F S E S 1. r 2 r N. t = pe sin f. T S p : E S. U = -p S # E

Transcription

1 Coulomb s law: For charges q 1 and q separated by a distance r, the magnitude of the electric force on either F = (1.) 1 ƒq 1 q ƒ 4pP 0 r charge is proportional to the product q 1 q and inversely proportional to r. The force on each charge is along the 1 line joining the two charges repulsive if q 1 and q have = * 10 9 N # m >C 4pP the same sign, attractive if they have opposite signs. In I 0 units the unit of electric charge is the coulomb, abbreviated C. (ee Examples 1.1 and 1..) When two or more charges each exert a force on a charge, the total force on that charge is the vector sum of the forces exerted by the individual charges. (ee Examples 1.3 and 1.4.) F on 1 q 1 q 1 F on 1 F 1 on q r r q F 1 on Electric field: Electric field E, a vector quantity, is the force per unit charge exerted on a test charge at any point. The electric field produced by a point charge is directed radially away from or toward the charge. (ee Examples ) E F 0 q 0 E 1 4pP 0 q r r N (1.3) (1.7) q E Electric dipoles: An electric dipole is a pair of electric charges of equal magnitude q but opposite sign, separated by a distance d. The electric dipole moment p has magnitude p = qd. The direction of p is from negative toward positive charge. An electric dipole in an electric field E experiences a torque T equal to the vector product of p and E. The magnitude of the torque depends on the angle f between p and E. The potential energy U for an electric dipole in an electric field also depends on the relative orientation of p and E. (ee Examples 1.13 and 1.14.) t = pe sin f T p : E U = -p # E (1.15) (1.16) (1.18) E F 5 qe p d q F 1 5 qe 1q d sin f f

2 Electric flux: Electric flux is a measure of the flow of electric field through a surface. It is equal to the product E = E cos f da L of an area element and the perpendicular component of E, integrated over a surface. (ee Examples.1.3.) = E da = E# da L L (.5) E f f A A A Gauss s law: Gauss s law states that the total electric flux through a closed surface, which can be written as the surface integral of the component of E normal to the surface, equals a constant times the total charge Q encl enclosed by the surface. Gauss s law is logically equivalent to Coulomb s law, but its use greatly simplifies problems with a high degree of symmetry. (ee Examples.4.10.) When excess charge is placed on a conductor and is at rest, it resides entirely on the surface, and E 0 everywhere in the material of the conductor. (ee Examples ) E = C E cos f da = C E da = C E # da = Q encl P 0 (.8), (.9) Outward normal to surface E f r R q E r da

3 Electric field of various symmetric charge distributions: The following table lists electric fields caused by several symmetric charge distributions. In the table, q, Q, l, and s refer to the magnitudes of the quantities. Point in Electric Field Charge Distribution Electric Field Magnitude ingle point charge q Distance r from q Charge q on surface of conducting sphere with radius R Outside sphere, r 7 R Infinite wire, charge per unit length l Inside sphere, r 6 R Distance r from wire Infinite conducting cylinder with radius R, charge per Outside cylinder, r 7 R unit length l Inside cylinder, r 6 R olid insulating sphere with radius R, charge Q distributed Outside sphere, r 7 R uniformly throughout volume Infinite sheet of charge with uniform charge per unit area s Inside sphere, r 6 R Any point E = E = E = 0 E = E = E = 0 E = E = E = 1 q 4pP 0 r 1 q 4pP 0 r 1 l pp 0 r 1 l pp 0 r 1 Q 4pP 0 r 1 Qr 4pP 0 R 3 s P 0 Two oppositely charged conducting plates with surface charge densities +s and - s Any point between plates E = s P 0 Charged conductor Just outside the conductor E = s P 0

4 Electric potential energy: The electric force caused by any collection of charges at rest is a conservative force. The work W done by the electric force on a charged particle moving in an electric field can be represented by the change in a potential-energy function U. The electric potential energy for two point charges q and q 0 depends on their separation r. The electric potential energy for a charge q 0 in the presence of a collection of charges q 1, q, q 3 depends on the distance from q 0 to each of these other charges. (ee Examples 3.1 and 3..) W ab = U a - U b 1 qq 0 U = 4pP 0 r (two point charges) (3.) (3.9) U = q 0 a q 1 + q + q 3 + 4pP 0 r 1 r r Á b 3 q 0 q i = (3.10) 4pP a 0 i r i ( q 0 in presence of other point charges) q 1 U 5 q q 1 q q 3 4pP 0 r 1 r r 3 q r 1 r q 3 r 3 q 0 Electric potential: Potential, denoted by V, is potential energy per unit charge. The potential difference between two points equals the amount of work that would be required to move a unit positive test charge between those points. The potential V due to a quantity of charge can be calculated by summing (if the charge is a collection of point charges) or by integrating (if the charge is a distribution). (ee Examples 3.3, 3.4, 3.5, 3.7, 3.11, and 3.1.) The potential difference between two points a and b, also called the potential of a with respect to b, is given by the line integral of E. The potential at a given point can be found by first finding E and then carrying out this integral. (ee Examples 3.6, 3.8, 3.9, and 3.10.) V = U 1 q = (3.14) q 0 4pP 0 r (due to a point charge) V = U 1 q i = (3.15) q 0 4pP a 0 i r i (due to a collection of point charges) 1 dq V = (3.16) 4pP 0 L r (due to a charge distribution) b b V a - V b = E# d l = E cos f dl L a L a (3.17) q 1 V q 1 q q 3 4pP 0 r 1 r r 3 q r 1 r q 3 r 3 P

