03 electric flux

8/16/2019 03 Electric Flux

1/31

R

2 R

a

a


2/31

Today…

• More on Electric Field: – Continuous Charge Distributions

• Electric Flux:

– Definition

– How to think about flux

1


3/31

Reminder: Lecture 2, ACT 3

• Consider a circular ring with total charge +Q.The charge is spread uniformly around the

ring, as shown, so there is λ = Q/2πR charge

per unit length.

• The electric field at the origin is

R

x

y

++++

+ +

+++

++

+ + + + ++++++

+

(a) zero R

πλ

πε

2

4

1

0

(b)2

04

1

R

Rλ π

πε (c)

But how would we calculate this??


4/31


5/31

AB

2) A finite line of positive charge is arranged as shown. What isthe direction of the Electric field at point A?

a) up b) down c) left d) right

e) up and left f) up and right

3) What is the direction of the Electric field at point B?

a) up b) down c) left d) righte) up and left f) up and right

L

Preflight 3:


6/31

Charge Densities

• How do we represent the charge “Q” on an extended object?

total charge

Q

small piecesof charge

dq

• Line of charge:λ = charge per

unit length

dq = λ dx

• Surface of charge:

σ = charge per

unit area

dq = σ dA

• Volume of Charge:

ρ = charge per

unit volume

dq = ρ dV


7/31

How We Calculate (Uniform) Charge

Densities:Take total charge, divide by “size”

Examples:10 coulombs distributed over a 2-meter rod.

10Cλ 5 C/m

2m= =

14 pC (pico = 10-12) distributed over a shell of radius 1 μm.

122

-6 2

14 10 C 14σ C/m

4π(10 m) 4π

−⋅= =

14 pC distributed over a sphere of radius 1 mm.

123 3

-3 343

14 10 C (3) 14ρ 10 C/m

π(10 m) 4π

−−⋅= = ⋅


8/31

Electric field from an infinite line charge

Approach:

“Add up the electric field contribution from each bit of

charge, using superposition of the results to get the final field.”

In practice:

• Use Coulomb’s Law to find the E-field per segment of charge

• Plan to integrate along the line…

– x: from −∞ to ∞ OR θ : from −π/2 to π/2

++++++++++++++++++++++++++

r E(r) = ?

Any symmetries ? This may help for easy cancellations

+++++++++++++++++++++++++++++

θ

x


9/31

Infinite Line of Charge

We need to add up the E-field

contributions from all

segments dx along the line.

Charge density = λ

++++++++++++++++ x

y

dx

r' r θ

dE


10/31


• To find the total field E, wemust integrate over allcharges along the line. If we

integrate over θ, we must write r’ and dq in terms of θ and d θ .

204

1

r

dq

dE ′=

πε

r E y

λ

πε 2

41

0=

0= x E

• The electric field due to dq is:

• Solution: After the appropriate

change of variables, we integrate andfind:

*the calculation is shown in the appendix

++++++++++++++++ x

y

dx

r' r

θ

dE


11/31


Conclusion:

• The Electric Field produced by an

infinite line of charge is:

- everywhere perpendicular to the line

- is proportional to the charge density

- decreases as

- next lecture: Gauss’ Law makes this

trivial!!

1r

++++++++++++++++ x

y

dx

r' r

θ

dE


12/31

Summary

Electric Field LinesElectric Field Patterns

Dipole ~ 1/ R3

Point Charge ~ 1/ R2

~ 1/ RInfinite

Line of Charge

Coming up:Electric field Flux

andGauss’ Law


13/31

The Story Thus Far

Two types of electric charge: opposite charges attract,

like charges repel

Coulomb’s Law:

Electric Fields• Charges respond to electric fields:

• Charges produce electric fields:

F qE =r r

2

q

E k r =

12212

21

12 ˆ

4

1

r r

QQ

F oε=

r


14/31


15/31

Lecture 3, ACT 1

•Examine the electric field

lines produced by the charges

in this figure.•Which statement is true?

(a) q1 and q2 have the same sign

(b) q1 and q2 have the opposite signs and q1 > q2(c) q1 and q2 have the opposite signs and q1 < q2

q1 q2


16/31

Lecture 3, ACT 1

Field lines start from q2 and terminate on q1.

This means q2 is positive; q1 is negative; so, … not (a)Now, which one is bigger ?

Notice along a line of symmetry between the two, that the E-fieldstill has a positive y component. If they were equal, it would be zero;

This indicates that q 2 is greater than q1

•Examine the electric field

lines produced by the charges

in this figure.•Which statement is true?

(a) q1 and q2 have the same sign

(b) q1 and q2 have the opposite signs and q1 > q2(c) q1 and q2 have the opposite signs and q1 < q2

q1 q2


17/31

Electric Dipole: Lines of Force

Consider imaginaryspheres centered on :

aa) +q (blue)

b

b) -q (red)c

c) midpoint (yellow)

• All lines leave a)

• All lines enter b)

• Equal amounts ofleaving and enteringlines for c)


18/31

Electric Flux

• Flux:Let’s quantify previous discussion about field-

line “counting”

Define: electric flux Φ through the closedsurface S

“S” is surfaceof the box•

S E S d E

rr


19/31

Flux

• How much of something is passing

through some surface

Ex: How many hairs passing through yourscalp.

• Two ways to define1.Number per unit area (e.g., 10 hairs/mm2)

This is NOT what we use here.

2.Number passing through an area of intereste.g., 48,788 hairs passing through my scalp.

This is what we are using here.


20/31

Electric Flux

∫

•

S E S d E

rr

•What does this new quantity mean?

