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Unformatted text preview: 165 Chapter 4: Electrostatics
Lesson #22
Chapter  Section: 41 to 43 Topics: Charge and current distributions, Coulomb's law Highlights: Maxwell's Equations reduce to uncoupled electrostatics and magnetostatics when charges are either fixed in space or move at constant speed. Line, surface and volume charge distributions Coulomb's law for various charge distributions Special Illustrations: Examples 43 and 44 CDROM Modules 4.14.5 CDROM Demos 4.14.8 166 Lesson #23
Chapter  Section: 44 Topics: Gauss's law Highlights: Gauss's law in differential and integral form The need for symmetry to apply Gauss's law in practice Coulomb's law for various charge distributions Special Illustrations: Example 46 CDROM Module 4.6 CDROM Demos 4.9 and 4.10 167 Lesson #24
Chapter  Section: 45 Topics: Electric potential Highlights: Concept of "potential" Relation to electric field Relation to charges Poisson's and Laplace's equations Special Illustrations: Example 47 168 Lesson #25
Chapter  Section: 46 and 47 Topics: Electrical materials and conductors Highlights: Conductivity ranges for conductors, semiconductors, and insulators Ohm's law Resistance of a wire Joule's law Special Illustrations: Example 49 Technology Brief on "Resistive Sensors" (CDROM) Resistive Sensors An electrical sensor is a device capable of responding to an applied stimulus by generating an electrical signal whose voltage, current, or some other attribute is related to the intensity of the stimulus. The family of possible stimuli encompasses a wide array of physical, chemical, and biological quantities including temperature, pressure, position, distance, motion, velocity, acceleration, concentration (of a gas or liquid), blood flow, etc. The sensing process relies on measuring resistance, capacitance, inductance, induced electromotive force (emf), oscillation frequency or time delay, among others. This Technology Brief covers resistive sensors. Capacitive, inductive, and emf sensors are covered separately (in this and later chapters). Piezoresistivity According to Eq. (4.70), the resistance of a cylindrical resistor or wire conductor is given by R = l/A), where l is the cylinder's length, A is its crosssectional area, and is the conductivity of its material. Stretching the wire by an applied external force causes l to increase and A to decrease. Consequently, R increases (A). Conversely, compressing the wire causes R to decrease. The Greek word piezein means to press, from which the term piezoresistivity is derived. This should not be confused with piezoelectricity, which is an emf effect (see EMF Sensors). 169 Lesson #26
Chapter  Section: 48, 49 Topics: Dielectrics, boundary conditions Highlights: Relative permittivity and dielectric strength Electrostatic boundary conditions for various dielectric and conductor combinations Special Illustrations: Example 410 170 Lesson #27
Chapter  Section: 410 Topics: Capacitance Highlights: Capacitor as "charge accumulator" General expression for C Capacitance of parallelplate and coaxial capacitors Joule's law Special Illustrations: Examples 411 and 412 Technology Brief on "Capacitive Sensors" (CDROM) Capacitive Sensors To sense is to respond to a stimulus (see Resistive Sensors). A capacitor can function as a sensor if the stimulus changes the capacitor's geometryusually the spacing between its conductive elementsor the dielectric properties of the insulating material situated between them. Capacitive sensors are used in a multitude of applications. A few examples follow. Fluid Gauge The two metal electrodes in (A), usually rods or plates, form a capacitor whose capacitance is directly proportional to the permittivity of the material between them. If the fluid section is of height Hf and the height of the empty space above it is (H Hf ), then the overall capacitance is equivalent to two capacitors in parallel: where w is the electrode plate width, d is the spacing between electrodes, and f and a are the permittivities of the fluid and air, respectively. 171 Lesson #28
Chapter  Section: 411 Topics: Energy Highlights: A charged capacitor is an energy storage device Energy density Special Illustrations: Technology Brief on "NonContact Sensors" (CDROM) NonContact Sensors Precision positioning is a critical ingredient of semiconductor device fabrication, as well as the operation and control of many mechanical systems. Noncontact capacitive sensors are used to sense the position of silicon wafers during the deposition, etching, and cutting processes, without coming in direct contact with the wafers. They are also used to sense and control robot arms in equipment manufacturing and to position hard disc drives, photocopier rollers, printing presses, and other similar systems. Basic Principle The concentric plate capacitor (A1) consists of two metal plates, sharing the same plane, but electrically isolated from each other by an insulating material. When connected to a voltage source, charges of opposite polarity will form on the two plates, resulting in the creation of electricfield lines between them. The same principle applies to the adjacentplates capacitor in (A2). In both cases, the capacitance is determined by the shapes and sizes of the conductive elements and by the permittivity of the dielectric medium containing the electric field lines between them. 172 Lesson #29
Chapter  Section: 412 Topics: Image method Highlights: Image method useful for solving problems involving charges next to conducting planes Remove conducting plane and replace with mirror images for the charges (with opposite polarity) Special Illustrations: Example 413 CDROM Demos 4.114.13 CHAPTER 4 173 Chapter 4
Sections 42: Charge and Current Distributions
Problem 4.1 A cube 2 m on a side is located in the first octant in a Cartesian coordinate system, with one of its corners at the origin. Find the total charge contained in the cube if the charge density is given by v xy2 e 2z (mC/m3 ). Solution: For the cube shown in Fig. P4.1, application of Eq. (4.5) gives
V
x 0 y 0 z 0 2 2 2 2z x 0 y 0 z 0 z 2m 2m 0 2m x y Figure P4.1: Cube of Problem 4.1. Problem 4.2 Find the total charge contained in a cylindrical volume defined by r 2 m and 0 z 3 m if v 20rz (mC/m3 ). Solution: For the cylinder shown in Fig. P4.2, application of Eq. (4.5) gives 3 2 2 r 0 0 z 0 3 r z Q z 0 0 r 0 2 10 3 2 20rz r dr d dz
2 3 480 (mC) 15C 1 2 3 x y e 12 Q v d V 2 2 2 xy2 e 2z dx dy dz e
4 8 1 3 2 62 mC 174 CHAPTER 4 z 3m 0 2m x 2m y Figure P4.2: Cylinder of Problem 4.2. Solution: For the cone of Fig. P4.3, application of Eq. (4.5) gives R 0 0 0 1 86 65 (mC) 128 3 2 2 3 2 5 R cos3 3 2 Q 0 0 2 4 2 R 0 10R2 cos2 R2 sin dR d d 4 2 Problem 4.3 Find the total charge contained in a cone defined by R 0 4, given that v 10R2 cos2 (mC/m3 ). 2 m and CHAPTER 4
z 175 2m /4 0 x y Figure P4.3: Cone of Problem 4.3. Solution: Solution: (a)
r 0 0 0 s0 a2 4 r 0 0 0 0 sin 2 2 2 Q s0 sin2 r dr d s0 a 2 r2 2 a 2 1 cos 2 2 (b) 0 0 d a2 s0 2 Q s ds a 2 s0 cos r dr d s0 r2 2 a 2 sin 0 Problem 4.5 Find the total charge on a circular disk defined by r (a) s s0 cos (C/m2 ), (b) s s0 sin2 (C/m2 ), (c) s s0 e r (C/m2 ), (d) s s0 e r sin2 (C/m2 ), where s0 is a constant. 5 5 5 a and z Q l dy 5 5 24y2 dy 24y3 3 5 Problem 4.4 If the line charge density is given by l charge distributed on the yaxis from y 5 to y 5. 24y2 (mC/m), find the total 2000 mC 2C 0 if: 176 (c) CHAPTER 4 r 0 0 0 (d) ^ Problem 4.6 If J y4xz (A/m2 ), find the current I flowing through a square with corners at 0 0 0 , 2 0 0 , 2 0 2 , and 0 0 2 . Solution: Using Eq. (4.12), the net current flowing through the square shown in Fig. P4.6 is z 2m J 2m x 0 y Figure P4.6: Square surface. y 0 x 0 z 0 S x 0 z 0 I J ds ^ y4xz 2 2 2 ^ y dx dz x z 2 2 e 1 a s0 1 a s0 1 e a r 0 s0 re r dr r 0 0 a Q a 2 s0 e r sin2 r dr d
2 0 sin2 d 1 a 2 2s0 1 e a 1 a 2s0 re Q s0 e r r dr d 2s0 a 2 a re r r dr e
r a 0 16 A CHAPTER 4 177 Solution: Using Eq. (4.12), we have R 5 0 0 Problem 4.8 An electron beam shaped like a circular cylinder of radius r 0 carries a charge density given by 0 (C/m3 v 1 r2 where 0 is a positive constant and the beam's axis is coincident with the zaxis. (a) Determine the total charge contained in length L of the beam. (b) If the electrons are moving in the zdirection with uniform speed u, determine the magnitude and direction of the current crossing the zplane. Solution: (a) (b) 0 Current direction is along ^ z. 2u0 u0 ln 1 r0 r 0 0 r0 2 I J ds u0 1 r2 r dr 1 r2 ^ z ^ zr dr d
2 r0 J v u ^ z u0 1 r2 (A/m2 ) (A) r 0 z 0 Q v d V r0 L 0 2r dr dz 2 r 0 z 0 1 r r0 r 20 L dr 0 L ln 1 2 0 1 r
r0 L 2 r0 5R cos S 0 0 2 100 I J ds ^ R 2 5 R ^ RR2 sin d d 314 2 Problem 4.7 If J ^ R5 R (A/m2 ), find I through the surface R 5 m. (A) 178 CHAPTER 4 Section 43: Coulomb's Law
Problem 4.9 A square with sides 2 m each has a charge of 40 C at each of its four corners. Determine the electric field at a point 5 m above the center of the square.
