chapter 21 electrical properties of matter

Physics Including Human Applications

Chapter 21- Electrical Properties of Matter

453

Chapter 21 ELECTRICAL PROPERTIES OF MATTER GOALS When you have mastered the contents of this chapter, you will be able to achieve the following goals: Definitions Define each of the following terms, and use it in an operational definition:

dielectric constant electrical field potential gradient potential difference

equipotential surfaces dipole capacitance of a capacitor

Coulomb's Law Apply the basic model of an electrostatic field, and use Coulomb's law to calculate the force on one, two, or three given point charges. Potential Gradient Apply the gradient to electrical field phenomena. Moving Charged Particles Explain the motion of charged particles in an electric field. Capacitance Solve capacitance and capacitor problems, including the use of a capacitor as a means of storing electrical energy. Applications of Electrostatics List a number of applications of electrostatic principles to daily living and to medical equipment.

PREREQUISITES Before beginning this chapter you should have achieved the goals of Chapter 2, Unifying Approaches, Chapter 4, Forces and Newton's Laws, Chapter 5, Energy, and Chapter 9, Transport Phenomena.



454

Chapter 21 ELECTRICAL PROPERTIES OF MATTER 21.1 Introduction You have observed and experienced many phenomena that are examples of the electrical nature of matter. During the winter you have experienced a shock by walking over a rug and then touching a metal object or by sliding across your automobile seat and touching the door handle. You have seen bolts of lightning. Clothes you take out of a clothes dryer often cling together due to static electricity. The focusing and imaging system of a television picture tube makes use of the force on electrons moving in an electric field. Your heart is an electric device that keeps its steady beat with electric synchronization. These examples all involve the electrical nature of matter and electric interactions. In this chapter, the electric field model will be introduced and used to explain electric phenomena. 21.2 Electrical Forces A dry glass rod after being rubbed with silk will pick up small bits of paper. A dry rubber rod, after being rubbed with cat's fur, will pick up small bits of paper and will attract the rubbed glass rod. An analysis of many experiments such as these has lead us to postulate that there are two kinds of electricity which we call positive and negative(Figure 21.1). Conventionally, positive electricity is defined as the electricity that appears on a glass rod when it is rubbed with silk, and negative electricity is defined as the electricity that appears on a hard rubber rod rubbed with cat's fur. If an object has equal amounts of positive electricity and negative electricity, it is said to be neutral. If an object has an excess of negative electricity, it is said to be negatively charged. If an object has a deficiency of negative electricity, it is said to be positively charged.



455

In 1909 Robert A. Millikan performed a series of experiments on charged oil drops. He reported that these drops always contained an amount of electrical charge that was an integer times a fundamental constant: total charge = Ne (21.1) where N is an integer, 0,ñ1,ñ2,ñ3,...;and e is the magnitude of fundamental charge, which we call the electronic charge. The best measurements of the size of the electronic charge now indicates that the unit of electric charge in the SI units, which is called a coulomb (C), is equal to a very large number of electron charges. 1 coulomb = 1 C = 6,241,450,300,000,000,000 e 1 C = 6.24 x 1018e (21.2) or e = 1.60 x 10-19 C (21.3) In the eighteenth century, experiments like those illustrated in Figure 21.2 had shown that like charges repel each other and unlike charges attract each other. The inverse square law for electrical forces was verified by the experiments of Charles Coulomb. According to Coulomb's law of electrical interaction, the magnitude of the electrical force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between the charges,

(21.4)

where q1 and q2 are the amounts of electric charge of the two point charges and where r is the distance between the point charges. This expression may be written as a vector equation as follows:

(21.5)

where F12 is the force acting on charge 2 because of the presence of charge q1 and is a vector of unit magnitude which points in the direction from charge 1 to charge 2 (Figure 21.3). If q1 andq2 are of like sign, Equation 21.5 has the force between the charge acting so as to push the two charges apart (repulsion). See Figure 21.3a and b. If the charges are of opposite sign, the force between the charges acts to push the charges together (attraction) as in Figure 21.3c and d. The value of the proportionality constant k which appears in Equation 21.5 depends upon the system of units that is being used and upon the medium in which the two charges are located. If the point charges are located in a vacuum and the force is measured in Newtons, the charges in Coulombs and the distance between the charges in meters, then k has the following value and units:

(21.6)



456

where ε0 is the permittivity of free space and has the numerical value of 8.85 x 10-12 C2/N-m2. For problems in this book you may use the value of k at 9.00 x 109 N- m2/C2 in SI units. If charges are located in a medium that has a dielectric constant ε, then Equation 21.5 must be replaced by the following equation,

(21.7)



457

Figure 21.3

Directions and magnitudes of forces between charges for repulsion and attraction.

