Chapter 21 Electrical Properties of Matter

Transcription

1 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 equipotential surfaces electrical field dipole potential gradient capacitance of a capacitor potential difference 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. Capitance 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. 169

2 Chapter 21 Electrical Properties of Matter OVERVIEW - The fundamental laws of electricity are abstract. As you look at this chapter you will find reference to "point" charges and "fields." These and other strange ideas have been introduced by scientists in an effort to explain the action in and around groups of charged particles interacting among themselves and with other groups of charges. Although some of the concepts seem difficult, most will be important throughout your study of electricity (e.g., force, work and voltage). SUGGESTED STUDY PROCEDURE - Begin your study of this chapter by reading the following Chapter Goals: Definitions, Coulomb's Law, Moving Charged Particles and Capacitance. In the first section of this Study Guide you will find a discussion of each of the terms listed under the goal of Definitions. Next, read text sections , 21.8 and Be careful to note the vector nature of the electrical force as illustrated in the Examples on page 476, whereas electrical potential is a scalar quantity see Figure Questions you encounter in your reading of the above-mentioned sections will be answered in the second section of this Study Guide chapter. Now turn to the end of the chapter and read the Chapter Summary and completesummary Exercises 1, 4, 5, 7-11, and Next, do Algorithmic Problems 1-6 and Exercises and Problems 2, 4, 5, 8, 10, and 15. For additional work with the concepts introduced in this chapter, see the Examples section of this Study Guide. Now you should be prepared to attempt the Practice Test provided in the last section of this chapter. When you finish the entire test, check your answers carefully against those given. If you have difficulty with any part of the test, refer back to a specific portion of the text or to a part of this Study Guide chapter for additional assistance. This study procedure is outlined below Chapter Goals Suggested Summary Algorithmic Exercises Text Readings Exercises Problems & Problems Definitions 21.1,21.4 1,4,5 1 Coulomb's Law ,10,&11 Moving Charged &15 2 2,4,5,10 Particles Capacitance 21.8,21.9 7,8,&16 3,4,5,6 8,15 Applications of 21.5, ,18 Electrostatics 170

3 DEFINITIONS DIELECTRIC CONSTANT - The ratio of the permitivity of a material to permitivity of a vacuum. Materials with high dielectric constants (like water), tend to be highly polar molecules. Thus when an external electric field is established, the material becomes polarized and acts to reduce the overall field within the substance. ELECTRIC FIELD - A property of space defined as the electric force per unit charge at each point. If the electric field is given for each point in space surrounding a charge distribution, the force on any single charge brought into that region can be found by multiplying the magnitude of the charge by the electric field value. The direction will be in the direction of the field if the charge is positive. POTENTIAL GRADIENT - The rate of change of potential in space; a vector quantity oriented in the direction of maximum change of potential. The potential gradient is identical in magnitude to the electric field in a region of space. POTENTIAL DIFFERENCE - If a charge is forced to move through a region of space against an electrical force, the work done in Joules per coulomb of charge is defined as the potential difference in volts. If a +1 coulomb charge is moved through a battery from the negative terminal to the positive terminal and 12 joules of work are done, then the battery has a potential difference of 12 volts. EQUIPOTENTIAL SURFACE - The surface defined by constant potential coordinates. Moving a charge along a line or surface of equipotential requires no work because no net voltage change occurs. DIPOLE - Equal but opposite charge distributions separated by some small distance. CAPACITANCE - Defined as the ratio of charge to voltage for a system of conductors 1 farad = coulomb/volt. Any arrangement of surfaces capable of storing electrical charges. 171