z P(0,0,5) R3 Q3(1,1,0) R4 R2 R1 Q2(1,1,0) y Q4(1,1,0) Q1(1,1,0) x Figure P4.9: Square with charges at the corners. Solution: The distance R between any of the charges and point P is CHAPTER 4 179 Problem 4.10 Three point charges, each with q 3 nC, are located at the corners of a triangle in the xy plane, with one corner at the origin, another at 2 cm 0 0 , and the third at 0 2 cm 0 . Find the force acting on the charge located at the origin. Solution: Use Eq. (4.19) to determine the electric field at the origin due to the other two point charges [Fig. P4.10]: y 2 cm Q R2 Q ^ R1 = x 2 cm ^ R2 = y 2 cm R1 Q 2 cm x Figure P4.10: Locations of charges in Problem 4.10. Problem 4.11 Charge q1 6 C is located at 1 cm 1 cm 0 and charge q2 is located at 0 0 4 cm . What should q2 be so that E at 0 2 cm 0 has no ycomponent? Solution: For the configuration of Fig. P4.11, use of Eq. (4.19) gives Employ Eq. (4.14) to find the force F qE ^ 202 2 x ^ y (N) E ^ 67 4 x ^ 1 3 nC x 0 02 4 0 02 3 3 nC ^ y 0 02 0 02 3 ^ y (kV/m) at R E R1 R2 3 R R3 ^ ^ ^ x y z5 27 3 2 5Q 5 ^ ^ z z 3 2 27 27 0 Q 40 Q 40 R 12 12 52 27. R3 R4 3 R R3 ^ ^ ^ ^ ^ ^ ^ ^ ^ x y z5 x y z5 x y z5 27 3 2 27 3 2 27 3 2 40 C 1 42 ^ 10 6 (V/m) z 51 2 (kV/m) 3 2 0 0 0 180
z 4 cm q2
^ ^ ^ y) R1 = x + y(21) = (x + ^ cm ^ ^ R2 = (y2  z4) cm CHAPTER 4 R2 0 1 cm x q1 1 cm R1 2 cm E1 y E2 Figure P4.11: Locations of charges in Problem 4.11. Problem 4.12 A line of charge with uniform density l 8 (C/m) exists in air along the zaxis between z 0 and z 5 cm. Find E at (0,10 cm,0). Solution: Use of Eq. (4.21c) for the line of charge shown in Fig. P4.12 gives 1 dl ^ l R 40 l R2 ^ ^ y 0 1 zz
z 0 6 71 86 ^ 10 y 4 47 3 ^ z 1 06 01 8 10 40 ^ y10z
2 ^ z z2 0 05 z 0 ^ y 321 4 8 10 6 1 40 0 05 ^ ^ y0 1 zz 2 01 z2 3 R E 2 dz 103 ^ z 76 2 If Ey 0, then q2 21 21 10 6 0 224 94 69 (C) 103 (V/m) ER ^ y2cm ^ ^ ^ ^ 1 6C x y 10 2 q2 y2 z4 10 2 2 3 2 4 2 10 20 10 2 3 2 1 ^ ^ x21 21 10 6 y 21 21 10 6 0 224q2 4 ^ z0 447q2 (V/m) CHAPTER 4
z 181 5 cm dz 0 10 cm
^ R' = y0.1 ^ zz  y x Figure P4.12: Line charge. Problem 4.13 Electric charge is distributed along an arc located in the xy plane and defined by r 2 cm and 0 4. If l 5 C/m), find E at 0 0 z and then evaluate it at (a) the origin, (b) z 5 cm, and (c) z 5 cm. 898 8 (a) At z (b) At z (c) At z ^ ^ x 1 6 y 0 66 (MV/m) 0, E ^ 81 4 y 33 7 z 226 (kV/m) ^ ^ x 5 cm, E ^ ^ ^ 5 cm, E x 81 4 y 33 7 z 226 (kV/m) 0 02 2 z2 3 2 ^ x0 014 ^ y0 006 0 ^ z0 78z l 4 l E 1 40 1 40 ^ R l dl R2 ^ ^ x0 02 cos y0 02 sin 2 0 02 z2 3 2 ^ zz 0 02 d (V/m) Solution: For the arc of charge shown in Fig. P4.13, dl r d ^ ^ ^ x0 02 cos y0 02 sin zz. Use of Eq. (4.21c) gives R 0 02 d and 182
z z CHAPTER 4 ^ R' =^ 0.02 + zz r /4
2 cm ^ r2 y r cm = ^ 0.02 m dz
x Figure P4.13: Line charge along an arc. Problem 4.14 A line of charge with uniform density l extends between z L 2 and z L 2 along the zaxis. Apply Coulomb's law to obtain an expression for the electric field at any point P r 0 on the xy plane. Show that your result reduces to the expression given by Eq. (4.33) as the length L is extended to infinity. Solution: Consider an element of charge of height dz at height z. Call it element 1. The electric field at P due to this element is dE 1 . Similarly, an element at z produces dE2 . These two electric fields have equal zcomponents, but in opposite ^ directions, and hence they will cancel. Their components along r will add. Thus, the net field due to both elements is ^ where the cos factor provides the components of dE 1 and dE2 along r. Our integration variable is z, but it will be easier to integrate over the variable from 0 to L 2 0 sin 1 r2 L 22 dE dE1 dE2 ^ r 2l cos dz 40 R2 ^ rl cos dz 20 R2 CHAPTER 4
z 183 L/2 dz 1 z R 2 0
r dE2 dE1 xy plane z 2 L/2 Figure P4.14: Line charge of length L. ^ r r2 L 2 2 For L r, L 2 1 r2 L 2 2 z 0 0 E dE dE 0 l cos d 20 r 0 l l ^ ^ sin 0 r r 20 r 20 r L 2 0 0 ^ r
0 l cos3 r sec2 d 20 r2 Hence, with R r cos , and z r tan and dz r sec 2 d, we have L 2 184 and l 20 r CHAPTER 4 Problem 4.15 Repeat Example 45 for the circular disk of charge of radius a, but in the present case assume the surface charge density to vary with r as where s0 is a constant. Solution: We start with the expression for dE given in Example 45 but we replace s with s0 r2 : 40 s0 h ^ z 20 r2
a To perform the integration, we use h dR Problem 4.16 Multiple charges at different locations are said to be in equilibrium if the force acting on any one of them is identical in magnitude and direction to the force acting on any of the others. Suppose we have two negative charges, one located at the origin and carrying charge 9e, and the other located on the positive xaxis at a distance d from the first one and carrying charge 36e. Determine the location, polarity and magnitude of a third charge whose placement would bring the entire system into equilibrium. Solution: If F1 force on Q1 a2 h2 ^ z s0 h 20 a2 h2 h2 2h h h s0 h ^ z 20 a2 h2 1 2 E ^ z s0 h 20 a2 h2 1 2 2R dR 2r dr R2 r2 h2 R2 0 h2 dR R2
a2 h2
1 2 E dE ^ z 2s0 r3 dr h2 3 2 r3 dr r2 h2 3 2 h s s0 r2 (C/m2 ) h2 dR R2 E ^ r (infinite line of charge) CHAPTER 4
Q1 = 9e x=0 x d (dx) Q3 Q2 = 36e 185 x Figure P4.16: Three collinear charges. then equilibrium means that The two original charges are both negative, which mean they would repel each other. The third charge has to be positive and has to lie somewhere between them in order to counteract their repulsion force. The forces acting on charges Q 1 , Q2 , and Q3 are respectively 40 R2 21 ^ R12 Q1 Q2 ^ R13 Q1 Q3 40 R2 31 40 R2 32 40 R2 23 Hence, equilibrium requires that Solution of the above equations yields Section 44: Gauss's Law
Problem 4.17 Three infinite lines of charge, all parallel to the zaxis, are located at the three corners of the kiteshaped arrangement shown in Fig. 429 (P4.17). If the Q3 4e x d 3 324e d2 9Q3 x2 324e d2 36Q3 d x2 9Q3 x2 36Q3 d x2 40 R2 13 2 F3 ^ R23 Q2 Q3 ^ x 9eQ3 40 x2 ^ x 36eQ3 40 d x 40 R2 12 2 F2 ^ R32 Q3 Q2 ^ x 324e2 40 d 2 ^ x 36eQ3 40 d x F1 ^ R21 Q1 Q2 ^ R31 Q1 Q3 ^ x 324e2 40 d 2 F1 F2 F3 ^ x 9eQ3 40 x2 F3 force on Q3 F2 force on Q2 186 CHAPTER 4 two right triangles are symmetrical and of equal corresponding sides, show that the electric field is zero at the origin.
