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Physics 2112 Unit 4: Gauss’ Law

Physics 2112 Unit 4: Gauss’ Law. Today’s Concepts: A) Conductors B) Using Gauss’ Law. Use #1: Determining E without calculus. E from infinite line of charge. We did this before with calculus. Remember?. Let’s do it again using Guass Law. Use #2: Determine Charge on Surfaces.

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Physics 2112 Unit 4: Gauss’ Law

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  1. Physics 2112 Unit 4: Gauss’ Law Today’s Concepts: A) Conductors B) Using Gauss’ Law

  2. Use #1: Determining E without calculus E from infinite line of charge. We did this before with calculus. Remember? Let’s do it again using Guass Law

  3. Use #2: Determine Charge on Surfaces E= 0 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 • Why? • Apply Gauss’ Law • Take Gaussian surface to be just inside conductor surface

  4. 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 ALWAYS TRUE!

  5. 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?

  6. Example 4.2 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. x R 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?

  7. Gauss’ Law ALWAYS TRUE! In cases with symmetry can pull E outside and get 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

  8. 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? The field points radially outward The field points radially inward The field is zero

  9. Gauss’ Law Symmetries ALWAYS TRUE! In cases with symmetry can pull E outside and get Spherical Cylindrical Planar

  10. 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 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

  11. 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? ρ/εo zero ρ(b3-a3)/(3εor2) none of the above

  12. 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?

  13. Quick Review: Infinite Sheet of Charge

  14. 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? Case A Case B They are the same

  15. Example 4.4 y • Point charge +3Qat center of neutral conducting shell of inner radius r1 and outer radius r2. • What is Efor the three regions of: • r < r1 • r1< r < r2 • r2 < r? r2 neutral conductor +3Q r1 x

  16. Calculation y Point charge +3Qat center of neutral conducting shell of inner radius r1 and outer radiusr2. What is E everywhere? r2 neutral conductor +3Q r1 x r<r1 r>r2 r1<r <r2

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