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Exercicios do Tipler Resolvidos (Volume 2, Capitulo 21 ao 41), Exercícios de Cultura

Volume 2, Capitulo 21 ao 41

Tipologia: Exercícios

2011
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Baixe Exercicios do Tipler Resolvidos (Volume 2, Capitulo 21 ao 41) e outras Exercícios em PDF para Cultura, somente na Docsity! = (o) Q (7) n ui [e [-5 [a] [ea (e) = (5) po o 7) o E ELA ITR Chapter 21 The Electric Field 1: Discrete Charge Distributions Conceptual Problems *1 •• Similarities: Differences: The force between charges and masses varies as 1/r2. There are positive and negative charges but only positive masses. The force is directly proportional to the product of the charges or masses. Like charges repel; like masses attract. The gravitational constant G is many orders of magnitude smaller than the Coulomb constant k. 2 • Determine the Concept No. In order to charge a body by induction, it must have charges that are free to move about on the body. An insulator does not have such charges. 3 •• Determine the Concept During this sequence of events, negative charges are attracted from ground to the rectangular metal plate B. When S is opened, these charges are trapped on B and remain there when the charged body is removed. Hence B is negatively charged and correct. is )(c 4 •• (a) Connect the metal sphere to ground; bring the insulating rod near the metal sphere and disconnect the sphere from ground; then remove the insulating rod. The sphere will be negatively charged. (b) Bring the insulating rod in contact with the metal sphere; some of the positive charge on the rod will be transferred to the metal sphere. (c) Yes. First charge one metal sphere negatively by induction as in (a). Then use that negatively charged sphere to charge the second metal sphere positively by induction. 1 Chapter 21 4 *12 • Determine the Concept We can use the rules for drawing electric field lines to draw the electric field lines for this system. In the field-line sketch to the right we’ve assigned 7 field lines to each charge q. 13 • Determine the Concept A positive charge will induce a charge of the opposite sign on the near surface of the nearby neutral conductor. The positive charge and the induced charge on the neutral conductor, being of opposite sign, will always attract one another. correct. is )(a *14 • Determine the Concept Electric field lines around an electric dipole originate at the positive charge and terminate at the negative charge. Only the lines shown in (d) satisfy this requirement. correct. is )(d *15 •• Determine the Concept Because θ ≠ 0, a dipole in a uniform electric field will experience a restoring torque whose magnitude is θsinxpE . Hence it will oscillate about its equilibrium orientation, θ = 0. If θ << 1, sinθ ≈ θ, and the motion will be simple harmonic motion. Because the field is nonuniform and is larger in the x direction, the force acting on the positive charge of the dipole (in the direction of increasing x) will be greater than the force acting on the negative charge of the dipole (in the direction of decreasing x) and thus there will be a net electric force on the dipole in the direction of increasing x. Hence, the dipole will accelerate in the x direction as it oscillates about θ = 0. 16 •• (a) False. The direction of the field is toward a negative charge. (b) True. (c) False. Electric field lines diverge from any point in space occupied by a positive charge. (d) True The Electric Field 1: Discrete Charge Distributions 5 (e) True 17 •• Determine the Concept The diagram shows the metal balls before they are placed in the water. In this situation, the net electric field at the location of the sphere on the left is due only to the charge –q on the sphere on the right. If the metal balls are placed in water, the water molecules around each ball tend to align themselves with the electric field. This is shown for the ball on the right with charge –q. (a) The net electric field r E net that produces a force on the ball on the left is the field r E due to the charge –q on the ball on the right plus the field due to the layer of positive charge that surrounds the ball on the right. This layer of positive charge is due to the aligning of the water molecules in the electric field, and the amount of positive charge in the layer surrounding the ball on the left will be less than +q. Thus, Enet < E. Because Enet < E, the force on the ball on the left is less than it would be if the balls had not been placed in water. Hence, the force will decrease when the balls are placed in the water. (b) When a third uncharged metal ball is placed between the first two, the net electric field at the location of the sphere on the right is the field due to the charge +q on the sphere on the left, plus the field due to the charge –Q and +Q on the sphere in the middle. This electric field is directed to the right. The field due to the charge –Q and +Q on the sphere in the middle at the location of the sphere on the right is to the right. It follows that the net electric field due to the charge +q on the sphere on the left, plus the field due to the charge –Q and +Q on the sphere in the middle is to the right and has a greater magnitude than the field due only to the charge +q on the sphere on the left. Hence, the force on either sphere will increase if a third uncharged metal ball is placed between them. Remarks: The reduction of an electric field by the alignment of dipole moments with the field is discussed in further detail in Chapter 24. Chapter 21 6 *18 •• Determine the Concept Yes. A positively charged ball will induce a dipole on the metal ball, and if the two are in close proximity, the net force can be attractive. *19 •• Determine the Concept Assume that the wand has a negative charge. When the charged wand is brought near the tinfoil, the side nearer the wand becomes positively charged by induction, and so it swings toward the wand. When it touches the wand, some of the negative charge is transferred to the foil, which, as a result, acquires a net negative charge and is now repelled by the wand. Estimation and Approximation 20 •• Picture the Problem Because it is both very small and repulsive, we can ignore the gravitational force between the spheres. It is also true that we are given no information about the masses of these spheres. We can find the largest possible value of Q by equating the electrostatic force between the charged spheres and the maximum force the cable can withstand. Using Coulomb’s law, express the electrostatic force between the two charged spheres: 2 2 l kQF = Express the tensile strength Stensile of steel in terms of the maximum force Fmax in the cable and the cross- sectional area of the cable and solve for F: A FS maxtensile = ⇒ tensilemax ASF = Equate these forces to obtain: tensile2 2 ASkQ = l Solve for Q: k ASQ tensilel= Substitute numerical values and evaluate Q: ( ) ( )( ) mC95.2 C/mN1099.8 N/m102.5m105.1m1 229 2824 = ⋅× ×× = − Q 21 •• Picture the Problem We can use Coulomb’s law to express the net force acting on the copper cube in terms of the unbalanced charge resulting from the assumed migration of half the charges to opposite sides of the cube. We can, in turn, find the unbalanced charge Qunbalanced from the number of copper atoms N and the number of electrons per atom. The Electric Field 1: Discrete Charge Distributions 9 Substitute for n to obtain: qkT PWE σ= (1) Substitute numerical values and evaluate E: ( )( )( )( ) ( )( )( ) N/C1041.2K300J/K1038.1C106.1 J/eV106.1eV1N/m10m10 6 2319 1925219 ×= ×× × = −− −− E (b) From equation (1) we see that: PE ∝ and 1−∝ TE i.e., E increases linearly with pressure and varies inversely with temperature. *23 •• Picture the Problem We can use Coulomb’s law to express the charge on the rod in terms of the force exerted on it by the soda can and its distance from the can. We can apply Newton’s 2nd law in rotational form to the can to relate its acceleration to the electric force exerted on it by the rod. Combining these equations will yield an expression for Q as a function of the mass of the can, its distance from the rod, and its acceleration. Use Coulomb’s law to relate the force on the rod to its charge Q and distance r from the soda can: 2 2 r kQF = Solve for Q to obtain: k FrQ 2 = (1) Apply to the can: ατ I=∑ mass ofcenter αIFR = Because the can rolls without slipping, we know that its linear acceleration a and angular acceleration α are related according to: R a =α where R is the radius of the soda can. Because the empty can is a hollow cylinder: 2MRI = where M is the mass of the can. Substitute for I and α and solve for F to obtain: Ma R aMRF == 2 2 