y 2l l l x Figure P4.17: Kiteshaped arrangment of line charges for Problem 4.17. Solution: The field due to an infinite line of charge is given by Eq. (4.33). In the present case, the total E at the origin is ^ ^ The components of E1 and E2 along x cancel and their components along y add. ^ Also, E3 is along y because the line charge on the yaxis is negative. Hence, Problem 4.18 Three infinite lines of charge, l1 3 (nC/m), l2 3 (nC/m), and l3 3 (nC/m), are all parallel to the zaxis. If they pass through the respective points But cos R1 R2 . Hence, E ^ y l R 1 0 R1 R2 ^ y l 0 R2 0 E ^ y 2l cos 20 R1 ^ y 2l 20 R2 E E1 E2 E3 CHAPTER 4
y 187 (0,b) l3 R3 l2 E2 P
(a,0) E1 x E3 (0,b) l1 Figure P4.18: Three parallel line charges. 0 b , 0 0 , and 0 b in the xy plane, find the electric field at a 0 0 . Evaluate your result for a 2 cm and b 1 cm. Solution: l1 Components of line charges 1 and 3 along y cancel and components along x add. Hence, using Eq. (4.33), E ^ x3 2a 2 20 a b2 a2 b2 and R1 with cos a a2 b2 , 1 a 10
9 (V/m) E ^ x 2l1 cos 20 R1 ^ x l2 20 a E E1 l3 l1 E2 E3 l2 3 (nC/m) 3 (nC/m) 188 CHAPTER 4 Problem 4.19 A horizontal strip lying in the xy plane is of width d in the ydirection and infinitely long in the xdirection. If the strip is in air and has a uniform charge distribution s , use Coulomb's law to obtain an explicit expression for the electric field at a point P located at a distance h above the centerline of the strip. Extend your result to the special case where d is infinite and compare it with Eq. (4.25).
z dE2 dE1 P(0,0,h) 0 R y y 1 d 2 Figure P4.19: Horizontal strip of charge. Solution: The strip of charge density s (C/m2 ) can be treated as a set of adjacent line charges each of charge l s dy and width dy. At point P, the fields of line charge at distance y and line charge at distance y give contributions that cancel each other ^ ^ along y and add along z. For each such pair, dE ^ z 2s dy cos 20 R y s x For a 2 cm and b 1 cm, E ^ x 1 62 (kV/m) CHAPTER 4 189 0 to 0 0 (4.25). Problem 4.20 Given the electric flux density determine (a) v by applying Eq. (4.26), (b) the total charge Q enclosed in a cube 2 m on a side, located in the first octant with three of its sides coincident with the x, y, and zaxes and one of its corners at the origin, and (c) the total charge Q in the cube, obtained by applying Eq. (4.29). Solution: (a) By applying Eq. (4.26) (b) Integrate the charge density over the volume as in Eq. (4.27): (c) Apply Gauss' law to calculate the total charge from Eq. (4.29) x 2 z 0 y 0 z 0 y 0 2x y dz dy 2z 2y 2 2 1 2 y 2 2 y 0 z 0 x 2 Ffront ^ x2 x y ^ y 3x 2y 2 2 ^ x dz dy Q D ds Ffront Fback Fright Fleft V x 0 y 0 z 0 Ftop Fbottom Q D dV 2 2 2 0 dx dy dz v D 2y 2x x 3x y 2y D ^ x2 x y ^ y 3x For an infinitely wide sheet, 0 2 and E ^ z s , which is identical with Eq. 20 2y (C/m2 ) E dE ^ z ^ z 0 0 d 2 d 2 s 0 cos dy R h s 0 cos2 d 0 0 h cos2 s ^ 0 z 0 2 24 With R h cos , we integrate from y 0 to d 2, which corresponds to 0 sin 1 d 2 h2 d 2 2 1 2 . Thus, 190 CHAPTER 4 Solution: x 0 y 0 z 0 V z 0 y 0 x 0 Q D dV 2 2 2 y z dx dy dz 3 3 xy4 z4 16 xy3 z3 (a) From Eq. (4.26), v D x (b) Total charge Q is given by Eq. (4.27): Problem 4.21 Repeat Problem 4.20 for D ^ xxy3 z3 (C/m2 ). y3 z3 Thus Q D ds 24 x 0 z 0 z 0 8 4 12 0 0 0 dy dx 0 2 2 0 x 0 z 0 z 0 2 Fbottom ^ x2 x y ^ y 3x 2y 2 2 x 0 z 0 z 2 0 dy dx 0 2 2 x 0 z 0 z 2 ^ z dy dx Ftop ^ x2 x y ^ y 3x 2y ^ z dy dx 2 2 2 2 y 0 z 0 x 0 z 0 x 0 32 C 3x 2y dz dx 2 2 3 2 x z 2 2 x 0 z 0 y 0 Fleft ^ x2 x y ^ y 3x 2y ^ y dz dx 2 2 2 y 2 z 0 x 0 z 0 x 0 12 3x 2y dz dx 2 2 3 2 z x 2 2 4x x 0 z 0 y 2 Fright ^ x2 x y ^ y 3x 2y ^ y dz dx 2 2 x 0 z 0 y 0 z 0 y 0 2 2x y dz dy 2 2 2 zy 2 y 0 z 0 x 0 2 8 4 Fback ^ x2 x y ^ y 3x 2y 2 2 ^ x dz dy CHAPTER 4
(c) Using Gauss' law we have 191 S ^ ^ Note that D xDx , so only Ffront and Fback (integration over z surfaces) will contribute to the integral. Problem 4.22 Charge Q1 is uniformly distributed over a thin spherical shell of radius a, and charge Q2 is uniformly distributed over a second spherical shell of radius b, with b a. Apply Gauss's law to find E in the regions R a, a R b, and R b. ^ Solution: Using symmetry considerations, we know D RDR . From Table 3.1, ^ R2 sin d d for an element of a spherical surface. Using Gauss's law in ds R integral form (Eq. (4.29)), S where Qtot is the total charge enclosed in S. For a spherical surface of radius R, 0 0 From Eq. (4.15), we know a linear, isotropic material has the constitutive relationship D E. Thus, we find E from D. DR DR R2 2 cos 0 Qtot Qtot 4R2 ^ RDR 2 ^ RR2 sin d d D ds Qtot Qtot Thus Q D ds 32 0 0 0 0 0 32 C x 0 z 0 y 0 z 0 y 0 x 0 Fback ^ x dy dz ^ xxy3 z3 2 2 2 2 x 2 y 0 z 0 y 0 z 0 xy3 z3 xy z 3 3 2 2 dy dz y4 z4 2 16 2 z 0 y 0 x 2 Ffront ^ xxy3 z3 2 2 ^ x dy dz 2 32 dy dz 0 D ds Ffront Fback Fright Fleft Ftop Fbottom 192 (a) In the region R CHAPTER 4
a, 4R2 4R2 4R2 Problem 4.23 The electric flux density inside a dielectric sphere of radius a centered at the origin is given by where 0 is a constant. Find the total charge inside the sphere. Solution: 2 0 Problem 4.24 In a certain region of space, the charge density is given in cylindrical coordinates by the function: Apply Gauss's law to find D. Solution: v 50re r (C/m3 20 a3 sin d 20 a3 cos 0 S Q D ds 0 0 ^ ^ R0 R RR2 sin d d D ^ R0 R (C/m2 ), R a 40 a3 Qtot Q1 Q2 E ^ RER ^ R Q1 (c) In the region R b, Q2 (V/m) Qtot (b) In the region a R b, Q1 E ^ RER ^ RQ1 (V/m) (C) Qtot 0 E ^ RER ^ RQtot 0 (V/m) CHAPTER 4
z 193 L r Figure P4.24: Gaussian surface. Method 1: Integral Form of Gauss's Law Since v varies as a function of r only, so will D. Hence, we construct a cylinder of radius r and length L, coincident with the zaxis. Symmetry suggests that D has the ^ functional form D r D. Hence, S r Method 2: Differential Method with Dr being a function of r. 1 rDr r r 50re r D v D ^ r Dr D ^ rD ^ r 50 e r 2 1 r 1 r re r 100L r2 e r 21 e r 0 Q 2L 50re ^ r D ds D 2rL r D ds Q r dr 1 r 194 rDr r CHAPTER 4 Problem 4.25 An infinitely long cylindrical shell extending between r 1 m and r 3 m contains a uniform charge density v0 . Apply Gauss's law to find D in all regions. D ^ r Dr r 3m Dr 2rL v0 L 32 4v0 ^ r r For r 3 m, 12 8v0 L D ^ r Dr ^ r ^ r v0 L r2 1 2rL v0 r2 1 2r Dr 2rL v0 L r2 S ^ r Dr ds Q Solution: For r 1 m, D For 1 r 3 m, 0. 12 1 r 3m D ^ r rDr ^ r 50 e r 2 1 r 1 r re 50 2 1 e r 0 0 1 r rDr dr r rDr r 50r2 e r 50r2 e r dr r r2 e
