Contents

Applications of Electrostatics

The study of electrostatics has proven useful in many areas. This module covers just a few of the many applications of electrostatics.

The Van de Graaff Generator

Van de Graaff generators (or Van de Graaffs) are not only spectacular devices used to demonstrate high voltage due to static electricity—they are also used for serious research. The first was built by Robert Van de Graaff in 1931 (based on original suggestions by Lord Kelvin) for use in nuclear physics research. Figure 18.38 shows a schematic of a large research version. Van de Graaffs utilize both smooth and pointed surfaces, and conductors and insulators to generate large static charges and, hence, large voltages.

A very large excess charge can be deposited on the sphere, because it moves quickly to the outer surface. Practical limits arise because the large electric fields polarize and eventually ionize surrounding materials, creating free charges that neutralize excess charge or allow it to escape. Nevertheless, voltages of 15 million volts are well within practical limits.

Schematic of Van de Graaff generator is shown. The parts of the generators shown include a conductor, insulator, nonconductive belt, ion source, and experimental area.

Figure 18.38 Schematic of Van de Graaff generator. A battery (A) supplies excess positive charge to a pointed conductor, the points of which spray the charge onto a moving insulating belt near the bottom. The pointed conductor (B) on top in the large sphere picks up the charge. (The induced electric field at the points is so large that it removes the charge from the belt.) This can be done because the charge does not remain inside the conducting sphere but moves to its outside surface. An ion source inside the sphere produces positive ions, which are accelerated away from the positive sphere to high velocities.

Take-Home Experiment: Electrostatics and Humidity

Rub a comb through your hair and use it to lift pieces of paper. It may help to tear the pieces of paper rather than cut them neatly. Repeat the exercise in your bathroom after you have had a long shower and the air in the bathroom is moist. Is it easier to get electrostatic effects in dry or moist air? Why would torn paper be more attractive to the comb than cut paper? Explain your observations.

Xerography

Most copy machines use an electrostatic process called xerography—a word coined from the Greek words xeros for dry and graphos for writing. The heart of the process is shown in simplified form in Figure 18.39.

A selenium-coated aluminum drum is sprayed with positive charge from points on a device called a corotron. Selenium is a substance with an interesting property—it is a photoconductor. That is, selenium is an insulator when in the dark and a conductor when exposed to light.

In the first stage of the xerography process, the conducting aluminum drum is grounded so that a negative charge is induced under the thin layer of uniformly positively charged selenium. In the second stage, the surface of the drum is exposed to the image of whatever is to be copied. Where the image is light, the selenium becomes conducting, and the positive charge is neutralized. In dark areas, the positive charge remains, and so the image has been transferred to the drum.

The third stage takes a dry black powder, called toner, and sprays it with a negative charge so that it will be attracted to the positive regions of the drum. Next, a blank piece of paper is given a greater positive charge than on the drum so that it will pull the toner from the drum. Finally, the paper and electrostatically held toner are passed through heated pressure rollers, which melt and permanently adhere the toner within the fibers of the paper.

Four stages of xerography are shown. A positively charged aluminum drum is shown which is grounded. In second stage image is being transferred to it, creating positive image. In third stage, negatively charged toner is attached with the drum and in fourth stage, toner is pulled by the paper which is highly charged.

Figure 18.39 Xerography is a dry copying process based on electrostatics. The major steps in the process are the charging of the photoconducting drum, transfer of an image creating a positive charge duplicate, attraction of toner to the charged parts of the drum, and transfer of toner to the paper. Not shown are heat treatment of the paper and cleansing of the drum for the next copy.

Laser Printers

Laser printers use the xerographic process to make high-quality images on paper, employing a laser to produce an image on the photoconducting drum as shown in Figure 18.40. In its most common application, the laser printer receives output from a computer, and it can achieve high-quality output because of the precision with which laser light can be controlled. Many laser printers do significant information processing, such as making sophisticated letters or fonts, and may contain a computer more powerful than the one giving them the raw data to be printed.

A laser printer mechanism is shown. Laser beam produced from a computer, laser, or optics is incident on the drum containing some image.

Figure 18.40 In a laser printer, a laser beam is scanned across a photoconducting drum, leaving a positive charge image. The other steps for charging the drum and transferring the image to paper are the same as in xerography. Laser light can be very precisely controlled, enabling laser printers to produce high-quality images.

Ink Jet Printers and Electrostatic Painting

The ink jet printer, commonly used to print computer-generated text and graphics, also employs electrostatics. A nozzle makes a fine spray of tiny ink droplets, which are then given an electrostatic charge. (See Figure 18.41.)