Substitute for F in equation (1): k MarQ 2 = Chapter 21 10 Substitute numerical values and evaluate Q: ( ) ( )( ) nC141 C/mN1099.8 m/s1kg018.0m1.0 229 22 = ⋅× =Q 24 •• Picture the Problem Because the nucleus is in equilibrium, the binding force must be equal to the electrostatic force of repulsion between the protons. Apply 0=∑F r to a proton: 0ticelectrostabinding =− FF Solve for Fbinding: ticelectrostabinding FF = Using Coulomb’s law, substitute for Felectrostatic: 2 2 binding r kqF = Substitute numerical values and evaluate Felectrostatic: ( )( ) ( ) N230m10 C106.1C/mN1099.8 215 219229 binding = ×⋅× = − − F Electric Charge 25 • Picture the Problem The charge acquired by the plastic rod is an integral number of electronic charges, i.e., q = ne(−e). Relate the charge acquired by the plastic rod to the number of electrons transferred from the wool shirt: ( )enq −= e Solve for and evaluate ne: 12 19e 1000.5C101.6 C8.0 ×= ×− − = − = − µ e qn 26 • Picture the Problem One faraday = NAe. We can use this definition to find the number of coulombs in a faraday. Use the definition of a faraday to calculate the number of coulombs in a faraday: ( )( ) C1063.9C/electron106.1electrons1002.6faraday1 41923A ×=××== −eN The Electric Field 1: Discrete Charge Distributions 11 *27 • Picture the Problem We can find the number of coulombs of positive charge there are in 1 kg of carbon from , where nenQ C6= C is the number of atoms in 1 kg of carbon and the factor of 6 is present to account for the presence of 6 protons in each atom. We can find the number of atoms in 1kg of carbon by setting up a proportion relating Avogadro’s number, the mass of carbon, and the molecular mass of carbon to nC. Express the positive charge in terms of the electronic charge, the number of protons per atom, and the number of atoms in 1 kg of carbon: enQ C6= Using a proportion, relate the number of atoms in 1 kg of carbon nC, to Avogadro’s number and the molecular mass M of carbon: M m N n C A C = ⇒ M mNn CAC = Substitute to obtain: M emNQ CA6= Substitute numerical values and evaluate Q: ( )( )( ) C1082.4 kg/mol012.0 C101.6kg1atoms/mol106.026 71923 ×=××= − Q Coulomb’s Law 28 • Picture the Problem We can find the forces the two charges exert on each by applying Coulomb’s law and Newton’s 3rd law. Note that because the vector pointing from q ir ˆˆ 2,1 = 1 to q2 is in the positive x direction. (a) Use Coulomb’s law to express the force that q1 exerts on q2: 2,12 2,1 21 2,1 r̂F r qkq = r Substitute numerical values and evaluate 2,1F r : ( )( )( ) ( ) ( )iiF ˆmN0.24 m3 µC6µC4/CmN108.99 2 229 2,1 = ⋅× = r Chapter 21 14 Convert to : 4,3r r 4,3̂r ( ) ( ) ( ) ( ) ji ji r r r ˆ707.0ˆ707.0 m05.0m05.0 ˆm05.0ˆm05.0ˆ 22 4,3 4,3 4,3 += + + == r r Substitute numerical values and evaluate 4,3F r : ( )( ) ( ) ( ) ( ) ( )ji jiF ˆN1014.1ˆN1014.1 ˆ707.0ˆ707.0 m205.0 nC3nC3/CmN1099.8 55 2 229 4,3 −− ×−×−= + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −⋅×= r Substitute and simplify to find 4F r : ( ) ( ) ( ) ( ) ( ) ( )ji jiijF ˆN1010.2ˆN1010.2 ˆN1014.1ˆN1014.1ˆN1024.3ˆN1024.3 55 5555 4 −− −−−− ×+×= ×−×−×+×= r 31 •• Picture the Problem The configuration of the charges and the forces on q3 are shown in the figure … as is a coordinate system. From the geometry of the charge distribution it is evident that the net force on the 2 µC charge is in the negative y direction. We can apply Coulomb’s law to express 3,1F r and 3,2F r and then add them to find the net force on q3. The net force acting on q3 is given by: 3,23,13 FFF rrr += The Electric Field 1: Discrete Charge Distributions 15 Express the force that q1 exerts on q3: jiF ˆsinˆcos3,1 θθ FF −= r where ( )( )( ( ) ( ) ) N3.12 m0.08m0.03 C2C5C/mN1099.8 22 229 2 31 = + ⋅× = = µµ r qkqF and °=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = − 6.20 cm8 cm3tan 1θ Express the force that q2 exerts on q3: jiF ˆsinˆcos3,2 θθ FF −−= r Substitute for 3,1F r and 3,2F r and simplify to obtain: j j ijiF ˆsin2 ˆsin ˆcosˆsinˆcos3 θ θ θθθ F F FFF −= − −−= r Substitute numerical values and evaluate 3F r : ( ) ( ) j jF ˆN66.8 ˆ6.20sinN3.1223 −= °−= r *32 •• Picture the Problem The positions of the charges are shown in the diagram. It is apparent that the electron must be located along the line joining the two charges. Moreover, because it is negatively charged, it must be closer to the −2.5 µC than to the 6.0 µC charge, as is indicated in the figure. We can find the x and y coordinates of the electron’s position by equating the two electrostatic forces acting on it and solving for its distance from the origin. We can use similar triangles to express this radial distance in terms of the x and y coordinates of the electron. Express the condition that must be ee FF ,2,1 = Chapter 21 16 satisfied if the electron is to be in equilibrium: Express the magnitude of the force that q1 exerts on the electron: ( )2 1 ,1 m25.1+ = r ekqF e Express the magnitude of the force that q2 exerts on the electron: 2 2 ,2 r eqk F e = Substitute and simplify to obtain: ( ) 2 2 2 1 m25.1 r q r q = + Substitute for q1 and q2 and simplify: ( ) ( ) 0m25.1 m2361.2m4.1 122 =+ +− −− rr Solve for r to obtain: m0.4386 and m036.2 −= = r r Because r < 0 is unphysical, we’ll consider only the positive root. Use the similar triangles in the diagram to establish the proportion involving the y coordinate of the electron: m1.12 m2.036 m5.0 =e y Solve for ye: m909.0=ey Use the similar triangles in the diagram to establish the proportion involving the x coordinate of the electron: m1.12 m2.036 m1 =e x Solve for xe: m82.1=ex The coordinates of the electron’s position are: ( ) ( )m0.909m,1.82, −−=ee yx The Electric Field 1: Discrete Charge Distributions 19 Substitute numerical values to obtain: ( ) ( )( ) ( ) ( ) ( ) ( ) ⎥ ⎦ ⎤ +⎢ ⎣ ⎡ ⋅× =− ii i ˆm04.0 m04.0 ˆm08.0 m0.08 C5C2/CmN1099.8 ˆN7.19 3 2 3 229 Qµµ Solve for and evaluate Q2: C00.32 µ−=Q Remarks: An alternative solution is to equate the electrostatic forces acting on q2 when it is at (17.75 cm, 0). 35 •• Picture the Problem By considering the symmetry of the array of charges we can see that the y component of the force on q is zero. We can apply Coulomb’s law and the principle of superposition of forces to find the net force acting on q. Express the net force acting on q: qQqxQq ,45ataxis,on 2 °+= FFF rrr Express the force on q due to the charge Q on the x axis: iF ˆ2axis,on R kqQ qxQ = r Express the net force on q due to the charges at 45°: i iF ˆ 2 2 ˆ45cos22 2 2,45at R kqQ R kqQ qQ = °=° r Substitute to obtain: i iiF ˆ 2 21 ˆ 2 2ˆ 2 22 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ += += R kqQ R kqQ R kqQ q r 36 ••• Picture the P oblem Let the Hr + ions be in the x-y plane with H1 at (0, 0, 0), H2 at (a, 0, 0), and H3 at ( )0,23,2 aa . The N−3 ion, q4 in our notation, is then at ( )32,32,2 aaa where a =1.64×10−10 m. To simplify our calculations we’ll set N1056.8 922 −×== Cake . We can apply Coulomb’s law and the principle of superposition of forces to find the net force acting on each ion. Chapter 21 20 Express the net force acting on q1: 1,41,31,21 FFFF rrrr ++= Find 1,2F r : ( ) iirF ˆˆˆ 1,22 1,2 21 1,2 CCr qkq −=−== r Find 1,3F r : ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +−= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = = ji ji rF ˆ 2 3ˆ 2 1 ˆ 2 30ˆ 2 0 ˆ 1,32 1,3 13 1,3 C a aa C r qkqr Noting that the magnitude of q4 is three times that of the other charges and that it is negative, express 1,4F r : ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛+⎟ ⎠ ⎞ ⎜ ⎝ ⎛= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − −== kji kji kji rF ˆ 3 2ˆ 32 1ˆ 2 13 ˆ 3 2ˆ 32 ˆ 2 3 3 2 322 ˆ 3 20ˆ 32 0ˆ 2 0 3ˆ3 222 1,41,4 C a aaa C aaa aaa CC r The Electric Field 1: Discrete Charge Distributions 21 Substitute to find : 1F r k kji jiiF ˆ6 ˆ 3 2ˆ 32 1ˆ 2 13 ˆ 2 3ˆ 2 1ˆ 1 C C CC = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛+⎟ ⎠ ⎞ ⎜ ⎝ ⎛+ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +−−= r From symmetry considerations: kFFF ˆ6132 C=== rrr Express the condition that molecule is in equilibrium: 04321 =+++ FFFF rrrr Solve for and evaluate : 4F r ( ) k FFFFF ˆ63 3 13214 C−= −=++−= rrrrr The Electric Field *37 • Picture the Problem Let q represent the charge at the origin and use Coulomb’s law for E r due to a point charge to find the electric field at x = 6 m and −10 m. (a) Express the electric field at a point P located a distance x from a charge q: ( ) P,02 r̂E x kqx = r Evaluate this expression for x = 6 m: ( ) ( )( ) ( ) ( )i iE ˆN/C999 ˆ m6 C4/CmN1099.8m6 2 229 = ⋅× = µr (b) Evaluate E r at x = −10 m: ( ) ( )( ) ( ) ( ) ( )iiE ˆN/C360ˆ m10 C4/CmN1099.8m10 2 229 −=− ⋅× =− µr (c) The following graph was plotted using a spreadsheet program: Chapter 