22.1: a) C.Nm75.160cos)m(0.250N/C)14( 22  AE   b) As long as the sheet is flat, its shape does not matter. ci) The maximum flux occurs at an angle  0  between the normal and field. cii) The minimum flux occurs at an angle  90  between the normal and field. In part i), the paper is oriented to “capture” the most field lines whereas in ii) the area is oriented so that it “captures” no field lines. 22.2: a) nAAE ˆ wherecos AθEA    CmN329.36cosm)(0.1)CN104((back) ˆ ˆ CmN329.36cosm)(0.1)CN104(front)( ˆ ˆ 090cosm)(0.1)CN104(bottom)( ˆ ˆ CmN24)9.36(90cosm)(0.1)CN104(right)( ˆ ˆ 090cosm)(0.1)CN104(top)( ˆ ˆ CmN24)93690(cosm)(0.1)CN104(left)( ˆ ˆ 223 223 23 223 23 S 223 66 55 44 33 22 11       SS SS SS SS S SS . in in kn jn kn jn b) The total flux through the cube must be zero; any flux entering the cube must also leave it. 22.3: a) Given that lengthedge,, ˆ D ˆ C ˆ B AEkjiE    L, and .BL ˆ ˆ ˆ .BL ˆ ˆ ˆ .DL ˆ ˆ ˆ .CL ˆ ˆ ˆ .DL ˆ ˆ ˆ .CL ˆ ˆ ˆ 2 6 2 5 2 4 2 3 2 2 2 1 66 55 44 33 22 11       SS SS SS SS SS SS A A A A A A nEin nEin nEkn nEjn nEkn nEjn       b) Total flux    6 1 i 0 i 22.4: C.Nm16.670cos)m(0.240)CN0.75( 22  AE   22.5: a) C.Nm1071.2)2( 25 m)(0.400C/m)1000.6( 2 0 6 00     εε l r πε πrlAE   b) We would get the same flux as in (a) if the cylinder’s radius was made larger—the field lines must still pass through the surface. c) If the length was increased to m,800.0l the flux would increase by a factor of two: C.Nm105.42 25  22.6: a) C.Nm452C)1000.4( 2 0 9 01 1   εεq S b) C.Nm881C)1080.7( 2 0 9 02 2   εεq S c) C.Nm429C)10)80.700.4(()( 2 0 9 021 3   εεqq S d) C.Nm723C)10)40.200.4(()( 2 0 9 021 4   εεqq S e) C.Nm158C)10)40.280.700.4(()( 2 0 9 0321 5   εεqqq S f) All that matters for Gauss’s law is the total amount of charge enclosed by the surface, not its distribution within the surface. 22.7: a) C.Nm1007.4C)1060.3( 25 0 6 0   εεq b) C.106.90)CNm780( 92 000   εεqεq c) No. All that matters is the total charge enclosed by the cube, not the details of where the charge is located. 22.8: a) No charge enclosed so 0 b) C.Nm678 NmC108.85 C1000.6 2 2212 9 0 2       ε q c) C.Nm226 NmC108.85 C10)00.600.4( 2 2212 9 0 21         ε qq 22.9: a) Since E  is uniform, the flux through a closed surface must be zero. That is:   .00 00 1 ρdVρdVd εε q AE   But because we can choose any volume we want, ρ must be zero if the integral equals zero. b) If there is no charge in a region of space, that does NOT mean that the electric field is uniform. Consider a closed volume close to, but not including, a point charge. The field diverges there, but there is no charge in that region. 22.10: a) If 0ρ and uniform, then q inside any closed surface is greater than zero.   00 AE   d and so the electric field cannot be uniform, i.e., since an arbitrary surface of our choice encloses a non-zero amount of charge, E must depend on position. b) However, inside a small bubble of zero density within the material with density ρ , the field CAN be uniform. All that is important is that there be zero flux through the surface of the bubble (since it encloses no charge). (See Exercise 22.61.) 