Hafez a radi, john o rasmussen auth principles of physics for scientists and engineers 2 29

20.7 Exercises Fig 20.26 See Exercise (8) 693 y q1 a q4 x P q2 q3 a y q1 a q4 x P q2 q3 a y q1 a q4 x P q2 q3 a Section 20.3 Electric Field of an Electric Dipole (9) Two point charges q1 = −6 μC and q2 = +6 μC are placed at two vertices of an equilateral triangle, see Fig 20.27 If a = 10 cm, find the electric field at the third corner Fig 20.27 See Exercise (9) a a q1 a q2 (10) A proton and an electron form an electric dipole and are separated by a distance of 2a = × 10−10 m, see Fig 20.28 (a) Use exact formulas to calculate the electric field along the x-axis at x = −10a, x = −2a, x = −a/2, 694 20 Electric Fields x = +a/2, x = +2a, and x = +10a (b) Show that at both points x = ±10a, the approximate formula given by Eq 20.13 has a very close percentage difference from the exact value Fig 20.28 See Exercise (10) -e Electron y Proton +e x -a a = 10− 10 m +a (11) Rework the calculations of Exercise 10 but on the y-axis at y = −10 a, y = −2 a, y = −a/2, y = +a/2, y = +2a, and y = +10a In part (b), use Eq 20.17 Section 20.4 Electric Field of a Continuous Charge Distribution (12) A non-conductive rod of length L has a total negative charge −Q that is uniformly distributed along its length, see Fig 20.29 (a) Find the linear charge density of the rod (b) Use the coordinates depicted in the figure to prove that the electric field at point P, a distance a from the right end of the rod, has the same form as the one given by Eq 20.27 (c) When P is very far from the rod, i.e a L, show that the electric field reduces to the electric field of a point charge (i.e the rod would look like a point charge) (d) If L = 15 cm, Q = 25 μC, and a = 20 cm, find the value of the electric field at P y −Q P x L a Fig 20.29 See Exercise (12) (13) A non-conductive rod lies along the x-axis with one of its ends located at x = a and the other end located at ∞, see Fig 20.30 Starting from the definition of an electric field of a differential element on the rod, find the electric field at the 20.7 Exercises 695 origin if: (a) the rod carries a uniform positive linear charge density λ (b) the rod carries a positive varying linear charge density λ = λ◦ a/x y λ -x ∞ a Fig 20.30 See Exercise (13) (14) A uniformly charged ring of radius 15 cm has a total charge of 50 μC Find the electric field on the central perpendicular axis of the ring at: (a) cm, (b) cm, (c) 10 cm, and (d) 100 cm (e) What you observe about the values you just calculated? (15) A charged ring of radius R = 0.5 m has a gap d = 0.1 m, see Fig 20.31 Calculate the electric field at its center C if it carries a uniform charge q = μC Fig 20.31 See Exercise (15) q=1 C C d = 1m R = m (16) Figure 20.32 shows a non-conductive semicircular arc of radius R that consists of two quarters The semicircle has a uniform positive total charge Q along its right half, and a uniform negative total charge −Q along its left half Find the resultant electric field at the center of the semicircle Fig 20.32 See Exercise (16) -Q P R +Q R R 696 20 Electric Fields (17) Two non-conductive semicircular arcs, one of a uniform positive charge +Q and the other of a uniform negative charge −Q, form a circle of radius R, see Fig 20.33 Find the resultant electric field at the center of the circle, and compare it with the result of Exercise 16 Fig 20.33 See Exercise (17) -Q +Q P R R (18) If you consider a uniformly charged ring of total charge Q and a fixed radius R (as in Fig 20.18), then the graph of Fig 20.34 would map the electric field along the axis of such a ring as a function of z/R Show that the maximum √ √ electric field is Emax = 2k Q/3 3R2 and occurs at z = R/ Fig 20.34 See Exercise (18) E max E 10 z /R (19) An electron is constrained to move along the central axis of a ring of radius R that has a uniform positive charge q, see Fig 20.35 Show that when the position x of the electron is much less than the radius R (x R), the electrostatic force exerted on the electron can cause it to oscillate through the center of the ring with an angular frequency given by ω = kqe/mR3 , where e and m are the electronic charge and