5 Finding electric field from electric potential: If the potential V is known as a function of the coordinates x, y, and x = - 0V 0x E y = - 0V 0y E z = - 0V 0z E z, the components of electric field E at any point are (3.19) given by partial derivatives of V. (ee Examples 3.13 and 3.14.) E aın 0V (3.0) 0x N 0V 0V kn 0y 0z b (vector form)

6 Problems 71 the values of u until a selfare each attached to silk Another Another point point charge charge Q = Q = mc mc Two... point Two charges point charges q 1 and q 1 and Figure P1.76 q are q held are in held place in 4.50 place cm4.50 apart. cm apart. q ung from a common point of mass of mass 5.00 g5.00 is initially g is initially located located 3.00 cm from each of these charges a m = 8.00 g. The radius of 3.00 cm from each of these charges o the distance between the (Fig. (Fig. P1.76) P1.76) and released and released from rest. from rest. int charges. One sphere is You You observe observe that the that initial acceleration of Q is 34 m>s the initial acceleration of is 34 m>s 4.50 cm Q upward, parallel a different positive charge Q upward, parallel to the line connecting the two point o that when the spheres are charges. to the Find line q 1 and connecting q. the two point angle u = 0.0 with the 1.77 charges.. Three Find identical q 1 and point q. charges for each sphere when in Three identical point charges q that act on each sphere. q are placed at each of three corners of ctrostatic force that acts on a square of side L. Find the magnitude n in each thread. (c) Based and direction of the net force on a point charge -3q placed (a) at en, what can you say about the center of the square and (b) at the vacant corner of the square. your answers. (d) A small In each case, draw a free-body diagram showing the forces exerted pheres, allowing charge to on the -3q charge by each of the other three charges. other until the two spheres Three point charges are placed on the y -axis: a charge q removed. Each thread now at y = a, a charge -q at the origin, and a charge q at y = -a. vertical. Determine the uch an arrangement is called an electric quadrupole. (a) Find the ge on the pair of spheres is magnitude and direction of the electric field at points on the positive x-axis. (b) Use the binomial expansion to find an approximate inary table salt) is made up expression for the electric field valid for x W a. Contrast this cm 3.00 cm

7 1.76. I DENTIFY: For the acceleration (and hence the force) on Q to be upward, as indicated, the forces due to q 1 and q must have equal strengths, so q 1 and q must have equal magnitudes. Furthermore, for the force to be upward, q 1 must be positive and q must be negative. ET UP: ince we know the acceleration of Q, Newton s second law gives us the magnitude of the force on it. We can then add the force components using F = F cosθ + F cosθ = F cosθ. The electrical force on Q is given by Coulomb s law, EXECUTE: First find the net force: F Qq 1 1 Qq = 4π r 0 1 Qq Qq Qq 1 1 (for q 1 ) and likewise for q. F = ma= ( kg)(34 m/s ) = 1.6 N. Now add the force components, calling θ the angle between the line connecting q 1 and q and the line connecting q 1 and Q. F 16N. F = FQq cosθ + F cos cos 1 Qq θ = F Qq θ and F 1.08 N. 1 Qq = = = Now find the charges 1 cosθ 5cm. 300cm. by solving for q 1 in Coulomb s law and use the fact that q 1 and q have equal magnitudes but opposite signs. F Qq 1 1 Qq = 4π r r F Qq ( m) (1. 08 N) 1 and q1 = = Q ( N m /C )( C) 4π 0 8 = C. q = q1 = C. EVALUATE: imple reasoning allows us first to conclude that q 1 and q must have equal magnitudes but opposite signs, which makes the equations much easier to set up than if we had tried to solve the problem in the general case. As Q accelerates and hence moves upward, the magnitude of the acceleration vector will change in a complicated way I DENTIFY: Use Coulomb s law to calculate the forces between pairs of charges and sum these forces as

