Chapter 21: Gauss s Law

Chapter 21: Gauss s Law

Electric field lines Electric field lines provide a convenient and insightful way to represent electric fields. A field line is a curve whose direction at each point is the direction of the electric field at that point. The spacing of field lines describes the magnitude of the field. Where lines are closer, the field is stronger. Vector and field-line diagrams of a point-charge field Tracing the field of an electric dipole

Field lines for simple charge distributions There are field lines everywhere, so every charge distribution has infinitely many field lines. In drawing field-line diagrams, we associate a certain finite number of field lines with a charge of a given magnitude. In the diagrams shown, 8 lines are associated with a charge of magnitude q. Note that field lines of static charge distributions always begin and end on charges, or extend to infinity.

Counting field lines How many field lines emerge from closed surfaces surrounding charge? Count each field line crossing going outward as +1, each inward crossing as 1. You ll find that the number of field lines crossing any closed surface is proportional to the net charge enclosed.

Electric flux Electric flux quantifies the notion number of field lines crossing a surface. The electric flux through a flat surface in a uniform electric field depends on the field strength E, the surface area A, and the angle between the field and the normal to the surface. Mathematically, the flux is given by Here is a vector whose magnitude is the surface area A and whose orientation is normal to the surface. Here, E is assumed to be uniform!

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, so small that each patch is essentially flat and the field is essentially uniform over each. We then sum the fluxes over each patch. In the limit of infinitely many infinitesimally small patches, the sum becomes a surface integral:

Gauss s law The electric flux through any closed surface is proportional to the charge enclosed. The proportionality constant is 1/ε 0. Therefore O where the circle designates any closed surface, and the integral is taken over the surface. q enclosed is the charge enclosed by that surface. This statement is Gauss s law. It s true for any surface and charge anywhere in the universe. 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

Electric field of a 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 d A =4πr 2 E = 1 0 Q E = 1 4π 0 Q r 2

Clicker Question What is the net electric flux through the closed cylindrical surface shown? A: 0 B: positive C: negative

CT 27.7 Three closed spherical surfaces enclose a + point charge. Which surface has the largest 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 sphere? A: 0 everywhere inside B: non-zero everywhere in the sphere C: Not enough info given + University of Colorado, Boulder (2008)

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 flux doesn t.

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)

da A = r 2 AdΩ da B = r 2 BdΩ Q A Q B = da A da B = r2 A r 2 B de A Q A /r 2 A de A de B =1

CT 27.16 The total charge on the thin spherical shell is Q. (Uniformly spread out) Using Gauss's Law, find the E-field outside the sphere. + + + + + + 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)

Example: The field of a uniformly charged sphere INTERPRET: The situation has spherical symmetry. DEVELOP: Appropriate gaussian surfaces are spheres. EVALUATE: The flux becomes Outside the sphere, the enclosed charge is the total charge Q. Then O Thus the field outside a spherical charge distribution is identical to that of a point charge. Inside the sphere, the enclosed charge is proportional to the volume enclosed: q enclosed = (r 3 /R 3 )Q. Then 4πr 2 E = (r 3 /R 3 )Q, so

Line symmetry In line symmetry, the charge density depends only on the perpendicular distance from a line, the axis of symmetry. This requires a charge distribution that is infinitely long. However, line symmetry is a good approximation for finite cylindrical charge distributions with length much greater than diameter, at points close to the charge. Applying Gauss s law in line symmetry requires the use of a cylindrical gaussian surface. The flux through this gaussian surface is For a line charge, Thus E = Φ =2πrLE Q = λl λ 2π 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. Applying Gauss s law in plane symmetry requires the use of a gaussian surface that straddles the plane. Φ =2AE = 1 0 σa E = 1 2 0 σ

Gauss s law and conductors Conduction 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, a condition in which there is no net charge motion, 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 (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.

CT 27.19 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: Other Inner surface University of Colorado, Boulder (2008)

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

Summary Gauss s law is one of the four fundamental laws of electromagnetism. In terms of the field line representation of electric fields, Gauss s law expresses the fact that the number of field lines emerging from any closed surface is proportional to the net charge enclosed. Mathematically, Gauss s law states that the electric flux O through any closed surface is proportional to the net charge enclosed: O Gauss s law embodies the inverse-square dependence of the point-charge field, and is equivalent to Coulomb s law. Gauss s law is always true. It can be used to calculate electric fields in situations with sufficient symmetry: spherical symmetry, line symmetry, or plane symmetry. Gauss s law requires that any net charge on a charged conductor reside on the conductor surface, and that the electric field at the conductor surface be perpendicular to the surface.