mass, respectively (20) Two non-conductive rings having the same radius R are arranged with their central axes along a common horizontal line and separated by a distance of R, see Fig 20.36 Ring has a uniform positive charge q1 , while ring has a uniform positive charge q2 Given that the net electric field is zero at point P, which is at a distance R from ring and on the common central axis of the two rings, (a) find the ratio between the two charges (b) If only the sign of q1 20.7 Exercises 697 is reversed, is it possible to have a point on the common axis where the net electric field is zero? If so, where would it be? q Fig 20.35 See Exercise (19) x R -x -e F R x x Fig 20.36 See Exercise (20) Ring Ring q1 q2 C1 C2 P R R R 4R (21) A disk of radius R = cm has a surface charge density σ = μC/m2 on its surface Calculate the magnitude of the electric field at points on the central axis of the disk located at: (a) mm, (b) cm, (c) 10 cm, and (d) 100 cm (22) A disk of radius R has a charge Q that is uniformly distributed over its surface area Show that Eq 20.55 transforms to: E= Show that when a formula: 2kQ a 1− √ 2 R R + a2 R, the electric field approaches that of a point charge E≈k Q (a a2 R) You may use the binomial expansion (1 + δ)p ≈ + pδ when δ (23) Compare the obtained results of Exercise 21 to the near-field approximation E = σ/2 ◦ as well as to the point charge approximation E = k(π R2 σ )/a2 , and find which result(s) of Exercise 21 match the two approximations 698 20 Electric Fields (24) A disk of radius R has a surface charge density σ and an electric field of magnitude E◦ = σ/2 ◦ at the center of its surface, see Fig 20.37 At what distance z along the central axis of the disk is the magnitude of the electric field E equal to one-half of E◦? Fig 20.37 See Exercise (24) E σ E0 Charge per unit area σ ° σ z ° R (25) Find the electric field between two oppositely-charged infinite sheets of charge, each having the same charge magnitude and surface charge density σ, but opposite signs, see Fig 20.38 + Fig 20.38 See Exercise (25) + E + + Section 20.5 Electric Field Lines (26) (a) A negatively charged disk has a uniform charge per unit area Sketch the electric field lines in the plane of the plane of the disk passing through its center (b) Redo part (a) taking the disk to be positively charged (c) A negatively charged rod has a uniform charge per unit length Sketch the electric field lines in the plane of the rod (d) Three equal positive charges are placed at the corners of an equilateral triangle Sketch the electric field lines in the plane of the charges (e) An infinite linear rod has a uniform charge per unit length Sketch the electric field lines in the plane of the rod 20.7 Exercises 699 Section 20.6 Motion of Charged Particles in a Uniform Electric Field (27) An electron and a proton are released simultaneously from rest in a uniform electric field of 105 N/C Ignore the effect of the fields of the electron and proton on each other (a) Find the speed and kinetic energy of the electron 50 ns after it has been released (b) Repeat part (a) for the proton (28) Figure 20.39 shows two oppositely charged parallel plates that are separated by a distance d = 1.5 cm Each plate has a charge per unit area of magnitude σ = μC/m2 An electron is released from rest at t = from the negative plate (a) Calculate the electric field between the two plates (b) Ignoring the effect of gravity, find the resultant force exerted on the electron? (c) Find the acceleration of the electron (d) How long does it take the electron to strike the positive plate? (e) What is the speed and kinetic energy of the electron just before striking the positive plate? + Fig 20.39 See Exercise (28) E + t ° + - t - + d (29) In Exercise 28 assume that the electron is projected from the positive plate toward the negative plate with an initial speed v◦ at time t = The electron travels the distance d = 1.5 cm between the two plates and stops temporarily before hitting the negatively charged plate (a) Find the magnitude and direction of its acceleration (b) Find the value of the electron’s initial speed (c) Find the time before the electron stops temporarily (30) Two