8 CALC Positive charge Q is distributed uniformly along the positive y-axis between Figure P1.90 y y = 0 and y = a. A negative a point charge -q lies on the positive x-axis, a distance x from the Q origin (Fig. P1.90). (a) Calculate the x - and y -components of O the electric field produced by the q x charge distribution Q at points on the positive x-axis. (b) Calculate the x - and y -components of the force that the charge distribution Q exerts on q. (c) how that if x W a, F x -Qq>4pP 0 x and F y +Qqa>8pP 0 x 3. Explain why this result is obtained Acharged line like that shown in Fig. 1.4extends >

9 1.90. I DENTIFY: Use Eq. (1.7) to calculate the electric field due to a small slice of the line of charge and integrate as in Example Use Eq. (1.3) to calculate F. ET UP: The electric field due to an infinitesimal segment of the line of charge is sketched in Figure sinθ = x y + y cosθ = x x + y Figure 1.90 lice the charge distribution up into small pieces of length dy. The charge dq in each slice is dq = Q( dy/ a). The electric field this produces at a distance x along the x-axis is de. Calculate the components of de and then integrate over the charge distribution to find the components of the total field. 1 dq Q dy EXECUTE: de = 4 = π 0 x + y 4π 0a x + y Qx dy dex = de cosθ = 4 3/ π 0a ( x + y ) de E E x y y Q ydy = desinθ = 4 π 3/ 0a ( x + y ) a Qx a dy Qx 1 y Q 1 dex 4 π 0 3/ 0a Ñ = ( x + y ) 4π 0a x x + y 4π 0x x + a 0 a Q a ydy Q 1 Q 1 1 dey 4 π 0 3/ 0a Ñ = ( x + y ) 4π 0a x y 4π 0a + x x + a 0 = = = = = = (b) F = q0e qq 1 qq 1 1 F = qe = ; F = qe = x + a x + a x x y y 4π 0x 4π 0a x (c) For x a, 1/ a 1 1 a 1 a = + = = x a x x x x + x x 3 qq 1 1, qq a Fx F y + qqa 3 = 3 4π 0x 4π 0a x x x 8π 0x qq EVALUATE: For x a, Fy Fx and F Fx = 4π P x charge distribution Q acts like a point charge. IDENTIFY: Apply Eq. (1.9) from Example and F is in the x-direction. For x a the

10 .47. Concentric pherical hells. A small conducting spherical shell with inner Figure P.47 radius a and outer radius b is concentric with a larger conducting spherical shell with inner radius c and outer radius d (Fig. P.47). a The inner shell has total charge +q, and the outer shell has charge +4q. (a) Calculate b c the electric field (magnitude and direc- tion) in terms of q and the distance r from d the common center of the two shells for (i) r 6 a; (ii) a 6 r 6 b; (iii) b 6 r 6 c; (iv) c 6 r 6 d; (v) how your results in a graph of the radial component of r 7 d. E as a function of r. (b) What is the total charge on the (i) inner surface of the small shell; (ii) outer surface of the small shell; (iii) inner surface of the large shell; (iv) outer surface of the large shell?.48. Repeat Problem.47, but now let the outer shell have

11 .47. IDENTIFY: Apply Gauss s law to a spherical Gaussian surface with radius r. Calculate the electric field at the surface of the Gaussian sphere. (a) ET UP: (i) r < a: The Gaussian surface is sketched in Figure.47a. EXECUTE: Φ E = EA= E(4 π r ) Q encl = 0; no charge is enclosed Qencl Φ = says E 0 E(4 π r ) = 0 and E = 0. Figure.47a (ii) a< r < b: Points in this region are in the conductor of the small shell, so E = 0. (iii) ET UP: b< r < c: The Gaussian surface is sketched in Figure.47b. Apply Gauss s law to a spherical Gaussian surface with radius b< r < c. Figure.47c Qencl 6q Φ E = gives E(4 π r ) = 0 0 6q E = 4π 0r EXECUTE: Φ E = EA= E(4 π r ) The Gaussian surface encloses all of the small shell and all of the large shell, so Qencl =+ q+ 4q= 6 q.. ince the enclosed charge is positive the electric field is radially outward. The graph of E versus r is sketched in Figure.47d. EXECUTE: Φ E = EA= E(4 π r ) The Gaussian surface encloses all of the small shell and none of the large shell, so Qencl =+ q. Figure.47b Q q q Φ = gives (4 ) = so =. ince the enclosed charge is positive the electric field is E encl E π r E 0 0 4π 0r radially outward. (iv) c< r < d: Points in this region are in the conductor of the large shell, so E = 0. (v) ET UP: r > d: Apply Gauss s law to a spherical Gaussian surface with radius r > d, as shown in Figure.47c. Figure.47d (b) IDENTIFY and ET UP: Apply Gauss s law to a sphere that lies outside the surface of the shell for which we want to find the surface charge. EXECUTE: (i) charge on inner surface of the small shell: Apply Gauss s law to a spherical Gaussian surface with radius a< r < b. This surface lies within the conductor of the small shell, where E = 0, so Φ = 0. Thus by Gauss s law Q encl = 0, so there is zero charge on the inner surface of the small shell. E