4 ANSWERS TO QUESTIONS FOUND IN THE TEXT SECTION 21.3 Electrical Forces Acting on Moving Charges Examples-1. On particles of charge -e in an upward vertical electric field E the force is acting downward. The downward force has a constant magnitude and direction, ee downward. Since there is no horizontal force on the charges, if they enter the electric field with a constant horizontal velocity, they will continue to maintain their constant horizontal motion. In addition they will feel a constant downward force. To what other type of motion is this similar? If you combine Newton's Second Law with the equations given in Section 3.8 what do you notice? The motion of this particle in this electric field is exactly like idealized projectile motion of an object near the surface of the earth except the gravitational force is increased by the addition of the downward electric force of magnitude ee. The path of the projectile is a downward curved parabola. This is a good place to show the relative strength of the electric field when compared to the gravitational field near the surface of the earth. Let us consider two of the so-called fundamental particles, the electron and the proton. The electron is attributed a charge of x C and a mass of 9.1 x kg. The proton is ascribed a charge of +1.6 x C and a mass of 1.7 x kg. A very weak electric field is of the order of 1 V/m. Right now you are probably seated a few meters from a 110 volt electric line so you are sitting in an electric field on the order of 10 to 50 volts/meter, which you are not able to detect. So a low field of 1 V/m is well below a level of human detection, nevertheless the electric force on either an electron or a proton in such a field is many times larger than the gravitational force on these particles, for the electron - F e /F g = qe/mg = ((1.6 x C)(1 V/m))/((9.1 x kg) x (9.8 m/s 2 )) = (1.6 x N) / (8.9 x N) F e /F g = 1.8 x (1) The electric force is 18 billion times the gravitational force acting on an electron in a weak electric field near the surface of the earth! for the proton - F e /F g = qe/mg = ((1.6 x C)(1 V/m))/((1.7 x kg) x (9.8 m/s 2 ) = (1.6 x N) / (1.7 x N) F e /F g = 9.4 x 10 6 (2) The electric force on a proton in a weak electric field near the surface of the earth is almost 10 million times the gravitational force on the proton! So you can see why it is possible to neglect the gravitational aspects of problems that involve the electric forces on fundamental particles. The gravitational forces on fundamental particles are so small for experiments carried out on the surface of the earth that they can be neglected. 2. The speed of migration of a constituent of a fluid in an electric field will be determined by such properties as the electric charge of the constituent, the mass of the constituent, and the resistance to flow of the constituent through the fluid. 172

5 SECTION 21.4 Electric Potential Question - Let us begin by assuming that there exists an equipotential surface Vo and at some point A on that surface there is an electric field (vector)e which is not perpendicular to the surface at point A. Then we can resolve (vector)e into two components, one component E which is perpendicular to the surface and E T which is tangential to the surface. The tangential component of E is able to do work on an electric charge and move it from point A to another point along the surface, say point B. Since the tangential component of (vector)e did work on the charge to move it from point A to point B, then point B must have a different electric potential than point A. But A and B are both points on the same equipotential surface. We have reached a contradiction. Something we have assumed to be true must not be true. The item that turns out to be false is the existence of a non-zero tangential component (vector)e. Therefore E T must be zero, so (vector)e is perpendicular to the equipotential surface. EXAMPLES COULOMB'S LAW 1. Find the magnitude and direction of the total force acting on a +4.0 C point charge at the origin if a +16.0C point charge is located at 4.0 m along the x- axis and a +9.0 C charge is located at 3.0 m along the y-axis. What data are given? The numerical values of this problem are shown in the sketch below at points 1, 2, and 3: The three charges are point charges, i.e. have no spatial dimensions, so we can use Coulomb's law to calculate the force between any two of the charges. Then the total force on any particle can be found by superposition. What physics principles are involved? We need to use Coulomb's law to find the vector forces between the charges. Then we will use the rules of vector algebra to add the individual vectors to find the total force. What equations are to be used? Coulomb's Law F 12 = (kq 1 q 2 )/ (r 2 ) r 12 (21.5) 173

6 Algebraic Solutions Let us begin by finding the total force acting on a charge Q 2 located at point 2. F 12 = (kq 1 Q 2 )/ (r 12 ) 2 r 12 (negative y-direction) (3) F 32 = (kq 3 Q 2 )/ (r 32 ) 2 r 34 (negative x-direction) (4) F 2 = F 12 + F 32. This is in case the two vectors F 12 and F 32 are at an angle θ = 90 ø with respect to one another. So we must use the vector algebra to add them. The magnitude of F 2 = SQR RT (F F 322 ) if Φ 2 = angle of F 2 downward from the negative x- axis; then tan Φ 2 = ((kq 1 Q 2 )/(r 12 ) 2 )/ (kq 3 Q 2 )/(r 32 ) 2 ) =Q 1 (r 32 ) 2 /Q 3 (r 12 ) 2 (5) Now let us find the total force on a charge Q 1 located at point 1 in Figure 21-1 above. F 21 = (KQ 2 Q 1 )/ (r 21 ) 2 r 21 (positive y-direction) (6) F 31 = (KQ 3 Q 1 )/ (r 31 ) 2 r 32 (an angle β from the positive y-axis) (7) The two forces acting on charge Q 1 are not at right angles, but we can use the knowledge of force components to add them together. The forces F 21 and F 31 in Figure 21-3 are equivalent to the following three forces: 174

