Copyright © 2012 Pearson Education Inc. PowerPoint ® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Lectures by Wayne Anderson Chapter 22 Gauss’s Law

Goals for Chapter 22 To use the electric field at a surface to determine the charge within the surface To learn the meaning of electric flux and how to calculate it To learn the relationship between the electric flux through a surface and the charge within the surface

Goals for Chapter 22 To use Gauss’s law to calculate electric fields Recognizing symmetry Setting up two-dimensional surface integrals To learn where the charge on a conductor is located

Charge and electric flux Positive charge within the box produces outward electric flux through the surface of the box.

Charge and electric flux Positive charge within the box produces outward electric flux through the surface of the box. More charge = more flux!

Charge and electric flux Negative charge produces inward flux.

Charge and electric flux More negative charge – more inward flux!

Zero net charge inside a box Three cases of zero net charge inside a box No net electric flux through surface of box!

Zero net charge inside a box Three cases of zero net charge inside a box No net electric flux through surface of box!

Zero net charge inside a box Three cases of zero net charge inside a box No net electric flux through surface of box!

Zero net charge inside a box Three cases of zero net charge inside a box No net electric flux through surface of box!

What affects the flux through a box? Doubling charge within box doubles flux.

What affects the flux through a box? Doubling charge within box doubles flux. Doubling size of box does NOT change flux.

Uniform E fields and Units of Electric Flux  = E·A = EA cos(  °)   C · m 2 = Nm 2 /C For a UNIFORM E field (in space)

Calculating electric flux in uniform fields

Example Electric flux through a disk Disk of radius 0.10 m with n at 30 degrees to E, with a magnitude of 2.0 x 10 3 N/C. What is the flux?

Example Electric flux through a disk Disk of radius 0.10 m with n at 30 degrees to E, with a magnitude of 2.0 x 10 3 N/C. What is the flux?  = E·A = EA cos(30°) A =  r 2 = m 2  =54 Nm 2 /C

Example Electric flux through a disk Disk of radius 0.10 m with n at 30 degrees to E, with a magnitude of 2.0 x 10 3 N/C. What is the flux? What if n is perpendicular to E? Parallel to E?

Electric flux through a cube An imaginary cube of side L is in a region of uniform E. Find the flux through each side…

Electric flux through a sphere  = ∫ E·dA

Electric flux through a sphere  = ∫ E·dA E = kq/r 2 = 1/(4  0 ) q/r 2 and is parallel to dA everywhere on the surface  = ∫ E·dA = E ∫dA = EA

Electric flux through a sphere  = ∫ E·dA E = kq/r 2 and is parallel to dA everywhere on the surface  = ∫ E·dA = E ∫dA = EA For q = +3.0nC, flux through sphere of radius r=.20 m?

Gauss’ Law  = ∫ E·dA =q enc 00 S

Gauss’ Law  = ∫ E·dA =q enc 00 Electric Flux is produced by charge in space S

Gauss’ Law  = ∫ E·dA =q enc 00 You integrate over a CLOSED surface (two dimensions!) S

Gauss’ Law  = ∫ E·dA =q enc 00 E field is a VECTOR S

Gauss’ Law  = ∫ E·dA =q enc 00 Infinitesimal area element dA is also a vector; this is what you sum S

Gauss’ Law  = ∫ E·dA =q enc 00 Dot product tells you to find the part of E that is PARALLEL to dA at that point (perpendicular to the surface) S

Gauss’ Law  = ∫ E·dA =q enc 00 Dot product is a scalar: E·dA = E x dA x + E y dA y + E z dA z = |E||dA|cos  S

Gauss’ Law  = ∫ E·dA =q enc 00 The TOTAL amount of charge… S

Gauss’ Law  = ∫ E·dA =q enc 00 … but ONLY charge inside S counts! S

Gauss’ Law  = ∫ E·dA =q enc 00 The electrical permittivity of free space, through which the field is acting. S

Why is Gauss’ Law Useful? Given info about a distribution of electric charge, find the flux through a surface enclosing that charge. Given info about the flux through a closed surface, find the total charge enclosed by that surface. For highly symmetric distributions, find the E field itself rather than just the flux.

Gauss’ Law for Spherical Surface… Flux through sphere is independent of size of sphere Flux depends only on charge inside.  = ∫ E·dA = +q/  0

Point charge inside a nonspherical surface As before, flux is independent of surface & depends only on charge inside.

Positive and negative flux Flux is positive if enclosed charge is positive, & negative if charge is negative.

