Chapter 22 - Resolução Sears - Vol 3

GAUSS’S LAW 22 22.1. (a) IDENTIFY and SET UP: ϕE = ∫ E cos 𝜙dA, where 𝜙 is the angle between the normal to the sheet 𝑛 and the electric field E. EXECUTE: In this problem E and cos 𝜙 are constant over the surface so ϕE = E cos 𝜙A = E cos(24°)(14 N/C)(cos(69°))(0.250 m²) = 1.8 N·m²/C. (b) EVALUATE: ϕE is independent of the shape of the sheet as long as 𝜙 and E are constant at all points on the sheet. (c) EXECUTE: (i) ϕE = E cos 𝜙A = E (cos(0°)) A, largest for 𝜙 = 0°, so cos 𝜙 = 1 and ϕE = EA. (ii) ϕE is smallest for 𝜙 = 90°, so cos 𝜙 = 0 and ϕE = 0. EVALUATE: ϕE is 0 when the surface is parallel to the field so no electric field lines pass through the surface. 22.2 IDENTIFY: The field is uniform and the surface is flat, so ϕE = E A cos 𝜙. SET UP: 𝜙 is the angle between the normal to the surface and the direction of E, so 𝜙 = 70°. EXECUTE: ϕE = (7.50 N/C)(0.400 m)(0.600 m)(cos 70°) = 6.16 N·m²/C EVALUATE: If the field were perpendicular to the surface the flux would be ϕE = EA = 18.0 N·m²/C. The flux in this problem is much less than this because only the component of E perpendicular to the surface contributes to the flux. 22.3. (a) EXECUTE: The electric flux through an area is defined as the product of the component of the electric field perpendicular to the area times the area. (a) SET UP: In this case, the electric field is perpendicular to the surface of the sphere, so ϕE = EA = E(4π r²). EXECUTE: Substituting in the numbers gives ϕE = (1.25×10⁶ N/C)(4π)(0.150 m)² = 3.53×10¹ N·m²/C (b) IDENTIFY: We use the electric field due to a point charge. SET UP: E = kq/r² EXECUTE: Solving for q and substituting the numbers gives q = 4πkϵo E r² = q = 9.00×10⁹ N·m²/C² × (1.25×10⁶ N/C)(0.150 m)² = 3.13×10⁻⁶ C 22-2 Chapter 22 (b) Total flux: ϕ = ϕ₁ + ϕ₂ = (0.081 – 0.135)(N·C⁻¹·m²) = –0.054 N·m²/C. Therefore, q = ϵoϕ = –4.78×10⁻¹¹ C. EVALUATE: Flux is positive which E is directed out of the volume and negative when it is directed into the volume. 22.5 IDENTIFY: The flux through the curved upper half of the hemisphere is the same as the flux through the flat circle defined by the bottom of the hemisphere because every electric field line that passes through the flat circle also must pass through the curved surface of the hemisphere. SET UP: The electric field is perpendicular to the flat circle, so the flux is simply the product of E and the area of the flat circle of radius r. EXECUTE: ϕE = EA = Eπr² = πRE² EVALUATE: The flux would be the same if the hemisphere were replaced by any other surface bounded by the flat circle. 22.6 IDENTIFY: Use Eq.(22.3) to calculate the flux for each surface. SET UP: ϕE = EA cos 𝜙 where A ⊥ A. EXECUTE: (a) ϕE = 0, so ϕ = –(4×10¹ N/C)(0.10 m²)cos(90° – 36.9°) = –24 N·m²/C. ... (e) ϕE tops = –(4×10¹ N/C)(0.10 m²)cos(36.9°) – 24 N·m²/C. (b) ϕE front = –(4×10¹ N/C)(0.10 m²)cos 53° (c) ϕE back = –(4×10¹ N/C)(0.10 m²)cos 50° = –32 N·m²/C. (d) ϕE bottom = –(4×10¹ N/C)(0.10 m²)cos 59° – 32 N·m²/C: ... (j) ϕE back = –(4×10¹ N/C)(0.10 m²)cos 29° = –28 N·m²/C: 22.7. EXECUTE: The area of the curved part of the cylinder is A = 2πrl. The electric field is parallel to the end caps of the cylinder, so E·A = 0 for the ends and the flux through the cylinder end caps is zero. The electric field is normal to the curved surface of the cylinder and has the same magnitude E = λ/2πϵor at all points on this surface. Thus 𝜙 = 0° and ϕE = EA cos 𝜙 = EA = (λ/2πϵor) × [2πrl] = (6.00×10⁴ C/m)(0.400 m) 2.71×10⁻² N·m²/C ... λ/8.854×10¹² (C/N·m²)/(m) = 2.71×10¹ N·m²/C. (b) In the calculation in part (a) the radius r of the cylinder divided out, so the flux remains the same. = 2.71×10¹ N·m²/C. (e) eA = (6.00×10⁴ C/m)(0.800 m) 8.854×10¹² C/N·m² = 5.42×10¹ N· m²/C (notice the flux calculated in parts (b) and (c)). EVALUATE: The flux depends on the number of field lines that pass through the surface of the cylinder. 22.8 IDENTIFY: Apply Gauss’s law to each surface. SET UP: qenc is the algebraic sum of the charges enclosed by each surface. Flux out of the volume is positive and flux into the enclosed volume is negative. EXECUTE: (a) = q¹/qenc = 4.00×10⁻⁶ C/ϵo = 452 N·m²/C. (b) ϕE = q¹/q = 7.80×10⁻¹ C ϕ = 81 N·m²/C. (c) ϕE = (4.00 – 7.80)×10⁻160 N/ϵo d ϕE = 429 N·m²/C. (d) ϕE = q¹/q = (4.00 + 2.40)×10⁻¹ C/ϵo = 723 N·m²/C. (e) ϕE = q¹/q = 4.00 + 2.40)×10⁻¹ C(ϵo = –156 N·m²/C. EVALUATE: (f) All that matters for Gauss’s law is the total amount of charge enclosed by the surface, not its distribution within the surface. 22.9 IDENTIFY: Apply the results in Example 22.10 for the field of a spherical shell of charge. SET UP: Example 22.10 shows that E = 0 inside a uniform spherical shell and that E = kq/r² outside the shell. EXECUTE: (a) E = 0 (kq/r²) (b) r = 0.060 m and E = (8.99×10⁹ N·m²/C²)) × (3.75×10² N/C)(0.060 m)² =... N/C (c) r = 0.110 m and (8.99×10⁹ N·m²/C²) (E) λ = 8.99 (8.99×10⁹ N·m²/C²)(0.110 m) = 3.1×10⁻6 C. EVALUATE: Outside the shell the electric field is the same as if all the charge were concentrated at the center of the shell. But inside the shell the field is not the same as for a point charge at the center of the shell, inside the shell the electric field is zero. 22.10 IDENTIFY: Apply Gauss’s law to the spherical surface. SET UP: qenc is the algebraic sum of the charges enclosed by the sphere. EXECUTE: (a) The charge enclosed so ϕ = qt/ϵo... 0.0130 x C (b) ϕE = 8.85×10¹¹ C/N·m² = Note: N·m²/C = –678 N·m²/C. (c) ϕE = q¹/q = q₁ (4,400 – 6x3.00)×10⁴ C/N·m² 8.85×10¹¹ C/N·m² EVALUATE: Negative flux corresponds to flux directed into the enclosed volume. The net flux depends only on the total charge enclosed by the surface and is not affected by any charges outside the enclosed volume. 