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# Chapter 22 - PowerPoint PPT Presentation

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

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### Chapter 22

Gauss’s Law

• 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

• Recognizing symmetry

• Setting up two-dimensional surface integrals

• To learn where the charge on a conductor is located

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

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

• Negative charge produces inward flux.

• More negative charge – more inward flux!

• Three cases of zero net charge inside a box

• No net electric flux through surface of box!

• Three cases of zero net charge inside a box

• No net electric flux through surface of box!

• Three cases of zero net charge inside a box

• No net electric flux through surface of box!

• Three cases of zero net charge inside a box

• No net electric flux through surface of box!

• Doubling charge within box doubles flux.

• Doubling charge within box doubles flux.

• Doubling size of box does NOT change flux.

• F= E·A = EA cos(q°)

• [F] = N/C ·m2 = Nm2/C

For a UNIFORM E field (in space)

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?

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

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?

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

F = ∫ E·dA

• = ∫ E·dA

E = kq/r2

= 1/(4pe0) q/r2 and is parallel to dAeverywhere on the surface

• = ∫ E·dA = E ∫dA = EA

• = ∫ E·dA

E = kq/r2 and is parallel to dAeverywhere on the surface

• = ∫ E·dA = E ∫dA = EA

For q = +3.0nC, flux through sphere of radius r=.20 m?

• = ∫ E·dA =qenc

e0

S

• = ∫ E·dA =qenc

e0

S

Electric Flux is produced by charge in space

• = ∫ E·dA =qenc

e0

S

You integrate over a CLOSED surface (two dimensions!)

• = ∫ E·dA =qenc

e0

S

E field is a VECTOR

• = ∫ E·dA =qenc

e0

S

Infinitesimal area element dA is also a vector; this is what you sum

• = ∫ E·dA =qenc

e0

S

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

S

Dot product is a scalar: E·dA =

ExdAx +EydAy + EzdAz = |E||dA|cosq

• = ∫ E·dA =qenc

e0

S

The TOTAL amount of charge…

• = ∫ E·dA =qenc

… but ONLY charge inside S counts!

e0

S

• = ∫ E·dA =qenc

e0

S

The electrical permittivity of free space, through which the field is acting.

• 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.

• Flux through sphere is independent of size of sphere

• Flux depends only on charge inside.

• F= ∫ E·dA = +q/e0

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

• Flux is positive if enclosed charge is positive, & negative if charge is negative.

• What is the flux through the surfaces A, B, C, and D?

• 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?

FB = -q/e0

FA = +q/e0

• 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?

FB = -q/e0

FA = +q/e0

FC = 0 !

FD = 0 !!

• Under electrostatic conditions, any excess charge on a conductor resides entirely on its surface.

• Under electrostatic conditions, E field inside a conductor is 0!

• 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 ???

• 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 conductoris 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!

• 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 cylindricalGaussian 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

dl

dq

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?

• dqgoes 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) /e0Gauss’ 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

• Example 22.7 for an infinite plane 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

• 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?

• 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

• E field between oppositely charged parallel conducting plates.

• Superposition of TWO E fields, from each plate…

• 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

• E field both inside and outside a sphere uniformly filled with charge.

• 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!

• E field insidea sphere uniformly filled with charge.

Find Qenclosed to start!

r = Qtotal/Volume (if uniform)

NB!!

Could be non-uniform!r(r) = kr2

• E field insidea sphere uniformly filled with charge.

Find Qenclosed in sphere of radius r <R?

Qenclosed = (4/3 pr3 ) x r

• E field insidea sphere uniformly filled with charge.

• Find Flux through surface?

• E field uniform, constant at any r

• E is parallel to dA at surfaceso…

• F= ∫ E·dA = E(4pr2)

• E field insidea 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)

• E field both inside and outside a sphere uniformly filled with charge.

• E = 0 inside conductor…

• Empty cavity inside conductor has no field, nor charge on inner surface

• Isolated charges inside cavity “induce” opposite charge, canceling field inside the conductor!

• Conceptual Example 22.11

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

• 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