Lecture 04

Week of February 17 05

Electric Field 1

Lecture 04

The Electric Field

Chapter 22 - HRW

Electric Field 2Week of February 17 05

Physics 2049 News

WebAssign was due today Another one is posted for

Friday You should be reading

chapter 22; The Electric Field.This is a very important concept. It is a little “mathy”

There will be a QUIZ on Friday.Material from chapters 21-22.Studying Works!


This is WAR

You are fighting the enemy on the planet Mongo.

The evil emperor Ming’s forces are behind a strange green haze.

You aim your blaster and fire … but ……

Ming themerciless

this guy is

MEAN!


Nothing Happens! The Green thing is a Force Field!

The Force may not be with you ….


Side View

TheFORCE FIELD

Force

Positiono

|Force| Big!


Properties of a FORCE FIELD

It is a property of the position in space.

There is a cause but that cause may not be known.

The force on an object is usually proportional to some property of an object which is placed into the field.


EXAMPLE: The Gravitational Field That We Live In.

m M

mgMg


The gravitational field: g

The gravitational field strength is defined as the Force per unit mass that the field creates on an object

This becomes g=(F/m)=(mg/m)=g

The field strength is a VECTOR. For this case, the gravitational

field is constant.magnitude=g (9.8 m/s) direction= down


Final Comment on Gravitational Field:

ggF

g m

m

massunit

orce

_

Even though we know what is causing the force, we really don’t usually think

about it.


Newton’s Law of Gravitation

R

MEarth

m


The Calculation

2

26

2411

2

2

/77.9

)104.6(

1061067.6

smg

x

xxg

R

M

m

R

mM

Earth

Earth

GF

g

GF


Not quite correct ….

Earth and the Moon (in background), seen from space)


More better …

FEarth

MEarth

m

Moon

Fmoon

mg


To be more precise …

g is caused byEarth (MAJOR)moon (small)Sun (smaller yet)Mongo (extremely teeny tiny)

g is therefore a function of position on the Earth and even on the time of the year or day.


The Electric Field E

In a SIMILAR WAYWe DEFINE the ELECTRIC FIELD

STRENGTH AS BEING THE FORCE PER UNIT CHARGE.

Place a charge q at a point in space.Measure (or sense) the force on the

charge – F Calculate the Electric Field by dividing

the Force by the charge,

Coulomb

Newtons

q

q

FE

EF



Electric Field Near a Charge


Two (+) Charges


Two Opposite Charges


A First Calculation

Q

r

q

A Charge

The spot where we wantto know the Electric Field

Place a “test chargeat the point and

measure the Force on it.


Doing itQ

r

q

A Charge

The spot where we wantto know the Electric Field

unit

unit

r

Qk

q

r

qQk

rF

E

rF

2

2

F


General-

unitjj

jjj

unit

unit

r

Qk

q

General

r

Qk

q

r

qQk

,2

2

2

rF

EE

rF

E

rF


Continuous Charge Distribution


ymmetry


Let’s Do it Real Time

Concept – Charge perunit length

dq= ds


The math

)sin(2

)cos(2

)cos()2(

)cos()2(

0

0

0

02

02

0

0

0

r

kd

r

kE

r

rdkE

r

dqkE

E

rdds

x

x

x

y

Why?


A Harder Problem

A line of charge=charge/length

setupsetup

dx

L

r

x

dE dEy


2/

02/322

2/

02/322

22

2

2

22

)(2

)(2

)()cos(

)(

)cos(

L

x

L

x

L

Lx

xr

dxkrE

xr

dxrkE

xr

r

xr

dxkE

(standard integral)


Completing the Math

r

kL

r

klE

Lr

L

Lrr

kLE

x

x

2

2

4

:line long VERY a oflimit In the

4

:nintegratio theDoing

22

22

1/r dependence


Dare we project this??

Point Charge goes as 1/r2

Infinite line of charge goes as 1/r1

Could it be possible that the field of an infinite plane of charge could go as 1/r0? A constant??


The Geometry

Define surface charge density=charge/unit-area

dq=dA

dA=2rdr

(z2+r2)1/2

dq= x dA = 2rdr


(z2+r2)1/2

R

z

z

rz

rdrzkE

rz

z

rz

drrk

rz

dqkdE

02/322

2/1222222

2

2)cos(


(z2+r2)1/2

Final Result

0z

220

2E

,R

12

When

Rz

zEz


Look at the “Field Lines”


What did we learn in this chapter??

We introduced the concept of the Electric FIELDFIELD.We may not know what causes

the field. (The evil Emperor Ming)

If we know where all the charges are we can CALCULATE E.

