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Gauss' Law

Gauss' Law. c. a. b. Fundamental Law of Electrostatics. Coulomb’s Law Force between two point charges OR Gauss’ Law Relationship between Electric Fields and charges. Preflight 4:. dS. dS. 2. 1.

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Gauss' Law

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  1. Gauss' Law c a b

  2. Fundamental Lawof Electrostatics • Coulomb’s Law Force between two point charges OR • Gauss’ Law Relationship between Electric Fields and charges

  3. Preflight 4: dS dS 2 1 2) A positive charge is contained inside a spherical shell. How does the electric flux dФEthrough the surface element dSchange when the charge is moved from position 1 to position 2? a) dФEincreases b) dФEdecreases c) dФEdoesn’t change

  4. dS dS 2 1 3) A positive charge is contained inside a spherical shell. How does the flux ФEthrough the entire surface change when the charge is moved from position 1 to position 2? a) ФE increases b) ФE decreases c) ФE doesn’t change

  5. Gauss’ Law • Gauss’ Law(a FUNDAMENTAL LAW): The net electric flux through any closed surface is proportional to the charge enclosed by that surface. • How do we use this equation?? • The above equation is ALWAYS TRUE but it doesn’t look easy to use. • It is very useful in finding E when the physical situation exhibits massive SYMMETRY.

  6. To solve the above equation forE, you have to be able to CHOOSE A CLOSED SURFACE such that the integral is TRIVIAL. • (1) Direction: surface must be chosen such that E is known to be either parallel or perpendicular to each piece of the surface; • If then • If then • (2) Magnitude: surface must be chosen such that E has the same value at all points on the surface when E is perpendicular to the surface. Gauss’ Law…made easy

  7. Gauss’ Law…made easy • With these two conditions we can bring E outside of the integral…and: • Note that is just the area of the Gaussian surface over which we are integrating. Gauss’ Law now takes the form: • This equation can now be solved for E (at the surface) if we know qenclosed(or for qenclosed if we know E).

  8. z c y b you may use different E’s for different surfaces of your “object” a x Geometry and Surface Integrals • If E is constant over a surface, and normal to it everywhere, we can take E outside the integral, leaving only a surface area z R R L

  9. E R +Q • Why? • E normal to every point on the surface • E has same value at every point on the surface • can take E outside of the integral! • Therefore, ! • Gauss’ Law • We are free to choose the surface in such problems… we call this a “Gaussian” surface Þ Þ Gauss Þ Coulomb • We now illustrate this for the field of the point charge and prove that Gauss’ Law implies Coulomb’s Law. • Symmetry Þ E-field of point charge is radial and spherically symmetric • Draw a sphere of radius R centered on the charge.

  10. What is the magnitude of the electric field due to a solid sphere of radius a with uniform charge density r (C/m3)? r a r Þ Gauss’ Law same as point charge! Uniform charged sphere • Outside sphere: (r>a) • We have spherical symmetry centered on the center of the sphere of charge • Therefore, choose Gaussian surface = hollow sphere of radiusr

  11. r a r E a r Uniform charged sphere • Outside sphere: (r > a) • Inside sphere: (r < a) • We still have spherical symmetry centered on the center of the sphere of charge. • Therefore, choose Gaussian surface = sphere of radiusr Gauss’ Law But, Thus:

  12. Charges on a conductor only reside on the surface(s)! + + + + + + + + Conducting sphere Gauss’ Law and Conductors • We know that E=0 inside a conductor (otherwise the charges would move). • But since  .