21 24 (c) Apply Coulomb’s law to obtain: ( ) ( ) ( ) ( ) jj ˆmN60.1ˆ m0.03 nC4 2 −=− −kq Solve for and evaluate q: ( )( ) ( )( ) nC40.0 nC4/CmN1099.8 m0.03mN60.1 229 2 −= ⋅× −=q 40 • Picture the Problem We can compare the electric and gravitational forces acting on an electron by expressing their ratio. We can equate these forces to find the charge that would have to be placed on a penny in order to balance the earth’s gravitational force on it. (a) Express the magnitude of the electric force acting on the electron: eEFe = Express the magnitude of the gravitational force acting on the electron: gmF eg = Express the ratio of these forces to obtain: mg eE F F g e = Substitute numerical values and evaluate Fe/Fg: ( )( ) ( )( ) 12 231 19 1069.2 m/s9.81kg109.11 N/C150C101.6 ×= × × = − − g e F F or ( ) ge FF 121069.2 ×= , i.e., the electric force is greater by a factor of 2.69×1012. (b) Equate the electric and gravitational forces acting on the penny and solve for q to obtain: E mgq = Substitute numerical values and evaluate q: ( )( ) C1096.1 N/C150 m/s9.81kg103 4 23 − − ×= × =q The Electric Field 1: Discrete Charge Distributions 25 41 •• Picture the Problem The diagram shows the locations of the charges q1 and q2 and the point on the x axis at which we are to find .E r From symmetry considerations we can conclude that the y component of E r at any point on the x axis is zero. We can use Coulomb’s law for the electric field due to point charges to find the field at any point on the x axis and to find the force on a charge q EF rr q= 0 placed on the x axis at x = 4 cm. (a) Letting q = q1 = q2, express the x- component of the electric field due to one charge as a function of the distance r from either charge to the point of interest: iE ˆcos2 θr kq x = r Express for both charges: xE r iE ˆcos2 2 θr kq x = r Substitute for cosθ and r, substitute numerical values, and evaluate to obtain: ( ) ( )( )( ) ( ) ( )[ ] ( )i iiiE ˆkN/C4.53 ˆ m0.04m0.03 m0.04nC6/CmN108.992ˆm04.02ˆm04.02 2322 229 32 = + ⋅× === r kq rr kq x r (b) Apply to find the force on a charge q EF rr q= 0 placed on the x axis at x = 4 cm: ( )( ) ( )i iF ˆN0.69 ˆkN/C4.53nC2 µ= = r *42 •• Picture the Problem If the electric field at x = 0 is zero, both its x and y components must be zero. The only way this condition can be satisfied with the point charges of +5.0 µC and −8.0 µC are on the x axis is if the point charge of +6.0 µC is also on the x axis. Let the subscripts 5, −8, and 6 identify the point charges and their fields. We can use Coulomb’s law for E r due to a point charge and the principle of superposition for fields to determine where the +6.0 µC charge should be located so that the electric field at x = 0 is zero. Chapter 21 26 Express the electric field at x = 0 in terms of the fields due to the charges of +5.0 µC, −8.0 µC, and +6.0 µC: ( ) 0 0 C6C8C5 = ++= − µµµ EEEE rrrr Substitute for each of the fields to obtain: 0ˆˆˆ 82 8 8 62 6 6 52 5 5 =++ − − − rrr r kq r kq r kq or ( ) ( ) 0ˆˆˆ 2 8 8 2 6 6 2 5 5 =−+−+ − − iii r kq r kq r kq Divide out the unit vector to obtain: î 02 8 8 2 6 6 2 5 5 =−− − − r q r q r q Substitute numerical values to obtain: ( ) ( ) 0 cm4 86 cm3 5 22 6 2 = − −− r Solve for r6: cm38.26 =r 43 •• Picture the Problem The diagram shows the electric field vectors at the point of interest P due to the two charges. We can use Coulomb’s law for E r due to point charges and the superposition principle for electric fields to find PE r . We can apply EF rr q= to find the force on an electron at (−1 m, 0). (a) Express the electric field at (−1 m, 0) due to the charges q1 and q2: 21P EEE rrr += The Electric Field 1: Discrete Charge Distributions 29 and ( ) 2322 2 ax kqxEx + = (b) For a,x << x2 + a2 ≈ a2, so: ( ) 3232 22 a kqx a kqxEx =≈ For a,x >> x2 + a2 ≈ x2, so: ( ) 2232 22 x kq x kqxEx =≈ (c) .2by given be wouldfield Its .2 magnitude of charge single a be appear to wouldby separated charges the, For 2x kqEq aax x = >> Factor the radicand to obtain: 23 2 2 2 12 − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ += x axkqxEx For a << x: 11 2 2 ≈+ x a and [ ] 2232 22 x kqxkqxEx == − *45 •• Picture the Problem The diagram shows the electric field vectors at the point of interest P due to the two charges. We can use Coulomb’s law for E r due to point charges and the superposition principle for electric fields to find PE r . We can apply EF rr q= to find the force on a proton at (−3 m, 1 m). Chapter 21 30 (a) Express the electric field at (−3 m, 1 m) due to the charges q1 and q2: 21P EEE rrr += Evaluate : 1E r ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( jiji jirE ˆkN/C544.0ˆkN/C908.0ˆ514.0ˆ857.0kN/C06.1 m3m5 ˆm3ˆm5 m3m5 C4/CmN1099.8ˆ 2222 229 P1,2 1.P 1 1 −+=+−−= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + +− + −⋅× == µ r kqr ) Evaluate 2E r : ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( jiji jirE ˆkN/C01.1ˆkN/C01.2ˆ447.0ˆ894.0kN/C25.2 m2m4 ˆm2ˆm4 m2m4 C5/CmN1099.8ˆ 2222 229 P2,2 P2, 2 2 −+−=−−= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + −+− + ⋅× == µ r kqr ) Substitute and simplify to find PE r : ( ) ( ) ( ) ( ) ( ) ( ) ji jijiE ˆkN/C55.1ˆkN/C10.1 ˆkN/C01.1ˆkN/C01.2ˆkN/C544.0ˆkN/C908.0P −+−= −+−+−+= r The magnitude of is: PE r ( ) ( ) kN/C90.1 kN/C55.1kN/C10.1 22P = +=E The Electric Field 1: Discrete Charge Distributions 31 The direction of is: PE r °=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − = − 235 kN/C10.1 kN/C55.1tan 1Eθ Note that the angle returned by your calculator for ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − −− kN/C10.1 kN/C55.1tan 1 is the reference angle and must be increased by 180° to yield θE. (b) Express and evaluate the force on a proton at point P: ( ) ( ) ( )[ ] ( ) ( )ji jiEF ˆN1048.2ˆN10.761 ˆkN/C55.1ˆkN/C10.1C106.1 1616 19 P −− − ×−+×−= −+−×== rr q The magnitude of F r is: ( ) ( ) N1004.3N1048.2N10.761 16216216 −−− ×=×−+×−=F The direction of F r is: °=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ×− ×− = − − − 235 N1076.1 N1048.2tan 16 16 1 Fθ where, as noted above, the angle returned by your calculator for ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ×− ×− − − − N1076.1 N1048.2tan 16 16 1 is the reference angle and must be increased by 180° to yield θE. 46 •• Picture the Problem In Problem 44 it is shown that the electric field on the x axis, due to equal positive charges located at (0, a) and (0,−a), is given by ( ) .2 2322 −+= axkqxEx We can identify the locations at which Ex has it greatest values by setting dEx/dx equal to zero. (a) Evaluate dx dEx : Chapter 21 34 Factor a2 from the denominator to obtain: 23 2 2 2 2 1 2 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + −= a xa xkqFx For x << a: x a kqFx 3 22 −= i.e., the bead experiences a linear restoring force. (b) Express the period of a simple harmonic oscillator: k' mT π2= Obtain k′ from our result in part (a): 3 22 a kqk' = Substitute to obtain: 2 3 3 2 2 2 2 2 kq ma a kq mT ππ == Motion of Point Charges in Electric Fields 49 • Picture the Problem We can use Newton’s 2nd law of motion to find the acceleration of the electron in the uniform electric field and constant-acceleration equations to find the time required for it to reach a speed of 0.01c and the distance it travels while acquiring this speed. (a) Use data found at the back of your text to compute e/m for an electron: C/kg1076.1 kg109.11 C106.1 11 31 19 ×= × × = − − em e (b) Apply Newton’s 2nd law to relate the acceleration of the electron to the electric field: ee m eE m Fa == net Substitute numerical values and evaluate a: ( )( ) 213 31 19 m/s1076.1 kg109.11 N/C100C101.6 ×= × × = − − a The Electric Field 1: Discrete Charge Distributions 35 field. electric theopposite iselectron an of onaccelerati theofdirection The (c) Using the definition of acceleration, relate the time required for an electron to reach 0.01c to its acceleration: a c a vt 01.0==∆ Substitute numerical values and evaluate ∆t: ( ) s170.0 m/s101.76 m/s1030.01 213 8 µ= × × =∆t (d) Find the distance the electron travels from its average speed and the elapsed time: ( )[ ]( ) cm5.25 s170.0m/s10301.00 821 av = ×+= ∆=∆ µ tvx *50 • Picture the Problem We can use Newton’s 2nd law of motion to find the acceleration of the proton in the uniform electric field and constant-acceleration equations to find the time required for it to reach a speed of 0.01c and the distance it travels while acquiring this speed. (a) Use data found at the back of your text to compute e/m for an electron: C/kg1058.9 kg1067.1 C106.1 7 27 19 ×= × × = − − pm e Apply Newton’s 2nd law to relate the acceleration of the electron to the electric field: pp m eE m Fa == net Substitute numerical values and evaluate a: ( )( ) 29 72 19 m/s1058.9 