The dielectric constant accounts for the contribution of the molecular charges in the medium in the determination of the electric field inside such materials. For many gases the charge distribution of the molecules is symmetrical around the center of the molecule. For gases with this spherical charge symmetry, e is about 1. Materials which show a charge distribution that is not symmetric but is instead characterized by the separation of its negative and positive charge centers, are called polar materials. Polar molecules are electrically neutral but their nonsymmetric charge distribution results in dielectric constants that are much greater than 1. A simple example of such a distribution would be equal volumes of positive and negative spherical charges with their centers slightly displaced from each other. Since electrical charges interact with each other across some distance, they represent an example of systems that interact-at-a-distance. An electric charge moving on one side of the room seems to cause changes in the motion of an electrical charge on the other side of the room. We then construct the model of an electric force field that fills all the space in the vicinity of an electrical charge. Since there are electrical charges in all molecules, the universe is more or less filled with electric fields. We construct electric fields that obey linear relationships. We can use the principle of superposition to solve problems that involve the interaction-at-a-distance of many different charges. In such cases we solve the problem for each individual charge as if the others did not exist, and then add up all of the individual results. Hence, in any region where an electric force is acting upon an electric charge at rest, we say there exists an electric field. The direction of the field is defined as the direction of the force that would act upon a positive charge at that location in space. The magnitude of the field at that point is given by the force divided by the charge. The electric field E at a point is given by E = F/q (21.8) where F is the force acting on a charge q at that location. The electric field is a vector quantity defined as the force upon a unit charge. The SI units for E are newtons per coulomb. In the region where there is a single positive point charge q, the electric field is given by,



458

(21.9)

where is a vector of unit magnitude that points in a radial direction away from the charge q. If there are several point charges in the region, the resultant electric field is the vector sum of the separate fields resulting from each individual charge.

Figure 21.4

Determination of electric fields.

EXAMPLE In the diagram shown in Figure 21.4, determine the field at B resulting from a charge of +1.6 x 10-12 C at A, 4 cm north of B, and a charge of -1.8 x 10-12 C 3 cm east of B. The field at B resulting from the charge at A is a repulsive force whose magnitude is given by,

and which is directed south. The field at B resulting from the charge at C is an attractive field whose magnitude is given by,

directed to the east.

tan θ = 9/18 = 1/2 θ = 26.6° south of east. A charge of +3 C located at point B would experience a force of 60.3 N acting 26.6° south of east. A charge of -2 C at point B would experience a force of 40.2 N acting 26.6° north of west.



459

21.3 Electrical Forces Acting on Moving Charges Now consider the behavior of a charged particle free to move in an electric field. We first investigate the case of a charged body in a uniform electric field. Suppose we have chargeq in a field of E. The force acting on the charge is q times E, F = qE (21.10) in the direction of E. In an earlier chapter on Newton's laws (Chapter 4) we learned that a constant force acting on a free body produces an acceleration given by Newton's second law, a = F/m =qE/m (21.11) where, in this case, the force is given by qE. We can apply the equations for uniformly accelerated motion to this situation. So, if we start from rest at time t = 0, at any later time t the velocity is given by the acceleration multiplied by the time, v = at = (qE/m)t (21.12) The displacement y is given by one-half the acceleration times the square of the time, y = 1/2 at2 = 1/2 (qE/m)t2 (21.13) The square of the velocity is proportional to the product of the magnitudes of the field and the displacement, v2 = 2 | a | | y | = 2q | E | | y |/m (21 .14) The kinetic energy acquired after moving a distance y is given by the work done in moving the charge, or the product of the force times the distance, KE = 1/2 mv2 = q | E | | y | (21.15) Thus, we see that the motion of charged bodies can be influenced by the application of electric fields. There are many applications of this principle. EXAMPLES 1. An instrument that makes use of this phenomenon is the oscilloscope. Consider the

case of a beam of particles of charge -e moving with a velocity v in a horizontal direction. They enter a vertical electric field E in the upward direction. What happens to the particles? What is the shape of their path in the electric field?

2. Electrophoresis is a process that is used to separate different types of molecules. It is very useful in separating proteins and amino acids for further analysis. The electrophoresis process is based on the different motions of charged particles (ions) in certain solvents (called buffers) under the influence of applied electric fields. Different molecular constituents in the solution migrate at different speeds. For example, when blood is subjected to this process, serum albumin and three globular proteins are separated according to their speed of migration in the applied electric field. What properties of the constituents would be involved in determining the speed of migration in an electric field?