r r CHAPTER 4
z 195 1m L 3m r Figure P4.25: Cylindrical shell. Problem 4.26 If the charge density increases linearly with distance from the origin such that v 0 at the origin and v 40 C/m3 at R 2 m, find the corresponding variation of D. Solution: v 0 a 0 v R a bR 196 CHAPTER 4 Applying Gauss's law to a spherical surface of radius R,
S V R 0 Section 45: Electric Potential with 2 V R3 At x a 2, R1 a 2 a 5 2 Q 20 2 a 2 5a R3 x R1 x a 2 a 2 2 a 2 a 2 2 2 0 55Q 0 a V Q 40 R1 Q 40 R2 Q 40 R3 Q 40 R4 Q 20 Solution: R1 R2 and R3 R4 . 1 R1 1 R3 Problem 4.27 A square in the xy plane in free space has a point charge of corner a 2 a 2 and the same at corner a 2 a 2 and a point charge of each of the other two corners. (a) Find the electric potential at any point P along the xaxis. (b) Evaluate V at x a 2. D DR 5R2 (C/m2 ) ^ ^ R DR R 5R2 (C/m2 ) DR 4R2 20R 4R2 dR D ds v d V v R Hence, b 20. 20R (C/m3 ) 80 R4 4 Q at Q at v 2 2b 40 CHAPTER 4
y 197 Q a/2 R3 a/2 a/2 R4 Q R1 P(x,0) x R2 Q a/2 Q Figure P4.27: Potential due to four point charges. Problem 4.28 The circular disk of radius a shown in Fig. 47 (P4.28) has uniform charge density s across its surface. (a) Obtain an expression for the electric potential V at a point P 0 0 z on the zaxis. (b) Use your result to find E and then evaluate it for z h. Compare your final expression with Eq. (4.24), which was obtained on the basis of Coulomb's law. Solution: (a) Consider a ring of charge at a radial distance r. The charge contained in width dr is dq s 2r dr 2s r dr The potential at P is dq 40 R 2s r dr 40 r2 z2 s 2 r 20 1 2 The potential due to the entire disk is 0 0 0 1 2 V dV z2 1 2 a2 z2 1 2 a s 20 a r dr 2 r z2 a dV s 20 z 198
z CHAPTER 4 E P(0,0,h)
h s dq = 2 s r dr r a dr y a x Figure P4.28: Circular disk of charge. a2 z2 Problem 4.29 A circular ring of charge of radius a lies in the xy plane and is centered at the origin. If the ring is in air and carries a uniform density l , (a) show that the electrical potential at 0 0 z is given by V l a 20 a2 z2 1 2 , and (b) find the corresponding electric field E. Solution: (a) For the ring of charge shown in Fig. P4.29, using Eq. (3.67) in Eq. (4.48c) gives 0 a2 z2 z2 V 00z 1 40 2 l a d l a 20 a2 Point 0 0 z in Cartesian coordinates corresponds to r z coordinates. Hence, for r 0, 0 z in cylindrical l 0 a2 r2 2ar cos z2 V R 1 40 l dl R 1 40 2 l a d The expression for E reduces to Eq. (4.24) when z h. E V ^ x V x ^ y V y ^ z V z ^ z s 1 20 (b) z CHAPTER 4
z 199 z R' = a2 + z2  0 a y d'
a x l dl' = a d' Figure P4.29: Ring of charge. (b) From Eq. (4.51),
3 2 Problem 4.30 Show that the electric potential difference V12 between two points in air at radial distances r1 and r2 from an infinite line of charge with density l along l 20 ln r2 r1 . the zaxis is V12 Solution: From Eq. (4.33), the electric field due to an infinite line of charge is Hence, the potential difference is r2 r2 Problem 4.31 Find the electric potential V at a location a distance b from the origin in the xy plane due to a line charge with charge density l and of length l. The line charge is coincident with the zaxis and extends from z l 2 to z l 2. V12 E dl r1 r1 ^ r l ^ r dr 20 r l r2 ln 20 r1 E ^ rEr ^ r l 20 r E V ^ z z2 1 2 l a 2 a 20 z ^ z l a z 20 a2 z2 (V/m) 200
z l/2 dz l z R' V(b) b CHAPTER 4 R' z2 + b2 = y l/2 Figure P4.31: Line of charge of length . Solution: From Eq. (4.48c), we can find the voltage at a distance b away from a line of charge [Fig. P4.31]: Hence, E 0 80 10 40 23 2 ^ R 0 ^ ^1 4 Again using Eq. (4.56) to find E at R 2 m and 90 , we have (mV/m) q 0 80 E qd 2 4 mV/m at 40 10 3 80 10 3 80 d 10 2 0 E qd ^ R2 cos 40 R3 Solution: For R 0 , E 1 m and 4 mV/m, we can solve for q using Eq. (4.56): ^ sin (C) Problem 4.32 For the electric dipole shown in Fig. 413, d 1 cm and E (mV/m) at R 1 m and 0 . Find E at R 2 m and 90 . l l 2 l z2 b2 4 V b 1 4 dz l l 2 l dl R l 4 l ln 4 l 2 4b2 l 2 4b2 CHAPTER 4 201 Problem 4.33 For each of the following distributions of the electric potential V , sketch the corresponding distribution of E (in all cases, the vertical axis is in volts and the horizontal axis is in meters): Solution:
V 30 3 30 5 8 11 13 16 x E 10 x 10 (a)
V 4 3 6 9 12 15 x 4 E 4.20 3 6 9 12 x 15 4.20 (b) 202
V 4 CHAPTER 4 3 6 9 12 15 x 4 E 2.6 3 6 9 12 15 x 2.6 (c) Figure P4.33: Electric potential distributions of Problem 4.33. Problem 4.34 Given the electric field Solution: A B 4 4 0 VAB ^ R ^ z 2 18 ^ z dz z2 z 0. Hence, 0 18 ^ z dz z2 2 ^ z 18 ^ z dz z2 4V ^ Along zdirection, R ^ z and E ^ z 18 for z z2 ^ 0, and R VAB VA VB E dl ^ z and E ^ z 18 for z2 find the electric potential of point A with respect to point B where A is at B at 4 m, both on the zaxis. 2 m and E ^ 18 R 2 R (V/m) CHAPTER 4 203 A z = 2m B z = 4m Figure P4.34: Potential between B and A. Problem 4.35 An infinitely long line of charge with uniform density l 9 (nC/m) lies in the xy plane parallel to the yaxis at x 2 m. Find the potential VAB at point A 3 m 0 4 m in Cartesian coordinates with respect to point B 0 0 0 by applying the result of Problem 4.30. Solution: According to Problem 4.30, where r1 and r2 are the distances of A and B. In this case, 12 VAB 9 10 9 2 8 85 10 ln Hence, r2 2m 2 17 r1 3 2 2 42 V l r2 ln 20 r1 17 m 117 09 V 204
z 4m A(3, 0, 4) CHAPTER 4 r1 r2 B y 2m Problem 4.36 The xy plane contains a uniform sheet of charge with s1 0 2 (nC/m2 and a second sheet with s2 0 2 (nC/m2 ) occupies the plane z 6 m. Find VAB , VBC , and VAC for A 0 0 6 m , B 0 0 0 , and C 0 2 m 2 m . Solution: We start by finding the E field in the region between the plates. For any point above the xy plane, E1 due to the charge on xy plane is, from Eq. (4.25), In the region below the top plate, E would point downwards for positive s2 on the top plate. In this case, s2 s1 . Hence, ^ Since E is along z, only change in position along z can result in change in voltage. 0 0 VAB ^ z 6 s 1 ^ z dz 0 s1 z 0 6 6s1 0 6 0 2 10 9 8 85 10 12 E E1 E2 ^ z s 1 20 ^ z s 2 20 E1 ^ z s 1 20 ^ z 2s1 20 ^ z s 1 0 135 59 V 3m x Figure P4.35: Line of charge parallel to yaxis. CHAPTER 4
