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Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
Ch23 gauss
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Ch23 gauss

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  • 1. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Ch 23 Gauss’ Law (HRW p605-619) Carl Friedrich Gauss was a German mathematician and physicist (1777 - 1855). His law concerns a closed hypothetical surface enclosing a charge distribution. Gauss’ Law relates the electric field at points on a closed Gaussian surface to the net charge enclosed by that surface (details to follow!) or If we know the electric field on a closed surface, we can determine the enclosed charge. Ch23-1/60 Flux Consider a wide air stream of uniform velocity v and a small square loop of area A. Let Φ represent the volume flow rate(vol/unit time) at which air passes thru the loop. This rate depends on the angle between v and A. 90 deg at θ, then component define area perpendicular to plane vector A | to plane parallel=>0 Ch23-2/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 1
  • 2. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Flux Consider a wide air stream of uniform velocity v and a small square loop of area A. θ θ Serway Fig24.2 Ch23-3/60 Flux The rate of air flow through the area A is an example of a flux - a volume flux in this case. If we consider the area vector of the loop as having magnitude = the area and direction normal to the surface in the direction of the flow, then the above eqn can be written: where θ is the angle between A and v Ch23-4/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 2
  • 3. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Flux of an Electric Field We want to define the flux of an electric field. Consider Gaussian surface - made up of small squares ∆A. - represent by an area vector ∆A -magnitude is ∆A -direction normal to surface (always outwards). Small area=> E constant over ∆A. E and ∆A are at angle θ Define flux of electric field for the Gaussian surface as Ch23-5/60 Flux of an Electric Field To make definition exact, let ∆A shrink to a differential limit dA. The sum becomes an integral. This tells us that the electric flux through a Gaussian surface is given by integrating the product over the surface. The loop on integral => over entire (closed) surface Ch23-6/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 3
  • 4. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Sample problem 23-1 Gaussian surface is a cylinder of radius R immersed in a uniform electric field E, with the axis parallel to the field. What is the flux Φ of the electric field through the closed surface. Here, and almost everywhere else in this course, the surface integrals are VERY SIMPLE. Take them one step at a time. Ch23-7/60 Sample problem 23-1(cont) On the left cap (a) angle between E and dA is 180o and E is uniform. Ch23-8/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 4
  • 5. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Sample problem 23-1 On the right cap (c) angle between E and dA is 0o and E is uniform. On cylinder, the angle is always 90o Ch23-9/60 Sample problem 23-1(cont) Adding up the contribution from each surface Congratulations, we have just worked out our first surface integral! What are the units of Φ? Ch23-10/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 5
  • 6. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Question What is the electric flux through a flat shape of area A lying in the xy-plane if the electric field is N/C ? (a) 0 N.m2/C (b) Aa N.m2/C (c) Ab N.m2/C (d) Ac N.m2/C (e) A(a2 + b2 +c2)1/2 N.m2/C Ch23-11/60 Question What is the electric flux through a flat shape of area A lying in the xy-plane if the electric field is N/C ? (a) 0 N.m2/C (b) Aa N.m2/C (c) Ab N.m2/C (d) Ac N.m2/C (e) A(a2 + b2 +c2)1/2 N.m2/C Ans: (d) Ac N.m2/C since the area vector is Ak and hence the other dot products in Φ are 0 Ch23-12/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 6
  • 7. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Question For this Gaussian cube of face area A, immersed in a uniform electric field E, along the Z axis, in terms of E and A, what is the electric flux through (a) The front face (in xy plane)? (b) The rear face? (c) The top face (d) The whole cube? Ch23-13/60 Question For this Gaussian cube of face area A, immersed in a uniform electric field E, along the Z axis, in terms of E and A, what is the electric flux through (a) The front face (in xy plane)?Ans: EA θ = 0o cos0o= 1 (b) The rear face? Ans: -EA θ = 180o (c) The top face Ans: 0 θ = 90o (d) The whole cube? Ans: 0 opposites cancel Ch23-14/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 7
  • 8. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Sample problem 23-2 A non-uniform electric field passes through the Gaussian cube above (E in N, x in m). What is the electric flux through the right face, the left face, the top face? Ch23-15/60 Sample problem 23-2 (cont) Key idea: we get the flux through the surface by integrating the scalar product over each face. Right face: an area vector is always perpendicular to the surface and points to the exterior. Thus dA on the right face points along the x-axis. Ch23-16/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 8