E is a VECTOR.The equation for E is the same

as for the force on a charge from Coulomb’s Law but divided by the “q of the test charge”.


What else did we learn in this chapter?

We introduced continuous distributions of charge rather than individual discrete charges.

Instead of adding the individual charges we must INTEGRATE the (dq)s.

There are three kinds of continuously distributed charges.


Kinds of continuously distributed charges Line of charge

or sometimes = the charge per unit length.

dq=ds (ds= differential of length along the line)

Area = charge per unit area dq=dA dA = dxdy (rectangular coordinates) dA= 2rdr for elemental ring of charge

Volume =charge per unit volume dq=dV dV=dxdydz or 4r2dr or some other

expressions we will look at later.


The Sphere

dqr

thk=dr

dq=dV= x surface area x thickness= x 4r2 x dr


Summary

222

,2

2

2

)()()(

r

rdsk

r

rdAk

r

rdVk

r

Qk

q

General

r

Qk

q

r

qQk

unitjj

jjj

unit

unit

E

rF

EE

rF

E

rF

(Note: I left off the unit vectors in the lastequation set, but be aware that they should

be there.)


To be remembered …

If the ELECTRIC FIELD at a point is E, then

E=F/q (This is the definition!)

Using some advanced mathematics we can derive from this equation, the fact that:

EF q


Example:

(2,8)? coordinate at the

placed isit if experience charge

coulomb 0.5 a wouldforceWhat

3xE

:expression by thegiven is

space ofregion ain field electric The

2


Solution

Newtons 6F

or

(N/C) 12 C 0.5 qE force The

matter.t doesn' coordinatey The

)/(12433xE

one!easy an is This2

CN


In the Figure, particle 1 of charge q1 = -9.00q and

particle 2 of charge q2 = +2.00q are fixed to an x

axis.

(a) As a multiple of distance L, at what coordinate on the axis is the net electric field of the particles zero?[1.89] L(b) Plot the strength of the electric field as a function of position (z).

q1 = -9q q2=+2q


Let’s do it backwards…

for this. EXCEL use sLet' brackets.

inquantity plot thejust and of

function a as at thislook can We

9

)1(

2)(

9

)(

2)(

Let

9

)(

2)(

...Then L.an greater th isthat

coordinate a hasit ofheck for the

whichposition xarbitrary an Take

222

222

22

L

qkxE

LLLqkxE

αLx

x

q

Lx

qkxE


EXCEL

a First Term Second Term Sum

-3 0.13 -1.00 -0.88

-2.9 0.13 -1.07 -0.94

-2.8 0.14 -1.15 -1.01

-2.7 0.15 -1.23 -1.09

-2.6 0.15 -1.33 -1.18

-2.5 0.16 -1.44 -1.28

-2.4 0.17 -1.56 -1.39

-2.3 0.18 -1.70 -1.52

-2.2 0.20 -1.86 -1.66

-2.1 0.21 -2.04 -1.83

-2 0.22 -2.25 -2.03

-1.9 0.24 -2.49 -2.26

-1.8 0.26 -2.78 -2.52

ETC ….


Bracket

-100.00

-50.00

0.00

50.00

100.00

-2 -1.8 -1.6 -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

alphaalpha=1.89

??


The mystery solved!!!

.

9

)(

2)(

0For x

9

)(

2)(

LFor x

9

)(

2)(

LFor x

22

22

22

x

q

Lx

qkxE

x

q

Lx

qkxE

x

q

Lx

qkxE

BE CAREFULL!BE CAREFULL!


In the Figure, the four particles are fixed in place and have charges q1 = q2 = +5e, q3 = +3e,

and q4 = -12e. Distance d = 9.0 mm. What is

the magnitude of the net electric field at point P due to the particles?



Figure 22-34 shows two charged particles on an x axis, q = -3.20 10-19 C at x = -4.20 m and q = +3.20 10-19 C at x = +4.20 m.

(a) What is the magnitude of the net electric field produced at point P at y = -5.60 m?[7.05e-11] N/C(b) What is its direction?[180]° (counterclockwise from the positive x axis)


Figure 22-40 shows two parallel nonconducting rings arranged with their central axes along a common line. Ring 1 has uniform charge q1 and radius R; ring 2 has

uniform charge q2 and the same radius R. The rings

are separated by a distance d = 3.00R. The net electric field at point P on the common line, at distance R from ring 1, is zero. What is the ratio q1/q2?

[0.506]


In the Figure, eight charged particles form a square array; charge q = +e and distance d = 1.8 cm. What are the magnitude and direction of the net electric field at the center?

Lecture 04

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