z 205 s2=  0.2 (nC/m2) A 6m C (0, 2, 2) s1= 0.2 (nC/m2) B 0 y x Figure P4.36: Two parallel planes of charge. The voltage at C depends only on the zcoordinate of C. Hence, with point A being at the lowest potential and B at the highest potential, Section 47: Conductors
Problem 4.37 A cylindrical bar of silicon has a radius of 4 mm and a length of 8 cm. If a voltage of 5 V is applied between the ends of the bar and e 0 13 (m2 /V s), h 0 05 (m2 /V s), Ne 1 5 1016 electrons/m3 , and Nh Ne , find (a) the conductivity of silicon, (b) the current I flowing in the bar, (c) the drift velocities ue and uh , (d) the resistance of the bar, and (e) the power dissipated in the bar. VAC VBC 2 VAB 6 VAB VBC 135 59 45 20 V 3 135 59 45 20 90 39 V 206 Solution: (a) Conductivity is given in Eq. (4.65), CHAPTER 4 (b) Similarly to Example 4.8, parts b and c, (c) From Eqs. (4.62a) and (4.62b), (d) To find the resistance, we use what we calculated above, Solution: (a) Conductivity is given in Eq. (4.65), (b) Similarly to Example 4.8, parts b and c, (c) From Eqs. (4.62a) and (4.62b), uh h E 02 5 0 08 12 5 E E (m/s) ue e E 04 5 0 08 E E 25 E E (m/s) I JA EA 23 5V 0 08 4 10 3 2 7 225 (mA) Ne e Nu u e 24 1019 0 4 02 16 10 19 23 Problem 4.38 Repeat Problem 4.37 for a bar of germanium with e h 0 2 (m2 /V s), and Ne Nh 2 4 1019 electrons or holes/m3 . 0 4 (m2 /V s), (S/m) (e) Power dissipated in the bar is P IV 5V 1 36 A 6 8 W R V I 5V 1 36 A 3 68 (M) uh h E 0 05 ue e E 0 13 5 E E 8 125 (m/s) 0 08 E E E 5 E 3 125 (m/s) 0 08 E E I JA EA 4 32 10 4 5V 0 08 4 10 3 2 1 36 (A) 15 1016 0 13 0 05 1 6 Ne e Nh h e 10
19 4 32 10 4 (S/m) CHAPTER 4
(d) To find the resistance, we use what we calculated above, 207 Problem 4.39 A 100mlong conductor of uniform cross section has a voltage drop of 4 V between its ends. If the density of the current flowing through it is 1 4 10 6 (A/m2 ), identify the material of the conductor. Solution: We know that conductivity characterizes a material: Problem 4.40 A coaxial resistor of length l consists of two concentric cylinders. The inner cylinder has radius a and is made of a material with conductivity 1 , and the outer cylinder, extending between r a and r b, is made of a material with conductivity 2 . If the two ends of the resistor are capped with conducting plates, show that the resistance between the two ends is R l 1 a2 2 b2 a2 . Solution: Due to the conducting plates, the ends of the coaxial resistor are each uniform at the same potential. Hence, the electric field everywhere in the resistor will be parallel to the axis of the resistor, in which case the two cylinders can be considered to be two separate resistors in parallel. Then, from Eq. (4.70), or Problem 4.41 Apply the result of Problem 4.40 to find the resistance of a 20cmlong hollow cylinder (Fig. P4.41) made of carbon with 3 10 4 (S/m). Solution: From Problem 4.40, we know that for two concentric cylinders, 1 a2 a2 R l 2 b2 () 1 a2 a2 R l 2 b2 () Rinner Router 1 R 1 1 1 A 1 l1 2 A 2 l2 1 a2 l 2 b2 l a2 From Table B2, we find that aluminum has 35 10 7 (S/m). J E 14 106 (A/m2 ) 4 (V) 100 (m) 35 107 (S/m) (e) Power dissipated in the bar is P IV 5V 7 225 mA 36 125 (mW) R V I 5V 7 225 mA 0 69 (k) 208 CHAPTER 4 3 cm 2 cm Carbon Figure P4.41: Cross section of hollow cylinder of Problem 4.41. Problem 4.42 A 2 10 3 mmthick square sheet of aluminum has 5 cm 5 cm faces. Find: (a) the resistance between opposite edges on a square face, and (b) the resistance between the two square faces. (See Appendix B for the electrical constants of materials). Solution: (a) 35 3 R 2 10 6 107 2 5 10 22 8 p (b) Now, l 2 10 3 mm and A 5 cm 35 7 5 cm 25 10 R 5 10 2 107 1 10 14 (m) 3 m2 . l 5 cm A For aluminum, 35 107 (S/m) [Appendix B]. 2 10
3 5 cm mm R l A 10 10 2 10 6 1 10 7 3 104 0 02 2 m2 R For air 1 0 (S/m), 2 3 104 (S/m); hence, 02 0 03 2 42 (m) CHAPTER 4 209 Section 49: Boundary Conditions
^ ^ ^ Problem 4.43 With reference to Fig. 419, find E 1 if E2 x3 y2 z2 (V/m), 20 , 2 180 , and the boundary has a surface charge density 1 s 3 54 10 11 (C/m2 ). What angle does E2 make with the zaxis? ^ ^ Solution: We know that E1t E2t for any 2 media. Hence, E1t E2t x3 y2. ^ ^ ^ Also, D1 D2 n s (from Table 4.3). Hence, 1 E1 n 2 E2 n s which gives Problem 4.44 An infinitely long conducting cylinder of radius a has a surface charge density s . The cylinder is surrounded by a dielectric medium with r 4 and contains no free charges. If the tangential component of the electric field in the ^ cos2 r2 , find s . region r a is given by Et Solution: Let the conducting cylinder be medium 1 and the surrounding dielectric medium be medium 2. In medium 2, ^1 2 cos2 r with Er , the normal component of E2 , unknown. The surface charge density is related to Er . To find Er , we invoke Gauss's law in medium 2: 2 sin cos rEr 2 sin cos r rEr dr r 1 dr r2 1 rEr cos2 r r2 Integrating both sides with respect to r, which leads to 2 sin cos r2 1 rEr r r 1 r or 1 cos2 r2 D2 0 E2 ^ rEr 0 ^ E2 z 2 E2 cos 9 4 4cos cos 1 Hence, E1 ^ x3 ^ y2 ^ z20 (V/m). Finding the angle E2 makes with the zaxis: 2 17 61 E1z s 2 E2z 1 3 54 10 20 11 18 2 2 3 54 10 11 2 8 85 10 12 18 20 (V/m) 210 or CHAPTER 4 Hence, ^ where n2 is the normal to the boundary and points away from medium 1. Hence, ^ ^ n2 r. Also, D1 0 because the cylinder is a conductor. Consequently, ^ r 2 E 2 ^ r D2 Problem 4.45 A 2cm conducting sphere is embedded in a chargefree dielectric ^ ^ medium with 2r 9. If E2 R 3 cos 3 sin (V/m) in the surrounding region, find the charge density on the sphere's surface. Solution: According to Eq. (4.93), ^ Problem 4.46 If E R150 (V/m) at the surface of a 5cm conducting sphere centered at the origin, what is the total charge Q on the sphere's surface? ^ Solution: From Table 43, n D1 D2 s . E2 inside the sphere is zero, since we assume it is a perfect conductor. Hence, for a sphere with surface area S 4a 2 , D1R s E1R s 0 Q S0 270 cos (C/m2 ) ^ ^ R 2 R 3 cos s ^ R D2 ^ In the present case, n2 ^ R and D1 0. Hence,