7 Thus the magnitude of the total force = SQR RT((F 31 sin β) 2 + (F 21 + F 31 cos β) 2 ) (8) if Φ 1 is the angle from the positive y-axis then tan Φ 1 = (F 31 sinβ)/(f 21 + F 31 cosβ) = [((kq 3 Q 1 )/(r 31 ) 2 ) x sinβ]/ (kq 2 Q 1 /(r 21 ) 2 + kq 3 Q 1 /(r 31 ) 2 cosβ) tan Φ 1 = (Q 3 sinβ/(r 31 ) 2 ) / ((Q 2 /(r 21 ) 2 ) + (Q 3 cosβ/r 31 ) 2 ) This expression can be rewritten to eliminate the angle b if you wish. The size of β is determined by the distances r 21 and r 31 in the Figure 21-1, so sinβ = r 31 /((SQR RT)(r r 212 ) and cosβ = r 21 /((SQR RT)(r r 21 2 ) These expressions can be put into Equation (9) to find tan Φ 1 in terms of Q 2, Q 3, r 21, and r 31. Notice how messy the algebra becomes for forces which are not at right angles or colinear to one another. In a similar way we can find the total force of the charge Q 3. We resolve the forces into components, add the components, and use vector addition to find the magnitude of the total force. We use trigonometry to determine the direction of the total force. At this point we think it is more instructive to proceed with the actual numerical solution than to go farther with the more general algebraic solutions. Numerical Solutions Before we begin to add the forces let us calculate all the forces and their directions using Coulomb's Law: Force on charge Q 1 from Q 2 = F 21 = ((9.0 x 10 9 Nm 2 /C)(9.0C)(4.0C))/(3.0m) 2 r 21 F 21 = 3.6 x N in the positive y-direction. Force on charge Q 2 from Q 1 = F 12 = 3.6 x N in the negative y-direction Force on charge Q 3 from Q 2 = F 23 = ((9.0 x 10 9 Nm 2 /C)(16.0C)(4.0C))/(4.0m) 2 F 23 = 3.6 x N in the positive y-direction Force on charge Q 2 from Q 3 = F 32 = 3.6 x N in the negative y-direction The distance between charges Q 1 and Q 3 = SQR RT((3.0) 2 + (4.0) 2 ) = 5.0 m Force on Q 1 from Q 3 = F 31 = ((9.0 x 10 9 Nm 2 /C)(16.0C)(9.0C))/(5.0m) 2 N F 31 = 5.2 x N at an angle β from the positive y-axis where tan β = 4.0/3.0; β = 53 ø Force on Q 3 from Q 1 = F 13 = 5.2 x at an angle β from the negative y axis, or an angle a from the positive x-axis; α = 37 ø. 175

8 Now we can use vector addition and trigonometry to find the numerical values for the magnitudes of the three total forces F 1,F 2 and F 3 and the three angles Φ 1, Φ 2, and Φ 3. Magnitude of F 1 = SQR RT((3.6 x N x N cos53 ø ) 2 + (5.2 x N sin 53 ø ) 2 )) = SQR RT((3.6 x x ) 2 + (4.2 x ) 2 N) = SQR RT((6.7 x ) 2 + (4.2 x ) 2 ) = SQR RT( ) x N F 1 = 7.9 x N angle Φ 1 = tan -1 (4.2 x 1010/6.7 x 1010) = 32 ø Magnitude of F 2 = SQR RT((3.6 x ) 2 + (3.6 x ) 2 ) = 5.1 x N angle Φ 2 = tan -1 (3.6 x /3.6 x ) = 45 ø Magnitude of F 3 = SQR RT ((3.6 x x cos37 ø ) 2 + (5.2 x sin37 ø ) 2 ) = SQR RT((3.6 x x ) 2 + (3.1 x ) 2 ) = SQR RT((6.8 x ) 2 + (3.1 x ) 2 ) = SQR RT( ) x N F 3 = 7.5 x N Φ 2 = tan -1 (3.1 x / 6.8 x ) = 25 ø Thinking about the answer Consider the three charges as rigidly attached to one another by massless rods so that they form a rigid body. What will the total force on the body be from the sum of the forces F 1, F 2, and F 3? Can you think of a way to answer this question without using algebraic or numerical computations? (Hint: See Section 4.2 in the textbook) 176