460

21.4 Electrical Potential Consider two points, A and B, in an electric field. A test charge q is moved from A to B, and we measure the work done in moving q from A to B without any acceleration. The electric potential difference between A and B, VAB, is defined as the work done per unit charge. So, VAB = VB - VA = W / Q or W =qVAB (21.16)

W may be positive, zero, or negative. If W is positive, B is said to be at a higher potential than A, and an external agent is required to move a positive charge q from A to B (Figure 21.5a) . If W is 0, A and B are at the same potential (moving a charge at right angles to the electric field lines, for example, as in Figure 21.5b. If W is negative, B is said to be at a lower potential than A and the work is done by the electric field (moving a positive charge from A to a point B closer to a negative charge as in Figure 21.5c. In the SI units, W is in joules, q in coulombs, and the potential is expressed in joules per coulomb, which is called volts (V). One volt equals an energy of one joule expended on one coulomb. The difference in potential between the two points is given by Equation 21.16. If one selects point A as a reference point and assigns it a value of zero potential, then any other potential, such as that at any point P, is given by Vp = Wp /q (21.16) whereWp represents the work required to take charge q from A to the point P under consideration. Usually the potential is taken to be zero at some point an infinite distance away. An electric charge of positive q will gain kinetic energy if it is released at an electric potential V and moves to zero potential. The work done on the charge is W = Vq . Since work is done on the charge, it will gain kinetic energy, KE = 1/2 mv2 = qV (21.17) The velocity of a charged particle, which starts from rest and undergoes a change of electric potential V is given by

(21.18) Since the electric field was developed as a linear model for interaction-at-a-distance,



461

the principle of superposition holds for potential, and you can find the resultant potential at a given point due to the separate individual charges by adding the individual potentials, Vtotal = V1 + V2 + V3 ... + Vn It can be shown that the potential due to a point charge at a distance r from the charge q is given by V = k(q / r) (21.19a) (See Equation 21.40.) Hence the potential at a given point produced by a number of point charges is given by V = k (q1 / r1 + q2 / r2 +q3 / r3 ... + qn / rn ) (21.19b) EXAMPLE Find the potential at B resulting from a charge of + 1.6 x 10-12 C at A, 4 cm north of B, and a charge of - 1.8 x 10-12 C at C, 3 cm east of B. See Figure 21.4. The potential at B due to the charge A is VA = kqA / rA = 9 x 109 x 1.6 x 10-12 / 4 x 10-2 = 0.36 V The potential at B due to charge C is VC = kqC /rC = 9 x 109( -1.8 x 10-12 / 3 x 10-2) = -0.54 V V at B = VA + VC = 0.36 + (-0.54) = -0.18 V where the potential at infinity is taken as zero. If no work is required to move a charge from one point to another, the two points are at the same potential and are said to be on an equi-potential surface. Let us consider the case of two large, parallel, flat conducting plates separated by a distance d with plate A charged positively, and plate B charged negatively. The field is directed from plate A to plate B, perpendicular to the plates, and is uniform (Figure 21.6). The work required to take a positive charge q from plate B to plate A is w = FD = + (-qEd) (21.20) because the force F must be equal but opposite to the electric field qE in order to move the charge from plate B to plate A. We can also express the work in terms of the potential difference between plates B and A, w = (VA -VB)q (21.21) We can set Equations 21.20 and 21.21 equal to each other and solve for E E = - (VA -VB )/ d (21.22)



462

The magnitude of the electric field E is the ratio of the change in potential (VA -VB) to the change in distance d. In other words, the electric field is a potential gradient, and the negative sign indicates that the electric field is oppositely directed to the potential gradient (that is, E is directed from high to low potential). In the case of parallel plates one notes that the direction of the electric field is perpendicular to the equipotential surfaces. This result was derived for a special case; however, it can be shown that the electric field is always equal to the negative of the potential gradient:

