Physics 212 Lecture 4 Gauss Law Physics 212

Physics 212 Lecture 4 Gauss’ Law Physics 212 Lecture 4, Slide 1

Gauss’ Law In general, the integral to calculate flux is difficult. But in cases with symmetry (spheres, cylinders, etc. ) we can find a closed surface where, is constant everywhere on the surface. For such cases, 19 Physics 212 Lecture 4, Slide 2

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

Examples with Symmetry Spherical 21 Cylindrical Planar Physics 212 Lecture 4, Slide 4

Checkpoint 4 In which case is E at point P the biggest? A) A B) B C) the same Recall that for an infinite plane with charge density (Coul/m 2), E = /2 0 27 Physics 212 Lecture 4, Slide 5

Superposition ! NET + Case A 34 + - + Case B Physics 212 Lecture 4, Slide 6

Checkpoint 1 The E field from a charged cube is not constant 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 Physics 212 Lecture 4, Slide 7

Conductors = charges free to move 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 Physics 212 Lecture 4, Slide 8

E = 0 inside conductors Charges reside on surfaces of conductors. Induced Charges Begin with a neutral conductor: Q=0 No charge inside or on the surface Now bring a positive charge +Q near the conductor. This induces (-) 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 + + + - +Q Physics 212 Lecture 4, Slide 9

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 Qouter Q D) inner = +Q/2, outer = -Q/2 E) inner = +Q, outer = -Q Qinner 10 Physics 212 Lecture 4, Slide 10

Checkpoint 3 26 Physics 212 Lecture 4, Slide 11

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 Physics 212 Lecture 4, Slide 12

Checkpoint 2 31 Physics 212 Lecture 4, Slide 13

Calculation y r 2 neutral conductor +3 Q r 1 x Point charge +3 Q at center of neutral conducting shell of inner radius r 1 and outer radius r 2. What is E everywhere? 36 A Magnitude of E is function of r. A Direction of E is along B Magnitude of E is function of (r-r 1). B Direction of E is along C Magnitude of E is function of (r-r 2). C Direction of E is along D None of the above Physics 212 Lecture 4, Slide 14

Calculation y r 2 neutral conductor +3 Q E is a function of r. Direction of along. r 1 x r < r 1 40 is Use Gauss’ Law with a spherical surface Centered on the origin to determine E(r). r 1 < r 2 r > r 2 A A B B C C Physics 212 Lecture 4, Slide 15

r 2 neutral conductor +3 Q r < r 1 r > r 2 r 1 < r 2 What is the induced surface charge density at r 1 ? A Gauss’ Law: r 2 B C 44 +3 Q r 1 Similarly: Physics 212 Lecture 4, Slide 16

y r 2 Now suppose we give the conductor a charge of -Q a) What is E everywhere? b) What are charge distributions at r 1 and r 2? -Q conductor +3 Q r 1 + + r 2 x + + + +3 Q + + + r < r 1 46 r > r 2 A A B B C C r 1 + -3 Q + + +2 Q + + r 1 < r 2 Physics 212 Lecture 4, Slide 17