Flux. Flux = Φ = ρva. This is the same as asking What is the flux of water through the rectangle? The answer depends on:

Transcription

1 Ch. 21: Gauss s Law Gauss s law is an alternative description of Coulomb s law that allows for an easier method of determining the electric field for situations where the charge distribution is contains symmetry.

2 Flux Gauss s law relates something called electric flux with a charge distribution. The concept of electric flux is tricky so first let s consider an analogy. Consider a flowing river, and placing a wire rectangle in the water. We can ask How much water flows through the rectangle each second? This is the same as asking What is the flux of water through the rectangle? The answer depends on: density of water velocity flow of water orientation of the rectangle. If rectangle is perpendicular to velocity: Flux = Φ = ρva

3 Flux Suppose you rotate the rectangle. Does the flux change? Sure it does! Φ=ρvA cos φ = ρva = ρv A

4 Electric flux The Electric Flux is defined as the amount of an electric field passing through a surface The electric field plays the same role as the electric field isn t flowing ) For a uniform electric field and a rectangular surface: Φ E = EAcos φ = E A ρv (note though that

5 Clicker question University of Colorado, Boulder (2008)

6 Electric flux with curved surfaces and nonuniform fields When the surface is curved or the field is nonuniform, we calculate the flux by dividing the surface into small patches da, so small that each patch is essentially flat and the field is essentially uniform over each. The flux of each patch is dφ = E da The total flux is given by summing up the contribution of each patch: In the limit of infinitely many infinitesimally small patches, the sum becomes a surface integral: Φ E = surface E da

7 Gaussian Surfaces A Gaussian surface is any mathematical closed surface that fully encloses a 3D volume. Valid examples Not Valid We will look at the electric flux through Gaussian surfaces.

8 By convention, electric fields going from the inside to the outside of the surface produces a positive flux, and fields going from the outside to the inside produces a negative flux Now let s look at the relationship between the electric flux through a Gaussian surface and the charge distribution.

9 Clicker Question A dipole moment is situated inside a Gaussian surface having the shape of a cuboid (rectangular box). The net electric flux through this surface is 1. positive 2. negative 3. zero 4. depends how and exactly where the dipole is situated inside

10 Clicker Question Field lines due to a point-charge are shown below, as well as a cylindrical Gaussian surface.what is the net electric flux through the surface? A: 0 B: positive C: negative

11 Gauss s law The electric flux through any closed surface is proportional to the charge enclosed. E da = q enclosed 0 0 = 1 4πk = C 2 /Nm 2 The contour integral is over the Gaussian surface; q enclosed is the charge enclosed by that surface. Gauss s law is true for any surface and charge distribution Gauss s law is equivalent to Coulomb s law; both describe the inverse square dependence of the point-charge field. One of the four fundamental laws of electromagnetism

12 Gauss s Law and Point Charge Gaussian shell of radius r with charge in center Spherical symmetry allows one to claim E is constant (in magnitude) on Gaussian surface E is also parallel to the normal everywhere on surface E da =4πr 2 E E = k Q r 2 E da =4πkQ = 1 0 Q This holds even if charge is not in center of sphere.

13 CT 27.7 Three closed surfaces (2 spheres and a cube) enclose a point charge. Which surface has the largest electric flux through it? A: Small (pink) cube B: middle (yellow) sphere C: larger (green) sphere D: all 3 are same E: Need more info green yellow pink University of Colorado, Boulder (2008)

14 CT 27.6b A single charge Q is just outside a spherical Gaussian surface. What is the electric field inside the Gaussian surface? A: 0 everywhere inside B: non-zero everywhere in the sphere C: Not enough info given University of Colorado, Boulder (2008)

15 Using Gauss s law Gauss s law is always true. But it s only useful for calculating the electric field in situations with sufficient symmetry: Spherical symmetry Line symmetry Plane symmetry Gauss s law is always true, so it holds in both situations shown. Both surfaces surround the same net charge, so the flux through each is the same. But only the left-hand situation has enough symmetry to allow the use of Gauss s law to calculate the field. The electric fields differ in the two situations, even though the fluxes are the same

16 CT 27.6 A spherical shell has a uniform positive charge density on its surface. (There are no other charges around) What is the electric field inside the sphere? A: E=0 everywhere inside B: E is non-zero everywhere in the sphere C: E=0 only at the very center, but non-zero elsewhere inside the sphere. D: Not enough info given University of Colorado, Boulder (2008)

17 Q A = σa A = σr 2 AdΩ Q B = σa B = σr 2 BdΩ dω =2π sin θdθ E A Q A /r 2 A E A = E B E B Q B /r 2 B

18 CT The total charge on the thin spherical shell is Q. (Uniformly spread out) Which statement is true outside the spherical shell?! A) E = B)! E is not Q 4"# 0 r 2 ˆ r Q 4"# 0 r 2 ˆ r University of Colorado, Boulder (2008)

