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

To use Gauss’s law to calculate electric fields 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 For a UNIFORM E field (in space) F= E·A = EA cos(q°) [F] = N/C · m2 = Nm2/C

Calculating electric flux in uniform fields

Calculating electric flux in uniform fields

Calculating electric flux in uniform fields

Example 22.1 - 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 103 N/C. What is the flux?

Example 22.1 - 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 103 N/C. What is the flux? F = E·A = EA cos(30°) A = pr2 = 0.0314 m2 F =54 Nm2/C

Example 22.1 - 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 103 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 F = ∫ E·dA

Electric flux through a sphere = ∫ E·dA E = kq/r2 = 1/(4pe0) q/r2 and is parallel to dA everywhere on the surface = ∫ E·dA = E ∫dA = EA

Electric flux through a sphere = ∫ E·dA E = kq/r2 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 =qenc e0 S

Electric Flux is produced by charge in space Gauss’ Law = ∫ E·dA =qenc e0 S Electric Flux is produced by charge in space

You integrate over a CLOSED surface (two dimensions!) Gauss’ Law = ∫ E·dA =qenc e0 S You integrate over a CLOSED surface (two dimensions!)

Gauss’ Law = ∫ E·dA =qenc e0 S E field is a VECTOR

Infinitesimal area element dA is also a vector; this is what you sum Gauss’ Law = ∫ E·dA =qenc e0 S Infinitesimal area element dA is also a vector; this is what you sum

= ∫ E·dA =qenc e0 Gauss’ Law Dot product tells you to find the part of E that is PARALLEL to dA at that point (perpendicular to the surface)

= ∫ E·dA =qenc e0 Gauss’ Law Dot product is a scalar: E·dA = ExdAx + EydAy + EzdAz = |E||dA|cosq

The TOTAL amount of charge… Gauss’ Law = ∫ E·dA =qenc e0 S The TOTAL amount of charge…

… but ONLY charge inside S counts! Gauss’ Law … but ONLY charge inside S counts! = ∫ E·dA =qenc e0 S

= ∫ E·dA =qenc e0 Gauss’ Law The electrical permittivity of free space, through which the field is acting.

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. F= ∫ E·dA = +q/e0

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.

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

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? FA = +q/e0

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? FB = -q/e0 FA = +q/e0

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? FB = -q/e0 FA = +q/e0 FC = 0 !

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? FB = -q/e0 FA = +q/e0 FC = 0 ! FD = 0 !!

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, any excess charge on a conductor resides entirely on its surface. Under electrostatic conditions, E field inside a conductor is 0! WHY ???

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! Assume the opposite! IF E field inside a conductor is not zero, then … E field WILL act on free charges in conductor Those charges WILL move in response to the force of the field They STOP moving when net force = 0 Which means IF static, THEN no field inside conductor!

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

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

Example 22.6: Field of a line charge E around an infinite positive wire of charge density l? Charge/meter = l ·E = ? You know charge, you WANT E field (Gauss’ Law!) Choose Gaussian Surface with symmetry to match charge distribution to make calculating ∫ E·dA easy!

Example 22.6: Field of a line charge E around an infinite positive wire of charge density l? 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 l? Look at ∫ E·dA; remember CLOSED surface! Three components: the cylindrical side, and the two ends. Need flux across each!

Example 22.6: Field of a line charge E around an infinite positive wire of charge density l? 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 l? 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 l? E is parallel to dA everywhere on the cylinder, so E ◦ dA = EdA = ∫ E·dA= ∫ EdA

Example 22.6: Field of a line charge E around an infinite positive wire of charge density l? dA rdq dq dl r The integration over area is two-dimensional |dA| = (rdq) dl

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

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

Example 22.6: Field of a line charge E around an infinite positive wire of charge density l? Limits of integration? dq goes from 0 to 2p dl goes from 0 to l (length of cylinder) = ∫ E·dA = E ∫ ∫ rdqdl

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

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

Example 22.6: Field of a line charge E around an infinite positive wire of charge density l? So… q/e0 = (l l) /e0 Gauss’ Law gives us the flux F = E(2pr) l = q/e0 = (l l) /e0

Example 22.6: Field of a line charge E around an infinite positive wire of charge density l? Don’t forget E is a vector! And… E = (l l) /e0 r = (l / 2pr e0) r (2pr) l Unit vector radially out from line

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

Field of a sheet of charge Example 22.7: What is E field a distance x away for an infinite plane sheet of charge with Q coulombs/sq. meter ? Find Q enclosed to start! Qenc = sA Where s is the charge/area of the sheet

Field of a sheet of charge Example 22.7: What is E field a distance x away for an infinite plane sheet of charge with Q coulombs/sq. meter ? Flux through entire closed surface of the cylinder? = ∫ E·dA = E1A (left) + E2A (right) (only) Why no flux through the cylinder?

Field of a sheet of charge Example 22.7: What is E field a distance x away for an infinite plane sheet of charge with Q coulombs/sq. meter ? Flux through entire closed surface of the cylinder? = ∫ E·dA = Qenc / e0 = 2EA = Qenc / e0 = sA / e0 So E = s / 2e0 for infinite sheet

Ex. 22.8 Field between two parallel conducting plates E field between oppositely charged parallel conducting plates.

Field between two parallel conducting plates Superposition of TWO E fields, from each plate…

Field between two parallel conducting plates Superposition of 2 fields from infinite sheets with same charge! For EACH sheet, E = s / 2e0 And s is the same in magnitude, directions of E field from both sheets is the same, so Enet = 2(s / 2e0) = s / e0

Ex. 22.9 - A uniformly charged insulating sphere E field both inside and outside a sphere uniformly filled with charge.

Ex. 22.9 - A uniformly charged insulating sphere E field both inside and outside a sphere uniformly filled with charge. E field outside a sphere uniformly filled with charge? EASY Looks like a point=charge field!

A uniformly charged insulating sphere E field inside a sphere uniformly filled with charge. Find Qenclosed to start! Start with charge density function r r = Qtotal/Volume (if uniform) NB!! Could be non-uniform! r(r) = kr2

A uniformly charged insulating sphere E field inside a sphere uniformly filled with charge. Find Qenclosed in sphere of radius r <R? Qenclosed = (4/3 pr3 ) x r

A uniformly charged insulating sphere E field inside a sphere uniformly filled with charge. Find Flux through surface? E field uniform, constant at any r E is parallel to dA at surface so… F= ∫ E·dA = E(4pr2)

A uniformly charged insulating sphere E field inside a sphere uniformly filled with charge. F= ∫ E·dA = E(4pr2) = (4/3 pr3 ) x r And E = rr/3e0 (for r < R) or E = Qr/4pR3e0 (for r < R)

Ex. 22.9 - A uniformly charged insulating sphere E 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

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 http://www.youtube.com/watch?v=FvtfE-ha8dE