• The integral is over a CLOSED SURFACE• Since is a SCALAR product, the electric flux is a SCALAR

quantity

• The integration vector is normal to the surface and points OUT

of the surface. is interpreted as the component of E which isNORMAL to the SURFACE

• Therefore, the electric flux through a closed surface is the sum of

the normal components of the electric field all over the surface.

• The sign matters!!Pay attention to the direction of the normal component as it

penetrates the surface… is it “out of” or “into” the surface?

• “Out of” is “+” “into” is “-”

S d E rr

•

S d E rr

•

S d r


21/31

How to think about flux

• We will be interested in netflux in or out of a closed

surface like this box

• This is the sum of the flux

through each side of the box

– consider each side separately

• Let E-field point in y-direction

– then and are parallel and

• Look at this from on top

– down the z-axis

Er

Sr

2w ES Errr

=

surface area vector:

yw

y AreaS

ˆ

ˆ

2=

•

r

w

“S” is surfaceof the box

x y

z


22/31

How to think about flux

• Consider flux through twosurfaces that “intercept

different numbers of field

lines”

– first surface is side of box from

previous slide

– Second surface rotated by an

angle θ

case 1

case 2

θ

case 1

case 2

y E E o ˆr

y E E o ˆr

2w

E-field surface area E S

2w E o

θ cos2 ⋅w E o

Case 2 issmaller!

Flux:

2w


23/31

The Sign Problem

• For an open surface wecan choose the direction ofS-vector two different ways – to the left or to the right

– what we call flux would bedifferent these two ways

– different by a minus sign

rightleft

A differential surfaceelement, with its vector • For a closed surface we

can choose the directionof S-vector two differentways

– pointing “in” or “out” – define “out” to be correct

– Integral of E dS over a closedsurface gives net flux “out,”

but can be + or -

2


24/31

E

1

2

Wire loops (1) and (2) have the

same length and width, but

differences in depth.

Wire loops (1) and (2) are

placed in a uniform electric

field as shown. Compare

the flux through the two

surfaces.

a) Ф1 >Ф2 b) Ф1 =Ф2

c) Ф1


25/31


26/31

Lecture 3, ACT 2

2A •Imagine a cube of side a positioned in aregion of constant electric field as shown

•Which of the following statements

about the net electric flux Φ E throughthe surface of this cube is true?

(a) Φ E = 0 (b) Φ E ∝2 a2 (c) Φ E∝ 6 a

2

a

a

2B R

2 R

• Consider 2 spheres (of radius R and 2 R)drawn around a single charge as shown.

– Which of the following statementsabout the net electric flux through the

2 surfaces (Φ2 R and Φ R) is true?

(a) Φ R < Φ2 R (b) Φ R = Φ2 R (c) Φ R > Φ2 R


27/31

Lecture 3, ACT 2

2A •Imagine a cube of side a positioned in aregion of constant electric field as shown

•Which of the following statements

about the net electric flux Φ E throughthe surface of this cube is true?

(a) Φ E = 0 (b) Φ E ∝2 a2 (c) Φ E∝ 6 a

2

a

a

• The electric flux through the surface is defined by: ∫ •≡Φ S d E rr

∫ • S d E

rr

• on the bottom face is negative. ( dS is out; E is in)∫ • S d E

rr

• on the top face is positive. ( dS is out; E is out)

• Therefore, the total flux through the cube is:

2 20 0 E dS Ea Eatop sides bottomΦ ≡ • = Φ +Φ +Φ = − + =

rr

˜

• is ZERO on the four sides that are parallel to the electric field.∫ • S d E rr


28/31

2B R

2 R

• Consider 2 spheres (of radius R and 2 R)drawn around a single charge as shown.

– Which of the following statementsabout the net electric flux through the

2 surfaces (Φ2 R

and Φ R

) is true?

(a) Φ R < Φ2 R (b) Φ R = Φ2 R (c) Φ R > Φ2 R

Lecture 3, ACT 2

•Look at the lines going out through each circle -- each circle has the

same number of lines.•The electric field is different at the two surfaces, because E isproportional to 1 / r 2, but the surface areas are also different. Thesurface area of a sphere is proportional to r 2.

•Since flux = , the r 2 and 1/ r 2 terms will cancel, and the two

circles have the same flux!

•There is an easier way. Gauss’ Law states the net flux is proportionalto the NET enclosed charge. The NET charge is the SAME in bothcases.

•But, what is Gauss’ Law ??? --You’ll find out next lecture!

∫ •

S

S d E rr


29/31

Summary

• Electric Fields of continuous charge distributions

•Electric Flux:

– How to think about flux: number of field lines

intercepting a surface, perpendicular to that surface

++++++++++++++++++++++++++ r

E( r) = r

r

ˆ

2

1

0

λ

πε

∫

•

S E S d E

rr

• Next Time: Gauss’ Law


30/31

Appendix


We use Coulomb’s Law to find dE:

2

04

1

r

dqdE

′

=

πε

What is r’ in terms of r ?

What is dq in terms of dx?

d dq xλ =

θ cos

r r =′

( )2

0 cos/4

1

θ

λ

πε r

dxdE =

Therefore,

2

2

0

cos

4

1

r

dxdE

θ λ

πε =

++++++++++++++++ x

y

dx

r' r θ

dE


31/31

x and θ are not independent!

x = r tan θ

dx = r sec2 θ dθ

r

d dE

θ λ

πε 04

1=

++++++++++++++++ x

y

dx

r' r θ

dE

We are dealing with too

many variables. Wemust write the integral in

terms of only one

variable θ or x). We will

use θ.

We still have x and θ variables.


03 electric flux

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