oppositely charged horizontal plates are separated by a distance d = cm and each has a length L = cm, see Fig 20.40 The electric field between the plates is uniform and has a magnitude E = 102 N/m An electron is projected between the plates with a horizontal initial speed of v◦ = 106 m/s as shown Assuming this experiment is conducted in a vacuum, where will the electron strike the upper plate? (31) Repeat Exercise 30 when a proton replaces the electron 700 20 Electric Fields Fig 20.40 See Exercise (30) y - + + + + + t d/2 e- ° x x E L (32) To prevent the Electron in exercise 30 from striking the upper plate, its initial horizontal speed is increased to v◦ = × 106 m/s, see Fig 20.41, and it then strikes a screen at a distance D = 30 cm (a) What is the acceleration of the electron in the region between the two plates? (b) Find the time when the electron leaves the two plates (c) What is the vertical position of the electron just before leaving the region between the two plates? (d) Find the electron’s total vertical distance just before hitting the screen y y2 + + + t + + d/ - t1 - E e- y1 D h y1 x L D Fig 20.41 See Exercise (32) (33) Repeat Exercise 32 when a proton replaces the electron 21 Gauss’s Law Although Coulomb’s law is the governing law in electrostatics, its form does not always simplify calculations in situations involving symmetry In this chapter, we introduce Gauss’s law as an alternative method for calculating electric fields of certain highly symmetrical charge distribution systems In addition to being simpler than Coulomb’s law, Gauss’s law permits us to use qualitative reasoning 21.1 Electric Flux → Consider a uniform electric field E penetrating a small area A oriented perpendicularly to the field as shown in Fig 21.1 Recall from Sect 20.5 that the number of electric field lines per unit area (measured in a plane perpendicular to the lines) is → proportional to the magnitude of E Therefore, the total number of lines penetrating the surface is proportional to EA This product is called the electric flux1 E Thus: E The SI units for E = EA (21.1) is newton-meters square per coulomb (N.m2 /C) Spotlight Electric flux is proportional to the number of electric field lines penetrating a certain area → If the area in Fig 21.1 is tilted by an angle θ with respect to E, the flux through it (the number of electric lines) will decrease To visualize this, Fig 21.2 shows an The word “flux” comes from the Latin word meaning “to flow” H A Radi and J O Rasmussen, Principles of Physics, Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_21, © Springer-Verlag Berlin Heidelberg 2013 701 702 21 Gauss’s Law area A that makes an angle θ with the field The number of lines that cross the area → A is equal to the number that cross the area A , which is perpendicular to E and hence A = A cos θ Thus, the flux through A, E (A), is equal to the flux through A , E (A ) But according to Eq 21.1, the flux through A is defined as Therefore, the flux through A is: E (A) = E (A E (A ) = EA = EA cos θ Area A ) = EA (21.2) Side view E Area A E → Fig 21.1 Electric field lines representing a uniform electric field E that penetrates an area A perpendicularly (shown both in 3D and as a side view) The electric flux E through this area is EA E Normal θ Area A θ Area A' Area A' θ E Area A Side view → Fig 21.2 Electric field lines representing a uniform electric field E penetrating an area A that is at an angle θ with the field (both three dimensional and side views are displayed) Since the flux through A is the same as through A , the flux through A is E = EA cos θ → If we define a vector area A whose magnitude represents the surface area and whose direction is defined to be perpendicular to the surface area as in Fig 21.3, then Eq 21.2 can be written as: → E → = E • A = EA cos θ (21.3) The flux through a surface of area A has a maximum value EA when the surface is perpendicular to the field (i.e when θ = 0◦ ), and is zero when the surface is parallel to the field (i.e when θ = 90◦ ) 708 21 Gauss’s Law Solution: The electric field inside the conductor must be zero If this is not the case, the field would exert forces on the free electrons and