Chapter 23 Gauss Law 23 1 What is

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Chapter 23. Gauss’ Law 23. 1. What is Physics? 23. 2. Flux 23. 3. Flux of an Electric Field 23. 4. Gauss' Law 23. 5. Gauss' Law and Coulomb's Law 23. 6. A Charged Isolated Conductor 23. 7. Applying Gauss' Law: Cylindrical Symmetry 23. 8. Applying Gauss' Law: Planar Symmetry 23. 9. Applying Gauss' Law: Spherical Symmetry

What is Physics? • Gaussian surface is a hypothetical (any imaginary shape) closed surface enclosing the charge distribution. • Gauss' law relates the electric fields at points on a (closed) Gaussian surface to the net charge enclosed by that surface.

Gaussian surface Let us divide the surface into small squares of area ΔA, each square being small enough to permit us to neglect any curvature and to consider the individual square to be flat. We represent each such element of area with an area vector • Magnitude is the area ΔA. • Direstion is perpendicular to the Gaussian surface and directed away from the interior of the surface.

Flux The rate of volume flow through the loop is :

Flux of an Electric Field • The electric field for a surface is • The electric field for a gaussian surface is The electric flux Φ through a Gaussian surface is proportional to the net number of electric field lines passing through that surface. SI Unit of Electric Flux: N·m 2/C

Problem 1 The drawing shows an edge-on view of two planar surfaces that intersect and are mutually perpendicular. Surface 1 has an area of 1. 7 m 2, while surface 2 has an area of 3. 2 m 2. The electric field E in the drawing is uniform and has a magnitude of 250 N/C. Find the electric flux through (a) surface 1 and (b) surface 2.

Sample Problem 2 Figure 23 -4 shows a Gaussian surface in the form of a cylinder of radius R immersed in a uniform electric field E , with the cylinder axis parallel to the field. What is the flux Φ of the electric field through this closed surface?

Sample Problem 3 A nonuniform electric field given by pierces the Gaussian cube shown in Fig. (E is in newtons per coulomb and x is in meters. ) What is the electric flux through the right face, the left face, the top face, and the Gaussian surface?

Gauss’ Law For a point charge:

Gauss’ Law For charge distribution Q: The electric flux through a Gaussian surface times by ε 0 ( the permittivity of free space) is equal to the net charge Q enclosed : • The net charge qenc is the algebraic sum of all the enclosed charges. • Charge outside the surface, no matter how large or how close it may be, is not included in the term qenc.

Check Your Understanding The drawing shows an arrangement of three charges. In parts (a) and (b) different Gaussian surfaces are shown. Through which surface, if either, does the greater electric flux pass?

Sample Problem Figure 23 -7 shows five charged lumps of plastic and an electrically neutral coin. The cross section of a Gaussian surface S is indicated. What is the net electric flux through the surface if q 1=q 4=3. 1 n. C, q 2=q 5=-5. 9 n. C, and q 3=-3. 1 n. C?

A Charged Isolated Conductor • If an excess charge is placed on an isolated conductor, that amount of charge will move entirely to the surface of the conductor. None of the excess charge will be found within the body of the conductor. • For an Isolated Conductor with a Cavity, There is no net charge on the cavity walls; all the excess charge remains on the outer surface of the conductor

The External Electric Field of a Conductor If σ is the charge per unit area, according to Gauss' law,

Sample Problem Figure 23 -11 a shows a cross section of a spherical metal shell of inner radius R. A point charge of q is located at a distance R/2 from the center of the shell. If the shell is electrically neutral, what are the (induced) charges on its inner and outer surfaces? Are those charges uniformly distributed? What is the field pattern inside and outside the shell?

Applying Gauss' Law: Cylindrical Symmetry Figure 23 -12 shows a section of an infinitely long cylindrical plastic rod with a uniform positive linear charge density λ.

Applying Gauss' Law: Planar Symmetry Figure 23 -15 shows a portion of a thin, infinite, nonconducting sheet with a uniform (positive) surface charge density σ

Applying Gauss' Law: Spherical Symmetry • A shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell's charge were concentrated at the center of the shell. • If a charged particle is located inside a shell of uniform charge, there is no electrostatic force on the particle from the shell.

Any spherically symmetric charge distribution with the volume charge density ρ • For r>R, the charge produces an electric field on the Gaussian surface as if the charge were a point charge located at the center, • For r<R, the electric field is

Checkpoint The figure shows two large, parallel, nonconducting sheets with identical (positive) uniform surface charge densities, and a sphere with a uniform (positive) volume charge density. Rank the four numbered points according to the magnitude of the net electric field there, greatest first.

Conceptual Questions (1) Two charges, +q and –q, are inside a Gaussian surface. Since the net charge inside the Gaussian surface is zero, Gauss’ law states that the electric flux through the surface is also zero; that is Φ=0. Does the fact that Φ =0 imply that the electric field E at any point on the Gaussian surface is also zero? Justify your answer. (2) The drawing shows three charges, labeled q 1, q 2, and q 3. A Gaussian surface is drawn around q 1 and q 2. (a) Which charges determine the electric flux through the Gaussian surface? (b) Which charges produce the electric field at the point P? Justify your answers.

(3) A charge +q is placed inside a spherical Gaussian surface. The charge is not located at the center of the sphere. (a) Can Gauss’ law tell us exactly where the charge is located inside the sphere? Justify your answer. (b) Can Gauss’ law tell us about the magnitude of the electric flux through the Gaussian surface? Why?