As you interact with the simulation, try to answer the following questions:

1. 1. Increase the applied voltage while keeping the capacitor geometry fixed. How do the stored charge and the electric field between the plates change?

 

1. 1. Increase the plate area while keeping all other variables constant. How does the capacitance change? How does the amount of stored charge change for the same applied voltage?

 

1. 1. Decrease the separation between the plates. What happens to the capacitance? What happens to the electric field if the applied voltage remains constant?

 

1. 1. Insert different dielectric materials between the plates. How does the dielectric affect the capacitance and the amount of charge stored at the same applied voltage?

 

1. 1. Use the built-in meters to measure the voltage and electric field. How are these quantities related as you change the plate separation? Does your observation agree with the relationship [latex]E=V/d[/latex]?

 

1. 1. Based on your observations, explain why engineers often increase plate area, decrease plate separation, and use dielectric materials when designing practical capacitors.

 

After completing the exploration, compare your observations with the concepts presented in this chapter. Notice that a capacitor stores electrical energy by separating charge on two conducting plates. Its capacitance is determined by its geometry and the dielectric material between the plates, while the applied voltage determines how much charge and energy are stored.

 

Section Summary

 

• • A capacitor is a device that stores electrical energy by separating equal amounts of positive and negative charge on two conductors.

 

• • The amount of charge [latex]Q[/latex] that a capacitor stores depends on both the applied voltage and the capacitor's physical characteristics, including its plate area, plate separation, and the material between the plates.

 

• • The capacitance [latex]C[/latex] is defined as the amount of charge stored per unit voltage:
    [latex]C=\frac{Q}{V}[/latex]

    The SI unit of capacitance is the farad (F).

 

• • For an ideal parallel plate capacitor with air or vacuum between the plates,
    [latex]C=\epsilon_{0}\frac{A}{d}[/latex]

    where [latex]\epsilon_{0}[/latex] is the permittivity of free space.

 

• • If the space between the plates is completely filled with a dielectric material, the capacitance becomes
    [latex]C=\kappa\epsilon_{0}\frac{A}{d}[/latex]

    where [latex]\kappa[/latex] is the dielectric constant (relative permittivity) of the material.

 

• • A dielectric increases the capacitance by reducing the electric field inside the capacitor for a given amount of stored charge. Dielectrics also allow capacitors to operate safely at higher voltages because many have much greater dielectric strength than air.

 

• • The dielectric strength of a material is the maximum electric field it can withstand before it begins to break down and conduct electricity.

 

 

 

Conceptual Questions

 

 

 

14. 14. Does the capacitance of a device depend on the applied voltage? What about the charge stored in it?

 

14. 14. Use the characteristics of the Coulomb force to explain why capacitance should be proportional to the plate area of a capacitor. Similarly, explain why capacitance should be inversely proportional to the separation between plates.

 

14. 14. Give the reason why a dielectric material increases capacitance compared with what it would be with air between the plates of a capacitor. What is the independent reason that a dielectric material also allows a greater voltage to be applied to a capacitor? (The dielectric thus increases [latex]C[/latex] and permits a greater [latex]V[/latex].)

 

14. 14. How does the polar character of water molecules help to explain water’s relatively large dielectric constant? (Figure 15.7)

 

14. 14. Sparks will occur between the plates of an air-filled capacitor at lower voltage when the air is humid than when dry. Explain why, considering the polar character of water molecules.

 

14. 14. Water has a large dielectric constant, but it is rarely used in capacitors. Explain why.

 

14. 14. Membranes in living cells, including those in humans, are characterized by a separation of charge across the membrane. Effectively, the membranes are thus charged capacitors with important functions related to the potential difference across the membrane. Is energy required to separate these charges in living membranes and, if so, is its source the metabolization of food energy or some other source?

 

 

 

 

Figure 15.9: The semipermeable membrane of a cell has different concentrations of ions inside and out. Diffusion moves the [latex]\text{K}^+[/latex] (potassium) and [latex]\text{Cl}^-[/latex] (chloride) ions in the directions shown until the Coulomb force halts further transfer. This results in a layer of positive charge on the outside, a layer of negative charge on the inside, and thus a voltage across the membrane. The membrane is normally impermeable to [latex]\text{Na}^+[/latex] (sodium ions).