22.11 IDENTIFY: Apply Gauss’s law. EXECUTE: (a) Since E is uniform, the flux through a closed surface must be zero. That is: ∮ E·dA= 0, so qnet = 0. EXECUTE: In this region, E is uniform and the result does not change any volume we want, q must be zero if the integral equals zero. 22.12. IDENTIFY: Apply Gauss’s law. SET UP: Use a small Gaussian surface located in the region of interest. EXECUTE: (a) If ρ > 0 and uniform, then 𝜎 inside any closed surface is greater than zero. This implies ϕᵧ > 0, so ∮ E·dA= 0 and so the electric field cannot be uniform. That is, 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 charge 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 problem 22.6.) EVALUATE: In a region of uniform field, the flux through any closed surface is zero. 22.13. (a) IDENTIFY and SET UP: It is rather difficult to calculate the flux directly from ϕ = ∫ E·dA since the magnitude of E and its angle with dA varies over the surface of the cube. A much easier approach is to use Gauss’s law to calculate the total flux through the cube. Let the cube be the Gaussian surface. The charge enclosed is the point charge. EXECUTE: ϕE = qnet/ϵo 6.00×10⁻⁶ C 8.854×10¹¹ C/N·m² = 1.08×10⁴ N·m²/C. By symmetry the flux is the same through each of the six faces, so the flux through one face equivalent to:... (1.08×10⁴ N·m²/C) = 1.81×10³ N·m²/C. (b) EVALUATE: In part (a) the size of the cube did not enter into the calculations. The flux through one face depends only on the amount of charge at the center of the cube. So the answer to (a) would not change if the size of the cube were changed. 22.25. IDENTIFY: The magnitude of the electric field is constant at any given distance from the center because the charge density is uniform inside the sphere. We can use Gauss’s law to relate the field to the charge causing it. (a) SET UP: Gauss’s law tells us that E= qencl/ ε0A and the charge density is given by ρ = q/V = q/(4/3)πR3 EXECUTE: Solving for q and substituting numbers gives q = E ε0 ∙ 4πr2 = (8750 N/C)(7.40×10−12 C/N ∙ m2) ∙4.866×10−2 C. •Using the formula for charge density we get ρ = q V = (4.3)π R3 = ns 4x3 (4.366×10−2 C) 2.66×10−7 C/m3. (b) SET UP: Take a Gaussian surface of radius r = 0.200 m, concentric with the insulating sphere. The charge enclosed within this surface is qenc = ρ(4 r3)π 3 and we can treat this charge as a point-charge, using Coulomb’s law E= 1 4πε0 [e4t =qenc] . The charge beyond r = 0.200 m makes no contribution to the electric field. EXECUTE: First find the enclosed charge: qenc = [4r(25)π 3] [2.66×10−7 C/m3] [(0.240.200m)3] = 8.70×10−9 C Now treat this charge as a point-charge and use Coulomb’s law to find the field: E= (9.0×109 N ∙ m2/C2 8.70×10−9 C 4(0.200 m)2 =1.96×102 N/C EVALUATE: Outside this sphere, it behaves like a point-charge located at its center. Inside of it, at a distance r from the center, the field is due only to the charge between the center and r. 22.26. IDENTIFY: Apply Gauss’s law and conservation of charge SET UP: Use a Gaussian surface that lies wholly within the conducting material. EXECUTE: (a) Positive charge is attracted to the inner surface of the conductor by the charge in the cavity. Its field lines terminate on negative charge, inducing an equal negative charge there. Since ® = 0 inside a conductor and a Gaussian surface that lies wholly within the conductor must enclose zero net charge. (b) For a Gaussian surface that lies entirely inside the region between the outer surface of the conductor and the charge “left behind” as the +6.00 nC moved to the inner surface: nqa enc = +10.0 nc +5.00 nc =-6.00 nc =–1.00 nC. 22.27. EVALUATE: The electric field outside the conductor is due to the charge on its surface. IDENTIFY: Apply Gauss’s law to each surface. SET UP: The field is zero within to the plates. By symmetry the field is perpendicular to the plates outside the plates and can depend only on the distance from the plates. Flux into the enclosed volume is positive. EXECUTE: SE and 9, enclose no charge, so the flux is zero and, no electric field outside the plates is zero. Between the plates, g, shows that – EA – Qe/ε0 = 4 A/ε0 = En & Eb /ε0; EVALUATE: Our result for the field between the plates agrees with the result stated in Example 22.28. 22.28. IDENTIFY: Close to a finite sheet the field is the same as for an infinite sheet. Very far from a finite sheet the field is that of a point charge. SET UP: For an infinite sheet, E= σ 2ε0 . For a point charge, E= λ 4πε0 = q 4 2r2 EXECUTE: (a) At a distance of 0.1 mm from the center, the sheet appears “infinite,” so E= λ 2ε0 =(7.50×10−6 C 2(0.180 m) = 662 N/C. (b) At a distance of 100 m from the center, the sheet looks like a point, so E= 1 (4r2) = (7.50×10−6 C) (4π(100 m)2) = 6.75×10−3 N/C. (c) There would be no difference if the sheet was a conductor. The charge would automatically spread out evenly over both faces, giving half the charge density on either face as the insulator but the same electric field. Far away, they both look like points with the same charge. EVALUATE: The sheet can be treated as infinite at points where the distance to the sheet is much less than the dimensions of the edge of the sheet. The sheet can be treated as a point charge at points for which the distance to the sheet is much greater than the dimensions of the sheet. 22-8 Chapter 22 22.29. IDENTIFY: Apply Gauss’s I law to a Gaussian surface and calculate E. (a) SET UP: Consider the charge on a length l of the cylinder. This can be expressed as q = λl. But since the surface area is 2πRl it can also be expressed as q = σ(2πRl). These two expressions must be equal, so λ = σ2πR and λ = 2πRσ. (b) Apply Gauss’s law to a Gaussian surface that is a cylinder of length l, radius r, and whose axis coincides with the axis of the charge distribution, as shown in Figure 22.29. EXECUTE: qencl = σ(2πRl) ΦE = 2πrlE Figure 22.29 qencl 9 gives 2πrlE = [2rπRl] , E= qr EVALUATE: Example 22.6 shows that the electric field of an infinite line of charge is E= λ 2πε0r , σ = λ 2πRε0r = Eλ 2πε0r is the same as for an infinite line of charge that is along the axis of the cylinder. 22.30. IDENTIFY: The net electric field is the vector sum of the fields due to each of the four sheets of charge. SET UP: The electric field of a large sheet of charges is E= |σ 2ε0| . The field is directed away from a positive sheet and toward a negative sheet. EXECUTE: (a) At E3=C (5μ C/m2 ± 2μ C/m2)ε0 = ±14 ε0 2 μ c/m n2± 8.82ϯ×10−2 N/C to the left. E4 = 1 (5μ, C/m2 ± 2μ C/m2, ±4μ C m (m n; +5 c/m2) n3 = 9.395×10 N/C to the left. (c) E6=12 q0 q quq4 y° ez e. -E J. [l Z6 26 20 20 5 { (4μ C/m2 + 6μ C/m2-5μ C/m2-ln2 = 1.69×10] N/C to the left EVALUATE: The field at C is not zero. The pieces of plastic are not conductors. 