9 POTENTIAL GRADIENT 2. Use the potential gradient concept to estimate the sizes of the environmental electric fields in which you live, e.g. under an electric blanket, in a typical room of a house, under a high voltage (380,000 volts) transmission line. What data are given? Not much, three typical situations, with the voltage given in only one; V transmission lines = 380,000V. What data are implied? In the United States of America, except for certain high power electrical devices, all the electric potentials in buildings and homes are 110 volts. You need to know typical distances to calculate gradients. You are only a few centimeters from an electric blanket, a few meters from the electrical wires in a house, and a few tens of meters from a transmission line. What physics principles are involved? The basic concept of the electric fields as the gradient of the electric potential is all that is needed for this problem, see Section 21.4 in the textbook. What equation is to be used? E = -ΔV/Δs (21.23) Solutions Since we are only interested in the size of the electric field we will omit the negative sign. Electric Field under an Electric Blanket 110V / 7cm 20 V/cm 2000 V/m Electric Field in a House 110V / 3m 40 V/m Electric Field under a Transmission Line 380,000V / 10m V/m Thinking about the answers You see that you are always living in some manmade electric fields. What are the effects of these fields on living organisms? Little is really known. This is presently an area of active research. Why don't you plan a career to study this topic? CAPACITANCE 3. A 3.0 µf capacitor and a 6.0 µf capacitor are connected in parallel and that combination is connected in series to a 4.0 µf capacitor. This group of three capacitors is connected across a 24 V battery. a) Draw the circuit, labeling all the elements by their specified values. b) Calculate the equivalent capacitance of the 3.0 µf and 6.0 µf capacitors. c) Calculate the equivalent capacitance of the combination of three capacitors. d) Calculate the voltage across each capacitor and e) Calculate the energy stored in each capacitor. What data are given? This question can be answered by answering part (a) of this question. So let us draw the circuit diagram and label the parts. 177

10 What data are implied? It is assumed that these are ideal capacitors that obey the conditions necessary for the deviations in Section 21.9 of the textbook to be valid. What physics principles are involved? The concepts of conservation of energy and conservation of electric charge can be used to derive equations for combining various combinations of capacitors. What equations are to be used? For a general system of three capacitors. Combined in Series: V total = V 1 + V 2 + V 3 (21.31) 1/C total = 1/C 1 + 1/C 2 + 1/C 3 (21.33) Q total = Q 1 = Q 2 = Q 3 (21.30) Combined in Parallel: V total = V 1 = V 2 = V 3 (21.34) C total = C 1 + C 2 + C 3 (21.36) Q total = Q 1 + Q 2 + Q 3 (21.35) Energy Stored in a Capacitor: E = (½) CV 2 (21.39) Algebraic Solution Let us begin by combining capacitors C 2 and C 3 and replace them by one capacitor C 23, then C 23 = C 2 + C 3 (11) The circuit then looks as follows: We can combine C 1 and C 23 to obtain C 1/C = 1/C 1 + 1/C 23 ; C = (C 1 C 23 )/(C 1 + C 23 ) (12) V = V 1 + V 23 (13) The total charge Q is given by Q = VC = (V (C 1 C 23 ))/(C 1 + C 23 ) (14) The separate charges Q 1 and Q 23 will be equal to Q since C 1 and C 23 are in series. V 1 = Q 1 /C 1 = Q/C 1 = VC/C 1 = VC 23 /(C 1 + C 23 ) (15) V 23 = Q 23 /C 23 = Q/C 23 = VC/C 23 = VC 1 /(C 1 + C 23 ) (16) The energy stored in the two capacitors can also be calculated E 1 = (½) C 1 V 2 1 = (C 1 V 2 (C 23 ) 2 )/(2(C 1 + C 23 ) 2 ) (17) E 23 = (½) C 23 V 2 23 = (C 23 V 2 (C 1 ) 2 )/ (2(C 1 + C 23 ) 2 ) (18) 178