22.11: C.Nm1008.1C)1060.9( 26 0 6 0sides6   εεq But the box is symmetrical, so for one side, the flux is: .CNm1080.1 25 side1  b) No change. Charge enclosed is the same. 22.12: Since the cube is empty, there is no net charge enclosed in it. The net flux, according to Gauss’s law, must be zero. 22.13: 0encl εQ E  The flux through the sphere depends only on the charge within the sphere. nC3.19)CmN360( 2 00encl  εεQ E 22.14: a) .CN44.7 m)(0.550 C)1050.2( 4 1 4 1 m)0.1m450.0( 2 10 0 2 0     πεr q πε rE b) 0E  inside of a conductor or else free charges would move under the influence of forces, violating our electrostatic assumptions (i.e., that charges aren’t moving). 22.15: a) m.62.1 CN614 C)10180.0( 4 1 4 1|| 4 1 || 6 00 2 0     πεE q πε r r q πε E b) As long as we are outside the sphere, the charge enclosed is constant and the sphere acts like a point charge. 22.16: a) C.1056.7)m(0.0610)CN1040.1(/ 825 000   εEAεqεqEA b) Double the surface area: C.1051.1)m(0.122)CN1040.1( 725 0   εq 22.17: C.1027.3m)(0.160)CN1150(44 92 0 2 0 4 1 2 0   πεErπεqE r q πε So the number of electrons is: .1004.2 10 C1060.1 C1027.3 e 19 9      n 22.18: Draw a cylindrical Gaussian surface with the line of charge as its axis. The cylinder has radius 0.400 m and is 0.0200 m long. The electric field is then 840 N/C at every point on the cylindrical surface and directed perpendicular to the surface. Thus   )2)(())(( cylinder πrLEAEd sE   /CmN42.2m)(0.0200m)(0.400)(2N/C)840( 2  π The field is parallel to the end caps of the cylinder, so for them 0  sE   d . From Gauss’s law: C1074.3 ) C mN 2.42() mN C 10854.8( 10 2 2 2 12 0       E εq 22.19: r E λ 2 1 0   22.20: a) For points outside a uniform spherical charge distribution, all the charge can be considered to be concentrated at the center of the sphere. The field outside the sphere is thus inversely proportional to the square of the distance from the center. In this case: CN53 cm0.600 cm0.200 )CN480( 2           E b) For points outside a long cylindrically symmetrical charge distribution, the field is identical to that of a long line of charge: , 2 λ 0 rπε E  that is, inversely proportional to the distance from the axis of the cylinder. In this case CN160 cm0.600 cm0.200 )CN480(           E c) The field of an infinite sheet of charge is ;2/ 0 εσE  i.e., it is independent of the distance from the sheet. Thus in this case .CN480E 22.21: Outside each sphere the electric field is the same as if all the charge of the sphere were at its center, and the point where we are to calculate E  is outside both spheres. 21 and EE  are both toward the sphere with negative charge. sphere.chargednegatively thetoward,CN1006.8 CN10471.5 m)(0.250 C1080.3|| CN10591.2 m)(0.250 C1080.1|| 5 21 5 2 6 2 2 2 2 5 2 6 2 1 1 1          EEE k r q kE k r q kE 22.22: For points outside the sphere, the field is identical to that of a point charge of the same total magnitude located at the center of the sphere. The total charge is given by charge density  volume: C1060.1m)150.0)( 3 4 )(mCn50.7( 1033   πq a) The field just outside the sphere is CN4.42 m)(0.150 