12 (ii) charge on outer surface of the small shell: The total charge on the small shell is + q. We found in part (i) that there is zero charge on the inner surface of the shell, so all + q must reside on the outer surface. (iii) charge on inner surface of large shell: Apply Gauss s law to a spherical Gaussian surface with radius c< r < d. The surface lies within the conductor of the large shell, where E = 0, so Φ = 0. Thus by Gauss s law Q encl = 0. The surface encloses the + q on the small shell so there must be charge q on the inner surface of the large shell to make the total enclosed charge zero. (iv) charge on outer surface of large shell: The total charge on the large shell is + 4 q. We showed in part (iii) that the charge on the inner surface is q, so there must be + 6q on the outer surface. EVALUATE: The electric field lines for b< r < c originate from the surface charge on the outer surface of the inner shell and all terminate on the surface charge on the inner surface of the outer shell. These surface charges have equal magnitude and opposite sign. The electric field lines for r > d originate from the surface charge on the outer surface of the outer sphere. IDENTIFY: Apply Gauss s law. E

13 > charge -Q CALC Anonuniform, but spherically symmetric, distribution of charge has a charge density r1r given as follows: r1r = r r>r for r R r1r = 0 for r Ú R where r 0 = 3Q>pR 3 is a positive constant. (a) how that the total charge contained in the charge distribution is Q. (b) how that the electric field in the region r Ú R is identical to that produced by a point charge Q at r = 0. (c) Obtain an expression for the electric field in the region r R. (d) Graph the electric-field magnitude E as a function of r. (e) Find the value of r at which the electric field is maximum, and find the value of that maximum field > 1 > > >

14 .65. ρ() r = ρ (1 r/r) for r R where ρ = 3 Q/ πr. ρ () r = 0 for r R (a) IDENTIFY: The charge density varies with r inside the spherical volume. Divide the volume up into thin concentric shells, of radius r and thickness dr. Find the charge dq in each shell and integrate to find the total charge. ET UP: The thin shell is sketched in Figure.65a. Figure.65a EXECUTE: The volume of such a shell is dv = 4 π r dr. The charge contained within the shell is 0 dq = ρ() r dv = 4 πr ρ (1 r/ R) dr. The total charge Q in the charge distribution is obtained by integrating dq over all such shells into which the sphere can be subdivided: R R 3 π ρ 0 0 πρ0 0 Q= dq= 4 r (1 r/ R) dr = 4 ( r r / R) dr 3 4 R 3 4 r r R R Q= 4πρ = 4πρ = 4 πρ ( R /1) = 4 π (3 Q/ π R )( R /1) = Q, as was to be shown R 3 4R 0 (b) IDENTIFY: Apply Gauss s law to a spherical surface of radius r, where r > R. ET UP: The Gaussian surface is shown in Figure.65b. Q EXECUTE: encl Φ E = 0 Q E(4 π r ) = 0 To calculate the enclosed charge Q encl use the same technique as in part (a), except integrate dq out to r rather than R. (We want the charge that is inside radius r.) r 3 r r r Qencl = 4πr ρ dr = 4πρ0 r dr R 0 R Q 3 4 r encl 4πρ0 4πρ0 4πρ0r 0 r r r r 1 r = = = 3 4R 3 4R 3 4R 3 3 3Q r 1 r r r ρ0 = so Q 3 encl = 1Q Q = 3 4R 3 π R R R R 3 Thus Gauss s law gives (4 ) Q r r E π r =. 0 R R Qr 3r E = 4, r R 3 4π 0R R (d) The graph of E versus r is sketched in Figure.65d. Figure.65b Q E = ; same as for point charge of charge Q. 4π 0r (c) IDENTIFY: Apply Gauss s law to a spherical surface of radius r, where r < R: ET UP: The Gaussian surface is shown in Figure.65c. Figure.65c EXECUTE: Q encl Φ E = 0 Φ E = E(4 π r ) Figure.65d de (e) Where the electric field is a maximum, 0. dr = Thus d 3r 4r = 0 so 4 6 r/ R= 0 and r = R/3. dr R Q R 3 R Q At this value of r, E = 4. 3 = 3 3 4π R R 3π R 0 0 EVALUATE: Our expressions for E( r ) for r < R and for r > R agree at r = R. The results of part (e) for the value of r where E() r is a maximum agrees with the graph in part (d). = ρ π

