Lecture 2: May 20 2009 - Department of Physics & Astronomygernot/Physics2220/L2 Chapter23.pdfPhysics...

Physics for Scientists and Engineers II , Summer Semester 2009

Lecture 2: May 20th 2009

Physics for Scientists and Engineers II


Electric Field due to a Continuous Charge Distribution

• We can model a system of charges as being continuous (instead of discrete) if the distance between the charges is much smaller than the distance to the point where the electric field is calculated.

• Procedure: - Divide charge distribution into small charge elements ∆q.- Add contributions to E from all charge elements.

r̂

PE∆ r

∆q

∑ ∆≈i

ii

e rr

qkE ˆ2

rrdqkr

rqkE e

ii

i

i

qe

i

ˆˆ 220

lim ∫∑ =∆=→∆

rrqkE e ˆ2

∆=∆


Charge Density (a useful concept when calculating E from charge distribution)

dldqlQ

dAdqAQ

dVdqVQ

λλ

σσ

ρρ

=⇒≡

=⇒≡

=⇒≡

:l)length of line aon ddistributeuniformly is Q (ifdensity chargeLinear

:A) area of surface aon ddistributeuniformly is Q (ifdensity charge Surface

:V) volumeat throughouddistributeuniformly is Q (ifdensity charge Volume


Example: Electric Field due to a Uniformly Charged Rod

l

x

y

P

a

x dx

E

dq = λ dx

22 :dq from on toContributixdxk

xdqkdEE ee

λ==

( ))11

1 :rod) (entire dq all from 22

alaQk

alalQk

xk

xdxk

xdxkEE

ee

al

ae

al

ae

al

ae

+=

+−=

−===

+++

∫∫ λλλ

charge.point a of field the, 0 lFor :otice 2aQkEN e→→


Example: Electric Field due to a Uniformly Charged Rod…..this is harder….

l

x

y

P

a

x

Ed dq = λ dx

( ) ( )j

xa

dxakixa

dxxk

jr

dxakir

dxxk

jra

rdxki

rx

rdxk

jrdxki

rdxkEdE

ee

ee

ee

ee

ˆˆ

ˆˆ

ˆˆ

ˆcosˆsin :dq from on toContributi

23222

322

33

22

22

++

+−=

+−=

+−=

Θ+Θ−=

λλ

λλ

λλ

λλ

Θr


Example: Electric Field due to a Uniformly Charged Rod…..this is harder….

l

x

y

P

a

x

Ed dq = λ dx

( ) ( )j

xa

dxakixa

dxxkEdE eeˆˆ :dq from on toContributi

23222

322 ++

+−= λλ

Θr

( ) ( )

( ) ( )dx

xaakEdx

xa

xkE

jxa

dxakixa

dxxkE

l

ey

l

ex

l

ee

∫∫

∫

+=

+−=

++

+−=

0 2322

0 2322

0 23222

322

1

ˆˆ :Ppoint at field electric Total

λλ

λλ


….solving the integral for Ex

( )dx

xa

xkEl

ex ∫+

−=0 2

322λ

( )( ) 2

122

2122 :onSubstituti

xa

dxxduxau+

=⇒+=

( ) ( )

+−+−=

+−−=

−−=−=

++−=

++

∫∫

22

22

22

20 2

12222

11

1112222

laaala

lQk

laak

ukdu

ukdx

xa

xxa

kE

ee

la

ae

la

ae

l

ex

λ

λλλ


….solving the integral for Ey

( )dx

xa

akEl

ey ∫+

=0 2

322λ

that)know you toexpect t wouldn'(Icos

tan :onSubstituti 2 ΘΘ

=⇒Θ= dadxax

( ) ( )

[ ]

2222

max0

002

23

2

02

232

02

23222

sinsin

coscos

1

cos1

1

cos1

tan1

1costan

max

maxmax

maxmax

alaQk

all

laQk

laQk

ak

da

kda

k

da

kda

aa

akE

ee

ee

ee

eey

+=

+=

Θ=Θ=

ΘΘ=ΘΘ

Θ

=

ΘΘΘ+

=ΘΘΘ+

=

Θ

ΘΘ

ΘΘ

∫∫

∫∫

λ

λλ

λλ

a

l

22 al +Θmax

22maxsinal

l+

=Θ


….and the final result

22 alaQkE e

y+

=

+−+−=

22

22

laaala

lQkE ex

again)chargepointaof(fieldand0:

:rodshort very aoflimit theIn

20

lim aQkEE e

yxl

→→→

akE

akE eyex

l

λλ →−→∞→

and:

:rod longvery aoflimit theIn

lim


Visualizing Electric Fields with Electric Field Lines

• The electric field vector is always tangent to the electric field line.• The electric field line has a direction (indicated by an arrow). The direction is

the same as that of the electric field (same direction as force on a positive test charge).

• The number of lines per unit area through a normal plane (perpendicular to field lines) is proportional to the magnitude of the electric field in that region.

