A capacitor is a device designed to store electric charge in a controlled way. Capacitors appear in many everyday technologies (camera flashes, phone and computer electronics, medical equipment) and they are especially important when a device needs to store electrical energy and then release it quickly. In health and biomedical contexts, a familiar example is energy storage in 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 are built from two conducting parts that are very close to each other but do not touch. These conducting parts are often called “plates,” even when they are rolled or shaped in other ways. The key requirement is that they are separated by an insulating region so that charge cannot freely flow directly from one conductor to the other. This separation may be simply air in idealized examples, but in real capacitors it is usually an insulating material placed between conductors. (That insulating material is called a dielectric, discussed later in this section.)

When the terminals of a battery are connected to an initially uncharged capacitor, the battery’s chemical processes do work to separate charge. Electrons are pulled away from one conducting plate and pushed onto the other. As a result, equal amounts of charge build up on the two plates: one plate has [latex]+Q[/latex] and the other has [latex]-Q[/latex]. The capacitor as a whole remains electrically neutral, because the total charge is still zero, but we still say it “stores charge” because it holds separated charges on opposite plates. That charge separation 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 major factors:

• The applied voltage (how much “push” the battery provides to separate charge).
• The physical design of the capacitor (size, geometry, and what material separates the plates).

To build intuition, consider a system made of 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 models because it clearly shows how voltage, electric field, and stored charge are connected.

In Figure 15.2, the plates carry [latex]+Q[/latex] and [latex]-Q[/latex]. Electric field lines begin on the positive plate and end on the negative plate. In a simplified diagram we may draw one field line per unit charge as a visual aid. The crucial point is that the density of field lines is proportional to the amount of charge on the plates. More charge means more field lines and therefore a stronger electric field.

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

From the uniform-field relationship between voltage and electric field (for parallel plates),

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

This is true for all capacitors: the greater the voltage applied across a capacitor, the more charge it stores. However, the proportionality constant is not the same for all capacitors. Some designs can store much more charge than others at the same voltage.

We define the capacitance [latex]C[/latex] as the quantity that measures how much charge a capacitor can store per volt. The defining relationship is

This equation is not a special-case formula—it is the definition of capacitance. It highlights the two major influences on stored charge:

• [latex]V[/latex], the applied voltage (how strongly charge is separated), and
• [latex]C[/latex], the capacitor’s ability to store charge for a given voltage (set by geometry and materials).

Rearranging the definition gives the conceptual meaning of capacitance directly:

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

A farad is a very large unit. If a capacitor had a capacitance of 1 F, it would store 1 coulomb of charge with only 1 volt applied. In most electronics, typical capacitances are much smaller, ranging from picofarads to millifarads:

• [latex]1\ \text{pF} = 10^{-12}\ \text{F}[/latex] (very small, often in radio-frequency circuits)
• [latex]1\ \text{mF} = 10^{-3}\ \text{F}[/latex] (larger, often in power supply filtering)

Some common capacitor types are shown in Figure 15.3. The physical size of a capacitor does not always tell you its capacitance, because capacitance depends strongly on internal construction, plate area, plate separation, and dielectric material.

Parallel Plate Capacitor

The parallel plate capacitor shown in Figure 15.4 has two identical conducting plates, each of 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 (meaning we ignore any material placed between them). When a voltage [latex]V[/latex] is applied, charges separate and the capacitor stores [latex]+Q[/latex] on one plate and [latex]-Q[/latex] on the other.

We can predict how capacitance depends on geometry using physical reasoning. If the plates have a larger area, charges can spread out more, lowering repulsion between like charges on the same plate. That makes it easier to store more charge for the same voltage, so capacitance increases with [latex]A[/latex]. If the plates are closer together, opposite charges on facing plates attract more strongly. This also makes it easier to store charge at a given voltage, so capacitance increases as [latex]d[/latex] decreases.

For an ideal parallel plate capacitor (air or vacuum between plates), the capacitance is

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

Capacitance of a Parallel Plate Capacitor

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

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 related to why the farad is such a large unit. To get large capacitance with parallel plates, one generally needs a very large plate area and/or an extremely small separation distance.

Remember that this formula is most accurate when the plates are large compared with their separation and when edge effects (field spreading at the edges) can be neglected. In real devices, the general trend remains true: larger area and smaller 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 [latex]C[/latex], then use [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 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\ \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 (thin foils, rolled layers) and very thin separations, along with dielectric materials.

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 plates (because [latex]E=V/d[/latex]). That breakdown would allow charge to leak through the air as a spark or discharge.

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 ions inside the cell and an abundance of positive sodium ions ([latex]\text{Na}^+[/latex]) outside. When a nerve cell is stimulated, ion channels open and [latex]\text{Na}^+[/latex] moves into the cell, producing a rapid change in membrane potential that signals along the neuron.

The plasma membrane is extremely thin—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 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. The reason cells can sustain such strong fields is that the membrane is a specialized dielectric material and the separation distance is extraordinarily small. This example illustrates why capacitors are a useful model for cell membranes: they are thin insulating layers separating conducting fluids on either side.

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

If [latex]E[/latex] becomes large enough, the air molecules can ionize. Once enough ions and electrons are present, the air no longer behaves like an insulator; it becomes a conductor and allows charge to move through it. This is electrical breakdown, and it often appears as a spark or discharge. The maximum field a material can withstand before breaking down is called its dielectric strength. Dielectric strengths for several materials are listed in Table 15.1.

