Ch. 22: 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 contains symmetry.
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 = = va
Flux Suppose you rotate the rectangle. Does the flux change? Sure it does! = va cos = va? = ~v ~A
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: E = EAcos = ~ E ~A ~v (note though that
QuickCheck Clicker Question 27.3 The electric flux through the shaded surface is A. 0. B. 400 cos20 N m 2 /C. C. 400 cos70 N m 2 /C. D. 400 N m 2 /C. E. Some other value. Slide 27-43
Clicker question University of Colorado, Boulder (2008)
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 d = 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: E = I ~E ~ da
Gaussian Surfaces A Gaussian surface is any mathematical closed surface that fully encloses a 3D volume. Valid examples Not Valid
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 Increasing enclosed charge increases magnitude of flux.
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
Clicker Question A dipole moment is situated inside a spherical Gaussian surface 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
Clicker Question A dipole moment is situated inside a spherical Gaussian surface 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
Gauss s law The electric flux through any closed surface is proportional to the charge enclosed. I ~E da ~ = q enclosed 0 0 = 1 4 k =8.85 10 12 C 2 /Nm 2 The integral is over the Gaussian surface Gauss s law is equivalent to Coulomb s law; both describe the inverse square dependence of the point-charge field. Gauss s law is true for any surface and charge distribution
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)
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)
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 I E is also parallel to the normal everywhere on surface ~E ~ da =4 r 2 E I E = k Q r 2 ~E ~ da =4 kq = 1 0 Q
This holds even if charge is not in center of sphere. I ~E ~ da = Q enc 0 d =EdA = kq r 2 r2 d = kq d 0 = E 0 da 0 r 2 r2 cos 0 = kq r 02 r02 d =d ) Flux through S is the same as flux through S sin d d
Using Gauss s law Gauss s law is always true. It s only useful for calculating the electric field in situations with sufficient symmetry: Spherical symmetry Line symmetry Plane symmetry 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
CT 27.16 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 4πε 0 r 2 ˆ r Q 4πε 0 r 2 ˆ r University of Colorado, Boulder (2008)
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)
The field of a uniformly charged sphere The charge distribution has spherical symmetry, so the appropriate Gaussian surfaces are spherical shells. Outside the sphere, the enclosed charge is the total charge Q. = I I ~E ~ da =4 r 2 E ~E ~ da = q enclosed 0 2 4 p re= Qe 0, so E= =. 2 2 4p e 0r r Q Thus the field outside a spherical charge distribution is identical to that of a point charge. kq
The field of a uniformly charged sphere Inside the sphere: I ~E da ~ = q enclosed 0 I = ~E da ~ =4 r 2 E q enclosed = 4 r 3 3 r3 = Q R 4 r 2 E = 1 r 3 Q 0 R E = Qr 4 0 R 3 = kq r R 3
Electric Field Inside a Hydrogen Atom Electron cloud produces a charge density Q e = a 3 exp( 2r/a 0 ) Z 0 ~E da ~ = 1 Q enc 0 E(r) Including the proton s charge Q: 4 r 2 E = 1 Z r Q e 4 r 2 dr 0 4 r 2 E = Q 0 1 1 a 3 0 0 Z r 0 exp( 2r/a 0 )4 r 2 dr E = kq r 2 exp( 2r/a 0) 2(r/a 0 ) 2 2(r/a 0 )1 Comparison with E proton EêEproton 1.0 0.8 0.6 0.4 0.2 1 2 3 4 rêa
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 a 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
QuickCheck Clicker Question 27.13 A cylindrical Gaussian surface surrounds an infinite line of charge. The total flux F e through the Gaussian surface is A. 0. B. 2p rle C. p r 2 LE. D. 2p rle 2p r 2 E E. It will require an integration to find out. Slide 27-95
Cylindrical symmetry Use a cylindrical Gaussian surface. I ~E ~ da = q enc / 0 =2 rle q enc = L E = 2 0 r
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.
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: =2AE = 1 0 A E = 1 2 0 E is uniform! Does not decrease with distance from plane.
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).
The electric field must be perpendicular to the surface. Very near surface, E 1 0 Gauss s law requires that any free charge on a conductor resides on the surface.
Clicker Question Consider a conductor that has a net charge Q. Where in the conductor is there an excess charge density? 1. All in the interior of the conductor. 2. All on the surface of the conductor 3. Both in the interior and the surface
CT 27.19 Clicker Question 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)
Clicker Question 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 outer 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)
Clicker Question 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
Faraday Cage https://www.youtube.com/watch?v=z51wvandevw
Clicker Question: A HOLLOW copper sphere has no net charge. A charge, q, is in the hole, SHIFTED right a bit from the center. 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
Clicker Question: A HOLLOW copper sphere has no net charge. A charge, q, is in the hole, SHIFTED right a bit from the center. 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
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
Parallel Plate Capacitor What is the electric field between the plates? EA = A/ 0 E = / 0
2.18 Clicker Question: 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