15 CP In experiments in which atomic nuclei collide, head-on collisions like that described in Problem 3.8 do happen, but near misses are more common. uppose the alpha particle in Problem 3.8 was not aimed at the center of the lead nucleus, but had an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude L = p 0 b, where p 0 is the magnitude of the initial momentum of the alpha particle and b = 1.00 * 10-1 m. What is the distance of closest approach? Repeat for b = 1.00 * m and b = 1.00 * m CALC A hollow, thin-walled insulating cylinder of

16 3.89. IDENTIFY: Angular momentum and energy must be conserved. ET UP: At the distance of closest approach the speed is not zero. E = K + U. q1 =, e q = 8 e. EXECUTE: mv 1 b = mv r. E1 E = gives E 1 kq q 1 1= mv + r. 1 E 1 = 11 MeV = J. r is the b b kq1q distance of closest approach. ubstituting in for v = v1 we find E1= E1 +. r r r ( E ) r ( kq q ) r Eb = 0. For b = 10 m, r = m. For b = 10 m, r = m And for b = 10 m, 14 r = m. EVALUATE: As b decreases the collision is closer to being head-on and the distance of closest approach decreases. Problem 3.8 shows that the distance of closest approach is m when b = 0. I : Consider the potential due to an infinitesimal slice of the cylinder and integrate over the length 14

17 Capacitors and capacitance: Acapacitor is any pair of conductors separated by an insulating material. When the capacitor is charged, there are charges of equal magnitude Q and opposite sign on the two conductors, and the potential V ab of the positively charged conductor with respect to the negatively charged conductor is proportional to Q. The capacitance C is defined as the ratio of Q to V ab. The I unit of capacitance is the farad (F): 1 F = 1 C>V. Aparallel-plate capacitor consists of two parallel conducting plates, each with area A, separated by a distance d. If they are separated by vacuum, the capacitance depends only on A and d. For other geometries, the capacitance can be found by using the definition C = Q>V ab. (ee Examples ) C = Q V ab C = Q V ab =P 0 A d (4.1) (4.) 1Q Q Wire Plate a, area A d Potential Wire Plate b, area A difference 5 V ab Capacitors in series and parallel: When capacitors with capacitances C 1, C, C 3, Á are connected in series, the reciprocal of the equivalent capacitance C eq equals the sum of the reciprocals of the individual capacitances. When capacitors are connected in parallel, the equivalent capacitance C eq equals the sum of the individual capacitances. (ee Examples 4.5 and 4.6.) 1 = Á C eq C 1 C C 3 (capacitors in series) C eq = C 1 + C + C 3 + Á (capacitors in parallel) (4.5) (4.7) a 1Q Q 1Q Q C 1 V ab 5 V c b a C V ac 5 V 1 V cb 5 V V ab 5 V C 1 Q 1 C Q b

18 Energy in a capacitor: The energy U required to charge a capacitor C to a potential difference V and a charge Q is equal to the energy stored in the capacitor. This energy can be thought of as residing in the electric field between the conductors; the energy density u (energy per unit volume) is proportional to the square of the electric-field magnitude. (ee Examples ) U = Q C = 1 CV = 1 QV u = 1 P 0 E (4.9) (4.11) 1Q V E Q + Dielectrics: When the space between the conductors is A C = KC 0 = KP 0 filled with a dielectric material, the capacitance d =PA d increases by a factor K, called the dielectric constant (parallel-plate capacitor of the material. The quantity P=KP 0 is called the permittivity of the dielectric. For a fixed amount of charge filled with dielectric) (4.19) on the capacitor plates, induced charges on the surface of the dielectric decrease the electric field and potential u = 1 KP 0 E = 1 PE (4.0) difference between the plates by the same factor K. The surface charge results from polarization, a microscopic Q # encl-free (4.3) rearrangement of charge in the dielectric. (ee Example C KE da = P ) Under sufficiently strong fields, dielectrics become conductors, a situation called dielectric breakdown. The maximum field that a material can withstand without breakdown is called its dielectric strength. In a dielectric, the expression for the energy density is the same as in vacuum but with P 0 replaced by P=KP. (ee Example 4.11.) Gauss s law in a dielectric has almost the same form as in vacuum, with two key differences: E is replaced by KE and Q encl is replaced by Q encl-free, which includes only the free charge (not bound charge) enclosed by the Gaussian surface. (ee Example 4.1.) Dielectric between plates s s s + i s i s i s i s s 809

19 nected across the 960-V potential difference? In Fig. P4.57, C 1 = C 5 = 8.4 mf and C = C 3 = Figure P4.57 C 4 = 4. mf. The applied potential is V ab = 0 V. (a) What is 1 C 3 C the equivalent capacitance of a the network between points a and b? (b) Calculate the charge C C on each capacitor and the 5 C 4 b potential difference across each capacitor