22.31. IDENTIFY: Apply Gauss’s law and conservation of charge SET UP: E=n0 in a conducting material. EXECUTE: (4) Gauss’s law says ¾Q=0 inner surface, ec = 0 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 runs through it; there is electric field in the cavity. (d) In electrostatic situation E = 0 inside a conductor. A Gaussian sphere with the -Q charge at its center and radius greater than outer: of the conductor must enclose zero charge since the +Q charge and the -Q on the surface of the metal so there is no flux through it and E= 0 outside the metal. e) No, E=0 there. Yes, the charge has been shielded by the grounded conductor. There is nothing like positive and negative waves (the gravity force is always attractive), so this cannot be done for gravity. EVALUATE: Field lines within the cavity terminate on the charges induced on the inner surface. 22-9 Gauss’s Law 22.32. IDENTIFY and SET UP: Eq(22.23 ) to calculate the flux. Identify the direction of the normal unit vector A for each surface. EXECUTE: (a) E= BE + Q Dr: A=L3 face E5=i= k ΦE=E5·A=E(+i)· (+k)= 0 CL. face E3=i–k ΦE=E3·A=E{ –i) 4–i) (5–i) { = ) -E(k) =IDI} -E { Jk) – (k)= =d = C. face E3=i cos:ms = ak, ΦE=E3·A=E{ +i) - (i)· (k)=0 EII face-E, = a.(k) = =akie, face E5=ak6-6a ΦE=E5·A=I–i(-k) (k)=) =CADL2 CL, that a add the flux through each of the six faces: a= CEL2 + 6 DL + CEL2 + DL: = CL (7) EL=0 there at all net electric flux through all six sides is zero. EVALUATE: All electric field lines that enter one face of the cube leave through another face. No electric field lines terminate inside the cube and the net flux is zero. 22.33. IDENTIFY: Use Fig 22.3, to calculate the flux through each surface and use Gauss’s law to relate the net flux to the enclosed charge. SET UP: Flux into the enclosed volume is negative and flux out of the volume is positive. EXECUTE: (a) q= EA (25$caringl/C6.06) m2 = 750 N m2/C. (b) Since the field is parallel to the surface, ac, gx 0. (o) Choose the Gaussian surface to equtvale to volume's surface. Then 750 N m/C- EA q = Ed, E EA: -0.4X10(N crta) + 750= [IN] NC, in the positive 4-direction. Since q < 0 we must have some net flux flowing in so the flux is [-EA] on second face. EVALUATE: (d) q <0 but we have E pointing away from face L This is due to an external field that does not affect the flux but affects the value of E. 22.34. IDENTIFY: Apply Gauss’s law of a cube centered at the origin and with side length 2L. SET UP: The total surface area of a cube with side length 2L is 6(2L)2 = 24L2 . EXECUTE: (a) Te square is sketched in Figure 22.34. (b) Imagine a charge q at the center of a cube of edge length 2L Here the square is one 24th of the surface area of the imaginary cube, so it intercepts 1/24 of the flux. That is, Φ E. = q/ E0 EVALUATE: Calculating the flux directly from Eqa(22.5) would involve a complicated integral. Using Gauss’s law and symmetry considerations is much simpler. 22-10 Chapter 22 22.35. (a) IDENTIFY: Find the net flux through the parallelepiped surface and then use that in Gauss’s law to find the net charge within. Flux out of the surface is positive and flux into the surface is negative. SET UP: E1 gives flux out of the surface. See Figure 22.35a. Figure 22.35a EXECUTE: ϕ1 = EA1 A1 = (0.0600 m)(0.0500 m) = 3.00x10-3 m2 E1 = E0 cos60° = (2.50x104 N/C)(cos60°) E1 = 1.25x104 N/C ϕ1 = E1A1 = (1.25x104 N/C)(3.00x10-3 m2) = 37.5 N·m2/C SET UP: E2 gives flux into the surface. See Figure 22.35b. Figure 22.35b EXECUTE: ϕ2 = E2A2 A2 = 0.0600 m)(0.0500 m) x 3.00x10-2 m2 E2 = E0 cos60° = (7.00x104 N/C)(cos60°) E2 = 3.50x104 N/C ϕ2 = E2A2 = (3.50x104 N/C)(3.00x10-2 m2) = -105.0 N·m2/C The net flux = ϕ1 + ϕ2 = +37.5 N·m2/C - 105.0 N·m2/C = -67.5 N·m2/C. The net flux is negative (inward), so the net charge enclosed is negative. Apply Gauss’s law: ϕ = Qenc/ε0 Qenc = ϕε0 = (-67.5 N·m2/C)(8.85x10-12 C2/N·m2) = -5.98x10-10 C. (b) EVALUATE: If there were no charge within the parallelepiped the net flux would be zero. This is not the case, so there is charge within. The electric field lines that pass out through the surface of the parallelepiped must terminate (there are no monopoles); they must terminate on negative charge. 22.36. IDENTIFY: The α particle feels no force, this means the net electrical field due to the two distributions of charge is zero. SIGN UP: The fields can cancel only if one field is inward A and B shown in Figure 22.36, apply on these two charges. EXECUTE: Enet = Eline Estretch 30 μC/cm 0.16 m = 16 cm = l. The fields cancel 16 cm. from the line in regions A and B. EVALUATE: The result is independent of the distance between the line and the sheet. The electric field of an infinite sheet of charge is uniform, independent of the distance from the sheet. Gauss’s Law 22-11 22.37. (a) IDENTIFY: Apply Gauss’s law to a Gaussian cylinder of length l and radius r, where r > c, as shown in Figure 22.37b. Figure 22.37b EXECUTE: ϕ = E(2πrl) Qenc = λl (the charge on the length l of the inner conductor that is inside the Gaussian surface; the outer conductor carries no net charge). ϕ = Qenc/ε0 gives E(2πrl) = λl / ε0 E = λ / 2πrε0 The enclosed charge is positive so the direction of E is radially outward. (e) E = 0 within a conductor. Thus E = 0 for r < a; E = λ / 2πrε0 for a < r < b; E > 0 for b < r < c; E = λ / 2πrε0 for r > c. The graph of E versus r is sketched in Figure 22.37c. Figure 22.37c EVALUATE: Inside either conductor E = 0. Between the conductors and outside both conductors the electric field is the same as for a line of charge with linear charge density λ and l lying along the axis of the inner conductor. (d) IDENTIFY and SET UP: inner surface; Apply Gauss’s law to a Gaussian cylinder with radius r, where b < r < c. We know E on this surface; calculate Qenc. EXECUTE: This surface lies within the conductor of the outer cylinder, where E = 0, so ϕ = 0. Thus by Gauss’s law Qenc = 0. The surface encloses charge λl on the inner conductor, so it must enclose charge - λl on the inner surface of the outer conductor. The charge per unit length on the inner surface of the outer cylinder is - λ. outer surface: The outer cylinder carries no net charge. So if there is charge per unit length + λ on its inner surface there must be charge per unit length - λl on the outer surface. EVALUATE: The electric field lines between the conductors originate on the surface charge on the outer surface of the inner conductor and terminate on the surface charges on the inner surface of the outer conductor. These surface charges are equal in magnitude (per unit length) and opposite in sign. The electric field lines outside the outer conductor originate from the surface charge on the outer surface of the outer conductor. 