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Physics 2112 Unit 4: Gauss’ Law Today’s Concepts: A) Conductors B) Using Gauss’ Law Unit 4, Slide 1 Use #1: Determining E without calculus E from infinite line of charge. We did this before with calculus. Remember? E x ( P ) k f h * sec * sin 2 d ( h * tan ) 0 2 h 2 Let’s do it again using Guass Law Unit 4, Slide 2 Use #2: Determine Charge on Surfaces E 0 inside any conductor at equilibrium (If E ≠0, then charge feels force and moves!) Excess charge on conductor only on surface at equilibrium E0 Why? Apply Gauss’ Law Take Gaussian surface to be just inside conductor surface Q enc E dA 0 surface o Q enc 0 Unit 4, Slide 3 Gauss’ Law + Conductors + Induced Charges How Does This Work? Charges in conductor move to surfaces to make Qenclosed 0. We say charge is induced on the surfaces of conductors Q E d A en c su rfa ce o ALWAYS TRUE! Unit 4, Slide 4 Example 4.1 An infinite line of charge with linear density λ1 = -5.1μC/m is positioned along the axis of a neutral conducting shell of inner radius a = 2.7 cm and outer radius b = 5.0 cm and infinite length. A) What is λb, the linear charge density on the outer surface of the conducting shell ? B) What is Ex(R), the electric field at point R, located a distance dR = 0.9 cm from the origin and making an angle of 30o with respect to the y-axis as shown? Unit 4, Slide 5 Example 4.2 xR An infinite line of charge with linear density λ1 = -5.1μC/m is positioned along the axis of a thick conducting shell of inner radius a = 2.7 cm and outer radius b = 5.0 cm and infinite length. The conducting shell is uniformly charged with a linear charge density λ 2 = +2.0 μC/m. C) What is Ex(R), the electric field at point R, located a distance dR = 7.0 cm from the origin and making an angle of 30o with respect to the y-axis as shown? Unit 4, Slide 6 Gauss’ Law Q enc E d A ALWAYS TRUE! 0 In cases with symmetry can pull E outside and get E Q enc A 0 In General, integral to calculate flux is difficult…. and not useful! To use Gauss’ Law to calculate E, need to choose surface carefully! 1) Want E to be constant and equal to value at location of interest OR 2) Want E dot A 0 so doesn’t add to integral Unit 4, Slide 7 CheckPoint: Charged Conducting Sphere & Shell 1 A positively charged solid conducting sphere is contained within a negatively charged conducting spherical shell as shown. The magnitude of the total charge on each sphere is the same. Which of the following statements best describes the electric field in the region between the spheres? A. The field points radially outward B. The field points radially inward C. The field is zero Unit 4, Slide 8 Gauss’ Law Symmetries Q E d A enc ALWAYS TRUE! 0 In cases with symmetry can pull E outside and get E Spherical A 4 r E 2 Q enc 4 r 0 2 Cylindrical Planar A 2 rL A 2 r E 2 r 0 E Q enc A 0 2 2 0 Unit 4, Slide 9 CheckPoint: Gaussian Surface Choice You are told to use Gauss' Law to calculate the electric field at a distance R away from a charged cube of dimension a. Which of the following Gaussian surfaces is best suited for this purpose? A. B. C. D. E. a sphere of radius R+1/2a a cube of dimension R+1/2a a cylinder with cross sectional radius of R+1/2a and arbitrary length This field cannot be calculated using Gauss' law None of the above Electricity & Magnetism Lecture 4, Slide 10 CheckPoint: Charged Sphericlal Shell A charged spherical insulating shell has inner radius a and outer radius b. The charge density on the shell is ρ. What is the magnitude of the E-field at a distance r away from the center of the shell where r < a? A. B. C. D. ρ/εo zero ρ(b3-a3)/(3εor2) none of the above Electricity & Magnetism Lecture 4, Slide 11 Example 4.3 A charged spherical insulating shell has inner radius a and outer radius b. The charge density on the shell is ρ. What is the magnitude of the E-field at a distance r away from the center of the shell where a< r < b? Electricity & Magnetism Lecture 4, Slide 12 Quick Review: Infinite Sheet of Charge Unit 4, Slide 13 CheckPoint: Infinite Sheets of Charge In both cases shown below, the colored lines represent positive (blue) and negative (red) charged planes. The magnitudes of the charge per unit area on each plane is the same. In which case is the magnitude of the electric field at point P bigger? A. Case A B. Case B C. They are the same Unit 4, Slide 14 Example 4.4 y r2 Point charge 3Q at center of neutral conducting shell of inner radius r1 and outer radius r2. neutral conductor 3Q r1 x What is E for the three regions of: r < r1 r1< r < r2 r2 < r ? Electricity & Magnetism Lecture 4, Slide 15 Calculation y r2 Point charge 3Q at center of neutral conducting shell of inner radius r1 and outer radius r2. neutral conductor 3Q r1 Q enc E d A 0 x What is E everywhere? r < r1 r > r2 r1 < r < r2 E 1 3Q 4 0 r 2 E 0 Electricity & Magnetism Lecture 4, Slide 16