© 2012 Pearson Education, Inc. A spherical Gaussian surface (#1) encloses and is centered on a point charge +q. A second spherical Gaussian surface (#2) of the same size also encloses the charge but is not centered on it. Compared to the electric flux through surface #1, the flux through surface #2 is A.greater. B.B. the same. C. less, but not zero. D. zero. E. not enough information given to decide Q22.1 +q+q Gaussian surface #1 Gaussian surface #2

© 2012 Pearson Education, Inc. A spherical Gaussian surface (#1) encloses and is centered on a point charge +q. A second spherical Gaussian surface (#2) of the same size also encloses the charge but is not centered on it. Compared to the electric flux through surface #1, the flux through surface #2 is A.greater. B.B. the same. C. less, but not zero. D. zero. E. not enough information given to decide A22.1 +q+q Gaussian surface #1 Gaussian surface #2

Conceptual Example 22.4 What is the flux through the surfaces A, B, C, and D?

Conceptual Example 22.4 What is the flux through the surfaces A, B, C, and D?  A = +q/  0

Conceptual Example 22.4 What is the flux through the surfaces A, B, C, and D?  A = +q/  0  B = -q/  0

Conceptual Example 22.4 What is the flux through the surfaces A, B, C, and D?  A = +q/  0  B = -q/  0  C = 0 !

Conceptual Example 22.4 What is the flux through the surfaces A, B, C, and D?  A = +q/  0  B = -q/  0  C = 0 !  D = 0 !!

© 2012 Pearson Education, Inc. Two point charges, +q (in red) and –q (in blue), are arranged as shown. Through which closed surface(s) is the net electric flux equal to zero? Q22.2 A.surface A B.B. surface B C. surface C D. surface D E. both surface C and surface D

© 2012 Pearson Education, Inc. Two point charges, +q (in red) and –q (in blue), are arranged as shown. Through which closed surface(s) is the net electric flux equal to zero? A.surface A B.B. surface B C. surface C D. surface D E. both surface C and surface D A22.2

Applications of Gauss’s law Under electrostatic conditions, any excess charge on a conductor resides entirely on its surface. Under electrostatic conditions, E field inside a conductor is 0!

Applications of Gauss’s law Under electrostatic conditions, field outside spherical conductor looks just like a point charge!

© 2012 Pearson Education, Inc. A conducting spherical shell with inner radius a and outer radius b has a positive point charge Q located at its center. The total charge on the shell is – 3Q, and it is insulated from its surroundings. In the region a < r < b, A. the electric field points radially outward. B. the electric field points radially inward. C. the electric field is zero. D. not enough information given to decide Q22.3

© 2012 Pearson Education, Inc. A conducting spherical shell with inner radius a and outer radius b has a positive point charge Q located at its center. The total charge on the shell is – 3Q, and it is insulated from its surroundings. In the region a < r < b, A. the electric field points radially outward. B. the electric field points radially inward. C. the electric field is zero. D. not enough information given to decide A22.3

Example 22.6: Field of a line charge E around an infinite positive wire of charge density ? ·E = ? Charge/meter =

Example 22.6: Field of a line charge E around an infinite positive wire of charge density ? Imagine closed cylindrical Gaussian surface around the wire a distance r away…

Example 22.6: Field of a line charge E around an infinite positive wire of charge density ? Look at E·dA on the cylinder, and at the ends.

Example 22.6: Field of a line charge E around an infinite positive wire of charge density ? E is orthogonal to dA at the end caps. E is parallel (radially outwards) to dA on cylinder

Example 22.6: Field of a line charge E around an infinite positive wire of charge density ? E is constant in value everywhere on the cylinder at a distance r from the wire!

Example 22.6: Field of a line charge E around an infinite positive wire of charge density ? E is constant in value everywhere on the cylinder at a distance r from the wire!  = ∫ E·dA

Example 22.6: Field of a line charge E around an infinite positive wire of charge density ? The integration over area is two-dimensional dA = (rd  dl dl rd  r dd dA

Example 22.6: Field of a line charge E around an infinite positive wire of charge density ? E is constant in value everywhere on the cylinder at a distance r from the wire!  = ∫ E·dA = ∫ ∫ (E) · (rd  )dl

Example 22.6: Field of a line charge E around an infinite positive wire of charge density ? E is constant in value everywhere on the cylinder at a distance r from the wire!  = ∫ E·dA = ∫ ∫ (E) · (rd  )dl = E ∫ ∫ rd  dl

Example 22.6: Field of a line charge E around an infinite positive wire of charge density ? Limits of integration? d  goes from 0 to 2  dl goes from 0 to l (length of cylinder)  = ∫ E·dA = E ∫ ∫ rd  dl

Example 22.6: Field of a line charge E around an infinite positive wire of charge density ? Surface Area of cylinder (but not end caps, since net flux there = 0

Example 22.6: Field of a line charge E around an infinite positive wire of charge density ? E is constant in value everywhere on the cylinder at a distance r from the wire!  = ∫ E·dA = (E) x (Surface Area!) = E(2  r)l

Example 22.6: Field of a line charge E around an infinite positive wire of charge density ? How much charge enclosed in the closed surface?