a current would flow within the conductor Of course, there are no such currents in an isolated conductor in electrostatic equilibrium First, we draw a Gaussian surface surrounding the conductor’s cavity, close to → its surface, as seen in Fig 21.8 Since E = inside the conductor, then E = and hence according to Gauss’s law, no net charge would exist on the inner walls of the cavity Then we draw a Gaussian surface just inside the outer surface of the conductor → Since E = inside the conductor, then E = and according to Gauss’s law, no net charge will exist inside the Gaussian surface If the excess charge is not inside the Gaussian surface, it must then be outside that surface, on the conductor’s surface; see Fig 21.9 Fig 21.9 + q Conductor's outer surface + + Conductor's inner surface + + + Cavity Example 21.4 Using Gauss’s law, find the electric field just outside the surface of a conductor carrying a positive surface charge density σ Solution: Consider a small section of the conductor’s surface so as to neglect curvature Then construct a cylindrical Gaussian surface normal to the conductor as shown in Fig 21.10, where one end of the cylinder is inside the conductor while → the other end is outside Each end has an area A The electric field E inside the → conductor is zero, and the electric field E just outside the conductor’s surface 21.3 Applications of Gauss’s Law 709 must be perpendicular to the surface If this were not true, the component of the field along the surface of the conductor would force the free electrons to move, violating the conductor’s electrostatic equilibrium E =0 + + A + + + + + E =0 + + + + + Conductor E Gaussian Charge per cylinder Side view unit area σ A E σ Fig 21.10 To find the net flux through this cylindrical Gaussian surface, let us consider → each of the four faces of the cylinder (1) Because E = inside the conductor, then the flux through the end of the cylinder inside the conductor is E (1) = (2) For the same reason, the flux through the curved surface of the cylinder inside the con→ → ductor is E (2) = (3) Since E ⊥ dA along the curved surface of the cylinder → → outside the conductor, the flux there is also E (3) = (4) Since E // dA along the end of the cylinder that is outside the conductor, the flux there is E (4) = EA Thus, the net flux through the cylindrical Gaussian surface is E = EA Since qin = σ A, we can then find electric field just outside the surface of the conductor as follows: E = → → E • dA = qin ◦ ⇒ EA = σA ◦ ⇒ E= σ ◦ Example 21.5 Find the electric field due to an infinite plane sheet of charge with a uniform positive surface charge density σ → Solution: By symmetry, the electric field E outside the infinite plane sheet must be: (1) uniform, (2) perpendicular to the sheet, (3) of the same magnitude at all points equidistant from the sheet, and (4) in opposite direction on the other side of the sheet, see Fig 21.11 The choice that reflects that symmetry is a cylindrical 710 21 Gauss’s Law Gaussian surface normal to the sheet as shown in Fig 21.11, where one end of the cylinder (of area A) is to the right of the sheet while the other end is to the left of it + + + + + + Sheet of charge + + A A + E E E + + Charge per unit area σ + A E + + + A Gaussian cylinder Side view σ Fig 21.11 As in Example 21.4, to find the net flux through this cylindrical Gaussian → → surface, let us consider each of the four faces of the cylinder (1) Because E // dA through the left end of the cylinder, then the flux there is E (1) = EA (2) Because → → E ⊥ dA through the left curved surface of the cylinder then the flux there is → → E (2) = (3) For the same reason (E ⊥ dA ), and the flux through the right → → curved surface of the cylinder is E (3) = (4) Because E // dA through the right end of the cylinder, then the flux there is E (4) = EA Thus, the net flux through the Gaussian surface is E = 2EA Since qin = σ A, we then can find electric field as follows: E = → → E • dA = qin ◦ ⇒ EA = σA ◦ ⇒ E= σ ◦ Example 21.6 Two infinite conducting plates have a uniform surface charge density of magnitude σ on their faces The excess charge is positive on one plate and negative on the other, see Fig 