  13. Preflight 4: A blue sphere A is contained within a red spherical shell B. There is a chargeQAon the blue sphere and chargeQBon the red spherical shell. A B 7) The electric field in the region between the spheres is completely independent of QB the charge on the red spherical shell. True False

  14. (a)EA< EB (b)EA= EB (c)EA> EB 1B (b) s1= 0 (c) s1> 0 • What is the surface charge density s1 on the inner surface of the conducting shell in case A? (a) s1< 0 Consider the following two topologies: A) A solid non-conducting sphere carries a total charge Q = -3 mC distributed evenly throughout. It is surrounded by an uncharged conducting spherical shell. s2 s1 -|Q| E B) Same as (A) but conducting shell removed 1A • Compare the electric field at point X in cases A and B:

  15. (a) EA< EB (b) EA= EB (c) EA> EB • Consider the following two topologies: • A solid non-conducting sphere carries a total charge Q = -3mC distributed evenly throughout. It is surrounded by an uncharged conducting spherical shell. s2 s1 -|Q| E 1A • Compare the electric field at point X in cases A and B: • Select a sphere passing through the point X as the Gaussian surface. • How much charge does it enclose? • Answer:-|Q|, whether or not the uncharged shell is present. • (The field at point X is determined only by the objects with NET CHARGE.)

  16. (b)s1= 0 (c)s1> 0 (a)s1< 0 s2 Consider the following two topologies: A solid non-conducting sphere carries a total charge Q = -3 mC and is surrounded by an uncharged conducting spherical shell. B) Same as (A) but conducting shell removed s1 -|Q| E • What is the surface charge density s1on the inner surface of the conducting shell in case A? 1B • Inside the conductor, we know the field E= 0 • Select a Gaussian surface inside the conductor • Since E = 0 on this surface, the total enclosed charge must be 0 • Therefore, s1 must be positive, to cancel the charge -|Q| • By the way, to calculate the actual value: s1= -Q/ (4 pr12)

  17. y Er Er + + + + + + + + + + + + + + + + + + + + x + + + + + + + + + h • Apply Gauss’ Law: • On the ends, Þ • On the barrel, AND NOTE: we have obtained here the same result as we did last lecture using Coulomb’s Law. The symmetry makes today’s derivation easier. Infinite Line of Charge • Symmetry ÞE-field must be ^ to line and can only depend on distance from line • Therefore, CHOOSE Gaussian surface to be a cylinder of radius r and length h aligned with the x-axis.

  18. A line charge l (C/m) is placed along the axis of an uncharged conducting cylinder of inner radius ri = a, and outer radius ro = b as shown. • What is the value of the charge density so (C/m2) on the outer surface of the cylinder? s0=? b a l (c) (b) (a)

  19. View end on: so Draw Gaussian tube which surrounds only the outer edge 0 b • A line charge l (C/m) is placed along the axis of an uncharged conducting cylinder of inner radius ri = a, and outer radius ro = b as shown. • What is the value of the charge density so (C/m2) on the outer surface of the cylinder? s0=? b a l (c) (b) (a)

  20. Preflight 4: 5) Given an infinite sheet of charge as shown in the figure. You need to use Gauss' Law to calculate the electric field near the sheet of charge. Which of the following Gaussian surfaces are best suited for this purpose? Note: you may choose more than one answer a) a cylinder with its axis along the plane b) a cylinder with its axis perpendicular to the plane c) a cube d) a sphere

  21. Infinite sheet of charge, surface charge density s • Symmetry: • direction of E = x-axis A • Therefore, CHOOSE Gaussian surface to be a cylinder whose axis is aligned with the x-axis. • Apply Gauss' Law: • On the barrel, • On the ends, • The charge enclosed =s A Therefore, Gauss’ LawÞ Conclusion: An infinite plane sheet of charge creates a CONSTANT electric field . +s x E E

  22. - + s - + - + + - + - A + - + - + - - + - + - A + - + - 0 E Two Infinite Sheets(into the screen) s E=0 E=0 • Field outside must be zero. Two ways to see: • Superposition • Gaussian surface encloses zero charge • Field inside is NOT zero: • Superposition • Gaussian surface encloses • non-zero charge

  23. Gauss’ Law: Help for the Homework Problems • Spherical Symmetry: Gaussian surface = sphere of radius r LHS: RHS:q = ALL charge inside radiusr • Cylindrical symmetry: Gaussian surface = cylinder of radius r LHS: RHS:q = ALL charge inside radiusr, lengthL • Planar Symmetry: Gaussian surface = cylinder of area A LHS: RHS:q = ALL charge inside cylinder =s A (See also: mandatory on-line “tutorial exercises”) • Gauss’ Law is ALWAYS VALID! • What Can You Do With This? • If you have (a) spherical, (b) cylindrical, or (c) planar symmetry AND: • If you know the charge (RHS), you can calculate the electric field (LHS) • If you know the field (LHS, usually because E=0 inside conductor), you can calculate the charge (RHS).