kg1067.1 N/C100C101.6 ×= × × = − − a field. electric the ofdirection in the isproton a of onaccelerati theofdirection The Chapter 21 36 (b) Using the definition of acceleration, relate the time required for an electron to reach 0.01c to its acceleration: a c a vt 01.0==∆ Substitute numerical values and evaluate ∆t: ( ) s313 m/s1058.9 m/s1030.01 29 8 µ= × × =∆t 51 • Picture the Problem The electric force acting on the electron is opposite the direction of the electric field. We can apply Newton’s 2nd law to find the electron’s acceleration and use constant acceleration equations to find how long it takes the electron to travel a given distance and its deflection during this interval of time. (a) Use Newton’s 2nd law to relate the acceleration of the electron first to the net force acting on it and then the electric field in which it finds itself: ee m e m EFa rr r − == net Substitute numerical values and evaluate a : r ( ) ( )j ja ˆm/s1003.7 ˆN/C400 kg109.11 C101.6 213 31 19 ×−= × × −= − −r (b) Relate the time to travel a given distance in the x direction to the electron’s speed in the x direction: ns0.50 m/s102 m0.1 6 =× = ∆ =∆ xv xt (c) Using a constant-acceleration equation, relate the displacement of the electron to its acceleration and the elapsed time: ( ) ( )( ) ( ) j j ay ˆcm79.8 ˆns50m/s1003.7 221321 2 2 1 −= ×−= ∆=∆ ty rr i.e., the electron is deflected 8.79 cm downward. 52 •• Picture the Problem Because the electric field is uniform, the acceleration of the electron will be constant and we can apply Newton’s 2nd law to find its acceleration and use a constant-acceleration equation to find its speed as it leaves the region in which there is a uniform electric field. Using a constant-acceleration xavv ∆+= 220 2 The Electric Field 1: Discrete Charge Distributions 39 55 •• Picture the Problem We can use constant-acceleration equations to express the x and y coordinates of the electron in terms of the parameter t and Newton’s 2nd law to express the constant acceleration in terms of the electric field. Eliminating the parameter will yield an equation for y as a function of x, q, and m. We can decide whether the electron will strike the upper plate by finding the maximum value of its y coordinate. Should we find that it does not strike the upper plate, we can determine where it strikes the lower plate by setting y(x) = 0. Express the x and y coordinates of the electron as functions of time: ( )tvx θcos0= and ( ) 2210 sin tatvy y−= θ Apply Newton’s 2nd law to relate the acceleration of the electron to the net force acting on it: e y e y y m eE m F a == net, Substitute in the y-coordinate equation to obtain: ( ) 20 2sin tm eE tvy e y−= θ Eliminate the parameter t between the two equations to obtain: ( ) ( ) 222 0 cos2 tan x vm eE xxy e y θ θ −= (1) To find ymax, set dy/dx = 0 for extrema: extremafor0 cos tan 22 0 = −= x' vm eE dx dy e y θ θ Solve for x′ to obtain: y e eE vmx' 2 2sin20 θ= (See remark below.) Substitute x′ in y(x) and simplify to obtain ymax: y e eE vmy 2 sin220 max θ = Substitute numerical values and evaluate ymax: ( )( ) ( )( ) cm02.1N/C103.5C101.62 54sinm/s105kg109.11 319 22631 max =×× °×× = − − y and, because the plates are separated by 2 cm, the electron does not strike the upper plate. Chapter 21 40 To determine where the electron will strike the lower plate, set y = 0 in equation (1) and solve for x to obtain: y e eE vmx θ2sin 2 0= Substitute numerical values and evaluate x: ( )( ) ( )( ) cm07.4N/C105.3C106.1 90sinm/s105kg1011.9 319 2631 = ×× °×× = − − x Remarks: x′ is an extremum, i.e., either a maximum or a minimum. To show that it is a maximum we need to show that d2y/dx2, evaluated at x′, is negative. A simple alternative is to use your graphing calculator to show that the graph of y(x) is a maximum at x′. Yet another alternative is to recognize that, because equation (1) is quadratic and the coefficient of x2 is negative, its graph is a parabola that opens downward. 56 •• Picture the Problem The trajectory of the electron while it is in the electric field is parabolic (its acceleration is downward and constant) and its trajectory, once it is out of the electric field is, if we ignore the small gravitational force acting on it, linear. We can use constant-acceleration equations and Newton’s 2nd law to express the electron’s x and y coordinates parametrically and then eliminate the parameter t to express y(x). We can find the angle with the horizontal at which the electron leaves the electric field from the x and y components of its velocity and its total vertical deflection by summing its deflections over the first 4 cm and the final 12 cm of its flight. (a) Using a constant-acceleration equation, express the x and y coordinates of the electron as functions of time: ( ) tvtx 0= and ( ) 221,0 tatvty yy += Because v0,y = 0: ( ) tvtx 0= (1) and ( ) 221 taty y= Using Newton’s 2nd law, relate the acceleration of the electron to the electric field: e y e y m eE m Fa − == net The Electric Field 1: Discrete Charge Distributions 41 Substitute to obtain: ( ) 2 2 t m eE ty e y−= (2) Eliminate the parameter t between equations (1) and (2) to obtain: ( ) 222 0 42 x K eE x vm eE xy y e y −=−= Substitute numerical values and evaluate y(4 cm): ( ) ( )( )( )( ) mm40.6J1024 m0.04N/C102C101.6m04.0 16 2419 −= × ×× −= − − y (b) Express the horizontal and vertical components of the electron’s speed as it leaves the electric field: θcos0vvx = and θsin0vvy = Divide the second of these equations by the first to obtain: 0 11 tantan v v v v y x y −− ==θ Using a constant-acceleration equation, express vy as a function of the electron’s acceleration and its time in the electric field: tavv yyy += ,0 or, because v0,y = 0 0 net, v x m eE t m F tav e y e y yy −=== Substitute to obtain: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −= −− K xeE vm xeE y e y 2 tantan 12 0 1θ Substitute numerical values and evaluate θ : ( )( )( ) ( ) °−=⎥⎦ ⎤ ⎢ ⎣ ⎡ × ×× −= − − − 7.17 J1022 m0.04N/C102C101.6tan 16 419 1θ (c) Express the total vertical displacement of the electron: cm12cm4total yyy += Relate the horizontal and vertical distances traveled to the screen to the horizontal and vertical components of its velocity: tvx x∆= and tvy y∆= Chapter 21 44 p falls off as 1/r3. 60 •• Picture the Problem We can use its definition to find the molecule’s dipole moment. From the symmetry of the system, it is evident that the x component of the dipole moment is zero. Using its definition, express the molecule’s dipole moment: jip ˆˆ yx pp += r From symmetry considerations we have: 0=xp The y component of the molecule’s dipole moment is: ( )( ) mC1086.1 nm0.058C101.62 2 29 19 ⋅×= ×= == − − eLqLpy Substitute to obtain: ( )jp ˆmC1086.1 29 ⋅×= −r 61 •• Picture the Problem We can express the net force on the dipole as the sum of the forces acting on the two charges that constitute the dipole and simplify this expression to show that We can show that, under the given conditions, is also given by .ˆnet iF Cp= r netF r ( ) îpdxdEx by differentiating the dipole’s potential energy function with respect to x. (a) Express the net force acting on the dipole: qq +− += FFF rrr net Apply Coulomb’s law to express the forces on the two charges: ( )iEF ˆ1 axqCqq −−=−=− rr and ( )iEF ˆ1 axqCqq +=+=+ rr Substitute to obtain: ( ) ( ) ii iiF ˆˆ2 ˆˆ 11net CpaqC axqCaxqC == ++−−= r where p = 2aq. The Electric Field 1: Discrete Charge Distributions 45 (b) Express the net force acting on the dipole as the spatial derivative of U: [ ] i iiF ˆ ˆˆ net dx dEp Ep dx d dx dU x x xx = −−=−= r 62 ••• Picture the Problem We can express the force exerted on the dipole by the electric field as −dU/dr and the potential energy of the dipole as −pE. Because the field is due to a point charge, we can use Coulomb’s law to express E. In the second part of the problem, we can use Newton’s 3rd law to show that the magnitude of the electric field of the dipole along the line of the dipole a distance r away is approximately 2kp/r3. (a) Express the force exerted by the electric field of the point charge on the dipole: rF ˆ dr dU −= r where is a unit radial vector pointing from Q toward the dipole. r̂ Express the potential energy of the dipole in the electric field: 2r kQppEU −=−= Substitute to obtain: rrF ˆ2ˆ 32 r kQp r kQp dr d −=⎥⎦ ⎤ ⎢⎣ ⎡−−= r (b) Using Newton’s 3rd law, express the force that the dipole exerts on the charge Q at the origin: FF rr −=Qon or rr ˆˆon FF Q −= and FF Q =on Express in terms of the field in which Q finds itself: QFon QEF Q =on Substitute to obtain: 3 2 r kQpQE = ⇒ 3 2 r kpE = General Problems *63 • Picture the Problem We can equate the gravitational force and the electric force acting on a proton to find the mass of the