Example: Electric field lines of a point charge

+

N field lines

Surface density of field lines at an imagined sphere of radius r is

Electric field strength is proportional to

24 rN

π

2

1r


Visualizing Electric Fields with Electric Field Lines

• For a single positive point charge: Electric field lines go from the positive charge to infinity.

• For a single negative point charge: Electric field lines go come from infinity and end at the negative point charge.

• For multiple point charges: Lines can start at the positive charges and end at the negative charges.

• Electric field lines can never cross (think about why that is so).• For two unequal point charges of opposite sign with charges Q1 and Q2 , the number N1 of

field lines terminating at Q1 and the number N2 of field lines terminating at Q2 are related by the equation

1

2

1

2

QQ

NN =


Motion of a Charged Particle in a Uniform Electric Field

• Assume particle has charge q, mass m.• Particle experiences a force

• The force results in an acceleration (according to Newton’s second law):

• For positive charges: Acceleration is in the same direction as electric field.• For negative charges: Acceleration is in a direction opposite to the electric field.• A uniform electric field will cause a constant acceleration of the particle.

You can use equations of motion for constant acceleration.

• Work is done on the particle by the electric force as the particle moves.

EqF e =

mEq

mFa e ==

xFW e ∆•=


Example (similar to Ex. 23.10 in book)

- - - - - - - - - -

+ + + + + + + + + +

E-?=iv

Electron: m = 9.11x10-31 kg ; q = 1.60x10-19 CElectric Field: E = 800 N/C

L = 0.100 m

The electron leaves the electric field at an angle of Θ = 65 degrees.Q1: What was the initial velocity of the electron?Q2: What is the final velocity of the electron (magnitude)?Q3: How low would the electric field have to be so that the net force on the electron is zero?Q4: Were we justified in neglecting the gravitational force in Q1 and Q2?

Θ


Θ===== tan;

:1Question

i

fy

i

eyfy v

vvLtt

mEqt

mFtav

( ) sm

kg

mCNC

LmEqv

vL

mEqv

vL

mEqv

i

ii

ify

631

19

105.2deg65tan1010.9

100.08001060.1

tan

tan

×=⋅×

⋅

−⋅×−

=Θ

=⇒

=Θ⇒=⇒

−

−

( ) sms

mvv if

66

101.665cos

105.2

cos

:2Question

×=°

×=

Θ=


force. nalgravitatio n thelarger thamuch iselectronan on force electric thefields, electric smallextremely for except Yes,

:4Question

106.5106.1

1010.98.9

:3Question

1119

312

CN

C

kgsm

qgmEg

mqE −

−

−

×=×

×⋅==⇒=


Gauss’s Law – An alternative procedure to calculate electric fields of highlysymmetric charge distributions

The concept of “Electric Flux”:

Area = A

E

area. lar toperpendicu being E and Econstant for :flux Electric E EA=Φ


E

⊥AE is lar toperpendicu Area

AE is lar toperpendicunot Area

Θ=⊥ cosAA

Θ

Θ Θ

Θ

The electric flux through the two surfaces is the same

Θ==Φ ⊥ cosAEAEE

Normal to green surface


Θ Θ

Θ

The electric flux through the two surfaces is the same

Θ==Φ ⊥ cosAEAEE

Normal to green surface

To calculate the flux through a randomly oriented area you need to know the anglebetween the electric field and the normal to the area.


∫

∑•=Φ

∆•≈Φ

∆•=Θ∆=∆Φ

surfaceE

iiiE

iiiiiE

AdE

AE

AEAE

:segments surface smallmally infinitesi oflimit in the ....and

:surface entiregh flux throu Electric

cos:element surfacegh flux throu Electric

How to treat situations where the electric field is not constant over the area?

• Divide area into small areas over which E is constant.• Calculate flux for each small area.• Add fluxes up.

Area vector:magnitude = areadirection = perpendicular to area

iEiΘ iA

“surface integral”


∫∫ =•=Φ dAEAdE nE : surface closedgh Flux throu

Flux through a closed surface:

•Convention: Area vectors always point outwards.Field lines that cross from the inside to the outside of the surface :(positive flux because cos Θ is positive)Field lines that cross from the outside to the inside of the surface:(negative flux because cos Θ is negative)

°<Θ 90

°<Θ<° 18090


Example: Cube in a uniform field

EdA2

dA1

dA3

dA4

dA5

dA6

00cos180cos

0000

22

2121

21

654321

=+−=+−=°+°=

++++•+•=

•+•+•+•+•+•=•=Φ

∫∫∫∫

∫∫

∫∫∫∫∫∫∫

ELELdAEdAEdAEdAE

AdEAdE

AdEAdEAdEAdEAdEAdEAdEE

Lecture 2: May 20 2009 - Department of Physics & Astronomygernot/Physics2220/L2 Chapter23.pdfPhysics...

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Transcript of Lecture 2: May 20 2009 - Department of Physics & Astronomygernot/Physics2220/L2 Chapter23.pdfPhysics...