7.12a, b The two plates are brought together as shown in Fig 21.12c Find the electric field to the left and right of the plates in each part of the figure Solution: The charge on the positively charged conductor in Fig 21.21a will spread over its two faces each with a surface density of magnitude σ From 21.3 Applications of Gauss’s Law 711 Example 21.4, the electric field just outside each of these surfaces would be directed away from the two faces and would have a magnitude of: E+ = Conductor σ′ + + + + (a) ◦ Conductor Conductor - - - - + + E+ σ E+ σ′ E− σ′ (b) E− E =0 σ′ + + + + + + σ = 2σ ′ E Conductor - E =0 (c) Fig 21.12 We have a similar situation in Fig 21.12b, except that the electric field is directed toward the two faces and has a magnitude given by: E− = σ ◦ When we bring the two plates together, the excess charge on one plate attracts the excess of charge on the other, and all the excess charge moves onto the inner surfaces of the plates, see Fig 21.12c The magnitude of the new surface charge density of the inner surfaces is σ = 2σ Thus, the electric field between the plates is to the right with a magnitude: E= σ ◦ (Between the plates) The electric field on the outer sides of the two plates of Fig 21.12c is zero since no charge is left on those sides Example 21.7 Two infinitely long nonconductive sheets are aligned in parallel, see Fig 21.13a Each sheet has a fixed uniform charge One sheet is positively charged and has a surface charge density of magnitude σ+ = 6.5 µC/m2 The other sheet is negatively charged with |σ− | = 4.5 µC/m2 Find the electric field: (a) to the left (L) of the sheets, (b) between (B) the sheets, and (c) to the right (R) of the sheets 712 21 Gauss’s Law σ+ L + + + + + + B - σ− σ+ E+ E− R L (a) + + + + + + E+ E− B (b) - σ− E+ σ+ EL E− R L + + + + + + EB B - σ− ER R (c) Fig 21.13 Solution: For items of fixed charge, we can calculate the electric field of the items in the same manner as if each item were isolated, i.e by adding the fields algebraically using the superposition principle Thus, the magnitude of the electric field due to the positive sheet is: E+ = σ+ 6.5 × 10−6 C/m2 = = 3.67 × 105 N/C ◦ 2(8.85 × 10−12 C2 /N.m2 ) Also, the magnitude of the electric field due to the negative sheet is: E− = σ− 4.5 × 10−6 C/m2 = = 2.54 × 105 N/C ◦ 2(8.85 × 10−12 C2 /N.m2 ) The directions of the fields to the left (L), between (B), and right (R) of the sheets are shown in Fig 21.13b The resultant field depends on the values of E+ and E− Since E+ > E− , then EL is: EL = E+ − E− = 3.67 × 105 N/C − 2.54 × 105 N/C = 1.13 × 105 N/C (to the left) The field to the right of the sheets ER has the same magnitude but is directed to the right Between the sheets, we have: EB = E+ + E− = 3.67 × 105 N/C + 2.54 × 105 N/C = 6.21 × 105 N/C (to the right) Example 21.8 Using Gauss’s law, find the electric field at a distance r from a long thin rod that has a uniform charge per unit length λ Solution: By symmetry, the electric fields outside the rod are radial and lie in a plane perpendicular to the rod Additionally, the field has the same magnitude 21.3 Applications of Gauss’s Law 713 at all points at the same radial distance from the rod This suggests that we can construct a cylindrical Gaussian surface of an arbitrary radius r and height Such a cylinder would have its ends perpendicular to the rod as shown in Fig 21.14 + + + r Gaussian cylinder Top view dA + r E E + + Fig 21.14 We divide the flux into two cases: (1) The flux through the two ends of the Gaussian → → → cylinder is zero because E is parallel to these surfaces, i.e E ⊥ A (2) The flux through the curved surface of the Gaussian cylinder can be obtained by taking into → → → → account that E = constant and E is parallel to dA , i.e E • dA = E dA Therefore, → → E • dA = E dA = E dA = EA, where A is the area of the curved cylinder and is given by A = 2π r The net charge inside the Gaussian cylinder is qin = λ We can now use Gauss’s law to find the electric field as follows: E = → → E • dA = qin ◦ ⇒ → → E • dA = E dA = E = E(2π r ) = Then: E= 2π dA λ ◦ λ λ = 2k r r ◦ This relation was derived in Chap 20 using Coulomb’s law (Eq 20.36) Example 21.9 A solid