proton under the given condition. (a) Express the condition that must be satisfied if the net force on the eg FF = Chapter 21 46 proton is zero: Use Newton’s law of gravity and Coulomb’s law to substitute for Fg and Fe: 2 2 2 2 r ke r Gm = Solve for m to obtain: G kem = Substitute numerical values and evaluate m: ( ) kg1086.1 kg/mN1067.6 C/mN1099.8C106.1 92211 229 19 − − − ×= ⋅× ⋅× ×=m (b) Express the ratio of Fe and Fg: 2 p 2 2 2 p 2 2 Gm ke r Gm r ke = Substitute numerical values to obtain: ( )( ) ( )( ) 36 2272211 219229 2 p 2 1024.1 kg1067.1kg/mN1067.6 C106.1C/mN1099.8 ×= ×⋅× ×⋅× = −− − Gm ke 64 •• Picture the Problem The locations of the charges q1, q2 and q2 and the points at which we are calculate the field are shown in the diagram. From the diagram it is evident that E r along the axis has no y component. We can use Coulomb’s law for E r due to a point charge and the superposition principle to find E r at points P1 and P2. Examining the distribution of the charges we can see that there are two points where E = 0. One is between q2 and q3 and the other is to the left of q1. The Electric Field 1: Discrete Charge Distributions 49 electrons: Relate N to Avogadro’s number, the mass of the copper penny, and the molecular mass of copper: M m N N = A ⇒ M mNN A= Relate ne to the free charge Q to be removed from the penny: [ ]enQ −= e ⇒ e Qn − =e A A meN QM M mN e Q f −=−= Substitute numerical values and evaluate f: ( )( ) ( )( )( ) %1029.31029.3mol106.02C101.6g3 g/mol5.63C15 79 12319 −− −− ×=×=×× − −= µf (b) Use Coulomb’s law to express the force of repulsion between the two pennies: ( ) 2 2 e 2 2 r enk r kqF == Substitute numerical values and evaluate F: ( )( ) ( ) ( ) N4.32 m25.0 C106.11038.9/CmN1099.8 2 219213229 = ××⋅× = − F 67 •• Picture the Problem Knowing the total charge of the two charges, we can use Coulomb’s law to find the two combinations of charge that will satisfy the condition that both are positive and hence repel each other. If just one charge is positive, then there is just one distribution of charge that will satisfy the conditions that the force is attractive and the sum of the two charges is 6 µC. (a) Use Coulomb’s law to express the repulsive force each charge exerts on the other: 2 2,1 21 r qkqF = Express q2 in terms of the total charge and q1: 12 qQq −= Chapter 21 50 Substitute to obtain: ( ) 2 2,1 11 r qQkqF −= Substitute numerical values to obtain: ( ) ( )[ ] ( )2 2 11 229 m3 C6/CmN108.99mN8 qq −⋅×= µ Simplify to obtain: ( ) ( ) 0C01.8C6 2121 =+−+ µµ qq Solve to obtain: C01.2andC99.3 21 µµ == qq or C99.3andC01.2 21 µµ == qq (b) Use Coulomb’s law to express the attractive force each charge exerts on the other: 2 2,1 21 r qkqF −= Proceed as in (a) to obtain: ( ) ( ) 0C01.8C6 2121 =−−+ µµ qq Solve to obtain: C12.1andC12.7 21 µµ −== qq 68 •• Picture the Problem The electrostatic forces between the charges are responsible for the tensions in the strings. We’ll assume that these are point charges and apply Coulomb’s law and the principle of the superposition of forces to find the tension in each string. Use Coulomb’s law to express the net force on the charge +q: qq FFT 421 += Substitute and simplify to obtain: ( ) ( ) ( ) 2 2 221 3 2 42 d kq d qkq d qkqT =+= Use Coulomb’s law to express the net force on the charge +4q: qq FFT 22 += Substitute and simplify to obtain: ( )( ) ( ) ( ) 2 2 222 9 2 442 d kq d qkq d qqkT =+= The Electric Field 1: Discrete Charge Distributions 51 *69 •• Picture the Problem We can use Coulomb’s law to express the force exerted on one charge by the other and then set the derivative of this expression equal to zero to find the distribution of the charge that maximizes this force. Using Coulomb’s law, express the force that either charge exerts on the other: 2 21 D qkqF = Express q2 in terms of Q and q1: 12 qQq −= Substitute to obtain: ( ) 2 11 D qQkqF −= Differentiate F with respect to q1 and set this derivative equal to zero for extreme values: ( )[ ] ( )[ ] extremafor0 1 112 11 1 2 1 = −+−= −= qQq D k qQq dq d D k dq dF Solve for q1 to obtain: Qq 211 = and QqQq 2112 =−= To determine whether a maximum or a minimum exists at Qq 211 = , differentiate F a second time and evaluate this derivative at Qq 211 = : [ ] ( ) . oftly independen 0 2 2 1 2 1 1 22 1 2 q D k qQ dq d D k dq Fd < −= −= . maximizes 2121 FQqq ==∴ *70 •• Picture the Problem We can apply Coulomb’s law and the superposition of forces to relate the net force acting on the charge q = −2 µC to x. Because Q divides out of our equation when F(x) = 0, we’ll substitute the data given for x = 8.0 cm. Using Coulomb’s law, express the net force on q as a function of x: ( ) ( ) ( )22 cm12 4 x Qkq x kqQxF − +−= Chapter 21 54 meter stick, and the mass required to maintain equilibrium when it is located either 25 cm to the right or to the left of the mid-point of the rigid stick. (a) Using Coulomb’s law, express the electric force between the two charges: 2 21 d qkqF = Substitute numerical values and evaluate F: ( )( ) ( ) N225.0 m1.0 C105C/mN1099.8 2 27229 = ×⋅× = − F (b) Apply the definition of torque to obtain: lF=τ Substitute numerical values and evaluate τ: ( )( ) ckwisecounterclo m,N113.0 m5.0N225.0 ⋅= =τ (c) Apply 0stickmeter theofcenter =∑τ to the meterstick: 0=− 'mglτ Solve for m: 'g m l τ = Substitute numerical values and evaluate m: ( )( ) kg0461.0m25.0m/s81.9 N113.0 2 ==m (d) Apply 0stickmeter theofcenter =∑τ to the meterstick: 0=+− 'mglτ Substitute for τ: 0=+− 'mgF ll Substitute for F: 02 21 =+− 'mg d 'qkq l where q′ is the required charge. Solve for q2′ to obtain: l l 1 2 2 kq 'mgdq = Substitute numerical values and evaluate q2′: ( ) ( )( )( ) ( )( )( ) C1003.5m5.0C105C/mN108.99 m25.0m/s81.9kg0461.0m1.0 7 7229 22 2 − − ×=×⋅× ='q The Electric Field 1: Discrete Charge Distributions 55 ic of forces to express the field at the rigin and use this equation to solve for Q. xpress the electric field at the origin due to the point charges Q: 74 •• Picture the Problem Let the numeral 1 refer to the charge in the 1st quadrant and the numeral 2 to the charge in the 4th quadrant. We can use Coulomb’s law for the electr field due to a point charge and the superposition o E ( ) ( ) ( )[ ] ( ) ( )[ ] ( ) i ijiji rrEEE ˆ ˆm8ˆm2ˆm4ˆm2ˆm4 ˆˆ0,0 333 0,22 0,2 0,12 0,1 21 xE= r kQ r kQ r kQ r kQ r kQ −=+−+−+−= +=+= rrr r is the distance from each charge to the origin and where ( ) 3 m8 r kQEx −= . Express r in terms of the coordinates , y) of the point charges: (x 22 yxr += Substitute to obtain: ( ) ( ) 2322 m8 yx kQEx + −= Solve for Q to obtain: ( ) ( )m8 2322 k yxEQ x += merical values and evaluate Q: Substitute nu ( ) ( ) ( )[ ] ( )( ) C97.4 m8/CmN108.99 m2m4kN/C4 229 2322 µ−= ⋅× + −=Q 75 •• Picture the Problem Let the numeral 1 denote one of the spheres and the numeral 2 the other. Knowing the total charge Q on the two spheres, we can use Coulomb’s law to fin the charge on each of them. A second application of Coulomb’s law when the spheres d arry the same charge and are 0.60 m apart will yield the force each exerts on the other. ss ach charge xerts on the other: c (a) Use Coulomb’s law to expre the repulsive force e e 2 2,1 21 r qkqF = Chapter 21 56 q2 in terms of the total charge nd q1: Express a 12 qQq −= Substitute to obtain: ( ) 2 2,1 11 r qQkqF −= ubstitute numerical values to obtain: S ( ) ( )[ ] ( )2 2 11 229 m6. C200/CmN108.99N201 qq −⋅×= µ implify to obtain the quadratic equation: 0 S ( ) ( ) 0C4810C200 2121 =+−+ µµ qq Solve to obtain: C172andC0.28 21q µµ == q or C0.28andC172 21 µµ == qq ss arge when 1 = q2 = 100 µC: (b) Use Coulomb’s law to expre the repulsive force each ch exerts on the other q 2 2,1 21 r qkqF = Substitute numerical values and evaluate F: ( )( ) ( ) N250 m6.0 C100/CmN108.99 2 2 229 =⋅×= µF 76 •• Picture the Problem Let the numeral 1 denote one of the spheres and the numeral 2 the other. Knowing the total charge Q on the two spheres, we can use Coulomb’s law to fin the charge on each of them. A second application of Coulomb’s law when the spheres d arry the same charge and are 0.60 m apart will yield the force each exerts on the other. s ach charge xerts on the other: c (a) Use Coulomb’s law to expres the attractive force e e 2 2,1 21 r qkqF −= rms of the total harge and q1: Express q2 in te c 12 qQq −= The Electric Field 1: Discrete Charge Distributions 59 *78 •• Picture the Problem Each sphere is in static equilibrium under the influence of the tensionT r , the gravitational force gF r , and the electric force . We can use Coulomb’s law to relate the electric force to the charge on each sphere and their separation and the conditions for static equilibrium to relate these forces to the charge on each sphere. EF r (a) Apply the conditions for static equilibrium