E = - ΔV / Δs (directed ⊥ to surfaces of constant potential) (21.23) where ΔV is the potential difference over the distance Δs. Also, the direction of the electric field is always perpendicular to the equipotential surfaces, directed from high to low potential. Questions Show that the electric field must always be perpendicular to the equipotential surfaces by assuming that E is not perpendicular to an equipotential surface and showing that this leads to a contradiction for an equipotential surface. 21.5 Gauss' Law Another formulation of the inverse square law for electric fields due to point charges is known as Gauss' law. First we must define the electric flux φE. To compute the electric flux through some small area ΔA, multiply the component of the electric field that is perpendicular to that area times the area ΔA; that is, ΔφE =E⊥ΔA(Figure 21.7). For a closed surface, Gauss' law states that the total flux passing through the area of that closed surface will be equal to the total electric charge enclosed divided by the permittivity of free space, ε0, φε =qtotal/ε0 (21.24) where φε is the total flux through the surface and qtotal is the total charge enclosed by the surface. The total flux, the sum of all the small fluxes ΔφE that pass through all the small areas ΔA needed to make up the total enclosed surface,

Therefore

(21.25)



463

EXAMPLES 1. Consider an isolated point charge. We know by symmetry that the electric field must be radial and depend only upon the distance R from the point charge. So we choose a spherical surface which has an area 4πR2 (Figure 21.8). Thus the electric flux is φE =E(4πR2) and φE =E(4πR2) = q/ε0 from Gauss' law. Solving for the value of the electric field we find:

(directed radially outward) (21.26)

This is Coulomb's law for a point charge q. Gauss' law is primarily useful when dealing with problems where we know from symmetry that the electric field is constant over a properly chosen surface around the given charge distribution. Gauss' law shows that the field outside any spherical distribution of charge is identical to that of a single point charge q total located at the center of the spherical distribution. 2. Let us apply Gauss' law to a long cylinder of positive charge neglecting end effects. (This is usually stated by saying the length is much greater than the radius of cylindrical charge.) Again the symmetry of the problem is such that we know there can only be an electric field pointing radially outward from the center of the cylinder. Therefore, if we pick a cylindrical surface surrounding the line charge (see Figure 21.9), the field is perpendicular to the surface and constant over the whole cylindrical area. Then from Gauss' law we get:



464

φE = E (area of cylindrical surface) = qtotal /ε. E = (2πRL) =qtotal / ε0

(21.27)

where qtotal/L is the linear charge density inside the cylinder. We see that a long line of charge produces a field that is proportional to 1/R as compared with 1/R2 for point or uniformly spherically distributed charges.

3. In the case of electrostatic equilibrium the electric field inside a conductor must be zero. If this were not the case, the free charge of the conductor would move under the influence of the electric field. Consequently, conductors are equipotential surfaces for electrostatic phenomena. We can use Gauss' law to show that the field inside an empty cavity in a conductor must be zero. Imagine a closed surface inside the cavity. From Gauss' law it follows that the field everywhere on the surface must be zero since there is no charge inside the surface. Thus, everywhere inside the cavity the electric field is zero. This result is completely independent of any charge on or outside the conductor. Such a space becomes an electrostatic shield, as motion of charge outside the conductor will have no affect on apparatus inside the cavity (Figure 21.10).

Figure 21.10

The field inside a hollow conductor is zero everywhere, and thus this space is shielded from effects due to charges outside of the conductor. This is the basis of electrostatic shielding that is used for many biological experimental set ups.

21.6 Electric Dipoles Recall that polar materials result from neutral charge distributions that have their



465

positive and negative charge centers separated by a distance we will call s. Such charge distributions are called electric dipoles (Figure 21.11). The product of the charge and the distance between charges qs gives the magnitude of the electric dipole and its vectors directed from the negative charge to the positive charge. It can be shown that the electric field due to a dipole is proportional to 1/r 3 where r is the distance from the dipole to the field point and r >> s. Substances with large molecular electric dipoles have high dielectric constants, for example, for water, ε = 80.36 at 20°C.

21.7 Electrostatic Applications The basic principles of electrostatic phenomena given above are sufficient to provide qualitative understanding of the electrical activity of the heart. One can ignore the electrical conductivity of the body and use Coulomb's law in order to show that the electrocardiogram (ECG) measures the potential distribution of a set of dipoles within the heart. See Footnote The dielectric constant of a material is a measure of the internal electric field for the material. An external electric field applied to a material with a high dielectric constant aligns the molecular dipoles inside the material producing a high internal electric field. The external field does work on the molecular dipoles, and this work is stored as potential energy of orientation within the material. The potential energy is a function of temperature. Can you propose a model to predict its qualitative temperature dependence? When the external field is removed, the molecular dipoles will gradually return to random orientations (what effect will temperature have on this process?) and produce very small depolarization currents in the materials. The dielectric constant of water is ~ 80. This high value is typical of biological cells and tissues. The interactions between such dielectric materials and externally applied fields are important areas of study in biophysics. The effectiveness of microwave ovens is a result of the resonant absorption of



