Gauss' law
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Gauss' law Presentation Transcript

  • 1. Electricity & Magnetism Maxwell’s Equations Faraday’s law of induction Gauss’ Law (Magnetism)Gauss’ Law (Electricity) There are 4 pillars that Ampere’s Law make up the foundation of Electricity & Magnetism. We’ll study each of these in varying degrees.
  • 2. First Pillar: Gauss’ LawKarl Fredrick Gauss (1777-1855) He was a contemporary of Charles Coulomb (1736-1806) Instead of finding the field from a single charge, Gauss found the field from a bunch of charges (charge distribution). Why is Gauss’ Law important?Specific GeneralCoulomb’s Law finds a Gauss’ Law finds a field/chargefield/charge from point charges. from any charged object. +Q +Q -Q +Q +Q
  • 3. What is Gauss’ Law? The electric field coming through a certain area is proportional to the charge enclosed. Gaussian Surface  An imaginary surface around a charge distribution (group of charges) arbitrarily chosen for its symmetry (so the Electric Field coming through the imaginary surface is fairly constant through all areas of the surface). Point Charge Wire Strange Shape Parallel Plates -Q -Q -Q -Q -Q +Q -Q (Capacitor)Examples: +Q + + + + + +Q +Q - - - - - Cylinder: Surface Area = 2πrh Strange Surface: Box: Sphere: Surface Area = L x W Calculus Surface Area = 4πr2
  • 4. What is Gauss’ Law? 2 The electric field coming through a certain area is proportional to the charge enclosed. 32. Electric Fields Quick reminders on Electric Field Lines 1. More field lines = stronger field. 2. Field lines always come out of the surface perpendicularly. 3. Out of +, into ‒ (show the direction a + charge will move)3. Charge enclosed The field is proportional to the charge inside the Gaussian Surface. More Field Lines = Stronger Field = Stronger Charge Inside.
  • 5. The electric field coming through a certain area isGauss’ Law: proportional to the charge enclosed. ∫EdA α Q How do we make this an equation? – Add a constant! ∫EdA = cQ c = 1/εo  remember this?!? Permitivity Constant εo = 8.85x10-12 Nm2/C  k = 1/4πεo = 8.99 x 109 Nm2/C2 Q ∫ EdA = ε o How much field through a certain area Rename this to be Electric Flux (ΦE)  how much field comes through a certain area.
  • 6. And finally… Gauss’ Law Summary The electric field coming through a certain area is proportional to the charge enclosed. Q Φ E = ∫ EdA = εo ΦE = Electric Flux (Field through an Area) E = Electric Field A = Area q = charge in object (inside Gaussian surface) εo = permittivity constant (8.85x 10-12)
  • 7. Sample Problem 1A Van de Graaff machine with a radius of 0.25 m has been charged up. What is the electric field 0.1 m away from the center of the sphere? Hint: Where are all the charges? On the outside. So how much charge is in the center? None. 0 Φ E = EA = =0 εo Since all the charge is on the surface, it proves there is no field inside a conducting surface!
  • 8. Sample Problem 2Find the electric field around a point charge, Q. Remember the area of a sphere (Gaussian Q Surface in this case) is 4πr2. EA = εo E (4πr ) = 2 Q +Q εo Q 1 E= → don t forget k = 4πε o r 2 4πε o kQ What does this look like? E= 2 Coulomb’s electric field for point charges! r
  • 9. Sample Problem 3A solid sphere of radius R = 40 cm has a total positivecharge of 26 μC uniformly distributed throughout itsvolume. Calculate the magnitude of the electric field atthe following distances from the center of the sphere. (a) 0 cm (b) 30 cm (c) 60 cm Inside sphere Q = 0 Still inside sphere ∴Q still = 0 Q Φ E = EA = εo Q Q Φ E = EA = Φ E = EA = Q εo εo E (4πr ) = 2 εo 0 0 Φ E = EA = =0 Φ E = EA = =0 26 x10 −6 εo εo E (4π 0.6 2 ) = 8.85 x10 −12 E = 6.49 x105 N / C
  • 10. Sample Problem 3A solid sphere of radius R = 40 cm has a total positivecharge of 26 μC uniformly distributed throughout itsvolume. Calculate the magnitude of the electric field atthe following distances from the center of the sphere. (a) 0 cm (b) 30 cm (c) 60 cm Inside sphere Q = 0 Still inside sphere ∴Q still = 0 Q Φ E = EA = εo Q Q Φ E = EA = Φ E = EA = Q εo εo E (4πr ) = 2 εo 0 0 Φ E = EA = =0 Φ E = EA = =0 26 x10 −6 εo εo E (4π 0.6 2 ) = 8.85 x10 −12 E = 6.49 x105 N / C