  24. σ1 σL σR E1 + EL - ER = 0 σ1 + σL - σR = 0 σL + σR = 0 (uncharged conductor) σ1 + 2σL = 0 σL = σR = A B C D Uncharged Conductor • Hints: • Assume σ is positive. If it’s negative, the answer will still work. • Assume to the right. • Use superposition, but keep signs straight • Think about which way a test charge would move. Sheets of Charge

  25. Summary • Gauss’ Law:Electric field flux through a closed surface is proportional to the net charge enclosed • Gauss’ Law is exact and always true…. • Gauss’ Law makes solving for E-field easy when the symmetry is sufficient • spherical, cylindrical, planar • Gauss’ Law proves that electric fields vanish in conductor • extra charges reside on surface

  26. Q2 Q1 R1 R2 Example 1: spheres • The inner sphere carries a charge Q1, and the spherical shell carries a charge Q2, such that Q2= -3Q1. • How is the charge distributed on the sphere? • How is the charge distributed on the spherical shell? • What is the electric field at r < R1? • Between R1 and R2? At r > R2? • What happens when you connect the two spheres with a wire? (What are the charges?) • A solid conducting sphere is concentric with a thin conducting shell, as shown A B C D

  27. Q2 Q1 R1 R2 A • How is the charge distributed on the sphere? * The electric field inside a conductor is zero. (A) By Gauss’s Law, there can be no net charge inside the conductor, and the charge must reside on the outside surface of the sphere + + + + + + + +

  28. Q2 Q1 R1 R2 B • How is the charge distributed on the spherical shell? * The electric field inside the conducting shell is zero. (B) There can be no net charge inside the conductor, therefore the inner surface of the shell must carry a net charge of -Q1, and the outer surface must carry the charge +Q1 + Q2, so that the net charge on the shell equals Q2. The charges are distributed uniformly over the inner and outer surfaces of the shell, hence and

  29. Q2 Q1 R1 R2 C • What is the Electric Field at r < R1? • Between R1 and R2? At r > R2? * The electric field inside a conductor is zero. (C)r < R1: Inside the conducting sphere (C) Between R1 and R2 : R1 < r < R2 Charge enclosed = Q1 (C) r > R2 Charge enclosed = Q1 + Q2

  30. - - - - - - - - - - - - - - - - - - - - - - D • What happens when you connect the two spheres with a wire? (What are the charges?) After electrostatic equilibrium is reached, there is no charge on the inner sphere, and none on the inner surface of the shell. The charge Q1 + Q2 on the outer surface remains. Also, for r< R2 and for r> R2

  31. Let’s try some numbers Q1 = 10mC R1 = 5cm Q2 = -30mC R2 = 7cm • sinner=-162 mC/m2 • souter = -325 mC/m2 • Electric field • r < 5cm: Er(r = 4cm) = 0 N/C • 5cm < r < 7cm: Er(r = 6cm) = 2.5 x 107 N/C • r> 7cm: Er(r = 8cm) = -2.81 x 107 N/C • Electric field • r > 7cm:Er(r = 9cm) = -2.22x 107 N/C B C D

  32. Example 2: Cylinders An infinite line of charge passes directly through the middle of a hollow, charged, CONDUCTING infinite cylindrical shell of radius R. We will focus on a segment of the cylindrical shell of length h. The line charge has a linear charge density l, and the cylindrical shell has a net surface charge density of stotal. stotal R l sinner souter h