sphere of radius R has a uniform volume charge density ρ and carries a total positive charge Q Find and sketch the electric field at any distance r away from the sphere’s center 714 21 Gauss’s Law Solution: We divide the solution to ≤ r ≤ R and r ≥ R (1) For ≤ r ≤ R When dealing with a spherically symmetric charge distribution, we chose a spherical Gaussian surface of radius r < R concentric with the charged sphere as shown in Fig 21.15 Fig 21.15 ρ R dA r E Gaussian sphere By symmetry, the magnitude of the electric field is constant everywhere on the → → spherical Gaussian surface and normal to the surface at any point, i.e E // dA Thus: → → E • dA = E dA = E dA = E(4π r ) It is important to notice that the volume, say V , of the Gaussian sphere encloses a net charge qin = ρV ; that is: qin = ρV = ρ( 43 π r ) We can now use Gauss’s law to find electric field as follows: E = → → E • dA = E= Then: qin ◦ ρ r ◦ ⇒ E(4π r ) = ρ( 43 π r ) ◦ (0 ≤ r ≤ R) Using the definition ρ = Q/( 43 π R3 ) and k = 1/(4π ◦ ), we get: E= Q 4π ◦R r=k Q r R3 (0 ≤ r ≤ R) 21.3 Applications of Gauss’s Law 715 (2) For r ≥ R Again, because the charge distribution is spherically symmetric, we can construct a Gaussian sphere of radius r > R concentric with the charged sphere, as shown in Fig 21.16 Fig 21.16 Q dA R Gaussian sphere → E r → Just as when r < R, E • dA = E(4π r ), but qin = Q Thus, we can use Gauss’s law to find the electric field as follows: E i.e.: = → → E • dA = E= 4π qin ⇒ ◦ Q Q =k 2 r r ◦ E(4π r ) = Q ◦ (r ≥ R) Notice that this is identical to the result obtained for a point charge Therefore, we conclude that the electric field outside any uniformly charged sphere is equivalent to that of a point charge located at the center of the sphere At r = R, the two cases give identical results E = kQ/R2 A plot of E versus r is shown in Fig 21.17 This figure shows the continuation of E and its maximum at r = R Fig 21.17 ρ R E=k E Q r2 Q E=k 3r R R r 716 21 Gauss’s Law Example 21.10 A thin spherical shell of radius R has a total positive charge Q distributed uniformly over its surface Find the electric field inside and outside the shell Solution: By symmetry, if any field exists inside the shell, it must be radial Let us construct a spherical Gaussian surface of radius r < R concentric with the shell, see the cross sectional view in Fig 21.18 Fig 21.18 + Q + R Gaussian sphere + Spherical + shell + r + + + Based on Gauss’s law, the lack of charge inside the surface indicates that → → E • dA = E(4π r ) = 0, or E = Accordingly, we conclude that there is no electric field inside a uniformly charged spherical shell Outside the shell, we construct a spherical Gaussian surface of radius r > R concentric with the charged shell as shown in Fig 21.19 Fig 21.19 Q + + r + Spherical shell Gaussian sphere + R + + → + + → Symmetry suggests that E = constant on that surface and E is parallel to dA , → → i.e E • dA = E(4π r ) Since the net charge qin inside the Gaussian surface is equal to the total charge Q on the shell, the shell is equivalent to a point charge located at the center That is: E= 4π Q Q =k 2 r r ◦ (r > R) 21.4 Conductors in Electrostatic Equilibrium 21.4 717 Conductors in Electrostatic Equilibrium Conductors contain free electrons that can move freely When there is no net motion of electrons within the conductor, the conductor is in electrostatic equilibrium and has the following four properties: (1) The electric field inside the conductor is zero (see Example 21.3) (2) The excess charge on an isolated conductor lies on its outer surface (see Example 21.3) (3) The electric field just outside a charged conductor at any point is perpendicular to its surface and has a magnitude E = σ/ ◦ , where σ is the surface charge density at that point (see Example 21.4) (4) The surface charge density is greatest at locations where the radius of curvature of the surface is smallest (see Chap 22) We can elaborate more about the first property by considering the conducting slab in electrostatic equilibrium on the left of Fig 21.20, where the free electrons are → uniformly distributed throughout the slab, i.e E int = When we place the slab in an → external electric