19 The field of a uniformly charged sphere The charge distribution has spherical symmetry, so the appropriate Gaussian surfaces are spheres. Outside the sphere, the enclosed charge is the total charge Q. Φ= E da =4πr 2 E E da = q enclosed πre= Qε0, so E= = πε 0r r Q Thus the field outside a spherical charge distribution is identical to that of a point charge. kq

20 The field of a uniformly charged sphere Inside the sphere: E da = q enclosed 0 Φ= E da =4πr 2 E q enclosed = ρ 4π r 3 3 r3 = Q R 4πr 2 E = 1 r 3 Q 0 R E = Qr 4π 0 R 3 = kq r R 3

21 Cylindrical symmetry The charge density depends only on the perpendicular distance from the cylindrical axis. This requires a charge distribution that is infinitely long. However, cylindrical symmetry is a good approximation for finite cylindrical charge distributions with length much greater than diameter, at points close to the charge. Example: A infinitely long line charge. By symmetry, E field must be radial Want a Gaussian surface that is perpendicular to field

22 Cylindrical symmetry Use a cylindrical Gaussian surface. Φ =2πrLE q enclosed = λl E da = q enc / 0 E = λ 2π 0 r

23 Plane symmetry In plane symmetry, the charge density depends only on the perpendicular distance from a plane (the plane of symmetry). This requires a charge distribution that extends infinitely in two directions. However, plane symmetry is a good approximation for finite slabs of charge whose thickness is much less than their extent in the other two dimensions, at points close to the charge. By symmetry, the electric field must be perpendicular to the plane of symmetry (say the z direction) and its magnitude can only depend on its distance from this plane of symmetry.

24 Example: Charged Plane Want Gaussian surface such that surface is either parallel to electric field (which contributes zero flux) or the surface is perpendicular to the electric field The Gaussian Pillbox Set up pillbox such that each end is the same distance from the plane. By symmetry each end must have the same electric flux: Φ =2AE = 1 0 σa E = σ E is uniform! Does not decrease with distance from plane.

25 2.18 Which is true about E at points on the imaginary dashed triangle? (I) (II) A. E is the same everywhere (uniform) in (I) ONLY B. E is uniform in (II) ONLY C. Uniform in both, but different for cases I and II D. Uniform in both, and same for cases I and II. E. E varies from point to point in both cases

26 Gauss s law and conductors Valence electrons in conductors are free to move, and they do so in response to an electric field. If a conductor is allowed to reach electrostatic equilibrium, then charges redistribute themselves to cancel the applied field inside the conductor. Therefore the electric field is zero inside a conductor (in electrostatic equilibrium).

27 The electric field must be perpendicular to the surface (a horizontal component would cause charges to accelerate/ move). Very near surface, E 1 0 σ Gauss s law requires that any free charge on a conductor reside on the conductor surface.

28 CT A negative point charge with charge -Q sits in the interior of a thick spherical (conducting) neutral metal shell. What is the total charge on the inner surface of the shell? -Q Outer surface Gaussian surface A: -Q B: Q C: 2Q D: zero E: Q/2 Inner surface University of Colorado, Boulder (2008)

29 A point charge Q sits outside a hollow spherical conducting shell. The electric field at point P inside of the shell is a) zero b) directed to the right c) directed to the left

30 Faraday Cage

31 2.30b A HOLLOW copper sphere has no net charge. A charge, q, is in the hole, SHIFTED right a bit from the center. (We are in static equilibrium.) What does the E field look like in the hole region? q A) Simple Coulomb field (straight away from q, right up to the wall) B) Complicated/ it s hard to compute

32 2.30b A HOLLOW copper sphere has no net charge. A charge, q, is in the hole, SHIFTED right a bit from the center. (We are in static equilibrium.) What does the E field look like outside the sphere? q A) Simple Coulomb field (radially outward from q ) B) Simple Coulomb field (radially outward from center of sphere) B) Complicated/ it s hard to compute

33 Clicker Question Consider two parallel, large conducting plates, each having a thickness a, oriented close to each other (this is a parallel plate capacitor). In the figure to the right, the plates are seen edge on. The left plate has a net charge Q and the right plate has a net charge Q. Q Q For the left plate, where does the charge reside? 1. All of the charge is on the right surface of the plate 2. All of the charge is on the left surface of the plate 3. Half of the charge is on each of the two plate surfaces. 4. The charge is distributed uniformly throughout the plate a

34 Parallel Plate Capacitor What is the electric field between the plates? EA = σ A/ 0 E = σ/ 0