Physics 212 Lecture 4

1. Physics 212 Lecture 4 Gauss’ Law

2. Gauss’ Law But in cases with symmetry (spheres, cylinders, etc.) we can find a closed surface where, is constant everywhere on the surface. For such cases, In general, the integral to calculate flux is difficult. 19

3. r L Example E field from an infinite line of charge with charge density  Choose cylinder of radius r, length L, centered on the line of charge. By symmetry, E field points radiallyoutward ( for positive  ) and so

4. Examples with Symmetry Spherical Cylindrical Planar 21

5. Checkpoint 4 Recall that for an infinite plane with charge density  (Coul/m2), E = /20 In which case is E at point P the biggest? A) A B) B C) the same 27

6. Superposition ! NET - + + + Case A Case B 34

7. Checkpoint 1 The E field from a charged cube is notconstant on any of these surfaces. Gauss’s law is true, but it does not help us to calculate E for this particular problem. (D) The field cannot be calculated using Gauss’ Law (E) None of the above How to calculate the field? Go back and find E by superposition. Add the contributions to E from each infinitesimal charge dq in the cube: 23

8. Conductors = charges free to move E = 0 Claim: E = 0 inside any conductor at equilibrium Why ? Charges in a conductor will move if E  0. They redistribute themselves until E = 0 and everything comes to equlibrium. Claim: Any excess charge in a conductor is on the surface (in equilibrium). Why? Take Gaussian surface (dashes) to be just inside conductor surface. • E = 0 everywhere inside conductor. • Use Gauss’ Law: 06

9. Induced Charges Begin with a neutral conductor: Q = 0 No charge inside or on the surface - - - + + + + Q E = 0 inside conductors  Charges reside on surfaces of conductors. Now bring a positive charge +Q near the conductor. Thisinduces (-) charge on one end and (+) charge on the other. The total charge on the conductor is still zero. The induced charge is on the surface of the conductor. 09

10. Charge in Cavity of Conductor A particle with charge +Q is placed in the center of an uncharged conducting hollow sphere. How much charge will be induced on the inner and outer surfaces of the sphere? A) inner = –Q, outer = +Q B) inner = –Q/2 , outer = +Q/2 C) inner = 0, outer = 0 D) inner = +Q/2, outer = -Q/2 E) inner = +Q, outer = -Q Qouter Q Qinner 10

11. Checkpoint 3 26

12. Checkpoint 3 What is direction of field OUTSIDE the red sphere? Using a Gaussian surface that enclosed both conducting spheres, the net enclosed charge will be zero. Therefore, the field outside will be zero. 29

13. Checkpoint 2 31

14. Calculation A B C D Magnitude of E is function of r. A B C D Direction of E is along Magnitude of E is function of (r-r1). Direction of E is along Magnitude of E is function of (r-r2). Direction of E is along None of the above. None of the above y neutral conductor r2 Point charge +3Q at center of neutral conducting shell of inner radius r1 and outer radius r2. What is E everywhere? +3Q r1 x 36

15. y Calculation r < r1 r1 < r < r2 r > r2 A B C A B C neutral conductor r2 E is a function of r . Direction of is along . +3Q r1 x Use Gauss’ Law with a spherical surface Centered on the origin to determine E(r). 40

16. Gauss’ Law: A B C r2 Similarly: +3Q r1 neutral conductor r2 r < r1 r > r2 +3Q r1 r1 < r < r2 What is the induced surface charge density at r1 ? 44

17. y + + + + r2 + + + + +3Q r1 + + +2Q + -3Q + + + + + + r < r1 + r1 < r < r2 A B C Now suppose we give the conductor a charge of -Q -Q • What is E everywhere? • What are charge distributions at r1 and r2? r2 conductor +3Q r1 x r > r2 A B C 46