C)10(1.06)/CmN109( 4 2 10229 2 0     rπε q E b) At a distance of 0.300 m from the center (double the sphere’s radius) the field will be 1/4 as strong: 10.6 CN c) Inside the sphere, only the charge inside the radius in question affects the field. In this case, since the radius is half the sphere’s radius, 1/8 of the total charge contributes to the field: CN2.21 m)(0.075 C)1006.1()8/1()C/mN109( 2 10229     E 22.23: The point is inside the sphere, so 3 / RkQrE  (Example 22.9) nC2.10 )m100.0( m)(0.220)CN950( 33  kkr ER Q 22.24: a) Positive charge is attracted to the inner surface of the conductor by the charge in the cavity. Its magnitude is the same as the cavity charge: nC,00.6 inner q since 0E inside a conductor. b) On the outer surface the charge is a combination of the net charge on the conductor and the charge “left behind” when the nC00.6 moved to the inner surface: nC.1.00nC6.00nC00.5 innertotouterouterinnertot  qqqqqq 22.25: 32 SandS enclose no charge, so the flux is zero, and electric field outside the plates is zero. For between the plates, 1 S shows that: . 000 εσEεAσεqEA  22.26: a) At a distance of 0.1 mm from the center, the sheet appears “infinite,” so:      .CN662 m)800.0(2 C1050.7 2 2 2 0 9 00 εAε q E ε q AEd AE   b) At a distance of 100 m from the center, the sheet looks like a point, so: .CN1075.6 m)(100 C)1050.7( 4 1 4 1 3 2 9 0 2 0      πεr q πε E c) There would be no difference if the sheet was a conductor. The charge would automatically spread out evenly over both faces, giving it half the charge density on any as the insulator 00 2 : :).( ε σ ε σ c Eσ  near one face. Unlike a conductor, the insulator is the charge density in some sense. Thus one shouldn’t think of the charge as “spreading over each face” for an insulator. Far away, they both look like points with the same charge. 22.27: a) .2 2 λπRσ L Q πRL Q A Q σ  b)   . 2 )2( 000 rε σR E ε πRLσ ε Q πrLEd AE c) But from (a), ,so,2λ 0 2 λ rπε ER   same as an infinite line of charge. 22.28: All the s'σ are absolute values. (a) at 0 1 0 4 0 3 0 2 2222 : ε σ ε σ ε σ ε σ A EA  left. thetoCN1082.2 )mC6mC4mC2mC5( 2 1 )( 2 1 5 2222 0 1432 0    μμμμ ε σσσσ ε E A (b) left. thetoCN1095.3 )mC5mC4mC2mC6( 2 1 )( 2 1 2222 5 2222 0 2431 00 2 0 4 0 3 0 1    μμμμ ε σσσσ εε σ ε σ ε σ ε σ E B (c) left thetoCN1069.1 )mC6mC4mC2mC5( 2 1 )( 2 1 2222 5 2222 0 1432 00 1 0 4 0 3 0 2    μμμμ ε σσσσ εε σ ε σ ε σ ε σ E C 22.29: a) Gauss’s law says +Q on inner surface, so 0E inside metal. b) The outside surface of the sphere is grounded, so no excess charge. c) Consider a Gaussian sphere with the –Q charge at its center and radius less than the inner radius of the metal. This sphere encloses net charge –Q so there is an electric field flux through it; there is electric field in the cavity. d) In an electrostatic situation 0E inside a conductor. A Gaussian sphere with the Q charge at its center and radius greater than the outer radius of the metal encloses zero net charge (the Q charge and the Q on the inner surface of the metal) so there is no flux through it and 0E outside the metal. e) No, 0E there. Yes, the charge has been shielded by the grounded conductor. There is nothing like positive and negative mass (the gravity force is always attractive), so this cannot be done for gravity. 