Dielectric strength is extremely important in practice because it sets a limit on how large a voltage can be applied across a capacitor for a given plate separation. Even if you design a capacitor with a large capacitance, you cannot increase the voltage indefinitely, because at some point the dielectric material between the plates will begin to conduct and the capacitor will discharge.

For example, consider the capacitor in Example 15.1, where the plate separation is [latex]d=1.00\ \text{mm}[/latex]. Using the dielectric strength of air ([latex]3\times10^{6}\ \text{V/m}[/latex]), the approximate voltage limit is

However, if we replace the air with Teflon, the dielectric strength is much larger ([latex]60\times10^{6}\ \text{V/m}[/latex]). For the same separation of [latex]1.00\ \text{mm}[/latex], the capacitor can withstand up to about [latex]60{,}000\ \text{V}[/latex] before breakdown. This means the same geometry can safely operate at a much higher voltage when filled with a strong dielectric.

This has two major consequences that work together:

(1) The dielectric increases the capacitance by a factor [latex]\kappa[/latex]. For Teflon, [latex]\kappa=2.1[/latex], so the capacitance becomes [latex]2.1[/latex] times larger than it was with air.

(2) The dielectric strength allows a much larger voltage to be applied without breakdown. Since the maximum charge stored is [latex]Q_{\max}=CV_{\max}[/latex], increasing both [latex]C[/latex] and [latex]V_{\max}[/latex] can dramatically increase the maximum charge and stored energy.

Using the capacitance from Example 15.1 for the air-filled capacitor ([latex]C_{\text{air}}=8.85\ \text{nF}[/latex]), the maximum charge for the Teflon-filled version is

This is about 42 times the charge that the same capacitor could store when air-filled at its breakdown limit. This example shows why practical capacitors almost always use a dielectric: it not only increases [latex]C[/latex], it also allows the capacitor to operate safely at higher voltages.

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 inside its atoms or molecules shift slightly in opposite directions. Even though the material remains electrically neutral overall, this tiny internal charge separation creates bound charges on the surfaces of the dielectric. Those bound charges affect the electric field inside the capacitor and the amount of free charge that can accumulate on the plates for a given applied voltage.

A good visual model is shown in Figure 15.5. The capacitor plates carry free charges [latex]+Q[/latex] and [latex]-Q[/latex]. The dielectric molecules between them become polarized: the side of each molecule closer to the positive plate becomes slightly negative, and the side closer to the negative plate becomes slightly positive. The resulting bound charge layer on the dielectric’s surfaces has the opposite sign to the nearby plate. Because opposite charges attract, this bound charge helps “pull” additional free charge onto the plates from the battery. The outcome is that for the same applied voltage, the capacitor ends up with more stored charge, meaning its capacitance is larger.

Another way to understand the same result is to focus on the electric field. With a dielectric present, some electric field lines terminate on the bound charges in the dielectric rather than traveling directly from one plate to the other. That means the net electric field between the plates is reduced compared with the vacuum case, even if the plates carry the same amount of free charge. Since the voltage is related to field by [latex]V=Ed[/latex], a smaller [latex]E[/latex] leads to a smaller voltage [latex]V[/latex] for the same [latex]Q[/latex]. Because capacitance is defined as [latex]C=Q/V[/latex], reducing [latex]V[/latex] while holding [latex]Q[/latex] fixed increases [latex]C[/latex]. In fact, the dielectric constant can be defined in terms of electric field as

where [latex]E_0[/latex] is the field that would exist with vacuum between the plates and [latex]E[/latex] is the field with the dielectric present.

A more modern description (developed further in Atomic Physics) treats electrons not as planets on fixed orbits but as probability clouds. Even so, the key idea remains: external electric forces distort the average charge distribution. The atom stays neutral overall, but it can exert forces because a nearby external charge may be closer to one side of the induced dipole than the other.

Some molecules are naturally polarized even without an external field. These are polar molecules. Water is the most important polar molecule in biology. A water molecule has two hydrogen atoms and one oxygen atom, arranged in a bent (“boomerang”) shape. Oxygen attracts electrons more strongly than hydrogen does, so the electron density is greater near the oxygen end. This makes the oxygen end slightly negative and the hydrogen ends slightly positive. Because water molecules already have a built-in separation of charge, they align relatively easily in electric fields and contribute strongly to polarization effects. That is why water has a large dielectric constant ([latex]\kappa\approx80[/latex]) in Table 15.1.

For health and bio majors, it is worth emphasizing that water’s polarity is one reason electric fields in biological environments behave differently than in air or vacuum. Polar water molecules can reduce (“screen”) electric fields from charged biomolecules, influence how ions move, and shape electrical behavior in tissues. This same molecular property helps explain why humid air breaks down more easily than dry air: water molecules in the air become polarized and facilitate charge motion and ionization, lowering the effective breakdown threshold.

Conceptual Questions

14. Does the capacitance of a device depend on the applied voltage? What about the charge stored in it?
15. 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.
16. 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].)
17. How does the polar character of water molecules help to explain water’s relatively large dielectric constant? (Figure 15.7)
18. 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.
19. Water has a large dielectric constant, but it is rarely used in capacitors. Explain why.
20. 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?