Example 22.6: Field of a line charge E around an infinite positive wire of charge density ? How much charge enclosed in the closed surface? Q(enc) = (charge density) x (length) =  l

Example 22.6: Field of a line charge E around an infinite positive wire of charge density ? So… q/  0 = (  l) /  0 Gauss’ Law gives us the flux = E(2  r) l = q/  0 = (  l) /  0

Example 22.6: Field of a line charge E around an infinite positive wire of charge density ? And… E = (  l) /  0 = ( / 2  r  0 )r (2  r) l Don’t forget E is a vector!

Field of a sheet of charge Example 22.7 for an infinite plane sheet of charge?

Field between two parallel conducting plates Example 22.8 for field between oppositely charged parallel conducting plates.

A uniformly charged insulating sphere Example 22.9 for field both inside and outside a sphere uniformly filled with charge.

Charges on conductors with cavities E = 0 inside conductor…

Charges on conductors with cavities Empty cavity inside conductor has no field, nor charge on inner surface

Charges on conductors with cavities Isolated charges inside cavity “induce” opposite charge, canceling field inside the conductor!

A conductor with a cavity Conceptual Example 22.11

© 2012 Pearson Education, Inc. A solid spherical conductor has a spherical cavity in its interior. The cavity is not centered on the center of the conductor. If a positive charge is placed on the conductor, the electric field in the cavity Q22.4 Conductor Cavity A. points generally toward the outer surface of the conductor. B. points generally away from the outer surface of the conductor. C. is zero. D. not enough information given to decide

© 2012 Pearson Education, Inc. A solid spherical conductor has a spherical cavity in its interior. The cavity is not centered on the center of the conductor. If a positive charge is placed on the conductor, the electric field in the cavity A. points generally toward the outer surface of the conductor. B. points generally away from the outer surface of the conductor. C. is zero. D. not enough information given to decide A22.4 Conductor Cavity

© 2012 Pearson Education, Inc. There is a negative surface charge density in a certain region on the surface of a solid conductor. Just beneath the surface of this region, the electric field A. points outward, toward the surface of the conductor. B. points inward, away from the surface of the conductor. C. points parallel to the surface. D. is zero. E. not enough information given to decide Q22.5

© 2012 Pearson Education, Inc. There is a negative surface charge density in a certain region on the surface of a solid conductor. Just beneath the surface of this region, the electric field A. points outward, toward the surface of the conductor. B. points inward, away from the surface of the conductor. C. points parallel to the surface. D. is zero. E. not enough information given to decide A22.5

© 2012 Pearson Education, Inc. For which of the following charge distributions would Gauss’s law not be useful for calculating the electric field? A. a uniformly charged sphere of radius R B. a spherical shell of radius R with charge uniformly distributed over its surface C. a right circular cylinder of radius R and height h with charge uniformly distributed over its surface D. an infinitely long circular cylinder of radius R with charge uniformly distributed over its surface E. Gauss’s law would be useful for finding the electric field in all of these cases. Q22.6

© 2012 Pearson Education, Inc. For which of the following charge distributions would Gauss’s law not be useful for calculating the electric field? A. a uniformly charged sphere of radius R B. a spherical shell of radius R with charge uniformly distributed over its surface C. a right circular cylinder of radius R and height h with charge uniformly distributed over its surface D. an infinitely long circular cylinder of radius R with charge uniformly distributed over its surface E. Gauss’s law would be useful for finding the electric field in all of these cases. A22.6

Testing Gauss’s law experimentally Faraday’s icepail experiment confirmed validity of Gauss’s law.

Testing Gauss’s law experimentally Faraday’s icepail experiment confirmed validity of Gauss’s law.

Testing Gauss’s law experimentally Faraday’s icepail experiment confirmed validity of Gauss’s law.

The Van de Graaff generator The Van de Graaff generator, shown in Figure below, operates on the same principle as in Faraday’s icepail experiment.

Electrostatic shielding A conducting box (a Faraday cage) in an electric field shields the interior from the field.

Electrostatic shielding A conducting box (a Faraday cage) in an electric field shields the interior from the field. See