  • 9. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Sample problem 23-2 (cont) Right surface (cont) which has x = 3.0 m everywhere =area=4m2 Ch23-17/60 Sample problem 23-2 (cont) Left face: dA now points along –ve x-axis x = 1 m. Follow same procedure and get Top face: the differential area vector points in +ve direction of y-axis so: Ch23-18/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 9
  • 10. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Gauss’ Law Gauss’ Law relates the net flux Φ of an electric field through an arbitrary closed surface (a Gaussian surface) to the net charge that is enclosed by the surface. or These equations hold only when the net charge is in vacuum (or air in practice). If qenc is +ve => net flux is out If qenc is –ve => net flux is in Ch23-19/60 Gauss’ Law (cont) Charge outside the Gaussian surface is not part of qenc, no matter how large or how close. The form and location of the charges inside the surface do not matter. The electric field in the equation is that from all charges, inside and outside the surface. But that is semantics, because the net flux from any charges outside the surface is zero (just consider the case with no charges inside, and Gauss’ Law tells us the net flux is zero). Ch23-20/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 10
  • 11. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Gauss’ Law (cont) What is the net flux over the 4 Gaussian surfaces in the regions around the dipole: +ve -ve 0 ? Surface S1 Flux outward everywhere => flux positive, as is the enclosed charge Surface S2 Flux inward everywhere => -ve flux and so is the enclosed charge Surface S3 No enclosed charge => net flux 0 -reasonable since in & out Surface S4 No net charge => net flux 0 -reasonable since lines out and in Ch23-21/60 Question The arrows indicate the direction of the field lines, and the numbers their magnitude in N.m2/C (all | to the surfaces). Do the Gaussian surfaces enclose net +ve, -ve or 0 charges? Ch23-22/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 11
  • 12. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Question The arrows indicate the direction of the field lines, and the numbers their magnitude in N.m2/C (all | to the surfaces). Do the Gaussian surfaces enclose net +ve, -ve or 0 charges? 1) Ans: 2 + 5 –3 +7 –4 -7 = 0 => no charge 2) Ans: -4 +10 –3 –6 +5 +3 = +5 => +ve charge 3) Ans: -7 –6 +8 –5 +5 +2 = -3 => -ve charge Ch23-23/60 Question What is the net flux through the Gaussian surface S (in 3dimensions) if: the coin is neutral q1 = q4 = +3.1 nC, q2 = q5 = -5.9 nC and q3 = -3.1nC ? Do q4 and q5 have an effect? What is the coin’s effect? What is the net flux thru S? Ch23-24/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 12
  • 13. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Question What is the net flux through the Gaussian surface S (in 3D) if the coin is neutral & q1 = q4 = +3.1 nC, q2 = q5 = -5.9 nC and q3 = -3.1nC ? Do q4 and q5 have an effect? Ans: no, since not enclosed What is the coin’s effect? Ans: none, since neutral What is the net flux thru S? Ch23-25/60 Gauss’ Law and Coulomb’s Law We derive Coulomb’s law from Gauss’ Law and symmetry considerations. Divide the surface of a Gaussian sphere centered on charge q into small areas dA. By definition, the area vector dA at any point is | to the surface and pointing outward. By symmetry, E is also | surface and pointing outward (+ve charge). Thus E.dA = EdA since cosθ = 1 Ch23-26/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 13
  • 14. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Gauss’ Law and Coulomb’s Law Gauss’ Law states which reduces to What is ? Which is just Coulomb’s Law, ie Gauss’ Law implies Coulomb’s Law Ch23-27/60 Finding electric field strength using Gauss’ Law Previous derivation - an example of finding an electric field strength, E, by applying Gauss’ Law. Major Point: We control how we define the surface To find E, construct a Gaussian surface which by symmetry makes the integration over the surface simple. Previous case made dot product simple, and constant E so resulting surface integral was simple. Gauss’ Law hold’s for any surface. If we had used a cube, none of the simplifications would have occurred. Ch23-28/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 14
  • 15. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons A charged isolated conductor Theorem: If an excess charge is placed on an isolated conductor, the charge will move entirely to the outer surface of the conductor. None of the excess charge will be found within the body of the conductor First we prove another important theorem: The electric field inside a conductor must be zero. I can guarantee the final exam will require one or the other of these theorems to be applied, most likely both. Ch23-29/60 A charged isolated conductor Theorem: The electric field inside a conductor is zero. Proof: Suppose a field exists => a force on the conduction (free) electrons => current would always exist within conductor. But there cannot be (or there is not) a perpetual current => field must be zero QED As conductor is charged, there is an internal electric field but added charge distributes itself so (a) there is no on-going current and hence (b) there is no internal electric field. Ch23-30/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 15