Once charged, the droplets can be directed, using pairs of charged plates, with great precision to form letters and images on paper. Ink jet printers can produce color images by using a black jet and three other jets with primary colors, usually cyan, magenta, and yellow, much as a color television produces color. (This is more difficult with xerography, requiring multiple drums and toners.)

An ink-jet printer mechanism is shown. Ink is projected from ink nozzle and passes through the charging electrodes moving through deflection plate and finally imprinting on paper.

Figure 18.41 The nozzle of an ink-jet printer produces small ink droplets, which are sprayed with electrostatic charge. Various computer-driven devices are then used to direct the droplets to the correct positions on a page.

Electrostatic painting employs electrostatic charge to spray paint onto odd-shaped surfaces. Mutual repulsion of like charges causes the paint to fly away from its source. Surface tension forms drops, which are then attracted by unlike charges to the surface to be painted. Electrostatic painting can reach those hard-to-get at places, applying an even coat in a controlled manner. If the object is a conductor, the electric field is perpendicular to the surface, tending to bring the drops in perpendicularly. Corners and points on conductors will receive extra paint. Felt can similarly be applied.

Smoke Precipitators and Electrostatic Air Cleaning

Another important application of electrostatics is found in air cleaners, both large and small. The electrostatic part of the process places excess (usually positive) charge on smoke, dust, pollen, and other particles in the air and then passes the air through an oppositely charged grid that attracts and retains the charged particles. (See Figure 18.42.)

Large electrostatic precipitators are used industrially to remove over 99% of the particles from stack gas emissions associated with the burning of coal and oil. Home precipitators, often in conjunction with the home heating and air conditioning system, are very effective in removing polluting particles, irritants, and allergens.

(a) Schematic of an electrostatic precipitator. Air is passed through grids of opposite charge. The first grid charges airborne particles, while the second attracts and collects them. (b) The dramatic effect of electrostatic precipitators is seen by the absence of smoke from this power plant.[alt]Schematic of an electrostatic precipitator is shown. Four filters are shown one after another. Air passes through initial filter, then through positively charged grid, then through the third grid which is negatively charged and finally through the final grid. The number of particles is shown decreasing as air passes through various filters.

Figure 18.42 (a) Schematic of an electrostatic precipitator. Air is passed through grids of opposite charge. The first grid charges airborne particles, while the second attracts and collects them. (b) The dramatic effect of electrostatic precipitators is seen by the absence of smoke from this power plant. (credit: Cmdalgleish, Wikimedia Commons)

Problem-Solving Strategies for Electrostatics

Examine the situation to determine if static electricity is involved. This may concern separated stationary charges, the forces among them, and the electric fields they create.
Identify the system of interest. This includes noting the number, locations, and types of charges involved.
Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful. Determine whether the Coulomb force is to be considered directly—if so, it may be useful to draw a free-body diagram, using electric field lines.
Make a list of what is given or can be inferred from the problem as stated (identify the knowns). It is important to distinguish the Coulomb force $F F$ from the electric field $E E$ , for example.
Solve the appropriate equation for the quantity to be determined (the unknown) or draw the field lines as requested.
Examine the answer to see if it is reasonable: Does it make sense? Are units correct and the numbers involved reasonable?

Integrated Concepts

The Integrated Concepts exercises for this module involve concepts such as electric charges, electric fields, and several other topics. Physics is most interesting when applied to general situations involving more than a narrow set of physical principles. The electric field exerts force on charges, for example, and hence the relevance of Dynamics: Force and Newton’s Laws of Motion. The following topics are involved in some or all of the problems labeled “Integrated Concepts”:

The following worked example illustrates how this strategy is applied to an Integrated Concept problem:

Example 18.5

Acceleration of a Charged Drop of Gasoline

If steps are not taken to ground a gasoline pump, static electricity can be placed on gasoline when filling your car’s tank. Suppose a tiny drop of gasoline has a mass of $4.00 \times 10^{–15} kg 4.00 \times 10^{–15}$ and is given a positive charge of $3.20 \times 10^{–19} C 3.20 \times 10^{–19}$ . (a) Find the weight of the drop. (b) Calculate the electric force on the drop if there is an upward electric field of strength $3.00 \times 10^{5} N/C 3.00 \times 10^{5}$ due to other static electricity in the vicinity. (c) Calculate the drop’s acceleration.

Strategy

To solve an integrated concept problem, we must first identify the physical principles involved and identify the chapters in which they are found. Part (a) of this example asks for weight. This is a topic of dynamics and is defined in Dynamics: Force and Newton’s Laws of Motion. Part (b) deals with electric force on a charge, a topic of Electric Charge and Electric Field. Part (c) asks for acceleration, knowing forces and mass. These are part of Newton’s laws, also found in Dynamics: Force and Newton’s Laws of Motion.