22.38. IDENTIFY: Apply Gauss’s law. SET UP: Use a Gaussian surface that is a cylinder of radius r, length l and that has the line of charge along its axis. The charge on length of the line of charge is q the charge of the tube is q = αl. Figure 22.38a. EXECUTE: (a)(i) For r < a, Gauss’s law gives E(2πrl) = αl / ε0 and E = α / 2πε0r. (ii) The electric field is zero because these points are within the conducting material. (iii) For r > b, Gauss’s law gives E(2πrl) = αl / ε0 and E = α / 2πε0r. The graph of E versus r is sketched in Figure 22.38. 22-12 Chapter 22 (b)(i) The Gaussian cylinder with radius r, for a < r < b, must enclose zero net charge, so the charge per unit length on the inner surface is −σ. (ii) Since the net charge per length for the tube is +σ and there is −σ on the inner surface, the charge per unit length on the outer surface must be +2σ. EVALUATE: For r > b the electric field is due to the charge on the outer surface of the tube. 22.39. (a) IDENTIFY: Use Gauss’s law to calculate E(r). (b) SET UP: For r < a Apply Gauss’s law to a cylindrical Gaussian surface of length l and radius r, where r < a, as sketched in Figure 22.39a. Figure 22.39a EXECUTE: ϕ = E(2πrl) Qenc = αl (the charge on length l of the line of charge) ϕ = Qenc/ε0 gives E(2πrl) = αl /ε0 E = α / 2πrε0 The enclosed charge is positive so the direction of E is radially outward. (ii) a < r < b; points in this region are within the conducting tube, so E = 0. (iii) SET UP: For r > b Apply Gauss’s law to a cylindrical Gaussian surface of length l and radius r, where r > b, as sketched in Figure 22.39b. Figure 22.39b EXECUTE: ϕ = E(2πrl) Qenc = αl (the charge on length l of the line of charge) − αl (the charge on length l of the tube) Thus Qenc = 0. ϕ = Qenc/ε0 gives E(2πrl) = 0 and E = 0. The graph of E versus r is sketched in Figure 22.39c. Figure 22.39c (b) IDENTIFY: Apply Gauss’s law to cylindrical surfaces that lie just outside the inner and outer surfaces of the tube. We know E so can calculate Qenc. (c) SET UP: inner surface. Apply Gauss’s law to a cylindrical Gaussian surface of length l and radius r, where a < r < b. Gauss’s Law 22-13 EXECUTE: This surface lies within the conductor of the tube, where E = 0, so Φe = 0. Then by Gauss’s law Qencl = 0. The surface encloses charge -aℓ on the line of charge so must enclose charge +aℓ on the inner surface of the tube. The charge per unit length on the inner surface of the tube is -a. (ii) outer surface The net charge per unit length on the tube is -a. We have shown in part (i) that this must all reside on the inner surface, so there is no net charge on the outer surface of the tube. EVALUATE: For r < a the electric field is due only to the line of charge. For r > b the electric field is the same as for a line of charge along its axis. The field’s of the line of charge and of the tube are equal in magnitude there is no net charge on any surface. For r < a the electric field lines originate on the line of charge and terminate on the surface charge on the inner surface of the tube. There is no electric field outside the tube and no surface charge on the outer surface of the tube. 22.40 IDENTIFY: Apply Gauss’s law. SET UP: Use a Gaussian surface that is a cylinder of radius r and length l, and that is coaxial with the cylindrical charge distributions. The volume of the Gaussian cylinder is πr²l and the area of its curved surface is 2πrl. The charge on a length l of the charge distribution is q = 2λ ,where λ = 2πεEl . EXECUTE: (a) For r < R, Qencl = ρπr²l and Gauss’s law gives E(2πrl) = Qencl so E = RρRl , radially outward. ε0 2πε0 (b) For r > R, Qencl = 2πRl² and Gauss’s law gives E(2πrl) = Qencl so E = 2, radially R 2πε0 2πε0 r outward. (c) At r = R, the electric field for BOTH regions is E = 2, so they are consistent. R (d) The graph of E versus r is sketched in Figure 22.40. EVALUATE: For r > R the field is the same as for a line of charge along the axis of the cylinder. 22.14 Chapter 22 22.42 IDENTIFY: Apply Gauss’s law. SET UP: Use a Gaussian surface that is a sphere of radius r and that is concentric with the conducting spheres. EXECUTE: (a) For r < a, E = 0, since these points are within the conducting material. For a < r < b, E = 4πq , since there is a q inside a radius r. 4πε0r² For b < r < c, E = 0, since these points are within the conducting material For r > c, E = 4πq, since again the total charge enclosed is +q. 4πε0r² (b) The graph of E versus r is sketched in Figure 22.42a. (c) Since the Gaussian sphere of radius r, for b < r < c, must enclose zero net charge, the charge on inner shell surface is +q. (d) Since the hollow sphere has no net charge and has charge -q on its inner surface, the charge on outer shell surface is -q. (e) The field lines are sketched in Figure 22.42b. Where the field is nonzero, it is radially outward. EVALUATE: The net charge on the inner solid conducting sphere is the surface of that sphere. The presence of the hollow sphere does not affect the electric field in the region r < b. -q, as on 22.43 IDENTIFY: Apply Gauss’s law and conservation of charge. SET UP: Use a Gaussian surface that is a sphere of radius r and that is concentric with the charge distributions. EXECUTE: (a) For r < R, E = 0, since these points are within the conducting material. For R < r < 2R, 4πQ, since the charge enclosed is Q. For r > 2R, E = 1 ( 2Q ), since the charge enclosed is 2Q. 4πε0r² 4πε0 (b) The graph of E versus r is sketched in Figure 22.43. EVALUATE: For r < 2R the electric field is unaffected by the presence of the charged shell. E Figure 22.42 22.15 Gauss’s Law (b) Since a Gaussian surface with radius r, for a < r < b, must enclose zero net charge, the total charge on the inner surface is -Q and the surface charge density on inner surface is -Q. 4πr² (c) Since the net charge on the shell is -3Q and there is -Q on the inner surface, there must be -2Q on the outer surface. The surface charge density on the outer surface is -2Q. 4πc² (d) The field lines and the locations of the charges are sketched in Figure 22.44a. (e) The graph of E versus r is sketched in Figure 22.44b. E Figure 22.44 EVALUATE: For r < a the electric field is due solely to the point charge Q. For r > b the electric field is due to the charge -2Q that is on the outer surface of the shell. 