  • 16. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons A charged isolated conductor Theorem: Excess charge placed on an isolated conductor will move entirely to the outer surface of conductor i.e., no excess charge will be found within body of conductor. Proof: Consider an isolated conductor held on insulating string with a Gaussian surface just below the surface of the conductor. E = 0 inside conductor (thm above) => => qenc=0 (by Gauss’ Law) => all charge must be on surface QED Ch23-31/60 A charged isolated conductor with a cavity Theorem: The walls of a cavity in a conductor have no net charge. Proof: Consider a Gaussian surface inside conductor but just surrounding the cavity. Again E = 0 everywhere inside conductor => => qenc=0 (by Gauss’ Law) => there is no charge on the inner surface/wall. So you are safe inside your car after a lightening strike! Ch23-32/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 16
  • 17. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons A charged isolated conductor: the external field Excess charge is on the outer surface, but unless conductor is spherical, the charge is not uniformly distributed. Variation in surface charge density σ (charge per unit area) makes determining E difficult in general, but it is easy very close to surface. Theorem: The electric field E at, or just outside, a conducting surface is | surface. Proof: Suppose not => there is a component of E along the conductor’s surface => there is a force on the surface charges => charges move until there are no forces along surface => at equilibrium, the electric field must be | to surface. QED Ch23-33/60 A charged isolated conductor: the external field Theorem: The magnitude of the field just outside a conducting surface is: where σ is surface charge density (C/m2). Proof Consider a small cylindrical Gaussian surface which cuts thru surface. Sum the flux thru the surface. Zero on interior end since E=0 inside Zero on side walls since E is | conductor’s surface => || to wall => E.dA = 0 Outer end: E | to end cap so Φ = EA => total flux = EA so Gauss’ Law=> Canceling A’s gives QED Ch23-34/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 17
  • 18. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Induced charges in a conducting shell Theorem: The charge induced on the inside of a neutral conducting shell equals the net charge inside the shell, but of opposite sign. Proof: Consider a Gaussian surface inside conducting shell, close to the inner surface (red line). E=0 in conductor => flux through Gaussian surface is zero. => net charge inside surface is 0 => charge on inner surface is equal and opposite contained charge. QED Ch23-35/60 Induced charges in a conducting shell (cont) In example, the interior charge is off centre and thus the induced +ve charge is not uniform. Much of it is close to the –ve charge. If the interior charge was centered, the induced charge would be uniform. Ch23-36/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 18
  • 19. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Induced charges in a conducting shell Theorem: Charge induced on the outside of a neutral conducting shell equals net charge inside the shell, with a distribution independent of that of interior charge. Proof: From previous theorem, charge on inner surface equals interior charge with opposite sign. Shell is neutral (i.e., net charge 0) => charge on outer surface is same as interior net charge. E=0 in conductor => distribution on outer surface is independent of interior distribution. QED Ch23-37/60 Induced charges in a conducting shell Theorem: If a charge is enclosed in a neutral conducting shell, then outside the shell, the system appears to be a point charge at the centre of the shell. Proof: The previous theorem states the same charge has been induced on the outer surface of the shell and is independent of the location of inner charge, i.e., uniform. Shell theorem states a uniform shell of charge acts as if all charge is at centre of shell. QED Ch23-38/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 19
  • 20. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Question A ball of charge of –50e is at the centre of a conducting shell of net charge –100e. What is the charge on (a) the shell’s inner surface? 0, +50e, -50e, -150e +100e (b) the shell’s outer surface? 0, +50e, -50e, -150e +100e Ch23-39/60 Question A ball of charge of –50e is at the centre of a conducting shell of net charge –100e. What is the charge on (a) the shell’s inner surface? 0, +50e, -50e, -150e +100e (b) the shell’s outer surface? 0, +50e, -50e, -150e +100e Ans (a): +50e to counter balance the enclosed charge Ans (b): -150e the original –100e plus the –50e lost to the inner surface. Ch23-40/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 20
