Gauss's Law

1. Ch. 24 Gauss's Law Electric Flux • Electric flux is the amount of electric field going across a surface • It is defined in terms of a direction, or normal unit vector, perpendicular to the surface • For a constant electric field, and a flat surface, it is easy to calculate • Denoted by E • Units of Nm2/C • When the surface is flat, and the fields are constant, youcan just use multiplication to get the flux • When the surface is curved, or the fields are not constant,you have to perform an integration

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5. Solve on Board

6. Total Flux Out of Various Shapes A point charge q is at the “center” of a (a) sphere (b) joined hemispheres (c) cylinder (d) cube. What is the total electric flux out of the shape? b a q q a q

7. Total Flux Out of a Cube A point charge q is at the “center” of a (a) sphere (b) joined hemispheres (c) cylinder (d) cube. What is the total electric flux out of the shape? What is the answer for the cube? A) < 4keq B) > 4keq C) = 4keq D) Insufficient information 2a q 2a 2a

8. Gauss’s Law • No matter what shape you use, the total electric flux out of a region containing a point charge q is 4keq = q/0. • Why is this true? • Electric flux is just measuring how many fieldlines come out of a given region • No matter how you distort the shape, the field linescome out somewhere • If you have multiple charges inside the region their effects add • However, charges outside the region do not contribute q q4 q3 q1 q2

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10. Using Gauss’s Law • Gauss’s Law can be used to solve three types of problems: • Finding the total charge in a region when you know the electric field outside that region • Finding the total flux out of a region when the charge is known • It can also be used to find the flux out of one side in symmetrical problems • In such cases, you must first argue from symmetry that the flux is identical through each side • Finding the electrical field in highly symmetrical situations • One must first use reason to find the direction of the electric field everywhere • Then draw a Gaussian surface over which the electric field is constant • Use this surface to find the electric field using Gauss’s Law • Works generally only for spherical, cylindrical, or planar-type problems

11. Using Gauss’s Law to find total charge A cube of side a has an electric field of constant magnitude |E| = E pointing directly out on two opposite faces and directly in on the remaining four faces. What is the total charge inside the cube? A) 6Ea20 B) – 6Ea20 C) 2Ea20D) – 2Ea20 E) None of the above a a a

14. Using Gauss’s Law to find flux A very long box has the shape of a regular pentagonal prism. Inscribed in the box is a sphere of radius R with surface charge density . What is the electric flux out of one lateral side of the box? Perspective view • The flux out of the end caps is negligible • Because it is a regular pentagon, the flux from theother five sides must be the same End view

15. Using Gauss’s Law to find E-field A sphere of radius a has uniform charge density  throughout. What is the direction and magnitude of the electric field everywhere? • Clearly, all directions are created equal in this problem • Certainly the electric field will point away from the sphere at all points • The electric field must depend only on the distance • Draw a sphere of radius r around this charge • Now use Gauss’s Law with this sphere r a Is this the electric field everywhere?

16. Using Gauss’s Law to find E-field (2) A sphere of radius a has uniform charge density  throughout. What is the direction and magnitude of the electric field everywhere? [Like example 24.3] • When computing the flux for a Gaussian surface, only include the electric charges inside the surface r a r/a

17. Electric Field From a Line Charge What is the electric field from an infinite line with linear charge density ? L r • Electric field must point away from the line charge, and depends only on distance • Add a cylindrical Gaussian surface with radius r and length L • Use Gauss’s Law • The ends of the cylinder don’t contribute • On the side, the electric field and the normal are parallel

18. Electric Field From a Plane Charge What is the electric field from an infinite plane with surface charge density ? • Electric field must point away from the surface, and depends only on distance d from the surface • Add a box shaped Gaussian surface of size 2dLW • Use Gauss’s Law • The sides don’t contribute • On the top and bottom, the electric field and the normal are parallel

20. Solve on Board

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22. Conductors and Gauss’s Law • Conductors are materials where charges are free to flow in response to electric forces • The charges flow until the electric field is neutralized in the conductor Inside a conductor, E = 0 • Draw any Gaussian surface inside the conductor In the interior of a conductor, there is no charge The charge all flows to the surface

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24. Electric Field at Surface of a Conductor • Because charge accumulates on the surface of a conductor, there can be electric field just outside the conductor • Will be perpendicular to surface • We can calculate it from Gauss’s Law • Draw a small box that slightly penetratesthe surface • The lateral sides are small and have no flux through them • The bottom side is inside the conductor and has no electric field • The top side has area A and has flux through it • The charge inside the box is due to the surface charge  • We can use Gauss’s Law to relate these

25. Where does the charge go? A hollow conducting sphere of outer radius 2 cm and inner radius 1 cm has q = +80 nC of charge put on it. What is the surface charge density on the inner surface? On the outer surface? A) 20 nC/cm2 B) 5 nC/cm2 C) 4 nC/cm2 D) 0E) None of the above 80 nC • The Gaussian surface is entirely contained in the conductor; therefore E = 0 and electric flux = 0 • Therefore, there can’t be any charge on the inner surface 1 cm • From the symmetry of the problem, the charge will be uniformly spread over the outer surface 2 cm cutaway view • The electric field: • The electric field in the cavity and in the conductor is zero • The electric field outside the conductor can be found from Gauss’s Law

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29. Sample problem An infinitely long hollow neutral conducting cylinder has inner radius a and outer radius b. Along its axis is an infinite line charge with linear charge density . Find the electric field everywhere. b a end-on view perspective view • Use cylindrical Gaussian surfaces when needed in each region • For the innermost region (r < a), the total charge comes entirely from the line charge • The computation is identical to before • For the region inside the conductor, the electric field is always zero • For the region outside the conductor (r > b), the electric field can be calculated like before • The conductor, since it is neutral, doesn’t contribute

30. Where does the charge go? (2) + – + – – + + – – + – + – + – + + + + – – – – + – + – + How can the electric field appear, then disappear, then reappear? + • The positive charge at the center attracts negative charges from the conductor, which move towards it • This leaves behind positive charges, which repel each other and migrate to the surface end-on view • In general, a hollow conductor masks the distribution of the charge inside it, only remembering the total charge • Consider a sphere with an irregular cavity in it cutaway view q