466

electromagnetic energy by molecular dipoles in the food being cooked in the oven. The frequency of these resonances is in the microwave region of the electromagnetic spectrum. Recent research shows that the mapping of absorption of microwaves by human tissue may be very useful in early detection of cancer. 21.8 Capacitance Electrical energy can be stored in a device called a capacitor. A capacitor can be made of two conducting plates separated by a nonconductor (dielectric). In an uncharged condition both plates are neutral. In a charged condition one plate is positively charged, and the other is negatively charged; that is, the two plates are at different potentials. The ratio of the charge on plates to the potential difference between the plates is defined as the capacitance. C = q / V (21.28) where q is the charge in coulombs, V is the potential in volts, and C is the capacitance in farads. The farad is a very large unit of capacitance, and the capacitance of typical devices is given in micro (10- 6) farads or pico (10-12) farads. The capacitance of a capacitor depends upon the area of the plates, the thickness, the properties of the dielectric, and the geometric configuration. For a parallel plate capacitor C = εε0 A /d (21.29) where C is in farads, ε is the dielectric constant, A is the area in square meters, d is the thickness in meters, and ε0 is the permittivity of free space (8.85 x 10-12 F/m). Note that the charge stored is the dielectric constant for a constant applied voltage. EXAMPLE What is the capacitance of an air capacitor whose parallel plates have an area of 1.00 cm2 and are spaced 1.00 mm apart? εair = 1.00 C = 1 x 1.00 x 10-4 x 8.85 x 10-12/1.00 x 10-3 = 8.85 x 10-13 F = 0.885 pF 21.9 Combinations of Capacitors In a variety of applications it is desirable to use a combination of capacitors. We can connect two charged capacitors together in two ways. We can connect the two oppositely charged plates or the two similarly charged plates.



467

When the unlike plates of charged capacitors are connected together, the arrangement is called a series connection (Figure 21.12). Charges will move from one capacitor to the other until the positive charge on one plate is equal to the negative charge on the plate connected to it, q1 = q2 = q3 = Q = total charge in the combination (21.30) In this configuration the potential across the total combination is the sum of the individual potentials, V = V1 + V2 + V3 (21.31) but the potential across a capacitor is defined as the ratio of the charge to the capacitance. We can make use of that definition to derive an expression for the effective capacitance of the three used in series,

(21.32)

By making use of Equation 21.30, we find that the effective capacitance C of a series configuration of capacitors is calculated by taking the reciprocal of the sum of the reciprocals of the individual capacitances. For three capacitors,C1, C2, andC3, 1/C = 1/C1 + 1/C2 + 1/C3 (21.33) The configuration in which the like plates of charged capacitors are connected together is called a parallel connection (Figure 21.13). In this configuration the charges rearrange themselves until the potential V is the same across all of the capacitors V = V1 = V2 = V3 (21.34) and the total charge in the combination is the sum of the individual charges, Q = q1 + q2 + q3 (21.35) We can use the definition of the charge on a capacitor, CV, to derive an equation for the effective capacitance for a number of capacitors connected in parallel,



468

CV = C1V1 +C2V2 +C3V3 = (C1 + C2 +C3)V C = C1 + C2 + C3 (21.36) The effective capacitance of a number of capacitors connected in parallel is the sum of the individual capacitances. The charged capacitor has the ability to do work, and thus possesses energy. The energy, which is stored in the dielectric, can be dissipated by connecting the two plates by conductor. If we do so a redistribution of charges occurs. The energy stored in a charged capacitor can be calculated by finding the increment of work required to charge a capacitor from zero to a small charge Δq, Δw = VΔq (21.37) So the total energy stored in the capacitor will be equal to the total work required to build up the charge to Q, w = Vave Q (21.38)

where .

(21.39)

where C is in farads, Q is in coulombs, V is in volts, and w is in joules. The capacitor is an important component of many electrical instruments and appliances. One example is the direct current defibrillator. Defibrillation is the application of an electric shock to the heart to stop the rapid, uncoordinated contractions of the heart muscle called fibrillation. In a defibrillator, the capacitor is usually of the order of 10 to 20mF, the charging potential may be of the order of a few thousand volts, and the energy delivered to the patient may be as much as 400 J. A capacitor is also used in the electrical circuit of an electrocardiograph. The capacitor is used to block any direct current component of the ECG signal. A third application is in a heart pacemaker. A pacemaker is an electronic device which provides a regular periodic electrical stimulus to the heart to make regular the rhythmic performance of the heart. The energy requirement for a pacemaker is many orders of magnitude smaller than that for a defibrillator.