r 2 cm ^ 3 sin ^ n2 D1 D2 s 80 sin cos (C/m2 ) a2 ^ ^ r r 0 r s r a r a 2 sin cos r2 ^ 1 2 cos2 r ^ n2 D1 According to Eq. (4.93), D2 s r a E2 ^ r 2 sin cos r2 ^ 1 2 cos2 r Er 2 sin cos r2 CHAPTER 4 211 Q ER S0 150 4 0 05 2 0 30 2 Problem 4.47 Figure 434(a) (P4.47) shows three planar dielectric slabs of equal thickness but with different dielectric constants. If E 0 in air makes an angle of 45 with respect to the zaxis, find the angle of E in each of the other layers.
z E0 45 0 (air) 1 = 30 2 = 50 3 = 70 0 (air) Figure P4.47: Dielectric slabs in Problem 4.47. Solution: Labeling the upper air region as region 0 and using Eq. (4.99), In the lower air region, the angle is again 45 . Sections 410 and 411: Capacitance and Electrical Energy
Problem 4.48 Determine the force of attraction in a parallelplate capacitor with A 5 cm2 , d 2 cm, and r 4 if the voltage across it is 50 V. 3 tan 1 tan 1 81 9 2 tan 1 tan 1 5 tan 71 6 3 7 tan 78 7 5 1 tan 1 tan 1 1 tan 0 0 2 tan 1 1 3 tan 2 2 3 tan 45 71 6 78 7 (C) 212 Solution: From Eq. (4.131), CHAPTER 4 Problem 4.49 Dielectric breakdown occurs in a material whenever the magnitude of the field E exceeds the dielectric strength anywhere in that material. In the coaxial capacitor of Example 412, (a) At what value of r is E maximum? (b) What is the breakdown voltage if a 1 cm, b 2 cm, and the dielectric material is mica with r 6? Solution: ^ (a) From Eq. (4.114), E rl 2r for a r b. Thus, it is evident that E is maximum at r a. (b) The dielectric breaks down when E 200 (MV/m) (see Table 42), or which gives l 200 MV/m 2 6 8 854 10 12 0 01 667 6 C/m). From Eq. (4.115), we can find the voltage corresponding to that charge density, Problem 4.50 An electron with charge Q e 1 6 10 19 C and mass 31 kg is injected at a point adjacent to the negatively charged plate in me 9 1 10 the region between the plates of an airfilled parallelplate capacitor with separation of 1 cm and rectangular plates each 10 cm 2 in area Fig. 433 (P4.50). If the voltage across the capacitor is 10 V, find (a) the force acting on the electron, (b) the acceleration of the electron, and (c) the time it takes the electron to reach the positively charged plate, assuming that it starts from rest. Solution: (a) The electric force acting on a charge Q e is given by Eq. (4.14) and the electric field in a capacitor is given by Eq. (4.112). Combining these two relations, we have F Qe E Qe V d 16 10 19 10 0 01 16 10 16 Thus, V 1 39 (MV) is the breakdown voltage for this capacitor. (N) V l b ln 2 a 667 6 C/m ln 2 12 8 854 10 12 F/m 1 39 2 60 10 2 E l 2r l 200 (MV/m) (MV) F ^ z ^ z 20 5 A E 2 2 10 4 50 0 02 2 ^ z 55 3 10 9 (N) CHAPTER 4
1 cm Qe 213 + V0 = 10 V Figure P4.50: Electron between charged plates of Problem 4.50. The force is directed from the negatively charged plate towards the positively charged plate. (b) F 1 6 10 16 a 1 76 1014 (m/s2 ) m 9 1 10 31 (c) The electron does not get fast enough at the end of its short trip for relativity to manifest itself; classical mechanics is adequate to find the transit time. From classical mechanics, d d0 u0t 1 at 2 , where in the present case the start position is d 0 0, 2 the total distance traveled is d 1 cm, the initial velocity u 0 0, and the acceleration is given by part (b). Solving for the time t, Calculate the electrostatic energy stored in the region and 0 z 3 m. 1m x E ^ x x2 2z ^ yx2 ^ zy Problem 4.51 In a dielectric medium with r 4, the electric field is given by z (V/m) 1 m, 0 t 2d a 2 0 01 1 76 1014 10 7 10 9 s 10 7 (ns) y 2 m, 214 Solution: Electrostatic potential energy is given by Eq. (4.124),
z 0 y 0 x 1 CHAPTER 4 Problem 4.52 Figure 434a (P4.52(a)) depicts a capacitor consisting of two parallel, conducting plates separated by a distance d. The space between the plates A1 1 2 A2 +  V d (a) + C1 C2 V (b) Figure P4.52: (a) Capacitor with parallel dielectric section, and (b) equivalent circuit. contains two adjacent dielectrics, one with permittivity 1 and surface area A1 and another with 2 and A2 . The objective of this problem is to show that the capacitance C of the configuration shown in Fig. 434a (P4.52(a)) is equivalent to two capacitances in parallel, as illustrated in Fig. 434b (P4.52(b)), with C C1 C2 4 62 10 9 (J) 40 2 1304 5 (4.132) x 1 y 0 40 2 2 5 x yz 5 2 2 3 z x y 3 4 3 z xy 3 1 y 12 1 2 z x 4 We x2 2z 2 x4 1 E 2 dV 2 V 2 3 2 1 y z 2 dx dy dz
3 z 0 CHAPTER 4
where 215 To this end, you are asked to proceed as follows: (a) Find the electric fields E1 and E2 in the two dielectric layers. (b) Calculate the energy stored in each section and use the result to calculate C 1 and C2 . (c) Use the total energy stored in the capacitor to obtain an expression for C. Show that Eq. (4.132) is indeed a valid result. Solution:
1 E1 E2 2 d + V  (c) Figure P4.52: (c) Electric field inside of capacitor. (a) Within each dielectric section, E will point from the plate with positive voltage to the plate with negative voltage, as shown in Fig. P452(c). From V Ed, (b) But, from Eq. (4.121), We We1 We2 A1 Hence C1 1 . Similarly, C2 d (c) Total energy is 2 A2 . d 2 A 2 1 CV 2 2 1 V2 1 A 1 2 d We1 1 C1V 2 2 We1 1 2 1 E 1 V 2 1 V2 1 A1 d 2 d2 E1 E2 V d 1 A1 1V 2 2 d C2 C1 1 A 1 d 2 A 2 d (4.133) (4.134) 216 Hence, 1 A 1 d 2 A 2 d CHAPTER 4 Problem 4.53 Use the result of Problem 4.52 to determine the capacitance for each of the following configurations: (a) conducting plates are on top and bottom faces of rectangular structure in Fig. 435(a) (P4.53(a)), (b) conducting plates are on front and back faces of structure in Fig. 435(a) (P4.53(a)), (c) conducting plates are on top and bottom faces of the cylindrical structure in Fig. 435(b) (P4.53(b)). Solution: (a) The two capacitors share the same voltage; hence they are in parallel. (b) 4 (c) C C3 3 2 0 12 10 12 F C2 3 2 C1 2 A1 r1 40 80 2 10 3 2 0 04 10 2 d 2 10 10 2 2 2 A2 r2 r1 20 2 40 4 10 3 2 2 10 d 2 10 2 10 2 2 2 r3 r2 A3 0 3 20 8 10 3 2 4 10 2 d 2 10 10 2 C1 C2 C3 0 22 10 12 F 1 C 12 C2 24 0 5 10 C1 1 A1 2 1 10 20 d 5 10 2 A2 3 2 10 2 40 d 5 10 2 C1 C2 0 5 10 12 F 4 0 8 0 10 2 2 F 0 06 C C2 C1 A1 5 1 10 4 20 50 10 2 d 2 10 2 A2 5 3 10 4 2 40 300 10 2 d 2 10 2 C1 C2 50 300 10 2 0 350 3 1 1 10