  • 21. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Exam problem, April 2005 -22.1nC 10 cm E = 21.0 kN/C 8 cm 3 cm 5 cm A spherical conductor of radius 3.00 cm has a charge of -2.21x10-8 C and is centered inside a hollow metal conducting spherical shell from 5.00 to 8.00 cm. The total charge on spherical shell is unknown. At a point 10.0 cm from the centre of the sphere and the shell, the electric field due to charges on these 2 conductors is 21.0 kN/C directed radially outward. 1) Using Gauss’ Law, determine the charge on inner surface of shell ? 2) In a similar fashion, use Gauss’ Law to find the charge on the outer surface of the shell ? 3) What are the magnitude and direction of the electric field at 2.00, 4.00 and 6.00 cm from the centre of the system ? Ch23-41/60 Exam problem, April 2005 Gaussian surface -22.1nC 10 cm E = 21.0 kN/C 8 cm 3 cm 5 cm 1) What is the charge on the inner surface of the shell ? Ans: +22.1 nC because an equal and opposite charge is induced on the inner surface. Proof: Consider a Gaussian surface as a sphere inside shell (red line). E is zero inside the conductor => flux integral is zero => by Gauss’ Law that enclosed charge = 0 =>induced charge on inner surface is equal and opposite. Ch23-42/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 21
  • 22. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Exam problem, April 2005 (cont) Gaussian 2) What is surface charge on outer surface of -22.1nC 10.0 cm E = 21.0 kN/C shell ? 8 cm 3 cm 5 cm Ans: Draw a Gaussian surface with radius 10.0 cm. By symmetry, E is the same everywhere on surface and parallel to dA so, using Gauss’ Law Other charges net to zero =>This is charge on outer surface . Ch23-43/60 Exam problem, April 2005 3) What is the magnitude and direction of the electric field at -22.1nC 10 cm E = 21.0 kN/C 2, 4.00 and 6 cm 8 cm 3 cm from the centre 5 cm of the system ? Ans: 2 cm is inside the inner conductor => E = 0. Ans: 4.00 cm is between the conductors. As before -ve sign => field is directed inwards. Ans: 6 cm is inside outer conductor => E = 0. Ch23-44/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 22
  • 23. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Gauss’ Law: Cylindrical Symmetry Given an infinitely long insulating rod with a uniform linear charge density λ. What is the magnitude of E at distance r? Select a Gaussian surface to match the symmetry of the situation: cylinder of length h and radius r, coaxial with the rod + 2 end caps. Symmetry => at all points on cylindrical surface, E has the same magnitude and is directed outwards (for +ve charge). Area of cylindrical wall = 2πrh. Flux of E thru cylindrical surface is Ch23-45/60 Gauss’ Law: Cylindrical Symmetry There is no flux through the end caps since E is parallel to these surfaces. Gauss’ Law gives E directed radially outwards for +ve charge and radially inward for –ve charge. Holds for a finite line of charge away from ends. Ch23-46/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 23
  • 24. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Gauss’ Law: Planar Symmetry: non-conducting sheet Uniform +ve surface charge density σ. What is E at distance r from sheet? Consider a small cylindrical Gaussian surface (with end caps of area A) which passes thru sheet and extends a distance r from the surface. Gauss’ Law gives: From symmetry we know E is | surface and | end caps and parallel the cylindrical sides. Hence and Note how much easier than previous derivation (Ch22-31)! Ch23-47/60 Gauss’ Law: Planar Symmetry: flat conducting plates Consider a single, infinite conducting plate. Since charge resides on surface, it resides equally on each surface to maximize distance between repelling charges. We know there will be an electric field | each surface with magnitude E = σ1 / εo (from Ch23-34) +ve charge => field is away from plate. Ch23-48/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 24
  • 25. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Conducting vs insulating sheet: where does the factor 2 come from? For a conductor, we have shown that near surface: For a plane of charge on an insulator, we have twice shown (Ch23-47,Ch22-31) that near the surface: How do we explain the factor 2 difference? Consider a conducting plane with a total surface charge of σ (C/m2) on both surfaces. By symmetry this charge will reside equally on each surface, σ1=σ/2. The electric field is E=σ1 /2 εo, consistent with the insulator equation where σ1&σ are the charge densities on 1 surface. Ch23-49/60 Gauss’ Law: Planar Symmetry: Two conducting plates Consider plate (b) with –ve excess charge with same surface charge density, σ1 = > same field strength as (a) but directed inwards. Now bring them close and parallel (c) . Opposite excess charges attract each other and all excess charges move onto inner surfaces with σ = 2σ1 Issue: How do we know that the attractive forces of the opposite charges dominate the repulsive? Ch23-50/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 25
