AP Physics C Mrs. Coyle. Electric Flux, The number of electric (flux) field lines which pass...

AP Physics CMrs. Coyle

Electric Flux, The number of electric (flux) field lines which pass through a given cross-sectional area A.

Which has the largest electric flux? Does flux depend on area?

Answer: A

Which has the largest electric flux? Does flux depend on direction?

Answer:1

The area vector is perpendicular to the surface A and has a magnitude equal to the area A.

A��������������

Electric Flux

Flux is a dot product and therefore a scalar

Units: ]=

[E]=

[A]= area m2 angle formed between E and the normal to the surface.

2Nm

CN

C

E dA cosEdA

When E is perpendicular to the loop?

= EA

Why?

What if you have a closed surface?

For a closed surface:

Gauss’s Law: the total electric flux through a closed surface is proportional to the enclosed charge.

Permittivity of free space:ε0 = 8.8542 x 10-12 C2 / (N m2)

0

in

surface

qE dA

Remember:Coulomb’s Law constant, k

9 2 2

0

19 10 /

4k x Nm C

Note: the dot product

0

cos in

surface

qEdA

Notes:The “Gaussian” surface can have any shape. You

choose the surface.The flux is positive if more electric field lines go

out of the surface and negative if more go in.The flux from a particular charge qin is the same

regardless of the shape or size of the surface.If the surface contains no charge, the flux

through it is zero. That means that every field line that enters the surface will also exit the surface.

More Notes on Gauss’s LawThe normal area vector is taken to point from

the inside of the closed surface out.

Gauss’s Law is also known as Maxwell’s first equation.

Gauss’s Law is second way used to find the electric field. What is the first way you learned?

Problem 1 The Electric Field due to a point charge. Prove E=kq/r2

What do you think would be your answer for E, if you chose a different Gaussian surface?

Problem 2: Show E= σ /(2 ε0) for a rectangular metal slab positively charged with a uniform surface charge density σ.

Problem 3: Field Due to a Spherically Symmetric Charge Distribution of an Insulatora)Outside the sphere( r>a)

in

2

2 2

4

4

Eo

eo

qd EdA

ε

EA E πr

Q QE k

πε r r

E A

Problem 3: Field Due to a Spherically Symmetric Charge Distribution of an Insulatorb)Inside the sphere( r<a)

qin < Q and since the volume charge density is uniform:

in

in2 34

Eo

eo

qd EdA

ε

q QE k r

πε r a

E A

3

3 3 34 4in

in

qQ rq Q

a r a

Problem 3: Field Due to a Spherically Symmetric Charge Distribution (Insulator)

2

Outside the sphere

e

QE k

r

3

Inside the sphere

e

QE k r

a

Problem 4: Field Due to a Thin Spherical Shell

For r > a, the enclosed charge is Q and E = keQ / r2

For r < a, the charge inside the surface is 0 and E = 0

Problem 5: Field due to a Line of Charge

Problem 5: Field Due to a Line of Charge

The flux through the ends of the gaussian cylinder is 0. Why?

End view

Problem 5: Field Due to a Line of Charge

in

2

22

Eo

o

eo

qd EdA

ε

λE πr

ε

λ λE k

πε r r

E A