20 4.57. (a) IDENTIFY: Replace series and parallel combinations of capacitors by their equivale ET UP: The network is sketched in Figure 4.57a. C1= C 5 = 84. µ F C = C3 = C 4 = 4. µ F Figure 4.57d = + + C C C C eq = + C 84. µ F 63. µ F eq C eq = 5. µ F Figure 4.57a EXECUTE: implify the circuit by replacing the capacitor combinations by their equiv are in series and can be replaced by C 34 (Figure 4.57b): Figure 4.57b C 34 CC 3 4 (4. µ F)(4. µ F) = = = 1. µ F C + C 4. µ F+ 4. µ F = + C34 C3 C4 1 C3+ C4 = C34 C3C4 C and C 34 are in parallel and can be replaced by their equivalent (Figure 4.57c): C34 = C + C34 C 34 = 4. µ F+ 1. µ F C 34 = 63. µ F EVALUATE: For capacitors in series the equivalent capacitor is smaller than any of those in series. For capacitors in parallel the equivalent capacitance is larger than any of those in parallel. (b) IDENTIFY and ET UP: In each equivalent network apply the rules for Q and V for capacitors in series and parallel; start with the simplest network and work back to the original circuit. EXECUTE: The equivalent circuit is drawn in Figure 4.57e. Figure 4.57e Q eq = C V Q1= Q5 = Q 34 = 550 µ C (capacitors in series have same charge) Q1 550 µ C V1 = = = 65 V C 84. µ F 1 Q5 550 µ C V5 = = = 65 V C5 84. µ F V 34 Q µ C = = = 87 V C 63. µ F 34 Now draw the network as in Figure 4.57f. eq Q eq = (. 5 µ F)(0 V) = 550 µ C V = V34 = V34 = 87 V capacitors in parallel have the same potential Figure 4.57c C1, C5 and C 34 are in series and can be replaced by C eq (Figure 4.57d): Figure 4.57f Q = CV = (4. µ F)(87 V) = 370 µ C Q34 = C34V 34 = (. 1µ F)(87 V) = 180 µ C Finally, consider the original circuit (Figure 4.57g).

21 34 = =. = Finally, consider the original circuit (Figure 4.57g). Q3 = Q4 = Q 34 = 180 µ C capacitors in series have the same charge Figure 4.57g V V 3 4 Q3 180 µ C = = = 43 V C 4. µ F 3 Q4 180 µ C = = = 43 V C 4. µ F 4 ummary: Q1= 550 µ C, V1= 65 V Q = 370 µ C, V = 87 V Q Q Q = 180 µ C, V = 43 V 3 3 = 180 µ C, V = 43 V 4 4 = 550 µ C, V = 65 V 5 5 EVALUATE: V3 V4 V and V1 V V5 Q = Q + Q and Q = Q + Q = + + = 0 V (apart from some small rounding error) I : We can make series and parallel combinations.

22 > Aparallel-plate capacitor is made from two plates Figure P cm on each side and 4.50 mm apart. Half of the space between these plates Plexiglas Air contains only air, but the other half is filled with Plexiglas of dielectric constant 3.40 (Fig. P4.7). An 18.0-V battery is connected across the plates. (a) What is the capacitance of this combination? (Hint: Can you think of this capacitor as equivalent to two capacitors in parallel?) (b) How much energy is stored in the capacitor? (c) If we remove the Plexiglas but change nothing else, how much energy will be stored in the capacitor? Aparallel-plate capacitor has square plates that are

23 4.7. IDENTIFY: The capacitor is equivalent to two capacitors in parallel, as shown in Figure 4.7. Figure 4.7 ET UP: Each of these two capacitors have plates that are 1.0 cm by 6.0 cm. For a parallel-plate A capacitor with dielectric filling the volume between the plates, C = K 0. For two capacitors in parallel, d C = C 1 + C. The energy stored in a capacitor is 1 U = CV. EXECUTE: (a) C = C 1 + C. C 1 A ( F/m)(0. 10 m)( m) 11 = = = F. d m C1= KC = (3. 40)( F) = F. C = C1+ C = F= 65. pf (b) U 1 CV 1 = = ( F)(18. 0 V) = J (c) Now C1= C and C = ( F) = F. U CV = = ( F)(18. 0 V) = J EVALUATE: The plexiglass increases the capacitance and that increases the energy stored for the same voltage across the capacitor IDENTIFY: The two slabs of dielectric are in series with each other. 11

