Gauss’ Law

Main Points of Chapter 17 Electric flux Gauss’ law Using Gauss’ law to determine electric fields

17-1 What Does Gauss’ Law Do? Consider the electric field lines associated with a point charge Q How many field lines pass through the tissue forming a sphere？ All the lines (12). If the sphere is displaced ? All the lines (12).

Suppose the tissue is some shape other than spherical, but still surrounds the charge. All the field lines still go through. Now, imagine the paper is crinkled and overlaps itself; how shall we deal with the lines that pierce the tissue three times? Notice that they go out twice and in once – if we subtract the “ins” from the “outs” we are left with one line going out, which is consistent with the other situations.

Like how many hairs passing through your scalp. In the electric field, the number of electric field lines through a curved surface A is defined as electric flux. The number of field lines penetrating a surface is proportional to the surface area, the orientation, and the field strength.

1. Electric flux for uniform electric field En We can define the electric flux through an infinitesimal area:   if then is normal to the surface A finite plane uniform electric field

2. General definition of electric flux 3. Electric flux through a closed surface The direction of unclosed surface: choose one side Outward: positive Closed surface: Inward: negative

The electric flux through a closed surface is the sum of the electric field lines penetrating the surface. If the electric flux is positive, there are net electric field lines “out of” the closed surface. If the electric flux is negative , there are net electric field lines “into” the closed surface. The closed surface does not have to be made of real matter. The imaginary surface is a Gaussian surface

Act Imagine a cube of side a positioned in a region of constant electric field as shown below. Which of the following statements about the net electric flux Fe through the surface of this cube is true? (a) FE = 0 (b) FE µ 2a2 (c) FE µ 6a2 a a

Look at the lines going out through each circle -- each circle has the Act Consider 2 spheres (of radius R and 2R) drawn around a single charge as shown. Which of the following statements about the net electric flux through the 2 surfaces (F2R and FR) is true? (a) FR < F2R (b) FR = F2R (c) FR > F2R Look at the lines going out through each circle -- each circle has the same number of lines. R 2R

Example: Calculate the electric flux of a constant electric field through a hemispherical surface of radius R whose circular base is perpendicular to the direction of the field. Solution Consider an infinitesimal strip at latitude q R dl r  

A2 Alternative Solution Consider a closed surface that consists of the hemisphere and the planar circle A1 A2 R

Act A nonuniform electric field given by What is the electric flux through the surface of the Gaussian cube shown in figure below ?

Act a point charge q is placed at the center of a sphere of radius of R. Calculate the electric flux through a circular plane of radius r as shown below. R r q S S’ the electric flux of the closed sphere the electric flux of any closed surface t surrounding the q?

17-2 Gauss’ Law 1. for a point charge + q Suppose a point charge q is at the center of a sphere，the electric flux of a spherical surface q r But the result would be the same if the surface was not spherical, or if the charge was anywhere inside it!

Suppose a point charge q is out of an arbitrary closed surface For the field of point charge, the total electric flux through a closed surface is only related to the total (net) electric charge inside the surface.

e of an arbitrary closed surface 2 for multiple point charges qi q2 qj e of an arbitrary closed surface P q1 qn

3 for continuous charge distributions We can quickly generalize this to any surface and any charge distribution; Gauss’ law The total electric flux through a closed surface is equal to the total (net) electric charge inside the surface, divided by . Gauss’s law provides a relationship between the electric field on a closed surface and the charge distribution within that surface.

Discussion Electric field Lines leave (+) charges with source Electric field Lines return to (-) charges with sink without source or sink Electric field Lines never discontinue in empty space is the electric field at each point on the surface produced by the total electric charge，e is connected to the algebraic sum of charges enclosed by the surface.

ACT If a closed surface surrounds a dipole, the net flux through the surface is zero. TRUE ACT If the net flux through a closed surface is zero, then there can be no charge or charges within that surface. FALSE

Act A full Gaussian surface encloses two of the four positively charged particles. a) Which of the particles contribute to the electric field at point P on the surface. b) Determine the electric flux through the surface. c) If I change the position of q4 outside the surface, Does Ep change? How about the electric flux ? + q4 + q3 a) The four particles + q1 P + q2 c) Ep changes doesn’t change

Act Consider a point charge q placed in the center of a cube. Determine the electric flux through each face of the cube . q S q If the charge is placed at a corner of the cube, what is the electric flux through surface S Put the charge in the middle of a larger cube

Example There is an electric field near Earth’s surface of about 100N/C that points vertically down. Assume this field is constant around Earth and that it is due to charge evenly spread on Earth’s surface. What is the total charge on the Earth? Solution choose a concentric sphere of radius R for Gaussian surface the total charge

4. Coulomb’s Law and Gauss’ Law Coulomb’s Law + Superposition of electric field Gauss’ Law Now we use Gauss’ Law to get the distribution of electric field for a point charge. Draw a concentric spherical Gaussian surface of radius r q Although E varies radially with distance from q, it has the same value everywhere on the spherical surface

Gauss’ law and Coulomb’s law , although expressed in different forms, are equivalent ways of describing the relation between electric charge and electric field in static situation. Gauss’ law is more general in that it is always valid. It is one of the fundamental laws of electromagnetism.