to the charged sphere: 0sinsin 2 2 E =−=−=∑ θθ Tr kqTFFx and ∑ =−= 0cos mgTFy θ Eliminate T between these equations to obtain: 2 2 tan mgr kq =θ Solve for q: k mgrq θtan= Referring to the figure, relate the separation of the spheres r to the length of the pendulum L: θsin2Lr = Substitute to obtain: k mgLq θθ tansin2= (b) Evaluate q for m = 10 g, L = 50 cm, and θ = 10°: ( ) ( )( ) C241.0 /CmN1099.8 10tanm/s81.9kg01.010sinm5.02 229 2 µ= ⋅× ° °=q Chapter 21 60 79 •• Picture the Problem Each sphere is in static equilibrium under the influence of the tensionT r , the gravitational force gF r , and the electric force . We can use Coulomb’s law to relate the electric force to the charge on each sphere and their separation and the conditions for static equilibrium to relate these forces to the charge on each sphere. EF r (a)Apply the conditions for static equilibrium to the charged sphere: 0sinsin 2 2 E =−=−=∑ θθ Tr kqTFFx and 0cos =−=∑ mgTFy θ Eliminate T between these equations to obtain: 2 2 tan mgr kq =θ Referring to the figure for Problem 80, relate the separation of the spheres r to the length of the pendulum L: θsin2Lr = Substitute to obtain: θ θ 22 2 sin4 tan mgL kq = or 2 2 2 4 tansin mgL kq =θθ (1) Substitute numerical values and evaluate : θθ tansin 2 ( )( ) ( )( )( ) 3 22 2229 2 1073.5 m1.5m/s9.81kg0.014 C75.0/CmN1099.8tansin −×=⋅×= µθθ Because : 1tansin 2 <<θθ θθθ ≈≈ tansin and 33 1073.5 −×≈θ Solve for θ to obtain: °== 3.10rad179.0θ The Electric Field 1: Discrete Charge Distributions 61 (b) Evaluate equation (1) with replacing q2 with q1q2: ( )( )( ) ( )( )( ) 33 22 229 2 1009.5 m1.5m/s9.81kg0.014 C1C5.0/CmN1099.8tansin θµµθθ ≈×=⋅×= − Solve for θ to obtain: °== 86.9rad172.0θ 80 •• Picture the Problem Let the origin be at the lower left-hand corner and designate the charges as shown in the diagram. We can apply Coulomb’s law for point charges to find the forces exerted on q1 by q2, q3, and q4 and superimpose these forces to find the net force exerted on q1. In part (b), we’ll use Coulomb’s law for the electric field due to a point charge and the superposition of fields to find the electric field at point P(0, L/2). (a) Using the superposition of forces, express the net force exerted on q1: 1,41,31,21 FFFF rrrr ++= Apply Coulomb’s law to express 1,2F r : ( ) ( ) jj rrF ˆˆ ˆ 2 2 3 1,23 1,2 12 1,22 1,2 12 1,2 L kqL L qqk r qkq r qkq =− − = == rr Apply Coulomb’s law to express 1,4F r : ( ) ( ) ii rrF ˆˆ ˆ 2 2 3 1,43 1,4 14 1,42 1,4 14 1,4 L kqL L qqk r qkq r qkq =− − = == rr Apply Coulomb’s law to express 1,3F r : ( ) ( )ji ji rrF ˆˆ 2 ˆˆ 2 ˆ 223 2 323 2 1,33 1,3 13 1,32 1,3 13 1,3 +−= −−= == L kq LL L kq r qkq r qkq rr Chapter 21 64 Using conservation of energy, relate the initial potential energy of the dumbbell to its kinetic energy when it is momentarily aligned with the electric field: 0=∆+∆ UK or, because Ki = 0, 0=∆+ UK where K is the kinetic energy when it is aligned with the field. Express the change in the potential energy of the dumbbell as it aligns with the electric field in terms of its dipole moment, the electric field, and the angle through which it rotates: ( )160cos coscos ff if −°= +−= −=∆ qaE pEpE UUU θθ Substitute to obtain: ( ) 0160cos =−°+ qaEK Solve for q: ( )°−= 60cos1aE Kq Substitute numerical values and evaluate q: ( )( )( ) C55.6 60cos1N/C600m0.3 J105 3 µ= °− × = − q *83 •• Picture the Problem The forces the electron and the proton exert on each other constitute an action-and-reaction pair. Because the magnitudes of their charges are equal and their masses are the same, we find the speed of each particle by finding the speed of either one. We’ll apply Coulomb’s force law for point charges and Newton’s 2nd law to relate v to e, m, k, and r. Apply Newton’s 2nd law to the positron: r vm r ke 2 1 2 2 2 = ⇒ 2 2 2mv r ke = Solve for v to obtain: mr kev 2 2 = 84 •• Picture the Problem In Problem 81 it was established that the period of an electric dipole in an electric field is given by .22 qEmaT π= We can use this result to find the frequency of oscillation of a KBr molecule in a uniform electric field of 1000 N/C. The Electric Field 1: Discrete Charge Distributions 65 Express the frequency of the KBr oscillator: ma qEf 2 2 1 π = Substitute numerical values and evaluate f: ( )( ) ( )( ) Hz1053.4 nm0.282kg101.4 N/C1000C101.62 2 1 8 25 19 ×= × × = − − π f 85 ••• Picture the Problem We can use Coulomb’s force law for point masses and the condition for translational equilibrium to express the equilibrium position as a function of k, q, Q, m, and g. In part (b) we’ll need to show that the displaced point charge experiences a linear restoring force and, hence, will exhibit simple harmonic motion. (a) Apply the condition for translational equilibrium to the point mass: 02 0 =−mg y kqQ Solve for y0 to obtain: mg kqQy =0 (b) Express the restoring force that acts on the point mass when it is displaced a distance ∆y from its equilibrium position: ( ) 2 00 2 0 2 0 2 0 2 y kqQ yyy kqQ y kqQ yy kqQF − ∆+ ≈ − ∆+ = because ∆y << y0. Simplify this expression further by writing it with a common denominator: 3 0 0 4 0 0 3 0 4 0 0 2 21 2 2 2 y ykqQ y yy ykqQy yyy ykqQyF ∆ −≈ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ∆ + ∆ −= ∆+ ∆ −= again, because ∆y << y0. From the 1st step of our solution: mg y kqQ =2 0 Chapter 21 66 Substitute to obtain: y y mgF ∆−= 0 2 Apply Newton’s 2nd law to the displaced point charge to obtain: y y mg dt ydm ∆−=∆ 0 2 2 2 or 02 0 2 2 =∆+ ∆ y y g dt yd the differential equation of simple harmonic motion with 02 yg=ω . 86 ••• Picture the Problem The free-body diagram shows the Coulomb force the positive charge Q exerts on the bead that is constrained to move along the x axis. The x component of this force is a restoring force, i.e., it is directed toward the bead’s equilibrium position. We can show that, for x << L, this restoring force is linear and, hence, that the bead will exhibit simple harmonic motion about its equilibrium position. Once we’ve obtained the differential equation of SHM we can relate the period of the motion to its angular frequency. Using Coulomb’s law for point charges, express the force F that +Q exerts on −q: ( ) 2222 xL kqQ xL QqkF + −= + − = Express the component of this force along the x axis: ( ) x xL kqQ xL x xL kqQ xL kqQFx 2322 2222 22 cos + −= ++ −= + −= θ The Electric Field 1: Discrete Charge Distributions 69 ∑ =−−= 011,11, EqgmTF yy Apply the conditions for static equilibrium to the charged sphere whose mass is m2: ( ) ( ) 0 sin sinsin sin 222 21 2 21 222 21 21 222 21 2, = + + ≈ + + = −=∑ θ θθ θ θθ θ T L qkq T LL qkq T r qkqFx and 022,22, =−−=∑ EqgmTF yy Express θ1 and θ 2 in terms of the components of T and : 1 r 2T r y x T T ,1 ,1 1 =θ (1) and y x T T ,2 ,2 2 =θ (2) Divide equation (1) by equation (2) to obtain: y y y x y x T T T T T T ,1 ,2 ,2 ,2 ,1 ,1 2 1 == θ θ because the horizontal components of and 1T r 2T r are equal. Substitute for T2,y and T1,y to obtain: Eqgm Eqgm 11 22 2 1 + + = θ θ Add equations (1) and (2) to obtain: ( ) ⎥⎦ ⎤ ⎢ ⎣ ⎡ + + ++ =+=+ EqgmEqgmL qkq T T T T y x y x 2211 2 21 2 21 ,2 ,2 ,1 ,1 21 11 θθ θθ Solve for θ1 + θ2: 3 2211 2 21 21 11 ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + + + =+ EqgmEqgmL qkqθθ Chapter 21 70 Substitute numerical values and evaluate 1 + θ2 and θ1/θ2: θ °==+ 4.16rad287.021 θθ and 34.1 2 1 = θ θ olve for θ1 and θ2 to obtain: °= 42.91θ and °= 98.61θ S 88 ••• Picture the Problem Each sphere is in static equilibrium under the influence of a tension, gravitational and Coulomb forc Let the mass of the sphere carrying the charge of 2.0 µC be m e. C s to e forces to the charges on the pheres. 