24 Current and current density: Current is the amount of charge flowing through a specified area, per unit time. The I unit of current is the ampere 11 A = 1 C>s. The current I through an area A depends on the concentration n and charge q of the charge carriers, as well as on the magnitude of their drift velocity v d. The current density is current per unit cross-sectional area. Current is usually described in terms of a flow of positive charge, even when the charges are actually negative or of both signs. (ee Example 5.1.) I = dq dt J nqvd = nƒqƒv d A (5.) (5.4) v d v d I v d v d v d v d E Resistivity: The resistivity r of a material is the ratio of the magnitudes of electric field and current density. Good conductors have small resistivity; good insulators have large resistivity. Ohm s law, obeyed approximately by many materials, states that r is a constant independent of the value of E. Resistivity usually increases with temperature; for small temperature changes this variation is represented approximately by Eq. (5.6), where a is the temperature coefficient of resistivity. r = E J r1t = r a1t - T 0 4 (5.5) (5.6) r 0 r O T 0 Metal: r increases with increasing T. lope 5 r 0 a T Resistors: The potential difference V across a sample of material that obeys Ohm s law is proportional to the current I through the sample. The ratio V>I = R is the resistance of the sample. The I unit of resistance is the ohm 11 Æ=1 V>A. The resistance of a cylindrical conductor is related to its resistivity r, length L, and cross-sectional area A. (ee Examples 5. and 5.3.) V = IR R = rl A (5.11) (5.10) Higher potential A I E L J V Lower potential I

25 > 1 > Circuits and emf: Acomplete circuit has a continuous current-carrying path. A complete circuit carrying a steady current must contain a source of electromotive force (emf) E. The I unit of electromotive force is the volt (1 V). Every real source of emf has some internal resistance r, so its terminal potential difference V ab depends on current. (ee Examples ) V ab = E - Ir (5.15) (source with internal resistance) I V ab 5 V a b V a + b r 5 V, E 5 1 V A a R 5 4 V b I Energy and power in circuits: Acircuit element with a potential difference V a - V b = V ab and a current I puts energy into a circuit if the current direction is from P = V ab I (general circuit element) (5.17) lower to higher potential in the device, and it takes P = V energy out of the circuit if the current is opposite. The ab I = I R = V ab R (5.18) power P equals the product of the potential difference (power into a resistor) and the current. A resistor always takes electrical energy out of a circuit. (ee Examples ) I V a a Circuit element V b b I Conduction in metals: The microscopic basis of conduction in metals is the motion of electrons that move freely through the metallic crystal, bumping into ion cores in the crystal. In a crude classical model of this motion, the resistivity of the material can be related to the electron mass, charge, speed of random motion, density, and mean free time between collisions. (ee Example 5.11.) E Net displacement

26 Resistors in series and parallel: When several resistors R 1, R, R 3, Á are connected in series, the equivalent resistance R eq is the sum of the individual resistances. The same current flows through all the resistors in a series connection. When several resistors are connected in parallel, the reciprocal of the equivalent resistance R eq is the sum of the reciprocals of the individual resistances. All resistors in a parallel connection have the same potential difference between their terminals. (ee Examples 6.1 and 6..) R eq = R 1 + R + R 3 + Á (resistors in series) 1 = R eq R 1 R R Á 3 (resistors in parallel) (6.1) (6.) Resistors in series a R 1 x R y R 3 b I Resistors in parallel a I R 1 R R 3 b I I Kirchhoff s rules: Kirchhoff s junction rule is based on conservation of charge. It states that the algebraic sum of the currents into any junction must be zero. Kirchhoff s loop rule is based on conservation of energy and the conservative nature of electrostatic fields. It states that the algebraic sum of potential differences around any loop must be zero. Careful use of consistent sign rules is essential in applying Kirchhoff s rules. (ee Examples ) a I = 0 a V = 0 (junction rule) (6.5) (loop rule) (6.6) At any junction: I E Loop I 1 Loop 1 + E Junction Loop 3 Around any loop: V 5 0 I I 1 1 I R A B

27 R-C circuits: When a capacitor is charged by a battery in series with a resistor, the current and capacitor charge are not constant. The charge approaches its final value asymptotically and the current approaches zero asymptotically. The charge and current in the circuit are given by Eqs. (6.1) and (6.13). After a time t = RC, the charge has approached within 1>e of its final value. This time is called the time constant or relaxation time of the circuit. When the capacitor discharges, the charge and current are given as functions of time by Eqs. (6.16) and (6.17). The time constant is the same for charging and discharging. (ee Examples 6.1 and 6.13.) Capacitor charging: q = CEA1 - e -t/rc B = Q f A1 - e -t/rc B i = dq = E dt R e-t/rc = I 0 e -t/rc Capacitor discharging: q = Q 0 e -t/rc i = dq = - Q 0 dt = I 0 e -t/rc RC e-t/rc (6.1) (6.13) (6.16) (6.17) i, q O + E i R 1q i q C q versus t i versus t t