  33. souter l R sinner stotal h A • How is the charge distributed on the cylindrical shell? • What issinner? • What issouter? • What is the electric field at r<R? • What is the electric field for r>R? B C

  34. What is sinner? A1 souter l R sinner stotal h • The electric field inside the cylindrical shell is zero. Hence, if we choose as our Gaussian surface a cylinder, which lies inside the cylindrical shell, we know that the net charge enclosed is zero. Therefore, there will be a surface charge density on the inside wall of the cylinder to balance out the charge along the line. • The total charge on the enclosed portion (of length h) of the line charge is: • Total line charge enclosed =l h • Therefore, the charge on the inner surface of the conducting cylindrical shell is • Qinner= -l h • The total charge is evenly distributed along the inside surface of the cylinder. Therefore, the inner surface charge densitysinneris justQinnerdivided by total area of the cylinder:sinner = -l h / 2p Rh = -l / 2p R • Notice that the result is independent ofh.

  35. A2 • What is souter? souter l R sinner stotal h • We know that the net charge density on the cylinder is stotal. The charge densities on the inner and outer surfaces of the cylindrical shell have to add up to stotal. Therefore, • souter =stotal – sinner = stotal +l /(2p R).

  36. B • What is the Electric Field at r<R? h Gaussian surface l r R h • Whenever we are dealing with electric fields created by symmetric charged surfaces, we must always first chose an appropriate Gaussian surface. In this case, for r <R, the surface surrounding the line charge is actually a cylinder of radius r. • Using Gauss’ Law, the following equation determines the E-field: • 2prhEr = qenclosed / eo • qenclosedis the charge on the enclosed line charge, which is lh, and (2prh) is the area of the barrel of the Gaussian surface. • The result is:

  37. C What is the Electric field for r>R? Gaussian surface l R stotal r h • As usual, we must first chose a Gaussian surface as indicated above. We also need to know the net charge enclosed in our Gaussian surface. The net charge is a sum of the following: • Net charge enclosed on the line:lh • Net charge enclosed within Gaussian surface, residing on the cylindrical shell:Q= 2pRhstotal • Therefore, net charge enclosed isQ + lh • The surface area of the barrel of the Gaussian surface is 2prh • Now we can use Gauss’ Law: 2prhE = (Q + lh) / eo • You have all you need to find the Electric field now. Solve for Erto find

  38. Let’s try some numbers R = 13 cmh = 168 cmstotal = 528 mC/m2l = 50mC/m A1 sinner = -61.21mC/m2 A2 souter = 589.2mC/m2 B • Electric Field (r<R); • Er(r = 5cm) = 1.798 x 107 N/C C • Electric Field (r>R); • Er(r = 20cm) = 4.328 x 107 N/C

  39. Example 3: planes Suppose there are infinite planes positioned at x1and x2. The plane at x1 has a positive surface charge density ofs1while the plane at x2has negative surface charge density of s2. Find: A the x-component of the electric field at a point x>x2 the x-component of the electric field at x1<x<x2 the x-component of the electric field at a point x<x1 B C y s 1 s 2 x1 x2 x

  40. Solutions A Ex(x>x2) We can use superposition to find . The E-field desired is to the right of both sheets. Therefore; y s 1 s 2 E2 E1 x1 x2 x

  41. B Ex(x1<x<x2) This time the point is located to the left ofs2and to the right of s1, therefore; y s 1 s 2 E2 E1 x1 x2 x

  42. C Ex(x<x1) When the point is located to the left of both sheets; y s 1 s 2 E1 E2 x1 x2 x

  43. Let’s add some numbers... x1= -2m x2= 2m s1= +2mC/m2 s2= -3mC/m2 A E1x= 1.130 x 105N/C Ex= -0.565 x 105N/C E2x= -1.695 x 105N/C B E1x= 1.130 x 105N/C Ex= 2.825 x 105N/C E2x= 1.695 x 105N/C C E1x= -1.130 x 105N/C Ex= 0.565 x 105N/C E2x= 1.695 x 105N/C

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