22.45 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) SET UP: r < a: The Gaussian surface is sketched in Figure 22.45a. EXECUTE: Φe = EA = E(4πr²) Qencl = 0; no charge is enclosed Qencl = 0 E(4πr²) = 0 and E = 0. Figure 22.45a (ii) a < r < b: Points in this region are in the conductor of the small shell, so E = 0. (iii) SET UP: b < r < c: The Gaussian surface is sketched in Figure 22.45b. Apply Gauss’s law to a spherical Gaussian surface with radius b < r < c. EXECUTE: Φe = EA = E(4πr²) The Gaussian surface encloses all of the small shell and none of the large shell, so Qencl = +2q. Φe = Qencl gives E(4πr²) = 2q, so E = 2q. Since the enclosed charge is positive the electric field is radially ε0 4πε0r² outward. (iv) c < r < d: Points in this region are in the conductor of the large shell, so E = 0. 22-16 Chapter 22 (v) Set Up: r > c: Apply Gauss’s law to a spherical Gaussian surface with radius r > d, as shown in Figure 22.45c. Execute: Φe = EA = E(4πr²) The Gaussian surface encloses all of the small shell and all of the large shell, so Qencl = +2q + 4q = 6q. Figure 22.45c Φe = Qencl gives E(4πr²) = 6q/ε₀ E = 6q/(4πε₀r²) Since the enclosed charge is positive the electric field is radially outward. The graph of E versus r is sketched in Figure 22.45d. E 6q/4πε0a² 2q/4πε0d² r Figure 22.45d (b) Identify and Set 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 Φe = 0. Thus by Gauss’s law 0en = 0, so there is zero charge on the inner surface of the small shell. (ii) charge on outer surface of the small shell: The total charge on the small shell is +2q. We found in part (i) that there is zero charge on the inner surface of the shell, so all +2q 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 Φe = 0. Thus by Gauss’s law Qencl = 0. The surface encloses the +2q on the small shell so there must be charge -2q 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 +4q. We showed in part (iii) that the charge on the inner surface is -2q. So there must be +6q on the outer surface. Evaluate: The electric field lines for 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. 22.46 Identify: Apply Gauss’s law. Set Up: Use a Gaussian surface that is a sphere of radius r and that is concentric with the charged shells. Execute: (a) (i) For r < a, E = 0, since the charge enclosed is zero. (ii) For a < r < b, E = 0, since the points are within the conducting material. (iii) For b < r < c, E = q/4πε₀r², outward, since charge enclosed is +2q. (iv) For c < r < d, E = 0, since the points are within the conducting material. (v) For r > d, E = 0, since the net charge enclosed is zero. The graph of E versus r is sketched in Figure 22.46. E q/4πε0b² q/4πε0c² r O c d Figure 22.46 22.47. Identify: Apply Gauss’s law Set Up: Use a Gaussian surface that is a sphere of radius r and that is concentric with the charged shells. Execute: (a) (i) For r < a, E = 0, since charge enclosed is zero. (ii) a < r < b, E = 0, since the points are within the conducting material. (iii) For b < r < c, E = 2q/4πε₀r², outward, since charge enclosed is +4q. (iv) For c < r < d, E = 0, since the points are within the conducting material. (v) For r > d, E = 4q/4πε0r2, inward, since charge enclosed is -2q. The graph of the radial component of the electric field versus r is sketched in Figure 22.47. When we use the convention that the radial field is positive if it is directed inward, the direction of Figure 22.47, then a positive radial electric field is negative. (b) small shell inner surface: Since a Gaussian surface with radius r, for a < r < b, must enclose zero net charge, the charge on this surface is zero. (ii) small shell outer surface: 4qa2. (iii) large shell inner surface: Since a Gaussian surface with radius r, a < r < d, must enclose zero net charge, the charge on this surface is -2q. (iv) large shell outer surface: Since there is -2q on the inner surface and the total charge on this conductor is -2q, the charge on this surface is zero. Evaluate: The outer shell has no effect on the electric field for r < c. For r > d the electric field is due only to the charge on the outer surface of the larger shell. 22-18 Chapter 22 22.48. Identify: Apply Gauss's law. Set Up: Use a Gaussian surface that is a sphere of radius r and that is concentric with the sphere and shell. The volume of the insulating shell is V = 4/3π[(2R)³ - R³] = 28πR³/3. Execute: (a) Zero net charge requires that -Q - 3Q/28πR³ = 0, so Q = 3Q/28πR³. (b) For r < R, E = 0 since this region is within the conducting sphere. For r > 2R, E = 0, since the net charge enclosed by the Gaussian surface with this radius is zero. For R < r < 2R, Gauss's law gives E(4πr²) = Qε₀ V + 4/3 (r³ - R³) and each r in this range is positive and the electric field is outward. E = Q/4πε₀(r³ - R³) Substituting ρ from part (a) gives E = 3Q/28πε₀R³. The net enclosed charge for each r in this range is positive and the electric field is outward. The graph is sketched in Figure 22.48. We see a discontinuity in going from the conducting sphere to the insulator due to the thin surface charge of the conducting sphere. But we see a smooth transition from the uniform insulator to the surrounding space. Evaluate: The expression for E within the insulator gives E = 0 at r = 2R. E Q/(28πε0R3)r R 2R r Figure 22.48 22.49 Identify: Use Gauss's law to find the electric field E produced by the shell for r < R and r > R and then use F = qE to find the force the shell exerts on the point charge. (a) Set Up: Apply Gauss’s law to a spherical Gaussian surface that has radius r > R and that is concentric with the shell, as sketched in Figure 22.49a. - q - Figure 22.49a Execute: Φe = E(4πr²) Qencl = -Q Φe = Qencl gives E(4πr²) = -Q/ε₀ The magnitude of the field is E = -Q/4πε₀r² and it is directed toward the center of the shell. Then F = qE = -qQ/4πε₀r² The magnitude of the field is E = directed toward the center of the shell. (This is because charge q is positive, while charge -Q is negative.) and E are in the same direction. (b) Set Up: Apply Gauss’s law to a spherical Gaussian surface that has radius r < R and that is concentric with the shell, as sketched in Figure 22.49b. Execute: Φe = E(4πr²) Qencl = 0 Φe = Qencl gives E(4πr²) = 0 Then E = 0 so F = 0. Evaluate: Outside the shell the electric field and the force it exerts is the same as for a point charge -Q located at the center of the shell. Inside the shell E = 0 and there is no force. 