field E ext as in the right part of Fig 21.20, the free electrons move to the left In time, more negative and positive charges accumulate on the left and right surfaces, respectively These two planes of charge create an increasing internal → → → electric field E int inside the conductor After awhile, E int will compensate E ext , → → → resulting in a zero net electric field inside the conductor, i.e E net = E ext − E int = The time to reach this new electrostatic equilibrium is of the order 10−6 s Before E net = After → Eext + + - E net = + + + - Eint → Fig 21.20 An external electric field E ext creates an internal electric field E int in the conductor such → that the net electric field E net is zero Example 21.11 A conducting sphere of radius R carries a net positive charge 2Q A conducting spherical shell of inner radius R1 (R1 > R) and outer radius R2 carries a net negative charge −Q This shell is concentric with the conducting sphere Find the 718 21 Gauss’s Law magnitude of the electric field at a distance r away from the common center when: (a) r < R, (b) R < r < R1 , (c) R1 < r < R2 , and (d) r > R2 Solution: The charge distributions under consideration are characterized by being spherically symmetrical around the common center c This suggests that a spherical Gaussian surface of radius r is to be constructed in each case such as S1 , S2 , S3 , and S4 that are displayed in Fig 21.21 In addition, we use the fact that the electric field inside a conductor is zero and all the excess charge will lie entirely on the outer surface of the isolated conductor S4 −Q S3 S2 2Q S1 R R1 R2 c Fig 21.21 (a) In this region the Gaussian sphere S1 of Fig 21.21 satisfies the condition r < R Because there is no charge inside the conductor in this region, i.e qin = 0; then, E1 = (b) In this region the Gaussian sphere S2 of Fig 21.21 satisfies the condi→ → tion R < r < R1 Because qin = 2Q inside this surface and because E • dA = E2 (4π r ), we can use Gauss’s law to find: E2 = 4π 2Q 2Q =k 2 r r ◦ (R < r < R1 ) 21.4 Conductors in Electrostatic Equilibrium 719 (c) In this region, the Gaussian sphere S3 of Fig 21.21 satisfies the condition R1 < r < R2 Because the electric field inside an equilibrium conductor is zero, i.e E3 = 0; then, based on Gauss’s law, the net charge qin must be zero From this argument, we find that an induced charge −2Q must be established on the inner surface of the shell to cancel the charge +2Q on the solid sphere In addition, because the net charge on the whole shell is −Q, we conclude that its outer surface must carry an induced charge +Q (d) In this region, the Gaussian sphere S4 of Fig 21.21 satisfies the condition r > R2 Because qin = 2Q − Q = Q inside this surface and because E4 (4π r ), we can use Gauss’s law to find: E4 = 4π Q Q =k 2 r r ◦ → → E • dA = (r > R2 ) Figure 21.22 shows a graphical representation of the variation of the electric field E with r In addition, the figure shows the final distribution of the charge on the two conductors Fig 21.22 E +Q 2Q E c −2Q 2Q r2 Q E4 = k r r E2 = k E R R1 R The expressions that we have arrived at for the electric fields established by simple charge distributions are presented in Table 21.1 720 21 Gauss’s Law Table 21.1 Electric fields due to simple charge distributions Charge distribution Electric field Single point charge q E =k Charge q uniformly distributed on the surface of a conducting sphere of radius R ⎧ q ⎨k r2 E= ⎩0 Charge q uniformly distributed with uniform charge density ρ over an insulating sphere of radius R ⎧ q ⎨k r E= ⎩k q r R3 Infinitely long thin rod of a uniform charge per unit length λ E = 2k Infinite plane sheet of charge of uniform surface charge density σ E= Conductor having surface charge density σ Two oppositely charged conducting plates with surface charge density of magnitude σ 21.5 E= E= q r2 r>0 λ r r ≥R r≤R Everywhere outside the plane ⎧ ⎨ σ ◦ Just outside the conductor Inside the conductor ⎧ ⎨ σ ⎩ r R (9) Figure 21.29 shows four closed surfaces S1 , S2 , S3 , and S4 and four point charges q, −q, 2q, and −2q (a) Find the electric flux through each surface (b) Would the electric field lines produced by the point charge −2q have an effect on the calculated fluxes? (c) Explain the reasoning behind your answer for (b) [...]