PHYS632_C2_23_Electric.ppt

8/15/2019 PHYS632_C2_23_Electric.ppt

1/28

Summer July 1

Lecture 2 Electric Fields Chp. 23• Cartoon - Analogous to gravitational field

• Opening Demo - Bending of water stream with charged rod• Warm-up problem

• Physlet

• opics

! "lectric field # $orce per unit Charge

! "lectric $ield %ines and "lectric $lu&

! "lectric field from more than ' charge ! "lectric Dipoles

! (otion of point charges in an electric field

! "&amples of finding electric fields from continuous charges

• %ist of Demos

! )an de *raaff *enerator+ wor,ings+lightning rod+ electroscope+ electric wind

! mo,e remover or electrostatic precipitator ! .elvin water drop generator

! ransparent C/ with visible electron gun

! $ield lines using felt+oil+ and '0 .) supply1


2/28

Summer July 2

he "lectric $ield

• Definition of the electric field1 Whenever charges are present and if 2 bring up another charge+ it will feela net Coulomb force from all the others1 2t is convenient to say that there is field there e3ual to the forceper unit positive charge1 E=F/q0. he direction of the electric field is along r and points in the direction a

positive test charge would move1 his idea was proposed by (ichael $araday in the '4506s1 he ideaof the field replaces the charges as defining the situation1 Consider two point charges7

303'

r


3/28

Summer July 3

he force per unit charge is E = F/q0

and then the electric field at r is E = kq1 /r 2 due to the point charge 3' 1

he units are 8ewton9Coulomb1 he electric field has direction and is a vector1

:ow do we find the direction1; he direction is the direction a unit positive test

charge would move1

3'

r E

< 303'

r

he Coulomb force is F= kq1q0 /r 2

2f 3' were positive


4/28

Summer July 4

Point negative charge

E= kq1 /r 2

3'

3'

r


5/28

Summer July 5

Electric Field Lines

Like charges (++) Opposite charges (+ -)

his is called an electric dipole1


6/28

Summer July 6

"lectric $ield %ines7 a graphic concept used

to draw pictures as an aid to develop

intuition about its behavior1

he te&t shows a few e&amples1 :ere are the drawing rules1

• "-field lines begin on < charges and end on - charges1 =or infinity>1

• hey enter or leave charge symmetrically1

•he number of lines entering or leaving a charge is proportional tothe charge

• he density of lines indicates the strength of " at that point1

• At large distances from a system of charges+ the lines becomeisotropic and radial as from a single point charge e3ual to the netcharge of the system1

• 8o two field lines can cross1

how a physlet ?1'1@+ ?1'1

how field lines using felt+oil+ and '0 .) supply


7/28

Summer July 7

ypical "lectric $ields =2 nits>

• ' cm away from ' nC of negative charge

! E = kq /r 2 = 1010 *10-9 / 10-4 =105 N /C

! N.m2 /C2 C / m2 = N/C

• $air weather atmospheric electricity # 100 N/C downward '00 ,m high in the ionosphere

• $ield due to a proton at the location of the electron in the :

atom1 he radius of the electron orbit is 01'0-'0 m1

! E = kq /r 2 = 1010 *1.6*10-19 / (0.5 *10-10 )2 = 4*1011 N /C

3

r

E

<

-r

:ydrogen atom

1

"arth

- - - - - - - - -

•"


8/28

Summer July 8

Example of field lines for a

uniform distribution of positive

charge on one side of a very large

nonconducting sheet.

:ow would the electric field change

if both sides were charged;

:ow would things change if the sheet

were conducting;

his is called a uniform electric field1


9/28

Summer July 9

(ethods of evaluating electric fields

• Direct evaluation from Coulombs %aw or brute force method

2f we ,now where the charges are+ we can find " from E = ∑ kqi /r i2.his is a vector e3uation and can be comple& and messy to

evaluate and we may have to resort to a computer1 he principleof superposition guarantees the result1

• 2nstead of summing the charge we can imagine a continuousdistribution and integrate it1 his distribution may be over a volume+a surface or Eust a line1

! E = ∫dE = ∫ kdqi /r 2 r where r is a unit vector directed from charge

d3 to the field point1

! d3 # ρd) + or d3 #σ dA+ or d3 # λ dl


10/28

Summer July 10

"&ample of finding electric field from two charges

$ind & and y components of electric field due to both charges

and add them up

We have 3'

#


11/28


12/28

Summer July 12

Example continued

&3'#'0 nc 3F #' nc

@

5

"y# '' < 51H # '@1H 89C

" -@14 89C

"

E = 14.6( )2 + −

4.8( )2 =15.4 N / C

(agnitude of electric field

E = E x2+ E y

2

φ1

φ1 = atan "y9" atan ='@1H9-@14># F14 deg

sing unit vector notation we can

also write the electric field vector as7

ρ E = −4.8 i

∧

+14.6 j∧


13/28

Summer July 13

Example of two identical

charges on the x axis. What

is the filed on the y axis?

"y#F" sin φ = 2∗6 ∗ 59 # 1F 89C

Ε &=0

θ

y

@ &3F #' nc

5

3F #' nc@

φ

5

" # '0'0 81mF9CF ' G'0-? C9=m>F # H 89C

θ5

y

@ &3F # -' nc 3F #' nc

@

φ

5

Ε y

=0

"F" cos φ = 2∗6 ∗ @9 # - ?1H 89C

Ε &

"y

Example of two opposite

charges on the x axis. What

is the filed on the y axis?