22.50 Identify: The method of Example 22.9 shows that the electric field outside the sphere is the same as for a point charge of the same charge located at the center of the sphere. Set Up: The charge of an electron has magnitude e = 1.60 x 10^-¹⁹ C. Gauss’s Law 22-19 Execute: (a) E = k |q| / r². For r = R = 0.150 m, E = 1150 N/C so |q| = (1150 N/C)(0.150 m)² = 2.88x10¯¹º C. 8.99×10⁹ N·m²/C² The number of excess electrons is 2.88×10¯¹ºC = 1.80×10¹⁰ electrons. 1.60×10¯¹⁹ C/electron (b: r = R + 0.100 m = 0.250 m . E = k |q| = (8.99×10⁹ N·m²/C²) 2.88×10¯¹º C = 414 N/C. (r + 0.250) m² (0.250) m² Evaluate: The magnitude of the electric field decreases according to the square of the distance from the center of the sphere. 22.51. Identify: The net electric field is the vector sum of the fields due to the sheet of charge on each surface of the plate. Set Up: The electric field due to the sheet of charge on each surface is E = σ/2ε₀ and is directed away from the surface. Execute: (a) For the conductor the charge absent on each surface produces fields of magnitude σ/2ε₀ and in the same direction, so the total field is twice this, or σ/ε₀. (b) At points inside the plate the fields at the sheets of charge on each surface are equal in magnitude and opposite in direction, so their vector sum is zero. At points outside the plate the fields produced by the sheets of charge are in the same direction so their magnitudes add, giving E = σ/ε₀. Evaluate: Gauss’s law can also be used to calculate the net charge enclosed, but it is less efficient in these regions. 22.52. Identify: Example 22.9 gives the expression for the electric field both inside and outside a uniformly charged sphere. Use F = eE to calculate the force on the electron. Set Up: The sphere has a radius R. Execute: (a) Only at r = 0 is the field inside a uniformly charged sphere. (b) At points inside the sphere, E = 4πε₀qr = F, = e 4πε₀e r. The minus sign indicates that F, is radially inward. For simple harmonic motion, F = -kx = -mw²x, where ω = (k/m) = 2πf . 22-20 Chapter 22 Execute: Consider the forces on electron 2. There is a repulsive force F₁₂ due to the other electron, electron 1. F₁₂ = − e² . 1 4πε₀ (2d)² The electric field inside the uniform distribution of positive charge is E = σr (Example 22.9), where q = +2e. 4πε₀R² At the position of electron 2, r = d. The force Fₓ exerted by the positive charge distribution is Fₓ = e( 2d)e σ 4πε₀ R³ and is attractive. The force diagram for electron 2 is given in Figure 22.53b. F₁₂ 4πε₀ R³ Fₓ 2d Z 2d Net force equals zero implies F₁₂ = Fₓ and 1 = 2d 4πε₀ 4πε₀R³ Thus 1/(4πε₀²) = 2d/R³ , so d² = R³/8 and d = R/2 . Evaluate: The electric field of the sphere is radially outward. It is zero at the center of the sphere and increases with distance from the center. The force this field exerts on one of the electrons is radially inward and increases as the electron gets further from the center. The potential energy of an electron in such a field is constant; it is infinite when r = 0 and decreases as d increases. It is reasonable therefore to consider d to be a value of r at which these forces balance. 22.54. Identify: Use Gauss’s law to find the electric field at a point inside and outside the slab of charge of charge. The charge is uniformly spread through the volume. Use then Gauss’s law to find the net electric field on the region where the forces of attraction and repulsion are equal. Set Up: Use a Gaussian surface that has one face of oval in the y-z plane at x = 0, and the other face is at the desired value x. The volume enclosed by such Gaussian surface is Δx. Execute: (a) The electric field inside the plate is zero by symmetry. There is no preferred direction in the y-z plane, so the electric field can only point in the x direction. But at the origin, neither the positive nor negative charged regions should suggest a net separation, so the net field must be zero. (b) For |x | ≤ d , Gauss’s law gives EA = Q = [ρ(x)x] = [ρ(Ax)] , with direction given by x i (away from the ε₀ ε₀ center of the slab). Note that this expression does give E = 0 at x = 0 and that only half the enclosed charge does not depend on x and is equal to ρA . For |x| , Gauss’s law gives EA = Q = ρA ε₀ ε₀ again with direction given by x i |x| i ε₀ Evaluate: At the surfaces of the slab, x = ±d . For these values of x the two expressions for E ± (for inside and outside the slab) give the same result. The charge per unit area of of the slab is given by ⍴/A = ρA(±d) and ρd = σ/2 . The result for E outside the slab can therefore be written as E = σ/2ε₀, and is the same as for a thin sheet of charge. 