469

ENRICHMENT 21.10 Calculus Derivations of Electrostatic Relationships In Chapter 5 on work and energy you learned that work is given by

Δw = F cosθΔ s or

where ds is the incremental element of displacement. Let us apply this to an electric field. The field is radial from a charge q (Figure 21.14). The line ab represents some arbitrary path between these two points. Consider some element of path where E makes an angle with the path. The work done to take unit charge along length ds is then E cosθ ds = dw. The integral of the portion

is the integral along the line from a to b. For the field in the vicinity of a point charge,

ds cos θ = dr where θ is the angle between ds and dr . Hence the work is equal to the integral of electric intensity along the line ab,

w = kq (1/rb -1/ra) =Vb -Va for a unit charge where Va = kq /ra Vb = kq /rb (21.40) If we choose ra = ∞ as reference point and Va = 0, then in general the potential due to a point q of a point r meters away is

(21.41)

To show that the work done in moving a unit charge around any closed path in an electrostatic field is zero, we proceed as follows: If a point charge q' is moved in an electric field, the force is q'E. The work for moving q' a distance ds is then



470

dw = (q'E cos θ) (ds) Again,

but ds cosθ = dr. Then,

where ra is the starting and ending point of the loop. In charging a capacitor, charge is transferred from a lower potential to a higher potential. This requires an expenditure of energy. The work done to transfer charge dq is dw = V Adq. But,V = q/C; so

Because Q = CV,

This result was introduced in Section 21.9 in an intuitive way.

SUMMARY Use these questions to evaluate how well you have achieved the goals of this chapter. The answers to these questions are given at the end of the summary with the number of the section where you can find related content material. Definitions 1. The electric field model for electrostatics is analogous to the gravitational field model.

Give the analogous electrostatic term for the given gravitational term: source of field: mass, ___________ distance dependence: 1/r2, ___________. 2. The dielectric constant for a material is defined by which of the following ratios (Ee =

external electric field in free space, Ei = internal electric field) a. Ee/Ei b. Ei/Ee c. qi/qe d. qe/qi



471

3. The relationship between the potential gradient and electric field can be written as gradient of V(r) equals a. energy b. E c. -E d. qtotal e. none of these 4. Equipotential surfaces are satisfied by the following a. conductors b. E = 0 surfaces c. potential gradient = 0 d. charges at rest e. all of these 5. The unit for potential differences is the volt; this is equivalent to: a. joule- second b. watt- second c. joule/second d. joule/coulomb e. joule- coulomb 6. The electric dipole consists of a neutral charge distribution with a separation between the centers of the positive and negative charge. The electric field of the dipole is proportional to (s = charge separation) a. s2 b. 1/s2 c. s d. 1/s3 e. 1/s 7. The capacitor is a device that stores energy in its electric field. This energy depends on the capacitor's a. geometry b. applied voltage c. total charge d. dielectric e. all of these 8. Capacitance has units of farads. One farad is equal to



472

a. volt/coulomb b. coulomb/volt c. volt x coulomb d. volt/meter e. none of these 9. By Coulomb's law the force on a unit positive charge half way between identical charges q (in terms of the force F due to one charge) is a. 2F b. F c. zero d. F/2 e. 4F 10. The magnitude of the force on a unit positive charge half way between +q and -q charge separated by a distance 2d is given as: a. k(q/d2) b. 2(kq/d2) c. kq/2d2 d. zero 11. The direction of the force in question 10 is toward a. +q b. origin c. -q d. undetermined Potential Gradient 12. The constant electric field in a region is known to be 10 N/C. The potential difference between two points 0.5 m apart is a. 10 V b. 5 V c. 20 V d. 0.05 V e. cannot tell from these data 13. The direction of the electric field is opposite the direction of the potential gradient where the potential gradient is: a. minimum b. zero c. maximum



473

d. unknown e. any value Moving Charged Particles 14. The energy gained by a particle of charge +q and mass m accelerating through a potential difference V is given by a. qV b. (qV)1/2 c. 2qV1/2/m d. V/q e. q/m 15. The energy gained by a charged particle passing through a potential difference V is independent of the particle's a. charge b. mass c. path length d. initial speed e. none of these Capacitance 16. The energy stored by a capacitor can be written in terms of C (capacitance), V (voltage), and q(charge) as a. 1/2 qV b. 1/2 CV2 c. 1/2 (q2/C) d. none of these e. 1/2 q2C Applications of Electrostatics 17. Gauss' law suggests that shielding equipment from electric fields can be accomplished by putting it inside a. a cavity in dielectric b. a vacuum c. a cavity in conductors d. none of these