17-3 Using Gauss’ Law to determine Electric Fields Gauss’s law is valid for any distribution of charges and for any closed surface. However, we can only calculate the field for several highly symmetric distributions of charge.

The steps of calculating the magnitude of the electric field using Gauss’ law Identify the symmetry of the charge distribution and the electric field it produces. *(2) Choose a Gaussian surface that is matched to the symmetry – that is, the electric field is either parallel to the surface or constant and perpendicular to it. (3) Calculate the algebraic sum of the charge enclosed by the Gaussian surface. (4) Find the electric field using Gauss’s law.

Example: Determine the electric field of an infinite uniformly charged line. The charge density per unit length:+ Solution

Solution we use as our Gaussian surface a coaxial cylinder surface with height l r l E r

Could we use Gauss’ law to find the field of a finite line of charge? What is the electric field of an infinite uniformly charged thin cylindrical shell? What is the electric field of an infinite uniformly charged cylinder?

Act Two long thin cylindrical shells of radius r1 and r2 are oriented coaxially. The charge density are . What is the electric field ? Solution

outside the sphere ( r > R ) Example: Determine the electric field both inside and outside a thin uniformly charged spherical shell of radius R that has a total charge Q. + Solution + + P take a concentric sphere of radius r for Gaussian surface r r R O + + + r E O From Gauss’ law outside the sphere ( r > R ) R inside the sphere( r < R ) on surface ( r = R ) E= ？

ACT A uniform spherical shell of charge of radius R surrounds a point charge at its center. The point charge has value Q and the shell has total charge -Q. The electric field at a distance R/2 from the center ________ A) does not depend on the charge of the spherical shell. B) is directed inward if Q>0. C) is zero. D) is half of what it would be if only the point charge were present. E) cannot be determined

outside the sphere ( r > R ) Act: A nonconducting spherical shell of radius R is uniformly charged with total charge of Q. We put point charges q1and q2 inside and outside spherical shell respectively . Determine the electric forces on the point charges. q1 q2 inside the sphere( r < R ) O r1 r2 outside the sphere ( r > R ) right Two shells law

Example: Find the electric field outside and inside a solid, nonconducting sphere of radius R that contains a total charge Q uniformly distributed throughout its volume. Solution take a concentric sphere of radius r for Gaussian surface r + + + r + + + + + R + outside the sphere ( r > R ) E inside the sphere ( r < R ) R O r

ACT Figure below shows four spheres, each with charge Q uniformly distributed through its volume. (a) Rank the spheres according to their volume charge density, greatest first. The figure also shows a point P for each sphere, all at the same distance from the center of the sphere. (b) Rank the spheres according to the magnitude of the electric field they produce at point P, greatest first. a, b, c, d a and b tie, c, d

Consider a point P in the cavity Example: A uniformly charged nonconducting sphere has volume charge density r. Material is removed from the sphere leaving a spherical cavity. Show that the electric field through the cavity is uniform and is given by where a is the distance of the centers of the two spheres Proof: Replace the sphere-with-cavity with two uniform spheres of equal positive and negative charge densities. Consider a point P in the cavity The e-Field produced by +r sphere at p P +r -r The e-Field produced by -r sphere at p

ACT A thin nonconducting uniformly charged spherical shell of radius R has a total charge of Q . A small circular plug is removed from the surface. (a) What is the magnitude and direction of the electric field at the center of the hole. (b) The plug is put back in the hole. Using the result of part (a), calculate the force acting on the plug. Solution (a) Replace the spherical shell-with-hole with a perfect spherical shell and a circle with equal negative surface charge densities. R Q

dq (b) R Q The “electrostatic pressure”( force per unit area) tending to expand the sphere

Example: Find the electric field outside an infinite, nonconducting plane of charge with uniform charge density s Solution To take advantage of the symmetry properties, we use as our Gaussian surface a cylinder with its axis perpendicular to the sheet of charge, with ends of area S. The charged sheet passes through the middle of the cylinder’s length, so the cylinder ends are equidistant from the sheet. x O Ex

ACT Nowhere inside (between the planes) an odd number of parallel planes of the same nonzero charge density is the electric field zero. TRUE

Act Consider two infinite parallel charged plates with surface charge density of s1and s2respectively. What is the electric field in the three regions. III I II x

Example: Find the electric field of a slab of nonconducting material forming an infinite plane .It has thickness d and carries a uniform positive charge density r  d Solution The Gaussian surface is shown in the figure Outside the slab S x x O Ex Inside the slab S d

*Find the distribution of charge according to the electric field （Differential form of Gauss’ Law ） Gauss’ law We now apply Gauss' divergence theorem If a closed surface S encloses a volume V then the surface integral of any vector over S is equal to the volume integral of the divergence of over V

As this is true for any closed surface the result requires that at every point in space This is Gauss' law for electric fields in differential form. It is the first of Maxwell's equations.

The formulas for the divergence of a vector Cartesian coordinates Spherical coordinates Cylindrical coordinates

Example The electric field of an atom is Find the distribution of charge of this atom Solution Atom is neutral

Alternative Solution The charge density should be r (r). We take a concentric spheres for Gaussian surface r r+dr

Summary of Chapter 17 Electric flux due to field intersecting a surface S: Gauss’ law relates flux through a closed surface to charge enclosed: Can use Gauss’ law to find electric field in situations with a high degree of symmetry