12 C C1 C2 F 10 12 F CHAPTER 4
1 cm 5 cm 2 cm 217 3 cm r = 2 r = 4 (a) r1 = 2mm r2 = 4mm r3 = 8mm 3 2 1 2 cm 1 = 80; 2 = 40; 3 = 20 (b) Figure P4.53: Dielectric sections for Problems 4.53 and 4.55. 218 CHAPTER 4 Problem 4.54 The capacitor shown in Fig. 436 (P4.54) consists of two parallel dielectric layers. We wish to use energy considerations to show that the equivalent capacitance of the overall capacitor, C, is equal to the series combination of the capacitances of the individual layers, C1 and C2 , namely where A 1 2 (a) C1 C2 + V  (b) Figure P4.54: (a) Capacitor with parallel dielectric layers, and (b) equivalent circuit (Problem 4.54). (a) Let V1 and V2 be the electric potentials across the upper and lower dielectrics, respectively. What are the corresponding electric fields E 1 and E2 ? By applying the appropriate boundary condition at the interface between the two dielectrics, obtain explicit expressions for E 1 and E2 in terms of 1 , 2 , V , and the indicated dimensions of the capacitor. (b) Calculate the energy stored in each of the dielectric layers and then use the sum to obtain an expression for C. C1 1 A d1 C2 2 A d2
d1 d2 + V  C C1C2 C1 C2 (4.136) CHAPTER 4 219 + d1 d2 1 1 E1 V1 + E2 V1  + V Figure P4.54: (c) Electric fields inside of capacitor. (c) Show that C is given by Eq. (4.136). Solution: (a) If V1 is the voltage across the top layer and V2 across the bottom layer, then V1 V2 E2 d1 d2 According to boundary conditions, the normal component of D is continuous across the boundary (in the absence of surface charge). This means that at the interface between the two dielectric layers, and or Hence, which can be solved for E1 : Similarly, d2 E2 d1 V 2 d1 1 E1 V 1 d2 2 V E1 d1 E2 d2 E1 d1 1 E 1 D1n D2n 2 E 2 1 E 1 d2 2 E1 V V1 V2 220 (b) 1 2 1 E 1 V 1 2 1 1 2 CHAPTER 4 2 (c) Multiplying numerator and denominator of the expression for C by A d 1 d2 , we have 1 A 2 A C1C2 d1 d2 C 1 A 2 A C1 C2 d1 d2 where Problem 4.55 Use the expressions given in Problem 4.54 to determine the capacitance for the configurations in Fig. 4.35(a) (P4.55) when the conducting plates are placed on the right and left faces of the structure. Solution: C C2 C1 A 2 5 10 4 20 200 10 2 1 77 d1 1 10 2 A 2 5 10 4 2 40 1 18 10 12 F d2 3 10 2 C1C2 1 77 1 18 10 12 0 71 10 12 F C1 C2 1 77 1 18 1 C1 1 A d1 C2 2 A d2 10 12 F C 1 2 A 1 2 Ad1 2 2 Ad2 2 1 2 d1 1 d2 2 But We 1 2 2 CV , hence, 2 d1 1 d2 2 d1 1 d2 2 We We1 We2 1 2 V 2 1 2 A 2 d1 1 d2 We2 1 2 2 E 2 V 2 2 1 2 2 V Ad2 2 d2 d1 1 1 2 Ad1 2 2 Ad2 2 1 1 d2 2 d1 2 1 2 V 2 2 2 2 Ad2 1 1 d2 2 d1 d1 2 We1 V 1 d2 2 Ad1 1 2 V 2 2 1 2 Ad1 2 2 d1 1 d2 CHAPTER 4
1 cm 5 cm 2 cm 3 cm 221 r = 2 r = 4 Figure P4.55: Dielectric section for Problem 4.55. Section 412: Image Method
Problem 4.56 With reference to Fig. 437 (P4.56), charge Q is located at a distance d above a grounded halfplane located in the xy plane and at a distance d from another grounded halfplane in the xz plane. Use the image method to (a) establish the magnitudes, polarities, and locations of the images of charge Q with respect to each of the two ground planes (as if each is infinite in extent), and (b) then find the electric potential and electric field at an arbitrary point P 0 y z .
z P(0, y, z) Q(0, d, d) y d d Figure P4.56: Charge Q next to two perpendicular, grounded, conducting half planes. Solution: (a) The original charge has magnitude and polarity Q at location 0 d d . Since the negative yaxis is shielded from the region of interest, there might as well be a conducting halfplane extending in the y direction as well as the y direction. This ground plane gives rise to an image charge of magnitude and polarity Q at location 222
z CHAPTER 4 Q d d Q P(y, z) y d Q d Q Figure P4.56: (a) Image charges. 0 d d . In addition, since charges exist on the conducting half plane in the z direction, an image of this conducting half plane also appears in the z direction. This ground plane in the xz plane gives rise to the image charges of Q at 0 d d and Q at 0 d d . (b) Using Eq. (4.47) with N 4, 1 1 x2 x2 x2 y2 y2 y2 2yd 1 2yd 2yd z2 z2 1 z2 2zd 2zd 2zd 2d 2 2d 2 2d 2 Q 4 1 z2 x2 y2 2yd 1 2zd 2d 2 (V) y d z d x2 2 2 x2 y d 2 z d 2 x2 y d z d 2 2 x2 y d 2 Q 4 1 1 ^ xx ^ zz d ^ xx ^ zz d z ^ xx ^ yy 1 ^ yy d ^ zz d ^ xx ^ yy 1 ^ yy d ^ zz d d V xyz Q 4 1 d 1 d 2 CHAPTER 4
From Eq. (4.51), 223 Problem 4.57 Conducting wires above a conducting plane carry currents I 1 and I2 in the directions shown in Fig. 438 (P4.57). Keeping in mind that the direction
I2 I1 (a) (b) Figure P4.57: Currents above a conducting plane (Problem 4.57). of a current is defined in terms of the movement of positive charges, what are the directions of the image currents corresponding to I1 and I2 ? Solution: (a) In the image current, movement of negative charges downward positive charges upward. Hence, image of I1 is same as I1 . x2 y d z d x2 y d z d 2 2 3 2 2 2 3 2 movement of ^ xx ^ yy d ^ zz d ^ xx ^ yy d ^ zz d x2 y d 2 z d 2 3 2 x2 y d 2 z d 2 3 2 Q 4 ^ xx ^ yy d ^ zz d ^ xx ^ yy d ^ zz x2 y d z d 2 2 x2 y d 2 1 1 z d 2 d y d z d x2 2 2 x2 y d 2 Q 4 E V 1 1 z d
2 (V/m) 224
+ q @ t=t1 I1 + q @ t=0 CHAPTER 4
I1 (image)  q @ t=0  q @ t=t1 Figure P4.57: (a) Solution for part (a). I1 @t=0 +q + q @ t=t1 q @t=0 I1  q @ t=t1 (image) Figure P4.57: (b) Solution for part (b). Problem 4.58 Use the image method to find the capacitance per unit length of an infinitely long conducting cylinder of radius a situated at a distance d from a parallel conducting plane, as shown in Fig. 439 (P4.58). Solution: Let us distribute charge l (C/m) on the conducting cylinder. Its image cylinder at z d will have charge density l . For the line at z d, the electric field at any point z (at a distance of d z from the center of the cylinder) is, from Eq. (4.33), E1 ^ z l 20 d z (b) In the image current, movement of negative charges to right positive charges to left. movement of CHAPTER 4
a 225 d V =0 Figure P4.58: Conducting cylinder above a conducting plane (Problem 4.58).