The following solutions to each part of the example illustrate how the specific problem-solving strategies are applied. These involve identifying knowns and unknowns, checking to see if the answer is reasonable, and so on.

Solution for (a)

Weight is mass times the acceleration due to gravity, as first expressed in

w = mg . w = mg . size 12{w= ital "mg"} {}

18.20

Entering the given mass and the average acceleration due to gravity yields

w = (4.00 \times 10^{- 15} kg) (9.80 {m/s}^{2}) = 3 . 92 \times 10^{- 14} N . w = (4.00 \times 10^{- 15} . size 12{w= \( "44" "." "00" times "10" rSup { size 8{ - "15"} } `"kg" \)  \( 9 "." "80""m/s" rSup { size 8{2} }  \) =3 "." "92" times "10" rSup { size 8{ - "14"} } N} {}

18.21

Discussion for (a)

This is a small weight, consistent with the small mass of the drop.

Solution for (b)

The force an electric field exerts on a charge is given by rearranging the following equation:

F = qE . F = qE . size 12{F= ital "qE"} {}

18.22

Here we are given the charge ( $3.20 \times 10^{–19} C 3.20 \times 10^{–19}$ is twice the fundamental unit of charge) and the electric field strength, and so the electric force is found to be

F = (3.20 \times 10^{- 19} C) (3 . 00 \times 10^{5} N/C) = 9 . 60 \times 10^{- 14} N . F = (3.20 \times 10^{- 19} . size 12{F= \( 3 "." "20" times "10" rSup { size 8{ - "19"} } C \)  \( 3 "." "00" times "10" rSup { size 8{5} } "N/C" \) =9 "." "60" times "10" rSup { size 8{ - "14"} } N} {}

18.23

Discussion for (b)

While this is a small force, it is greater than the weight of the drop.

Solution for (c)

The acceleration can be found using Newton’s second law, provided we can identify all of the external forces acting on the drop. We assume only the drop’s weight and the electric force are significant. Since the drop has a positive charge and the electric field is given to be upward, the electric force is upward. We thus have a one-dimensional (vertical direction) problem, and we can state Newton’s second law as

a = \frac{F_{net}}{m} . a = \frac{F_{net}}{m} . size 12{a= {  { ital "net"`F}  over  {m} } `} {}

18.24

where $F_{net} = F - w F_{net} = F - w size 12{F=F - w} {}$ . Entering this and the known values into the expression for Newton’s second law yields

\begin{array}{l} a & = & \frac{F - w}{m} \\ = & \frac{9.60 \times 10^{- 14} N - 3.92 \times 10^{- 14} N}{4.00 \times 10^{- 15} kg} \\ = & 14 . 2 {m/s}^{2} . \end{array} \begin{array}{l} a & = & \frac{F - w}{m} \\ = \frac{9.60 \times 10^{- 14}}{4.00 \times 10^{- 15}} \\ = 14.2 . \end{array} alignl { stack {
 size 12{a= {  {F - w}  over  {m} } }  {} #
 size 12{ {}= {  {9 "." "60" times "10" rSup { size 8{ - "14"} } N - 3 "." "92"` times "10" rSup { size 8{ - "14"} } N}  over  {4 "." "00" times "10" rSup { size 8{ - "15"} }  ital "kg"} } }  {} #
="14" "." 2m/s rSup { size 8{2} }  {}
} } {}

18.25

Discussion for (c)

This is an upward acceleration great enough to carry the drop to places where you might not wish to have gasoline.

This worked example illustrates how to apply problem-solving strategies to situations that include topics in different chapters. The first step is to identify the physical principles involved in the problem. The second step is to solve for the unknown using familiar problem-solving strategies. These are found throughout the text, and many worked examples show how to use them for single topics. In this integrated concepts example, you can see how to apply them across several topics. You will find these techniques useful in applications of physics outside a physics course, such as in your profession, in other science disciplines, and in everyday life. The following problems will build your skills in the broad application of physical principles.

Unreasonable Results

The Unreasonable Results exercises for this module have results that are unreasonable because some premise is unreasonable or because certain of the premises are inconsistent with one another. Physical principles applied correctly then produce unreasonable results. The purpose of these problems is to give practice in assessing whether nature is being accurately described, and if it is not to trace the source of difficulty.

Problem-Solving Strategy

To determine if an answer is reasonable, and to determine the cause if it is not, do the following.

Solve the problem using strategies as outlined above. Use the format followed in the worked examples in the text to solve the problem as usual.
Check to see if the answer is reasonable. Is it too large or too small, or does it have the wrong sign, improper units, and so on?
If the answer is unreasonable, look for what specifically could cause the identified difficulty. Usually, the manner in which the answer is unreasonable is an indication of the difficulty. For example, an extremely large Coulomb force could be due to the assumption of an excessively large separated charge.