1 = 0.01 kg, and that of the sphere carrying the charge of 1.0 µ be m2 = 0.02 kg. We can use Coulomb’ law to relate the Coulomb force to the charge on each sphere and their separation and the conditions for static equilibrium relate thes s Apply the conditions for static equilibrium to the charged sphere whose mass is m1: ( ) ( ) 0 sin sinsin sin 112 21 2 21 112 21 21 112 21 1, = + + −≈ + + −= +−=∑ θ θθ θ θθ θ T L qkq T LL qkq T r qkqFx and ∑ =−= 01,11, gmTF yy to the charged sphere whose mass is m2: Apply the conditions for static equilibrium ( ) ( ) 0 sin sinsin sin 222 21 2 21 222 21 21 222 21 2, = + + ≈ + + = −=∑ θ θθ θ θθ θ T L qkq T LL qkq T r qkqFx The Electric Field 1: Discrete Charge Distributions 71 and 02,22, =−=∑ gmTF yy a ter ponents of and Using the small-angle approximation sinθ ≈ tanθ ≈θ, express θ1 nd θ2 in ms of the com 1T r 2T r : y x T T ,1 ,1 1 =θ (1) and y x T T ,2 ,2 2 =θ (2) ation (1) by equation (2) obtain: Divide equ to y y yT ,2 x y x T T T T T ,1 ,2 ,2 ,1 ,1 2 1 == θ θ because the horizontal components of 1T r and 2T r are equal. Substitute for T2,y and T1,y to obtain: 1 2 2 1 m m = θ θ Add equations (1) and (2) to obtain: ( ) ⎥⎦ ⎤ +⎢ ⎣ ⎡ + = +=+ gmgmL qkq T T T T y x y x 21 2 21 2 21 ,2 ,2 ,1 ,1 21 11 θθ θθ Solve for θ1 + θ2: 3 21 2 21 21 11 ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ +=+ gmgmL qkqθθ and valuate θ1 + θ2 and θ1/θ2: Substitute numerical values e °==+ 4.28rad496.021 θθ and 2 1 2 1 = θ θ Solve for θ1 and θ2 to obtain: °= 47.91θ and °= 9.181θ Remarks: While the small angle approximation is not as good here as it was in the receding problems, the error introduced is less than 3%. p Chapter 21 74 ) Apply Newton’s 2nd law to the ring to obtain: (c ( ) yya kqQ dt ydm 23222 2 2 + −= or ( ) yyam kqQ dt yd 23222 2 2 + −= Factor the radicand to obtain: y mL kqQy ma kqQ 162 y a yma kqQ dt yd 33 23 2 2 3 2 2 1 2 −=−≈ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + −= provided y << a = L . Thus we have: /2 y mL kqQ dt yd 2 32 16 −= or 016 32 2 + y mL kqQ dt yd = simple harmonic motion. cy of the simple the differential equation of Express the frequen harmonic motion in terms of its angular frequency: π ω 2 =f From the differential equation describing the motion we have: 3 2 16 mL kqQ =ω and 3 16 2 1 mL kqQf π = Substitute numerical values and evaluate f: ( )( )( ) ( )( ) Hz37.9 m0.24kg0.03 C2C5/CmN1099.816 2 1 3 229 = ⋅× = µµ π f The Electric Field 1: Discrete Charge Distributions 75 92 ••• Picture the Problem The free body diagram shows the forces acting on the microsphere of mass m and having an excess charge of q = Ne when the electric field is downward. Under terminal-speed conditions the sphere is in equilibrium under the influence of the electric force eF r , its weight ,m and the drag force gr .dF r We can apply Newton’s 2nd law, under terminal-speed conditions, to relate the number of excess charges N on the sphere to its mass and, using Stokes’ law, find its terminal speed. (a) Apply to the microsphere: ∑ = yy maF ymaFmgF =−− de or, because ay = 0, 0terminald,e =−− FmgF Substitute for Fe, m, and Fd,terminal to obtain: 06 t =−− rvVgqE πηρ or, because q = Ne, 06 t 3 3 4 =−− rvgrNeE πηρπ Solve for N to obtain: eE rvgr N t 3 3 4 6πηρπ + = Substitute numerical values and evaluate gr ρπ 334 : ( ) ( )( N1018.7 m/s81.9kg/m1005.1 m105.5 15 233 37 3 43 3 4 − − ×= ×× ×= πρπ gr ) Substitute numerical values and evaluate t6 rvπη : ( )( ) ( ) N1016.2 m/s1016.1 m105.5sPa108.166 14 4 75 t − − −− ×= ×× ×⋅×= ππηrv Substitute numerical values in equation (1) and evaluate N: ( )( ) 3 V/m106C106.1 N1016.2N1018.7 419 1415 = ×× ×+× = − −− N (b) With the field pointing upward, the electric force is downward and the application of to ∑ = yy maF 0eterminald, =−− mgFF or 06 334t =−−− grNeErv ρππη Chapter 21 76 the bead yields: Solve for vt to obtain: r grNeE v πη ρπ 6 3 3 4 t + = Substitute numerical values and evaluate vt: ( )( ) ( ) ( )( ) ( )( ) m/s1093.1 m105.5sPa108.16 m/s81.9kg/m1005.1m105.5V/m106C106.13 4 75 23337 3 4419 t − −− −− ×= ×⋅× ××+×× = π πv *93 ••• Picture the Problem The free body diagram shows the forces acting on the microsphere of mass m and having an excess charge of q = Ne when the electric field is downward. Under terminal-speed conditions the sphere is in equilibrium under the influence of the electric force eF r , its weight ,m and the drag force g r .dF r We can apply Newton’s 2nd law, under terminal-speed conditions, to relate the number of excess charges N on the sphere to its mass and, using Stokes’ law, to its terminal speed. (a) Apply to the microsphere when the electric field is downward: ∑ = yy maF ymaFmgF =−− de or, because ay = 0, 0terminald,e =−− FmgF Substitute for Fe and Fd,terminal to obtain: 06 u =−− rvmgqE πη or, because q = Ne, 06 u =−− rvmgNeE πη Solve for vu to obtain: r mgNeEv πη6u − = (1) With the field pointing upward, the electric force is downward and the application of to the microsphere yields: ∑ = yy maF 0eterminald, =−− mgFF or 06 d =−− mgNeErvπη Solve for vd to obtain: r mgNeEv πη6d + = (2) 79 Chapter 22 The Electric Field 2: Continuous Charge Distributions Conceptual Problems *1 •• (a) False. Gauss’s law states that the net flux through any surface is given by insideS nnet 4 kQdAE πφ == ∫ . While it is true that Gauss’s law is easiest to apply to symmetric charge distributions, it holds for any surface. (b) True 2 •• Determine the Concept Gauss’s law states that the net flux through any surface is given by insideS nnet 4 kQdAE πφ == ∫ . To use Gauss’s law the system must display some symmetry. 3 ••• Determine the Concept The electric field is that due to all the charges, inside and outside the surface. Gauss’s law states that the net flux through any surface is given by insideS nnet 4 kQdAE πφ == ∫ . The lines of flux through a Gaussian surface begin on charges on one side of the surface and terminate on charges on the other side of the surface. 4 •• Picture the Problem We can show that the charge inside a sphere of radius r is proportional to r3 and that the area of a sphere is proportional to r2. Using Gauss’s law, we can show that the field must be proportional to r3/r2 = r. Use Gauss’s law to express the electric field inside a spherical charge distribution of constant volume charge density: A kQE inside4π= where 24 rA π= . Express Qinside as a function of ρ and r: 3 3 4 inside rVQ πρρ == Substitute to obtain: rk r rkE 3 4 4 4 2 3 3 4 πρ π πρπ == Chapter 22 80 *5 • (a) False. Consider a spherical shell, in which there is no charge, in the vicinity of an infinite sheet of charge. The electric field due to the infinite sheet would be non-zero everywhere on the spherical surface. (b) True (assuming there are no charges inside the shell). (c) True. (d) False. Consider a spherical conducting shell. Such a surface will have equal charges on its inner and outer surfaces but, because their areas differ, so will their charge densities. 6 • Determine the Concept Yes. The electric field on a closed surface is related to the net flux through it by Gauss’s law: 0insideS ∈== ∫ QEdAφ . If the net flux through the closed surface is zero, the net charge inside the surface must be zero by Gauss’s law. 7 • Determine the Concept The negative point charge at the center of the conducting shell induces a charge +Q on the inner surface of the shell. correct. is )(a 8 • Determine the Concept The negative point charge at the center of the conducting shell induces a charge +Q on the inner surface of the shell. Because a conductor does not have to be neutral, correct. is )(d *9 •• Determine the Concept We can apply Gauss’s law to determine the electric field for r < R1 and r > R2. We also know that the direction of an electric field at any point is determined by the direction of the electric force acting on a positively charged object located at that point. From the application of Gauss’s law we know that the electric field in both of these regions is not zero and is given by: 2n r kQE = A positively charged object placed in either of these regions would experience an attractive force from the charge –Q located at the center of the shell. correct. is )(b The Electric Field 2: Continuous Charge Distributions 81 *10 •• Determine the Concept We can decide what will happen when the conducting shell is grounded by thinking about the distribution of charge on the shell before it is grounded and the effect on this distribution of grounding the shell. The negative point charge at the center of the conducting shell induces a positive charge on the inner surface of the shell and a negative charge on the outer surface. Grounding the shell attracts positive charge from ground; resulting in the outer surface becoming electrically neutral. correct. is )(b 11 •• Determine the Concept We can apply Gauss’s law to determine the electric field for r < R1 and r > R2. We also know that the direction of an electric field at any point is determined by the direction of the electric force acting on a positively charged object located at that point. From the application of Gauss’s law we know that the electric field in the region r < R1 is given by 2n r kQE = . A positively charged object placed in the region r < R1 will experience an attractive force from the charge –Q located at the center of the shell. With the conducting shell grounded, the net charge enclosed by a spherical Gaussian surface of radius r > R2 is zero and hence the electric field in this region is zero. correct. is )(c 12 •• Determine the Concept No. The electric field on a closed surface is related to the net flux through it by Gauss’s law: 0insideS ∈== ∫ QEdAφ . φ can be zero without E being zero everywhere. If the net flux through the closed surface is zero, the net charge inside the surface must be zero by Gauss’s law. 13 •• False. A physical quantity is discontinuous if its value on one side of a boundary differs from that on the other. We can show that this statement is false by citing a counterexample. Consider the field of a uniformly charged sphere. ρ is discontinuous at the surface, E is not. Estimation and Approximation *14 •• Picture the Problem We’ll assume that the total charge is spread out uniformly (charge density = σ) in a thin layer at the bottom and top of the cloud and that the area of each Chapter 22 84 (b) Substitute numerical values and evaluate Ex at x = 6 m: ( ) ( )( )( )( ) N/C26.2 m5m6m6 nC17.5/CmN108.99m6 229 = − ⋅× =xE (c) Substitute numerical values and evaluate Ex at x = 9 m: ( ) ( )( )( )( ) N/C37.4 m5m9m9 nC17.5/CmN108.99m9 229 = − ⋅× =xE (d) Substitute numerical values and evaluate Ex at x = 250 m: ( ) ( )( )( )( ) mN/C57.2m5m502m502 nC17.5/CmN108.99m502 229 = − ⋅× =xE (e) Use Coulomb’s law for the electric field due to a point charge to obtain: ( ) 2x kQxEx = Substitute numerical values and evaluate Ex(250 m): ( ) ( )( ) ( ) mN/C52.2 m250 nC17.5/CmN108.99m250 2 229 = ⋅× =xE Note that this result agrees to within 2% with the exact value obtained in (d). 