14/28

Summer July 14

4 equal charges symmetrically spaced along a line. What is the field at

point P? (y and x = 0)

θ3

φ

y

&

3F 353' 3@

r 'r F

r 5r @

θ2θ1

θ4

E y = k qii=1

4

∑ cosθi / r i2

P

E y = k ∆qii=1

∞

∑ cosθi / r i2


15/28


16/28

Summer July 16

What is the electric field from an infinitely long wire

with linear charge density of +100 nC/m at a point 10

away from it. What do the lines of flux look like?

E y =2k λ

ysinθ0

y #'0 cm1

E y =

2 *1010 Nm

2*100*10

−9C / m

0.1m sin90 = 2 *10

4 N

/ C

ypical field for the electrostatic smo,e cleaner

"y


17/28

Summer July 17

"lectric field gradient

• When a dipole is an electric field that varies with position+ then

the magnitude of the electric force will be different for the two

charges1 he dipole can be permanent li,e 8aCl or water or

induced as seen in the hanging pith ball1 2nduced dipoles are

always attracted to the region of higher field1 "&plains why woodis attracted to the teflon rod and how a smo,e remover or

microwave oven wor,s1

• how smo,e remover demo1


18/28

Summer July 18

"lectrostatic smo,e precipitator model

• 8egatively charged central wire has electric field that varies as'9r =strong electric field gradient>1 $ield induces a dipole moment

on the smo,e particles1 he positive end gets attracted more to

the wire

• 2n the meantime a corona discharge is created1 his Eust means

that induced dipole moments in the air molecules cause them tobe attracted towards the wire where they receive an electron

and get repelled producing a cloud of ions around the wire1

• When the smo,e particle hits the wire it receives an electron

and then is repelled to the side of the can where it stic,s1

:owever+ it only has to enter the cloud of ions before it isrepelled1

• 2t would also wor, if the polarity of the wire is reversed


19/28


20/28

Summer July 20

Water =:FO> is a molecule that has a permanent dipole moment1

When a dipole is an electric field+ the dipole

moment wants to rotate to line up with

the electric field1 2t e&periences a tor3ue1

*2ven p # H1F & '0 - 50 C m And 3 # -'0 e and 3 #


21/28

Summer July 21

:FO in a niform "lectric $ield

here e&ist a tor3ue on the water molecule

o rotate it so that p lines up with E1

&

$%rqu! #%u& &'! c%m #τ

$ & sin θ + $=d-&>sin θ = $dsin θ =3"dsin θ = p"sin θ = p & E

Potential "nergy # # -W # -p"cosθ = − p Ε

2s a minimum when p aligns with "

τ = p & E


22/28

Summer July 22

(otion of point charges in electric fields

•When a point charge such as an electron is placed in an electricfield "+ it is accelerated according to 8ewton6s %aw7

# = F/m = qE/m %r ui%rm !!c&ric i!d"

# = F/m = m+/m = + %r ui%rm +r#,i&i%# i!d"

2f the field is uniform+ we now have a proEectile motion problem-

constant acceleration in one direction1 o we have parabolic

motion Eust as in hitting a baseball+ etc e&cept the magnitudes

of velocities and accelerations are different1

/eplace g by 3"9m in all e3uationsJ

For example, In y =1/2at2 we get y =1/2(qE/m)t2


23/28

Summer July 23

"&ample7 An electron is proEected perpendicularly to a

downward electric field of E= 2000 N/C with a horiKontalvelocity ,=106 m/". :ow much is the electron deflectedafter traveling ' cm1

ince velocity in & direction does not change+ &=d/, =10-2 /106 = 10-6 "!cso the distance the electron falls upward is

=1/2#&2 = 0.5*!E/m*&2 = 0.5*1.6*10-19*2*10 /10 - 0*(10-)2 = 0.016m

)

"

d

•e

"

•Demo ransparent C/ with electron gun


24/28

Summer July 24

Back to computing Electric Fields

• "lectric field due to a line of uniform charge

• "lectric field due to an arc of a circle of uniform charge1

• "lectric field due to a ring of uniform charge

• "lectric field of a uniform charged dis,

• 8e&t we will go on to another simpler method to calculate

electric fields that wor,s for highly symmetric situations using

*auss6s %aw1


25/28

Summer July 25

d" , d3 cos θ /r F

E y =

dE y

− L / 2

L / 2

∫ = 0

s#r θds#r dθ

E x = k λ rd θcosθ / r 2

− L / 2

L / 2

∫ = k λ / r d θcosθ−θ0

θ0

∫

d" , λ ds cos θ /r F

E x =2k λ

r

sinθ0

Field due to arc of charge

What is the field at the center

of a circle of charge;

Ans1 0

Fi d th l t i fi ld th i f if l h d i ith


26/28

Summer July 26

Find the electric field on the axis of a uniformly charged ring with

linear charge density λ = L9Fπ/.

E z

= dE cosθ

dE = k

dq

r 2 = k

λds

r 2

E z

= k λcosθ

r 2

ds∫

E z

= k λcosθ

r 2

2π R

d3 # λds

r F #KF


27/28


28/28

Summer July 28

Kelvin Water Drop Generator

Am. J. Phys. 68,1084(2000)

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PHYS632_C2_23_Electric.ppt

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