474

Answers

1. charge, 1/r2 (Section 21.2) 2. a (Section 21.2) 3. c (Section 21.4) 4. a,c (Section 21.4) 5. d (Section 21.4) 6. c (Section 21.6) 7. e (Sections 21.8 and 21.9) 8. b (Section 21.7) 9. c (Section 21.2)

10. b (Section 21.2) 11. c (Section 21.2) 12. b (Section 21.4) 13. e (Section 21.4) 14. a (Section 21.4) 15. b,c,d (Section 21.4) 16. a,c,b, (Section 21.9) 17. c (Section 21.5)

ALGORITHMIC PROBLEMS Listed below are the important equations from this chapter. The problems following the equations will help you learn to translate words into equations and to solve single-concept problems. Equations

(21.5)

(21.6)

(21.8, 21.9)

KE = 1/2 mv2 = q | E | | y | (21.15) W = qV (21.16) KE = qV (21.17)

E = - ΔV / Δs (directed ⊥ to surfaces of constant potential) (21.23)

(21.25)

C = q / V (21.28) C = εε0 A /d (21.29) 1/C = 1/C1 + 1/C2 + 1/C3 (series) (21.33) C = C1 + C2 + C3 (parallel) (21.36)



475

(21.39)

V = kq / r (21.19a) Problems 1. What is the electrostatic force between two ions in vacuum if one has a charge of 1.6 x

10-19 C and the other has a charge of 3.2 x 10-19 C and the charges are separated by a distance of 4 x 10-10 m?

2. What is the kinetic energy of an ion that has a charge of 1.6 x 10-19 C and is accelerated through a potential of 1.0 x 106 V?

3. Two parallel plates separated by 1.0 mm in a vacuum have a potential difference of 1000 V. What is the electric field of the capacitor?

4. A 2-mF capacitor is charged to a potential difference of 100 V. What is the charge on the capacitor?

5. What is the energy stored in the charged capacitor of problem 4? 6. A 2.0-µF and a 4.0-µF capacitor are connected in series. What is the capacitance of an

equivalent single capacitor? Answers

1. 2.88 x 10-9 N 2. 1.6 x 10-13 J 3. E = 1.0 x 106 N/C from positive to negative plate; E is a vector quantity perpendicular to the plates

4. 200 mC 5. 10-2 J 6. 1.3 µF

EXERCISES These exercises are designed to help you apply the ideas of a section to physical situations. When appropriate, the numerical answer is given in brackets at the end of the exercise. Section 21.2 1. Two point charges +4.00 C and -2.00 C are located along the x-axis at the origin and at

20 cm respectively. Sketch the electric field in the region of these charges. What is the field far, far away from the origin? Locate all the points on the x-axis where the electric field is zero. [2k/r2 N/C; x = 68.3 cm]

Section 21.3 2. A hydrogen ion of charge +e with a mass of 1.67 x 10-27 kg is initially at rest in an electric

field of 1.00 x 10-6 N/C. What is the velocity of the ion after 5 sec? How far has it traveled? What is its kinetic energy? [u=479 m/sec;s = 1.2 km; ΔK = 19.2 x 10-25 J]

Section 21.4 3. For the charges and locations given in exercise 1, find the location of zero potential

points on the x-axis. What is the potential far, far away from the origin? [40 cm, 13.3 cm, ~ 2k/r J/C]



476

4. In atomic and molecular experiments a unit of energy called an electron volt (eV) is used. One electron volt is the energy gained by a charge e as it changes its electric potential by one volt. Calculate the value of an electron volt in joules. [1 eV = 1.6 x 10-19 J]

5. In a given medical x-ray tube, electrons are accelerated through a potential of 10,000 V. How much energy does the accelerated electron have? What is the velocity of the electron after acceleration? [1.6 x 10-15 J or 104 eV, 5.9 x 107 m/sec]

Section 21.5 6. Given a hollow spherical conducting shell with a +Q charge at the center of its inner

cavity. Show that the charge on the inner surface of the conductor is -Q and that a charge of +Q is on the outer surface.