18 • Picture the Problem Let the charge densities on the two plates be σ1 and σ2 and denote the three regions of interest as 1, 2, and 3. Choose a coordinate system in which the positive x direction is to the right. We can apply the equation for E r near an infinite plane of charge and the superposition of fields to find the field in each of the three regions. The Electric Field 2: Continuous Charge Distributions 85 (a) Use the equation for E r near an infinite plane of charge to express the field in region 1 when σ1 = σ2 = +3 µC/m2: i ii EEE ˆ4 ˆ2ˆ2 21 1 21 σπ σπσπ σσ k kk −= −−= += rrr Substitute numerical values and evaluate :1E r ( )( ) ( )iiE ˆN/C1039.3ˆC/m3/CmN1099.84 522291 ×−=⋅×−= µπ r Proceed as above for region 2: 0ˆ2ˆ2 ˆ2ˆ2 212 21 =−= −=+= ii iiEEE σπσπ σπσπσσ kk kk rrr Proceed as above for region 3: ( )( ) ( )i i i iiEEE ˆN/C1039.3 ˆC/m3/CmN1099.84 ˆ4 ˆ2ˆ2 5 2229 213 21 ×= ⋅×= = +=+= µπ σπ σπσπσσ k kk rrr The electric field lines are shown to the right: (b) Use the equation for E r near an infinite plane of charge to express and evaluate the field in region 1 when σ1 = +3 µC/m2 and σ2 = −3 µC/m2: 0ˆ2ˆ2 ˆ2ˆ2 211 21 =−= −=+= ii iiEEE σπσπ σπσπσσ kk kk rrr Proceed as above for region 2: ( )( ) ( )i i i iiEEE ˆN/C1039.3 ˆC3/CmN1099.84 ˆ4 ˆ2ˆ2 5 229 211 21 ×= ⋅×= = +=+= µπ σπ σπσπσσ k kk rrr Chapter 22 86 Proceed as above for region 3: 0ˆ2ˆ2 ˆ2ˆ2 213 21 =−= −=+= ii iiEEE σπσπ σπσπσσ kk kk rrr The electric field lines are shown to the right: 19 • Picture the Problem The magnitude of the electric field on the axis of a ring of charge is given by ( ) ( ) 2322 axkQxxEx += where Q is the charge on the ring and a is the radius of the ring. We can use this relationship to find the electric field on the x axis at the given distances from the ring. Express E r on the axis of a ring charge: ( ) ( ) 2322 ax kQxxEx + = (a) Substitute numerical values and evaluate Ex for x = 1.2 cm: ( ) ( )( )( ) ( ) ( )[ ] N/C1069.4cm5.8cm2.1 cm2.1C75.2/CmN1099.8cm2.1 52322 229 ×= + ⋅× = µ xE (b) Proceed as in (a) with x = 3.6 cm: ( ) ( )( )( ) ( ) ( )[ ] N/C1013.1cm5.8cm6.3 cm6.3C75.2/CmN1099.8cm6.3 62322 229 ×= + ⋅× = µ xE (c) Proceed as in (a) with x = 4.0 m: ( ) ( )( )( ) ( ) ( )[ ] N/C1054.1cm5.8m4 m4C75.2/CmN1099.8m4 32322 229 ×= + ⋅× = µ xE (d) Using Coulomb’s law for the electric field due to a point charge, express Ex: ( ) 2x kQxEx = The Electric Field 2: Continuous Charge Distributions 89 We’re given that: 021 2∈= σxE Equate these expressions: ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + −= 22 0 12 4 ax xkσπ ε σ Simplify to obtain: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + −= 22 0 12 4 ax xkσπ ε σ or, because k = 1/4πε0, 22 1 2 1 ax x + −= Solve for x to obtain: 3 ax = 23 • Picture the Problem We can use ( ) 2322 ax kQxEx + = to find the electric field at the given distances from the center of the charged ring. (a) Evaluate Ex at x = 0.2a: ( ) ( ) ( )[ ] 2 2322 189.0 2.0 2.02.0 a kQ aa akQaEx = + = (b) Evaluate Ex at x = 0.5a: ( ) ( ) ( )[ ] 2 2322 358.0 5.0 5.05.0 a kQ aa akQaEx = + = (c) Evaluate Ex at x = 0.7a: ( ) ( ) ( )[ ] 2 2322 385.0 7.0 7.07.0 a kQ aa akQaEx = + = (d) Evaluate Ex at x = a: ( ) [ ] 22322 354.0 a kQ aa kQaaEx = + = Chapter 22 90 (e) Evaluate Ex at x = 2a: ( ) ( )[ ] 22322 179.02 22 a kQ aa kQaaEx = + = The field along the x axis is plotted below. The x coordinates are in units of x/a and E is in units of kQ/a2. -0.4 -0.2 0.0 0.2 0.4 -3 -2 -1 0 1 2 3 x /a E x 24 • Picture the Problem We can use ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + −= 22 12 Rx xkEx σπ , where R is the radius of the disk, to find the electric field on the axis of a disk charge. Express Ex in terms of ε0: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + − ∈ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + − ∈ = 22 0 22 0 1 2 1 4 2 Rx x Rx xEx σ π πσ (a) Evaluate Ex at x = 0.2a: ( ) ( ) 0 22 0 2 804.0 2.0 2.01 2 2.0 ∈ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − ∈ = σ σ aa aaEx The Electric Field 2: Continuous Charge Distributions 91 (b) Evaluate Ex at x = 0.5a: ( ) ( ) 0 22 0 2 553.0 5.0 5.01 2 5.0 ∈ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − ∈ = σ σ aa aaEx (c) Evaluate Ex at x = 0.7a: ( ) ( ) 0 22 0 2 427.0 7.0 7.01 2 7.0 ∈ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − ∈ = σ σ aa aaEx (d) Evaluate Ex at x = a: ( ) 0 22 0 2 293.0 1 2 ∈ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + − ∈ = σ σ aa aaEx (e) Evaluate Ex at x = 2a: ( ) ( ) 0 22 0 2 106.0 2 21 2 2 ∈ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − ∈ = σ σ aa aaEx The field along the x axis is plotted below. The x coordinates are in units of x/a and E is in units of .2 0∈σ 0.0 0.4 0.8 1.2 1.6 2.0 -3 -2 -1 0 1 2 3 x/R E x Chapter 22 94 and 03 222 =−+ xax Solve for x to obtain: 2 ax ±= as our candidates for maxima or minima. A plot of E, in units of kQ/a2, versus x/a is shown to the right. This graph shows that E is a minimum at 2ax −= and a maximum at 2ax = . -0.4 -0.2 0.0 0.2 0.4 -3 -2 -1 0 1 2 3 x/a E x 27 •• Picture the Problem The line charge and point (0, y) are shown in the diagram. Also shown is a line element of length dx and the field E r d its charge produces at (0, y). We can find dEx from E r d and then integrate from x = x1 to x = x2. Express the x component of E r d : ( ) dxyx xk dx yx x yx k dx yx kdEx 2322 2222 22 sin + −= ++ −= + −= λ λ θλ The Electric Field 2: Continuous Charge Distributions 95 Integrate from x = x1 to x2 to obtain: ( ) ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + + + −−= ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + + + −−= ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + −−= + −= ∫ 22 1 22 2 22 1 22 2 22 2322 11 1 2 1 2 1 yx y yx y y k yxyx k yx k dx yx xkE x x x x x λ λ λ λ From the diagram we see that: 22 2 2cos yx y + =θ or ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = − y x21 2 tanθ and 22 1 1cos yx y + =θ or ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = − y x11 1 tanθ Substitute to obtain: [ ] [ ]12 12 coscos coscos θθλ θθλ −= +−−= y k y kEx 28 •• Picture the Problem The diagram shows a segment of the ring of length ds that has a charge dq = λds. We can express the electric field E r d at the center of the ring due to the charge dq and then integrate this expression from θ = 0 to 2π to find the magnitude of the field in the center of the ring. (a) and (b) The field E r d at the center of the ring due to the charge dq is: ji EEE ˆsinˆcos θθ dEdE ddd yx −−= += rrr (1) The magnitude dE of the field at the center of the ring is: 2r kdqdE = Chapter 22 96 Because dq = λds: 2r dskdE λ= The linear charge density varies with θ according to λ(θ) = λ0 sin θ : 2 0 sin r dskdE θλ= Substitute rdθ for ds: r dk r rdkdE θθλθθλ sinsin 02 0 == Substitute for dE in equation (1) to obtain: j iE ˆsin ˆcossin 2 0 0 r dk r dkd θθλ θθθλ − −= r Integrate E r d from θ = 0 to 2π: j j j iE ˆ ˆ0 ˆsin ˆ2sin 2 0 0 2 0 20 2 0 0 r k r k d r k d r k λπ λπ θθλ θθλ π π −= −= − −= ∫ ∫ r . is magnitude its anddirection negative in the isorigin at the field The 0 r ky λπ 29 •• Picture the Problem The line charge and the point whose coordinates are (0, y) are shown in the diagram. Also shown is a segment of the line of length dx. The field that it produces at (0, y) is .E r d We can find dEy from E r d and then integrate from x = 0 to x = a to find the y component of the electric field at a point on the y axis. (a) Express the magnitude of the field E r d due to charge dq of the 2r kdqdE =
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