22.30: Given , ˆ )m)CN(00.3( ˆ )m)CN(00.5( kiE zx   edge length ,m300.0L ,m300.0L and .0 ˆ ˆ ˆ 11 1  A Ss nEjn   )CN(00.3( ˆ ˆ ˆ 21 2 A SS nEkn   zz )m)CN(27.0()m300.0)(m 2 .m)CN(081.0)m300.0)(m)CN(27.0( 2  .0 ˆ ˆ ˆ 33 3  A SS nEjn  ).0(0)m)CN(27.0( ˆ ˆ ˆ 44 4  zzA SS nEkn  xxA SS )m)N/C(45.0()m300.0)(m)CN(00.5( ˆ ˆ ˆ 2 5 55  nEin  ).m)CN(135.0()m300.0)(m)N/C(45.0( 2  ).0(0)m)CN(45.0( ˆ ˆ ˆ 66 6  xxA SS nEin  b) Total flux: CNm054.0m)CN()135.0081.0( 22 52  C1078.4 13 0   εq 22.31: a) b) Imagine a charge q at the center of a cube of edge length 2L. Then: ./ 0 εq Here the square is one 24th of the surface area of the imaginary cube, so it intercepts 1/24 of the flux. That is, .24 0  q 22.32: a) .CmN750)m0.6)(CN125( 22  EA b) Since the field is parallel to the surface, .0 c) Choose the Gaussian surface to equal the volume’s surface. Then: 750 – EA= ,CN577)750C1040.2( 0 8 m0.6 1 0 2    Eεq in the positive x -direction. Since 0q we must have some net flux flowing in so AEEA  on second face. d) 0q but we have E pointing away from face I. This is due to an external field that does not affect the flux but affects the value of E. 22.33: To find the charge enclosed, we need the flux through the parallelepiped: CmN5.3760cos)CN1050.2)(m0600.0)(m0500.0(60cos 24 11  AE CmN10560cos)CN1000.7)(m0600.0)(m0500.0(120cos 24 22  AE So the total flux is ,CmN5.67CmN)1055.37( 22 21  and .C1097.5)CmN5.67( 10 0 2 0   εεq b) There must be a net charge (negative) in the parallelepiped since there is a net flux flowing into the surface. Also, there must be an external field or all lines would point toward the slab. 22.34: The  particle feels no force where the net electric field is zero. The fields can cancel only in regions A and B. sheetline EE  00 22 λ ε σ rπε  cm16m16.0 )C/m100( C/m50 λ 2     r The fields cancel 16 cm from the line in regions A and B. [...]... in r 22. 60: a)  g  g 4πr 2  4πGM  g   22. 61: a) For a sphere NOT at the coordinate origin:    Q ρ 4πr 3 ρr  E r   r  b    4πr  2 E  encl  , ε0 ε0 3 3ε0 ˆ in the r  - direction    ρ(r  b ) E 3ε0 b) The electric field inside a hole in a charged insulating sphere is:        ρr ρ(r  b ) ρb  E hole  Esphere  E(a)    3ε0 3ε0 3ε0  Note that E is uniform 22. 62:... 2 , since 22. 38: a) r  a, E  1 Q 4 πε 0 r 2 , since the charge enclosed is Q a  r  b, E  0, since the –Q on the inner surface of the shell cancels the +Q at the center of the sphere 1 r  b, E   4πε 0 2Q , since the total enclosed charge is –2Q r2 Q b) The surface charge density on inner surface: σ   4πa 2 c) The surface charge density on the outer surface: σ   42Q2 πb d) e) 22. 39: a)(i)... conducting sphere—but we see a smooth transition from the uniform insulator to the outside 22. 43: a) The sphere acts as a point charge on an external charge, so: 1 F  qE  4πε 0 qQ , radially inward r2 (b) If the point charge was inside the sphere (where there is no electric field) it would feel zero force 22. 