7. Show that if a copper ball is given a charge Q, the entire charge resides on its outer surface.

Section 21.8 8. A parallel plate capacitor consists of two flat plates 20 cm square separated by a

dielectric 0.2 mm thick. a. Find the capacitance if the dielectric is air. b. Find capacitance if the dielectric is mica (dielectric constant =6). [a. C = 17.8 x 10-

10 F; b. 106.2x 10-10 F] Section 21.9 9. If one has three capacitors with capacitances of 0.5, 1.0, and 2.0µF, what capacitance

can be produced by connecting these in various parallel and series combinations? [all series: 0.29µF; all parallel: 3.5 µF; six other arrangements: 2.33 µF; 1.40 µF; 1.17 µF; 0.86 µF; 0.71 µF; 0.43 µF]

PROBLEMS The following problems may involve more than one physical concept. Where appropriate, the numerical answer is given in brackets at the end of the problem. 10. Find the force of attraction between an electron and a proton (hydrogen nucleus) at

a distance of 5.3 x 10-11 m. How does this compare with their gravitational attraction? [82 x 10-9 N, 3.6 x 10-47 N]

11. Given that the dielectric constant of sodium chloride is 6.12, find the force of attraction between a Na+ ion and Cl- ion in a salt crystal if the separation is 2.8 x 10-

10 m. Calculate the energy of the ionic bond in a vacuum and in water (e = 80). [0.48 x 10-9 N; evac = 8.2 x 10-19 J, ewater = evac/80]

12. A square ABCD is 10.0 cm on a side. A charge of 2.00 x 10-10 C is placed at B, and a charge of -3.00 x 10-10 C is placed at C. Find:

a. the field at D b. the potential at D [a. 216 N/C down to right making angle of 17° with CD; b. -14.3

volt] 13. If a charge of 5.00 x 10-10 C is moved from D in problem 12 to the center of the

square, how much work is done? [7.55 x 10-10 J]



477

14. Given that the capacitance of 1.0 cm2 of a cell membrane is 1.0 µF, find the number of ions necessary to charge the membrane to 70 mV (resting potential for an axon). Assume that the ions are singly charged with q = 1.6 x 10-19 C. [44 x 1010 ions]

15. An air capacitor (0.1 µF) is charged with 20 µC of charge. The separation of the plates is 1 mm.

a. Find the electric field between the plates of the capacitor. b. Find the energy needed to charge the capacitor as given. [a. E = 2 x 105 V/m; b.

w = 2x 10- 3 J] 16. The electrical potential difference between the inside and the outside of a heart

muscle is about 90 mV. If the wall of each cell is an insulating layer of thickness of 5.0 x 10-9 m, what is the electric field in the cell membrane? Compare this with other electrical fields, such as when an electrical breakdown occurs in air (104 V/cm). [18 x 106 V/m, 3 x 106 V/m]

17. A direct current defibrillator has a maximum energy output of 400 J. The capacitance of the capacitor is 20µF. What is the maximum charging potential required? What is the electrical charge impulse sent through the body for maximum charging potential? [6300 V, 0.13 C]

18. If the energy for each impulse of a pacemaker, 2.4 x 10-4 J, is stored in a capacitor charged to a potential of 6.0 V, what is the capacitance of the capacitor, and what is the charge of each impulse? If 70 impulses are given per minute, how much energy is needed per day? [13.3µF, 80 µC, 24 J]

19. Specially designed capacitors called ion chambers are used to detect ionizing radiation. The incoming radiation produces ion pairs in the capacitor. These pairs tend to neutralize the charged capacitor. The battery attached to the capacitor recharges the capacitor. Outline a method that could be used to determine the energy deposited in the capacitor.

20. A 3.0-µF and 6.0-µF capacitor are connected in series to a 120-V source of potential. The capacitors are disconnected and reconnected with positive plates together and negative plates together. What is: a. The original charge on each capacitor? b. The initial potential difference for each capacitor? c. The final charge on each capacitor? d. The final potential difference for each? e. The change in energy of the charged capacitors? How do you account for the

difference [a. Q3 = Q6 = 240 µC; b. V3> = 80 V, V6 = 40 V; c. Q3 = 160 µC, Q6 = 320 µC; d. V3 = V6 =

53V; e. 1600 µJ] 21. A 2.00-µF and 3.00-µF capacitor are connected in parallel across a 100-V line. They

are disconnected and reconnected with the positive plate of each capacitor connected to the negative plate of the other. What is the final charge on each capacitor and potential difference across each? What is the change in energy? Explain the difference. [Q2 = 40 µC, Q3 = 60 µC, V2 = V3 = 20 V; ΔE = 24.0 x 10-3 J]

FOOTNOTES 1) For a complete discussion of electrostatic phenomena as applied to electrocardiograms, see R. K. Hobbie, "The Electrocardiogram as an Example of Electrostatics," American Journal of Physics 41 (June 1973): 824-831.

chapter 21 electrical properties of matter

Documents