11 We have completed the calculations we can make for the simplified circuit shown in Figure We can now complete the calculations for the more complex circuit given in the problem. Since none of the values we have calculated for C1 will change if we consider the two parallel capacitors C2 and C3 which are equivalent to C23 we need not do any further calculations for C1. We do need to find the charge and energy stored for both C2 and C3. Since C2 and C3are in parallel we already know the voltage across them. V 2 = V 3 = V 23 = (VC 1 )/(C 1 + C 23 ) (19) where C 23 = C 2 +C 3 from Equation (11). The charge on each capacitor is given by: Q 2 = C 2 V 2 = (C 2 VC 1 )/(C 1 + C 23 ) (20) Q 3 = C 3 V 3 = (C 3 VC 1 )/(C 1 + C 23 ) (21) The energy stored in each capacitor can be calculated also E 2 = (½) C 2 (V 2 )2 = (½) C 2 x (V 2 (C 1 ) 2 )/ ((C 1 + C 23 ) 2 ) (22) E 3 = (½) C 3 (V 3 )2 = (½) C 3 x (V 2 (C 1 ) 2 )/ (C 1 + C 23 )) 2 (23) We are finished. We have an algebraic solution for every part of this question: part (b) is answered by Equation (11); part (c) is answered by Equation (12): part (d) is answered by Equations (15) and (19); part (e) is answered by Equations (17), (22) and (23). Numerical Solutions part (b) Equivalent capacitance of 6.0 µf and 3.0 µf in parallel C 23 = (3.0)+(6.0) µf = 9.0 µf part (c) Equivalent capacitance of all three capacitors 1/C = 1/C 1 + 1/C 23 = 1/(4.0µF) + 1/(9.0 mf); C = 2.8µF part (d) Q = CV = (2.8µF)(24V) = 6.7 x 10-5 C V 1 = Q/C 1 = (6.7 x 10-5 )/4 F = 17V V 23 = 7.4V = V 2 = V 3 part (e) E 1 = (½)C 1 (V 1 ) 2 = (½) (4.0)(17) 2 = 5.8 x 10-4 joules E 2 = (½)C 2 (V 2 ) 2 = (½) (6.0)(7.4) 2 = 1.6 x 10-4 joules E 3 = (½)C 3 (V 3 ) 2 = (½) (3.0)(7.4) 2 = 8.2 x 10-5 joules Thinking about the answers You will notice how keeping two significant figures in each of the answers leads to small peculiarities in some of the answers. We know that V 1 and V 23 should add up to 24V. If you add the numbers calculated separately you must add 17V and 7.4V. How can you check the answer to part (e)? The total energy in the three capacitors should be equal to the sum of the three energies. E total = ½ CV 2 = (½) (2.8mF)(24V) 2 = 8.0 x 10-4 joules add up the values obtained in part (e) 5.8 x x x 10-4 = 8.2 x 10-4 joules The small difference of 2 parts in 80 arises from the use of two significant figures in the calculations. 179

12 PRACTICE TEST 1. A proton (q p = +1.6 x coulomb) and an electron (q e = -1.6 x coulomb) are stationary and are separated by a distance of 5 x meters (typical hydrogen atom separation). A. Calculate the magnitude and the direction of the force exerted on both the proton and the electron due to the presence of the other. Show the direction of both of these forces on the diagram above. B. Sketch the electrostatic field which exists between the charges by drawing several lines of force between the charges in the diagram above. 2. An electron (q e = -1.6 x coulomb and m e = 9.11 x kg) is accelerated through a series of three sets of parallel plates. Each set of plates (A, B, and C) has a similar voltage differential (100 volts) but a progressively increasing plate separation (10.0 cm, 20.0 cm, and 30.0 cm). (See the diagram shown below). A. Predict the electron's velocity when it leaves plate set A, at point x. B. What change in electron velocity occurs between points x and x 1? C. Calculate the electron's energy at y and again at z. D. If a fourth set of plates were added, can you predict the final velocity of the electron? 180

13 3. Three capacitors are connected in a series arrangement as indicated by the diagram shown below. After the circuit was connected, a physics student finds that the lµf capacitor stores a charge of 27.5 µcoul. A. What is the charge stored on each of the other capacitors? B. Find the voltage of the battery. C. Calculate the voltage drop across the 2.00µF capacitor. D. How much energy is stored by the entire series configuration of capacitors? ANSWERS: x N(AH reaction) x 10 6 m/s, none, 8.3 x 10 6 m/s, 10.2 x 10 6 m/s, yes, 11.8 x 10 6 m/s 3. A) 27.5 µcoul, B) 50 volts, C) 13.8 volts, D) 688 x 10-6 Joules 181