R

2R

a

a

• Consider a circular ring with total charge +Q. The charge is spread uniformly around the ring, as shown, so there is λ = Q/2pR charge per unit length.

• The electric field at the origin is

R

x

y++++

+ +

+++

++

+ + + + ++++++

+

(a) zeroR

2

4

1

0

(b) 204

1

R

R

(c)

But how would we calculate this??

Electric Fieldsfrom

Continuous Charge Distributions

• Principles (Coulomb’s Law + Law of Superposition) remain the same.

Only change:

 

  

Examples: • line of charge• charged plates• electron cloud in atoms, … ++++++++++++++++++++++++++

rE(r) = ?

AB

2) A finite line of positive charge is arranged as shown. What is the 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) right

e) up and left f) up and right

L

Preflight 3:

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

total chargeQ

small piecesof chargedq

• Line of charge:l = charge per

unit lengthdq = l dx

• Surface of charge:s = charge per

unit area

dq = s dA

• Volume of Charge:

r= charge per unit volume

dq = r dV

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.12

2-6 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

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 q : from - /2p to+ /2p

++++++++++++++++++++++++++rE(r) = ?

Any symmetries ? This may help for easy cancellations

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

q

x

Infinite Line of Charge

We need to add up the E-field contributions from all segments dx along the line.

Charge density = l

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

y

dx

r'r

q

dE

Infinite Line of Charge

• To find the total field E, we must integrate over all charges along the line. If we integrate over q, we must write r’ and dq in terms of qand dq.

204

1

r

dqdE

rE y

24

1

0

0xE

• The electric field due to dq is:

•Solution: After the appropriate change of variables, we integrate

and find:

*the calculation is shown in the appendix

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

y

dx

r'r

q

dE

Infinite Line of Charge

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

q

dE

SummaryElectric Field Lines

Electric Field Patterns

Dipole ~ 1/R3

Point Charge ~ 1/R2

~ 1/RInfinite

Line of Charge

Coming up:Electric field Flux

and Gauss’ Law

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

2

qE k

r

12212

2112 ˆ

4

1r

r

QQF

o

The Story Thus Far

We want to be able to calculate the electric fields from various charge arrangements. Two ways:

2. Gauss’ Law: The net electric flux through any closed surface is proportional to the charge enclosed by that surface.

In cases of symmetry, this will be MUCH EASIER than the brute force method.

1. Brute Force: Add up / integrate contribution from each charge.

Often this is pretty difficult.

Ack!Ex: electron cloud around nucleus

2

• 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

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 q2 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

Electric Dipole: Lines of Force

Consider imaginary spheres centered on :

aa) +q (blue)

b

b) -q (red)c

c) midpoint (yellow)

• All lines leave a)• All lines enter b)• Equal amounts of

leaving and entering lines for c)

Electric Flux• Flux:

Let’s quantify previous discussion about field-line “counting”

Define: electric flux F E through the closed surface S

“S” is surfaceof the box

SE SdE

Flux

• How much of something is passing through some surface

Ex: How many hairs passing through your scalp.

• 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 interest

e.g., 48,788 hairs passing through my scalp.

This is what we are using here.

Electric Flux

S

E SdE

• 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 is NORMAL 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 “-”

SdE

SdE

Sd

How to think about flux• We will be interested in net

flux 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

E

S

2wESE

surface area vector:

yw

yAreaS

ˆ

ˆ2

w

“S” is surfaceof the box

xy

z

How to think about flux• Consider flux through two

surfaces that “intercept different numbers of field lines”– first surface is side of box from

previous slide– Second surface rotated by an

angle q

case 1

case 2

qcase 1

case 2

yEE o ˆ

yEE o ˆ

2w

E-field surface area E S

2wEo

cos2 wEo

Case 2 is smaller!

Flux:

2w

The Sign Problem• For an open surface we

can choose the direction of S-vector two different ways– to the left or to the right– what we call flux would be

different these two ways– different by a minus sign

rightleft

A differential surfaceelement, with its vector

•For a closed surface we can choose the direction of S-vector two different ways

–pointing “in” or “out”–define “out” to be correct–Integral of EdS over a

closed surface gives net flux “out,” but can be + or -

2

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 < Ф2

5)

Preflight 3:

6) A cube is placed in a uniform electric field. Find the flux through the bottom surface of the cube.

a) Фbottom < 0

b) Фbottom = 0

c) Фbottom > 0

Preflight 3:

Lecture 3, ACT 2

2A • Imagine a cube of side a positioned in a region of constant electric field as shown• Which of the following statements

about the net electric flux FE through the surface of this cube is true?

(a) FE = 0 (b) FE µ 2a2 (c) FE µ 6a2

a

a

2BR

2R

• Consider 2 spheres (of radius R and 2R) drawn around a single charge as shown.– Which of the following statements

about the net electric flux through the

2 surfaces (F2R and FR) is true?

(a) FR < F2R (b) FR = F2R (c) FR > F2R

2A • Imagine a cube of side a positioned in a region of constant electric field as shown• Which of the following statements

about the net electric flux FE through the surface of this cube is true?

(a) FE = 0 (b) FE µ 2a2 (c) FE µ 6a2

a

a

• The electric flux through the surface is defined by: SdE

SdE

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

SdE

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

• Therefore, the total flux through the cube is:

2 20 0E dS Ea Eatopsides bottom

˜

• is ZERO on the four sides that are parallel to the electric field. SdE

2BR

2R

• Consider 2 spheres (of radius R and 2R) drawn around a single charge as shown.– Which of the following statements

about the net electric flux through the

2 surfaces (F2R and FR) is true?

(a) FR < F2R (b) FR = F2R (c) FR > F2R

•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 is proportional to 1 / r 2, but the surface areas are also different. The surface 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 proportional to the NET enclosed charge. The NET charge is the SAME in both cases.

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

S

SdE

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

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

rE(r) = r

rˆ

2

1

0

S

E SdE

• Next Time: Gauss’ Law

AppendixInfinite Line of Charge

We use Coulomb’s Law to find dE:

204

1

r

dqdE

What is r’ in terms of r ?

What is dq in terms of dx?

d dq x

cosr

r

20 cos/4

1

r

dxdE

Therefore,

2

2

0

cos

4

1

r

dxdE

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

y

dx

r'r

q

dE

x and q are not independent!

x = r tan qdx = r sec2 q dq

r

ddE

04

1

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

y

dx

r'r

q

dE

We are dealing with too many variables. We must write the integral in terms of only one variable (qor x). We will use q.

We still have x and qvariables.

Infinite Line of Charge

Infinite Line of Charge

• Components:

sin4

1

0 r

ddEx

cos4

1

0 r

ddEy

• Integrate:

2/

2/ 0

cos4

1

r

ddEE yy

2/

2/ 0

sin4

1

r

ddEE xx

Ex

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

y

dx

r'r

q

dE qEy

Infinite Line of Charge

• Now

• The final result:

0

1

2yEr

0xE

/ 2

/ 2

sin 0d

/ 2

/ 2

cos 2d

Ex

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

y

dx

r'r

q

dE qEy