a l d z d a l Figure P4.58: (a) Cylinder and its image. ^ where z is the direction away from the cylinder. Similarly for the image cylinder at distance d z and carrying charge l , z The potential difference between the cylinders is obtained by integrating the total electric field from z d a to z d a: d a d a d z d z ^ z l 20 2 V E1 ^ E2 z dz 1 1 ^ z dz 1 E2 ^ z l 20 d z ^ z l 20 d 226
d a l 1 1 dz 20 d z d z d a l a ln d z ln d z d d a 20 l ln a ln 2d a ln 2d a 20 l 2d a ln 0 a CHAPTER 4 and the capacitance per unit length is Problem 4.59 A circular beam of charge of radius a consists of electrons moving with a constant speed u along the z direction. The beam's axis is coincident with the zaxis and the electron charge density is given by where c is a constant and r is the radial distance from the axis of the beam. (a) Determine the charge density per unit length. (b) Determine the current crossing the zplane. Solution: (a) 0 r 0 0 l v ds
a 2 cr2 r dr d v cr2 (c/m3 ) 2c 1 r4 4 a ca4 2 C C L 0 ln 2d a (C/m) a a For a length L, Q l L and C Q V l L l 0 ln 2d (C/m) ln a CHAPTER 4
(b) 227 a 0 Problem 4.60 A line of charge of uniform density l occupies a semicircle of radius b as shown in the figure. Use the material presented in Example 44 to determine the electric field at the origin.
z z l
y b x Solution: Since we have only half of a circle, we need to integrate the expression for dE1 given in Example 44 over from 0 to . Before we do that, however, we need to set h 0 (the problem asks for E at the origin). Hence, Problem 4.61 A spherical shell with outer radius b surrounds a chargefree cavity of radius a b. If the shell contains a charge density given by v0 v a R b R2 where v0 is a positive constant, determine D in all regions. E1 dE1 ^ ^ l b rb zh d 2 2 3 2 40 b h ^ r l d 40 b ^ r l dE1 40 b 0 h 0 2cu r 0 0 r3 dr a I J ds
2 J v u ^ z cr2 u (A/m2 ) ^ ^ z cur2 z r dr d cua4 2 l u (A) 228 CHAPTER 4 r3 v
a r1 r2 b Solution: Symmetry dictates that D is radially oriented. Thus, At any R, Gauss's law gives S S DR v0 R a R2 Hence, a R R a Q v dV (b) For a R b,
R v0 4R2 dR R2 R a 4v0 R a
R DR 0 R a b (a) For R a, no charge is contained in the cavity. Hence, Q DR 4R2 DR ^ ^ RDR R ds D ds Q Q Q Q 4R2 0, and D ^ R DR CHAPTER 4
(c) For R 229 b,
R a Problem 4.62 Two infinite lines of charge, both parallel to the zaxis, lie in the xz plane, one with density l and located at x a and the other with density l and located at x a. Obtain an expression for the electric potential V x y at a point P x y relative to the potential at the origin.
y P(x,y) r'' l
(a, 0) r' l
(a, 0) x Solution: According to the result of Problem 4.30, the electric potential difference between a point at a distance r1 and another at a distance r2 from a line charge of density l is l r2 V ln 20 r1 Applying this result to the line charge at x a, which is at a distance a from the origin: l a ln V r2 a and r1 r 20 r x a 2 y2 l ln 20 a Similarly, for the negative line charge at x l a ln V 20 r a, r2 a and r1 r x a 2 y2 l ln 20 a DR v0 b a R2 R b Q v dV 4v0 b b a 230 The potential due to both lines is CHAPTER 4 At the origin, V 0, as it should be since the origin is the reference point. The potential is also zero along all points on the yaxis (x 0). Problem 4.63 A cylindershaped carbon resistor is 8 cm in length and its circular cross section has a diameter d 1 mm. (a) Determine the resistance R. (b) To reduce its resistance by 40%, the carbon resistor is coated with a layer of copper of thickness t. Use the result of Problem 4.40 to determine t. Solution: (a) From (4.70), and using the value of for carbon from Appendix B,
2 (b) The 40%reduced resistance is: Using the result of Problem 4.40: 1 a2 a2 and Thus, the addition of a copper coating less than 0.1 m in thickness reduces the resistance by 40%. 0 00086 10 4 m t b a 5 00086 5 b 5 00086 10 4 m 10 4 0 086 m With 1 3 4 104 S/m (carbon), 2 5 10 4 m, and b unknown, we have 58 107 S/m (copper), a R l 2 b2 2 04 R 0 6R 06 34 2 04 3 3 2 2 R l A l d 2 8 10 104 10 2 34 1 mm 2 x a x a 2 y2 2 y2 V V V l ln 20 a ln a CHAPTER 4 231 Problem 4.64 A coaxial capacitor consists of two concentric, conducting, cylindrical surfaces, one of radius a and another of radius b, as shown in the figure. The insulating layer separating the two conducting surfaces is divided equally into two semicylindrical sections, one filled with dielectric 1 and the other filled with dielectric 2 .
2
b a 1 E + + l  V + (a) Develop an expression for C in terms of the length l and the given quantities. Solution: (a) For the indicated voltage polarity, the E field inside the capacitor exists in only the dielectric materials and points radially inward. Let E 1 be the field in dielectric 1 and E2 be the field in dielectric 2 . At the interface between the two dielectric sections, E1 is parallel to E2 and both are tangential to the interface. Since boundary conditions require that the tangential components of E 1 and E2 be the same, it follows that: ^ E1 E2 rE (b) Evaluate the value of C for a l 4 cm. 2 mm, b 6 mm, r1 2, r2 4, and 232 CHAPTER 4 At r a (surface of inner conductor), in medium 1, the boundary condition on D, as stated by (4.101), leads to or Similarly, in medium 2 Thus, the E fields will be the same in the two dielectrics, but the charge densities will be different along the two sides of the inner conducting cylinder. Since the same voltage applies for the two sections of the capacitor, we can treat them as two capacitors in parallel. For the capacitor half that includes dielectric 1 , we can apply the results of Eqs. (4.114)(4.116), but we have to keep in mind that Q is now the charge on only one half of the inner cylinder. Hence, Similarly, and (b) C 6 07 pF. 4 10 2 2 4 8 85 ln 6 2 C C1 C2 l 1 2 ln b a C2 2 l ln b a C1 1 l ln b a s2 2 E s1 ^ r 1 E 1 E D1 1 E 1 ^ n s1 ^ r s1 10 12 ...
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This note was uploaded on 02/09/2011 for the course EE 172 taught by Professor . during the Spring '10 term at UCLA.
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