44: a) ρinner  outer  q 3q  1  q  4 3 4 3   Vb  Va 3 πb  3 πa 4π  b3  a 3 ... dV  ε0 ε0 q 4π 3 q q(r 3  c 3 ) 3 E 4 r   (r  c ) ρouter, so E   ε0 3ε0 4πε0 r 2 4πε0 r 2 (d 3  c 3 )  q q  (v) r  d  E  d A    0  E  0 ε0 ε0 2 22. 45: a) a  r  b, E  b) r  c, E  1 4 πε0 1 2λ , radially outward, as in 22. 48 (b) 4 0 r 2λ r , radially outward, since again the charge enclosed is the same as in part (a) c) d) The inner and outer surfaces of the outer cylinder must... λ outer  λ α q αl  E 2πε0 r ε0 ε0 (ii) a  r  b, there is no net charge enclosed, so the electric field is zero 22. 46: a) (i) r  a, E (2πrl )  (iii) r  b, E (2πrl )  α q 2αl  E ε0 ε0 πε0 r b) (i) Inner charge per unit length is  α (ii) Outer charge per length is  2 22. 47: a) (i) r  a, E (2πrl )  q ε0  αl ε0 E α 2 πε 0 r , radially outward (ii) a  r  b, there is not net charge... Inner charge per unit length is   (ii) Outer charge per length is ZERO 22. 48: a) r  R, E (2πrl )  q ε0  ρπr 2 l ε0 E b) r  R, and λ  ρπR 2 , E (2πrl )  ρr 2ε 0 q ε0  , radially outward ρπR 2l ε0 E ρR 2 2 ε0 r c) r  R the electric field for BOTH regions is E  ρR 2ε 0  λ 2 πε0 r  2 kλ r , so they are consistent d) 22. 49: a) The conductor has the surface charge density on BOTH sides, so... undergo simple harmonic motion, because for r  R, F  1 r 2 , and is not linear 22. 51: The electrons are separated by a distance 2d , and the amount of the positive nucleus’s charge that is within radius d is all that exerts a force on the electron So: ke 2 Fe   Fnucleus  2ke 2 Rd3  d 3  R 3 / 8  d  R / 2 2 (2d ) 22. 52: a) Q(r )  Q   ρdV  Q  Q 3 πa0 e  2 r / a0 r 2 dr sin  dθ d  Q... E , ε0 ε0 ε0 ˆ again with direction given by x i | x| 22. 55: a) Again, E is zero at x  0 , by symmetry arguments b) x  d :   EA  x  d :   EA  Qencl ρ A  0 2 ε0 ε0 d Qencl ρ A  0 2 ε0 ε0 d x 2  x' dx'  0 d  x' 0 2 dx'  ρ0 Ax 3 ρ x3 x ˆ  E  0 2 , in i direction 2 3ε0 d 3ε0 d |x| ρ0 Ad ρd x ˆ  E  o , in i direction 3ε0 3ε0 | x| 22. 56: a) We could place two charges  Q on either side... 0.097 m 0.900  10 9 C Q2 22. 36: a) For r  a, E  0, since no charge is enclosed For a  r  b, E  q 1 4 πε 0 r 2 , since there is +q inside a radius r For b  r  c, E  0, since now the –q cancels the inner +q For r  c, E  q 1 4 πε 0 r 2 , since again the total charge enclosed is +q b) c) Charge on inner shell surface is –q d) Charge on outer shell surface is +q e) 22. 37: a) r  R, E  0, since... two electric fields cancel, so E  0 d) x  3R : now both spheres contribute fields pointing to the right:  1  Q 1 10Q ˆ Qˆ  E ( x  3R )   (3R) 2  R 2  i  4πε 9 R 2 i  4πε0   0 22. 64: (See Problem 22. 63 with Q   Q for terms associated with right sphere)  1 Q ˆ a) E ( x  0)   i 4πε0 4 R 2  ˆ 1  (Q 8) 1  Q 4Q  ˆ 1 17Q ˆ Q R b) E  x     ( R 2) 2  (3 R 2) 2  i  4πε  2 . )mC6mC4mC2mC5( 2 1 )( 2 1 5 222 2 0 1432 0    μμμμ ε σσσσ ε E A (b) left. thetoCN1095.3 )mC5mC4mC2mC6( 2 1 )( 2 1 222 2 5 222 2 0 2431 00 2 0 4 0 3 0. thetoCN1069.1 )mC6mC4mC2mC5( 2 1 )( 2 1 222 2 5 222 2 0 1432 00 1 0 4 0 3 0 2    μμμμ ε σσσσ εε σ ε σ ε σ ε σ E C 22. 29: a) Gauss’s law says +Q on