28 5.53. A 540-W electric heater is designed to operate from 10-V lines. (a) What is its resistance? (b) What current does it draw? (c) If the line voltage drops to 110 V, what power does the heater take? (Assume that the resistance is constant. Actually, it will change because of the change in temperature.) (d) The heater coils are metallic, so that the resistance of the heater decreases with decreasing temperature. If the change of resistance with temperature is taken into account, will the electrical power consumed by the heater be larger or smaller than what you calculated in part (c)? Explain. ection 5.6 Theory of Metallic Conduction

29 5.53. V IDENTIFY: P= I R= = VI. V = IR. R ET UP: The heater consumes 540 W when V = 10 V. Energy = Pt. EXECUTE: (a) (b) P = VI so V V (10 V) P = so R = = = 6. 7 Ω R P 540 W P 540 W I = = = 450. A V 10 V V (110 V) (c) Assuming that R remains 6. 7 Ω, P = = = 453 W. P is smaller by a factor of (110/10). R 6. 7 Ω EVALUATE: (d) With the lower line voltage the current will decrease and the operating temperature will decrease. R will be less than 6. 7 Ω and the power consumed will be greater than the value calculated in part (c).

30 + rather than 5.0 h. What differences in performance do you see? CP Consider the circuit shown in Fig. P5.83. The emf source has negligible internal resistance. The resistors Figure P5.83 R have resistances R 1 = 6.00 Æ and R = 4.00 Æ. The capacitor has capacitance C = 9.00 mf. When the capacitor is fully E R 1 C charged, the magnitude of the charge on its plates is Q = 36.0 mc. Calculate the emf E CP Consider the circuit shown in Fig. P5.84. The battery

31 5.83. IDENTIFY: No current flows through the capacitor when it is fully charged. ET UP: With the capacitor fully charged, EXECUTE: V R I = R ε + R 1. VR = IR and V = Q/ C. Q µ C VC = = = 400. V. VR = V 400 V C 900. µ F 1 C =. and V I = = = A. R Ω = IR = ( A)(4. 00 Ω ) =. 668 V. ε = V + V = 400. V V= 667. V. R1 R EVALUATE: When a capacitor is fully charged, it acts like an open circuit and prevents any current from flowing though it I : No current flows to the capacitors when they are fully charged. V R C

32 parts (a) and (b) for t between 0 and 0 s CP In the circuit shown in Fig. E6.43 both capacitors are initially charged to 45.0 V. (a) How long after closing the switch will the potential across each capacitor be reduced to 10.0 V, and (b) what will be the current at that time? Aresistor and a capacitor Figure E mf mf 30.0 V 50.0 V

33 6.43. IDENTIFY: The capacitors, which are in parallel, will discharge exponentially through the resistors. ET UP: ince V is proportional to Q, V must obey the same exponential equation as Q, V / trc 0. = V e The current is 0 / trc I = ( V / R) e. EXECUTE: (a) olve for time when the potential across each capacitor is 10.0 V: (b) 0 t = RC ln( V/ V ) = (80.0 Ω )(35.0 µ F) ln(10/45) = 410 µ s = 4.1 ms / trc 0 I = ( V / R) e. Using the above values, with V0 = 45.0 V, gives I = 0.15 A. EVALUATE: ince the current and the potential both obey the same exponential equation, they are both reduced by the same factor (0.) in 4.1 ms. I : In τ = RC use the equivalent capacitance of the two capacitors.

34 internal resistance (a) Find the current through the battery and each resistor in the circuit shown in Fig. P6.66. (b) What is the equivalent resistance of the resistor network? Figure P6.66 R V +R V 14.0 V R V R 5.00 V R V..

35 . +. = IDENTIFY: Apply the loop and junction rules. ET UP: Use the currents as defined on the circuit diagram in Figure 6.66 and obtain three equations to solve for the currents. EXECUTE: Left loop: 14 I1 ( I1 I) = 0 and 3I1 I = 14. Top loop: ( I I ) I I 0 I + 3I + I = = and 1 Bottom loop: ( I I1+ I) + ( I1 I) I = 0 and I + 3I1 4I = 0. olving these equations for the currents we find: I = I = 10.0 A; I = I = 6.0 A; I = I =.0 A. o the other currents are: (b) R eq V V = = = 140. Ω. I A 4 5 battery 1 R R 1 R 1 R 1 R 1 3 I = I I = 40A;. I = I I = 40A;. I = I I + I = 60A.. EVALUATE: It isn t possible to simplify the resistor network using the rules for resistors in series and parallel. But the equivalent resistance is still defined by V = IR eq. Figure 6.66