  • 26. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Gauss’ Law: Planar Symmetry: Two conducting plates Theorem: When two equally and B oppositely charged conducting plates are brought parallel to each other, charge lies entirely on the inner surfaces. A Proof: Consider Gaussian surface A (red lines) which just encloses both plates. Total charge enclosed is 0 since charges on two plates are equal and opposite => flux over the surface is zero. By symmetry, any exterior field is | surface => flux = AE But flux = 0 => E = 0. Now consider Gaussian surface B (blue line) which is a cylinder cutting either outer surface. If there were any charge on the outer surface, the field outside would be σ/εo, but it is 0 from above => σ = 0 on outer surfaces. Ch23-51/60 Gauss’ Law: Planar Symmetry: Two conducting plates Theorem: The electric field between two parallel, equally charged, conducting plates is E = 2σ1/εo where σ1 is original surface charge on each. Proof: The surface charge, σ is equal to 2σ1 because the charge from both sides of the plates is now on the inner surface as discussed before. Consider a cylindrical Gaussian surface as shown in red. As before, for a conducting surface, the electric field vector is | surface and given by σ/εo = 2σ1/εo QED Note how powerful Gauss’ Law is here. It gets the right answer without even enclosing the second plate! But the field has doubled. Ch23-52/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 26
  • 27. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Gauss’ Law: Planar Symmetry: Two conducting plates Summary First showed field near a single plate is | surface with E=σ1/εo Next, showed for two parallel plates that all charge move to the inner surface and E = 0 outside the conductor pair. Finally, showed the field between them is E = 2σ1/εo Note: Final charge distn is NOT sum of two original ones, but electric field vectors are (not a general result). Ch23-53/60 Gauss’ Law: Planar Symmetry: Two non-conducting plates Different from previous because charges cannot move as a result of each other. Use superposition to add up electric fields from each separately The fields on opposite sides of each plane are uniform and | surface with magnitude E = σ/2εo (from before). Between the planes the vectors add, on outside they partially cancel (exactly if equal charge densities) Ch23-54/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 27
  • 28. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Gauss’ Law: Spherical Symmetry: shell Use Gauss’ Law to prove two shell theorems stated previously. Theorem: A shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell’s charge were at centre of shell. Proof: Gaussian surface S2 surrounds a total charge q. Following the technique of the earlier derivation re Coulomb’s Law (Ch23-26) we can show that electric field vectors on S2 are radial, with magnitude => shell of charge has same effect as a charge q located at the centre of shell. QED Ch23-55/60 Gauss’ Law: Spherical Symmetry: shell Theorem: If a charged particle is located inside a shell of uniform charge there is no electrostatic force on the particle from the shell of charge. Proof: Consider a Gaussian surface S1 inside the shell of charge with no charge inside S1. By Gauss’ Law we can write Thus, either everywhere E = 0 or E is | to the area vector of surface S1 so E.dA = 0 => E tangential to the Gaussian surface. The latter is impossible since there is no charge for the electric field to start or end on while tangential to surface. Hence E=0 everywhere. QED So much simpler than a 3-D integral over surface! Ch23-56/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 28
  • 29. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Gauss’ Law: Spherical Symmetry: sphere Consider non-conducting sphere of charge rather than shell. A spherically symmetric charge distribution can be considered as a set of nested shells of charge. We will assume that the volume charge distribution ρ (C/m3) is a function of radius only, i.e. spherically symmetric. Case: r > R (outer radius of charge) Construct a spherical Gaussian surface through r. As before, E is radial with where the total charge q is given by The last equality holds only if ρ is constant with r. Thus once again, the symmetrical distn of charges is equivalent to a point charge at the centre of the sphere. Ch23-57/60 Gauss’ Law: Spherical Symmetry: uniform non-conducting sphere Case: r < R Construct a spherical Gaussian surface through r. The shell of charge outside r has no effect on the charge inside (as shown before). As before,the charge enclosed, q’ (see right for uniform case), sets up a field as if concentrated at the centre. The magnitude of the electric field scales linearly with r Ch23-58/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 29
  • 30. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Question Two large, parallel non-conducting sheets with identical +ve uniform surface charge densities and a sphere with a uniform +ve charge density. Rank the 4 numbered points according to the magnitude of the net electric field there, largest first. Ch23-59/60 Question Two large, parallel non-conducting sheets with identical +ve uniform surface charge densities and a sphere with a uniform +ve charge density. Rank the 4 numbered points according to the magnitude of the net electric field there, largest first. Ans: 3 and 4 tie as largest, then 2, then 1 Key ideas: E from the sheets is constant everywhere between them (0 actually), rest is 1/r2. Ch23-60/60Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 30

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