... 21 .3 Applications of Gauss’s Law 21 .3 707 Applications of Gauss’s Law Example 21 .2 Using Gauss’s law, find the electric field at a distance r from a positive point charge q, and compare it with Coulomb’s law Solution: We apply Gauss’s law to the spherical Gaussian surface of radius r in Fig 21 .5 From the symmetry of the problem, we know that at any point, the → electric field E is perpendicular to... one plate and negative on the other, see Fig 7.1 2a, b The two plates are brought together as shown in Fig 21 .12c Find the electric field to the left and right of the plates in each part of the figure Solution: The charge on the positively charged conductor in Fig 21 .2 1a will spread over its two faces each with a surface density of magnitude σ From 21 .3 Applications of Gauss’s Law 711 Example 21 .4,... whole surface dA θ E → Fig 21 .4 The differential surface vector area d A of magnitude dA and direction perpendicular to the → differential surface area and pointing outwards When the electric field E makes an angle θ with that → → differential surface area, the differential flux d E is E • d A → We start by considering a differential vector surface area dA to be normal to the surface and to point outwards... Section 21 .2 Gauss’s Law (8) A point charge q is located at the center of a charged ring of radius R The ring has a linear charge density λ, see Fig 21 .28 (a) Find the total electric flux through the Gaussian sphere S1 of radius r < R (b) Find the total electric flux through the Gaussian sphere S2 of radius r > R (9) Figure 21 .29 shows four closed surfaces S1 , S2 , S3 , and S4 and four point charges... law, find the electric field at a distance r from a long thin rod that has a uniform charge per unit length λ Solution: By symmetry, the electric fields outside the rod are radial and lie in a plane perpendicular to the rod Additionally, the field has the same magnitude 21 .3 Applications of Gauss’s Law 713 at all points at the same radial distance from the rod This suggests that we can construct a. .. entering the surface 21 .2 Gauss’s Law In this section, we introduce a new foundation of Coulomb’s law, called Gauss’s law This law can be used to take advantage of symmetry in the problem under consideration Central to Gauss’s law is a hypothetical closed surface called a Gaussian surface In Example 21 .1 we noticed that the flux over a sphere of radius r was equal to the charge q inside the sphere divided... = E (2 r ) = Then: E= 1 2 dA λ ◦ λ λ = 2k r r ◦ This relation was derived in Chap 20 using Coulomb’s law (Eq 20 .36) Example 21 .9 A solid sphere of radius R has a uniform volume charge density ρ and carries a total positive charge Q Find and sketch the electric field at any distance r away from the sphere’s center 714 21 Gauss’s Law Solution: We divide the solution to 0 ≤ r ≤ R and r ≥ R (1) For 0... surface of a conductor carrying a positive surface charge density σ Solution: Consider a small section of the conductor’s surface so as to neglect curvature Then construct a cylindrical Gaussian surface normal to the conductor as shown in Fig 21 .10, where one end of the cylinder is inside the conductor while → the other end is outside Each end has an area A The electric field E inside the → conductor is... surface Example 21 .1 Find the electric flux through a sphere of radius r enclosing at its center: (a) a positive charge q, and (b) a negative charge −q Solution: (a) When a positive charge q is at the center of such a sphere, the electric field would be directed outwards and be normal to the surface It would also have a constant magnitude, E = q/(4π ◦ r 2 ), see Fig 21 .5 Therefore: → → E • dA = E dA cos 0◦... to right along the positive x-axis If the magnitude of the field is 2 × 105 N/C, what flux passes through a circular loop of area 0.5 m2 if the normal to loop is: (a) in the positive x-direction, (b) in the negative x-direction, (c) in the positive y-direction, (d) in the negative y-direction, and (e) in a direction that makes an angle 60◦ from the x-axis (2) The maximum flux through a rectangle of