22.55. (a) Identify and set up: Consider the direction of the field for x slightly greater than and slightly less than zero. The slab is sketched in Figure 22.55a. Gauss’s Law 22-21 (b) Identify and Set Up: |x <| > d (outside the slab) Apply Gauss’s law to a cylindrical Gaussian surface whose axis is perpendicular to the slab and whose end caps have area A and are the same distance |x| > d from x = 0, as shown in Figure 22.55b. Execute: Φₑ = 2EA Figure 22.55b To find Qₑnc consider a thin disk at coordinate x and with thickness dx, as shown in Figure 22.55c. The charge within this disk is d𝑞 = ρ𝑉ⁿ = ρAd𝑥 = (ρA/d²) (x²/d²) Figure 22.55c The total charge enclosed by the Gaussian cylinder is Qenc = 2∫ (dq) _0 = ∫ (⍴A/d²) (x²/d²) x dx = Qₑnc = 2(⍴A/3d x³) = 2(⍴A/3d) Then Q = ⍴, 2EA = 2⍴A(1/3d) ε₀ ε₀ = E = ⍴d(2/3d) E is directed away from x = 0, so F = -⍴d |x(1/3)|i Identify and Set Up: |x| <d (inside the slab) Apply Gauss’s law to a cylindrical Gaussian surface whose axis is perpendicular to the slab and whose end caps have area A, and are the the same distance |x| < d from x = 0, as shown in Figure 22.55d Execute: Φₑ = 2EA Figure 22.55d Qₑ is found as above, but now the integral on dx is only from 0 to h, instead of 0 to d. Q, ᵤ ε₀ Q,ₑ = 2⍴A/3d Then Q = , gives 2EA = 2⍴A/x³d^|3|,/d ε₀ = E = ⍴x²/3dE E is directed away from x = 0, so E = -⍴d³x³(|x|/4)d Evaluate: Note that at E = 0 at x = 0 as stated in par (a). Note also that the expressions for |x| <d and |x| <d agree for x = d. 22.56. Identify: Apply F = qE to relate the force on q to the electric field at the location of q. Set Up: Flux is negative if the electric field is directed into the enclosed volume. For a sphere whose center is located by vector b, a point inside the sphere and located by F is located by the vector \( \vec{r} = \vec{r}' - \vec{b} \) relative to the center of the sphere, as shown in Figure 22.61. EXECUTE: Thus \( \vec{E} = \frac{\rho(\vec{r} - \vec{b})}{3\varepsilon_0} \) Figure 22.61 EVALUATE: When b = 0 this reduces to the result of Example 22.9. When \( \vec{F} = \vec{b} \), this gives E = 0, which is correct since we know that E = 0 at the center of the sphere. (b) IDENTIFY: The charge distribution can be represented as a uniform sphere with charge density \( +\rho \) and centered at the origin added to a uniform sphere with charge density \( -\rho \) and centered at \( \vec{r} = \vec{b} \). SET UP: \( \vec{E} = \vec{E}_{sphere} + \vec{E}_{hole} \) where \( \vec{E}_{sphere} \) is the field of a uniformly charged sphere with charge density \( \rho \) and \( \vec{E}_{hole} \) is the field of a sphere located at the hole and with charge density \( -\rho \). (Within the spherical hole the net charge density is \( \rho + (-\rho) = 0. \)) EXECUTE: \( \vec{E}_{hole} = \frac{-\rho \vec{r}}{3\varepsilon_0} \), where \( \vec{r} \) is a vector from the center of the sphere. \( \vec{E}_{hole} = \frac{-\rho(\vec{F} - \vec{b})}{3\varepsilon_0} \) at points inside the hole. Then \( \vec{E} = \frac{\rho(\vec{r} - \vec{b})}{3\varepsilon_0} - \frac{\rho(\vec{F} - \vec{b})}{3\varepsilon_0} = \frac{\rho b \vec{b}}{3\varepsilon_0} \) EVALUATE: \( \vec{E} \) is independent of r so is uniform inside the hole. The direction of \( \vec{E} \) inside the hole is in the direction of the vector \( \vec{b} \), the direction from the center of the insulating sphere to the center of the hole. 22.62 IDENTIFY: We first find the field of a cylinder of axis c, then take the field difference at the location of the hole. The difference between two electric fields: a solid cylinder on the axis c and related off-axis b, in the location of the hole. SET UP: Let α locate a point within the circle defining the relative axes, c = circle center, and let E locate the point relative to the axis of the hole. Let b locate the axis of the hole relative to the axis of the cylinder. As shown in Figure 22.62. F = r - b. Problem 23.48 shows that at points within a long insulating cylinder, \( \vec{E} = \frac{\rho \vec{r}}{2\varepsilon_0} \). EXECUTE: \( \vec{E}_{inside} = \frac{\rho \vec{r}}{2\varepsilon_0} - \frac{\rho(\vec{r} - \vec{b})}{2\varepsilon_0} = \frac{\rho b \vec{b}}{2\varepsilon_0} \) Note that \( \vec{E} \) is uniform. EVALUATE: If the hole is coaxial with the cylinder, b = 0 and \( \vec{E}_{hole} = 0 \). Figure 22.62 22-26 Chapter 22 22.63. IDENTIFY: The electric field at each point is the vector sum of the fields of the two charge distributions. SET UP: Inside a sphere of uniform positive charge, \( \vec{E} = \frac{\rho r}{3\varepsilon_0} \), \( \rho = \frac{Q}{\frac{4}{3} \pi R^3} \), so \( \vec{E}_{outside} = \frac{Q}{4\pi \varepsilon_0 r^2} \), directed away from the center of the sphere. Outside a sphere of uniform positive charge, \( \vec{E}_{outside} = \frac{Q}{4\pi \varepsilon_0 R^2} \), directed away from the center of the sphere. EXECUTE: (a) \( r = 0. \) This point is inside sphere 1 and outside sphere 2. The fields are shown in Figure 22.63a. \( \vec{E}_2 = \frac{Q}{4\pi \varepsilon_0 R^2}, \vec{E}_1 = 0, \) since \( r = 0. \) \( \vec{E}_x = \frac{Q}{4\pi \varepsilon_0 R^2}, \vec{E}_y = 0, \) so \( \vec{E} = \vec{E}_2 + \vec{E}_1 \) in the \( -x \)-direction. Thus \( E_x = \frac{Q}{16\pi \varepsilon_0 R^2} \). (b) \( r = R/2. \) This point is inside sphere 1 and outside sphere 2. Each field is directed away from the center of the cylinder that produces it. The fields are shown in Figure 22.63b. \( \vec{E}_1 = \frac{Q}{8\pi \varepsilon_0 R^2} \) with \( r = R \) so \( \vec{E}_2 = \frac{Q}{9\pi \varepsilon_0 R^2} \). \( E_x = \vec{E}_1 + \vec{E}_2 = \frac{Q}{9\pi \varepsilon_0 R^2} \) in the \( +x \)-direction and \( \vdash{E} = \frac{Q}{72\pi \varepsilon_0 R^2} \) (c) \( r = R \). This point is at the surface of each sphere. The fields have equal magnitudes and opposite directions, so E = 0. (d) \( r = 3R \). This point is outside both spheres. Each field is directed away from the center that produces it. The fields are shown in Figure 22.63c. \( \vec{E}_2 = \frac{Q}{4\pi \varepsilon_0 R^2}, \vec{E}_1 = \frac{Q}{36\pi \varepsilon_0 R^2} \) \( \vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{SQ}{18\varepsilon_0 R^2} = \frac{SQ}{18\varepsilon_0 R^2} \) in the \( +x \)-direction and \( \vdash{E} = \frac{SQ}{18\varepsilon_0 R^2} \) EVALUATE: The field of each sphere is radially outward from the center of the sphere. We must use the correct expressions for E(r) for each sphere, depending on whether the field point is inside or outside that sphere. 22.64. IDENTIFY: The net electric field at any point is the vector sum of the fields at each sphere. SET UP: Example 22.9 gives the electric field inside and outside a uniformly charged sphere. For the positively charged sphere the field is radially outward and for the negatively charged sphere the electric field is radially inward. EXECUTE: (a) At this point the field of the left-hand sphere is zero and the field of the right-hand sphere is toward the center of that sphere, in the +x-direction. This point is outside the right-hand sphere, a distance r = 2R from its center. \( \vec{E} = \frac{1}{4\pi \varepsilon_0} \frac{Q}{81R^2} \). (b) This point is inside the left-hand sphere, at r = R/2, and is outside the right-hand sphere, at r = 3R/2. Both fields are in the x-direction. \( \begin{aligned} \vec{E}_1 & = \frac{Q}{4\pi \varepsilon_0 R^2} \frac{r}{R} = \frac{Q}{4\pi \varepsilon_0 R^2} \frac{R}{3R} = \frac{Q}{4\pi \varepsilon_0 R^2}, \\ \vec{E}_2 & = \frac{40}{9\pi} \frac{Q}{R^2} = \frac{1}{9\pi} \frac{Q}{4\pi \varepsilon_0 R^2}, \\ \end{aligned} \) \( \vec{E} = \vec{E}_1 + \vec{E}_2 \) (c) This point is outside both spheres, at a distance r = R from their centers. Both fields are in the x-direction. \( \begin{aligned} \vec{E}_1 & = \frac{1}{4\pi \varepsilon_0} \frac{Q}{R^2}, \\ \vec{E}_2 & = \frac{1}{4\pi \varepsilon_0} \frac{3R}{9R^2}, \\ \vec{E} & = \frac{80}{18\pi} \frac{Q}{R^2} + \frac{Q}{4\pi \varepsilon_0 R^2} \end{aligned} \) (d) This point is outside both spheres, a distance r = 3R from the center of the left-hand sphere and a distance r = R from the center of the right-hand sphere. The field of the left-hand sphere is in the direction and the field of the right-hand sphere is in the -x-direction. \( \frac{1}{4\pi \varepsilon_0} \frac{Q}{181R^2} + \frac{80}{18\pi} \frac{Q}{R^2} \) EVALUATE: At all points on the x-axis the net field is parallel to the x-axis. 22.65. IDENTIFY: Let \( -Q \) be the electron charge contained within a spherical shell of radius r and thickness dr. Integrate r from 0 to r to find the total charge \( -Q \) within a shell of radius r. Apply Gauss’s law to a sphere of radius r to find the electric field E(r). SET UP: The volume of the spherical shell is \( dV = 4\pi r^2 dr \). EXECUTE: \( q(r) = -Q \int_{a}^{r} dv = -4Q \int_a^r \frac{1}{r^2} dr, \ q = -4Q [\frac{1}{4} - \underline{\frac{1}{r^2}}] = \hat{O}_i \) If r → ∞ → q(r) → \( \underline {0 \)} ; the total charge of the atom is zero. (b) The electric field is radially outward. Gauss’s law gives: \( E(4\pi r^2), \ q(r) = Q \end{bmatrix} \) The graph of E versus r is sketched in Figure 22.65. What is plotted is the scaled E, equal to \( E/E_{e-charging} \), versus scaled r, equal to \( r/a \) : \( \frac{QE}{\rho_{chg}} \) is the field of a point charge. EVALUATE: As r → 0 the field approaches that of the positive point charge (the proton). For increasing r the electric field falls to zero more rapidly than the \( 1/r^2 \end {for a point charge. \)} 22.66. IDENTIFY: The charge in a spherical shell of radius r and thickness dr is \( dq = \rho(4\pi r^2) dr \). Apply Gauss’s law. SET UP: Use a Gaussian surface that is a sphere of radius r. Let Q1 be the charge in the region r ≤ R/2 and let Q2 be the charge in the region where R/2 ≤ r ≤ R. Distribution of these contents holds: \( 4\pi \varepsilon_0(group \ of \, population) \) ) 22.26 Chapter 22 EXECUTE: (a) The total charge is Q = Q_1 + Q_2 , where Q_2 = \dfrac{-4 \pi (R^2) a xR^4}{6} and Q_1 = 4 \pi (x_0) \int^{R}_{R/2} r^2 | \dfrac{r}{x} | \, dr = \dfrac{xa R^3}{3} - \dfrac{xR^4}{16} = \dfrac{3 x R^3}{16} . Therefore, Q = \dfrac{15 a x R^4}{24} and \dfrac{8 Q}{5 \pi x R_0} . (b) For r \leq R/2 , Gauss 's law gives E4 \pi r^2 = \dfrac{4 \pi R^3 xa^3}{2 } and E = \dfrac{a}{3 x R_0} . For R / 2 \leq r \leq = R , E4 \pi r^2 = \dfrac{Q_1}{R^3}[(r^3 - R^3) (r - R) / 16)] = \dfrac{a}{15 \pi R^3} \, (64 x(r / R)^3 - 48x(r / R) = \dfrac{xQ}{15\pi R^3}[64x(r / R)^3 - 48x(R/K) - 1] . For r \geq R/2 , E4 \pi (E)^2 = Q_1 / 4 \pi x_0. E_{(c)} Q_{(4 Q)( 15)} = E_{0.267} . (d) For £ (R / 2) , F = - e^{x} = - \dfrac{8Q_{0}}{15 \pi x R_{0}^{2}} \cdot r \, so the restoring force depends upon displacement to the first power, and we have simple harmonic motion. (e) Comparing to r = - k \cdot x , k = \dfrac{8Q_0}{15 \pi R_{0} }, \sqrt{\dfrac{k}{m}}, \dfrac {15 \pi x_{0} \sqrt{m}}{m}, \dfrac{k}{m}, \, \dfrac{T_{0}}{2} = X_{0}^{m} \, \dfrac{15 \pi}{a} = 8Q_{0} \cdot T_{2} r =\sqrt{m} \cdot K{\cdot} . EVALUATE: (f) If the amplitude of oscillation is greater than R / 2 , the force is no longer linear in r , and is thus no longer simple harmonic. 22.67. IDENTIFY: The charge in a spherical shell of radius r and thickness dr is dq = \rho (r)x/3r^{2} dr . Apply Gauss’s law. Set Up : Use a Gaussian surface that is a sphere of radius r . Let Q_{e} be the charge in the region r \leq R / 2 and let Q_{e} be the charge in the region where R/2 \leq r \leq R . EXECUTE: (a) The total charge is Q = Q_1 + Q_2 , where Q_2 = \dfrac{4 \pi}{3} . \dfrac{3 a}{2} . \dfrac{x^3}{R^2} \int^{x} \, {R} \dfrac{6 \pi x}{5} - \dfrac{3}{32} \dfrac{5x^{2}}{4} = \dfrac{x}{8R} . \dfrac{3 x R^4}{3} . Q_2 = 4 \pi (1 - (xR)^2) r^2 dr = 4 \div 4 \pi x_{0}\dfrac{x^2}{R^2} , \dfrac{31/24}{16} \dfrac{47}{120} \dfrac{47}{230} \cdot Q_{0} - (D) - \dfrac{3 xR}{8R} , \dfrac{1}{ 3} = 3 \cdot \dfrac{3 x R^4}{48}\cdot a_{Q} \longrightarrow , \dfrac{3 x^3}{8R},\dfrac{A}{8Q}R,\dfrac{8a}{120 R^4}\cdot (\dfrac{1}{2}\cdot)(.\dfrac{Q}{(2xR)^2} 480Q\over 233(axR^3) (b) For r \leq R / 2 , Gauss's law gives E4 \pi x = \dfrac{4 \pi ax}{2} \cdot \dfrac{5 a}{18 x P R x} and E_{R} =\dfrac{x}{8}(x)= \dfrac{90 Q^2}{16 R^2} \text{For r/2 £ r ¿ R} E 4 \pi x \{(1 - (xR)^2 r)\}\,{dr} \cdot\ ,\ , \dfrac{Q}{8 x} = \dfrac{x special{R} }{24 \cdot \pi x} \dfrac{R^3}{P} \dfrac{3 \cdot 3 x\cdot}{4 \cdot \pi} \dfrac{49 x^4}{53 x =3 . 120}\equiv R \cdot = \frac{x}{18} \dfrac{x}{32} = ( x - \bigg[ {\frac{x}{639P(c)}} \dfrac{1.3}{[6]-ini\bigg] Multiply. \dfrac{E =\dfrac{x^9}{a}(24 \cdot x )^{3}\m, = ( 8 \over 3200 )£R,\ {6} \bigg.+ \dfrac{R kms.)}{8 \pi x} = P}\ 55 \cdot\over 240 \text{Since charge is enclosed.} 480Q\over 233axR_0 ^2 ,(c) The fraction of Q between R / 2 \leq r \leq R (D.), \dfrac{47}{120} \divon{\dfrac{3 x^8}{R^3}} 49\div(3907 \cdot (C = \dfrac{1}{2} \text{ (Div.blogspot.x.x.120) = \dfrac{d)iterator of electric field expressions above\over}\xi»,- 0 .7.5 } EVALUATE:(d,E_{0} = \frac{180}{2334 axR_{x}^{k}}.using\, checking\,math\endbut (d) \text{ HE = X_X_X_y}\over } .\ , \dfrac{x}{240 P L}\AM\ .=\( O {E O}\[f \]\{-195,15107\:\ ......... \dfrac{x}{7 \times x \sqrt{x(R)}} \dfrac{K}{^}...&\ =A_xR_0_