Physics for Scientists and Engineers II, Summer Semester 2009 Lecture 2: May 20 th 2009 Physics for Scientists and Engineers II

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

ii

i

e rr

qkE ˆ

2

rr

dqkr

r

qkE e

ii

i

i

qe

i

ˆˆ22

0lim

rr

qkE e ˆ

2


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

dldql

Q

dAdqA

Q

dVdqV

Q

: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 = dx

22 :dq from on toContributi

x

dxk

x

dqkdEE ee

)11

1 :rod) (entire dq all from

22

ala

Qk

alal

Qk

xk

x

dxk

x

dxkEE

ee

al

ae

al

a

e

al

a

e

charge.point a of field the, 0 lFor :otice2a

QkEN e


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

l

x

y

P

a

x

Ed dq = dx

j

xa

dxaki

xa

dxxk

jr

dxaki

r

dxxk

jr

a

r

dxki

r

x

r

dxk

jr

dxki

r

dxkEdE

ee

ee

ee

ee

ˆˆ

ˆˆ

ˆˆ

ˆcosˆsin :dq from on toContributi

23

2223

22

33

22

22

r


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

l

x

y

P

a

x

Ed dq = dx

j

xa

dxaki

xa

dxxkEdE ee

ˆˆ :dq from on toContributi2

3222

322

r

dx

xaakEdx

xa

xkE

jxa

dxaki

xa

dxxkE

l

ey

l

ex

l

ee

0 23

220 2

322

0 23

2223

22

1

ˆˆ :Ppoint at field electric Total


….solving the integral for Ex

dx

xa

xkE

l

ex

0 2

322

2

122

21

22 :onSubstitutixa

dxxduxau

22

22

22

20 2

122

22

11

1112222

laa

ala

l

Qk

laak

ukdu

ukdx

xa

x

xakE

ee

la

ae

la

a

e

l

ex


….solving the integral for Ey

dx

xa

akE

l

ey

0 2

322

that)know you toexpect t wouldn'(Icos

tan :onSubstituti2

da

dxax

2222

max0

002

23

2

02

23

20

22

3222

sinsin

coscos

1

cos1

1

cos

1

tan1

1

costan

max

maxmax

maxmax

ala

Qk

al

l

la

Qk

la

Qk

a

k

da

kd

a

k

da

kd

a

aa

akE

ee

ee

ee

eey

a

l

22 al max

22maxsinal

l


….and the final result

22 ala

QkE e

y

22

22

laa

ala

l

QkE ex

again)chargepointaof(fieldand0:

:rodshort very aoflimit theIn

20

lim a

QkEE e

yxl

akE

akE eyex

l

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 r

N

2

1

r


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

Q

Q

N

N


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

m

Eq

m

Fa

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 = 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?


tan;

:1Question

i

fy

i

eyfy v

v

v

Ltt

m

Eqt

m

Ftav

s

m

kg

mCN

CL

m

Eqv

v

L

m

Eqv

v

L

m

Eqv

i

ii

ify

631

19

105.2deg65tan1010.9

100.08001060.1

tan

tan

s

msmv

v if

6

6

101.665cos

105.2

cos

:2Question


force. nalgravitatio n thelarger thamuch is

electronan on force electric thefields, electric smallextremely for except Yes,

:4Question

106.5106.1

1010.98.9

:3Question

1119

312

C

N

C

kgs

m

q

gmEg

m

qE


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

The concept of “Electric Flux”:

Area = A

E

area. lar toperpendicu being E and Econstant for

:flux Electric E EA


E

AE is lar toperpendicu Area

AE is lar toperpendicunot Area

cosAA

The electric flux through the two surfaces is the same

cosAEAEE

Normal to green surface


The electric flux through the two surfaces is the same

cosAEAEE

Normal to green surface

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


surface

E

i

iiE

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

iE

iiA

“surface integral”


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 is positive)Field lines that cross from the outside to the inside of the surface: (negative flux because cos is negative)

90

18090


Example: Cube in a uniform field

E

dA2

dA1

dA3

dA4

dA5

dA6

00cos180cos

0000

22

2121

21

654321

ELELdAEdAEdAEdAE

AdEAdE

AdEAdEAdEAdEAdEAdEAdEE

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Physics for Scientists and Engineers II, Summer Semester 2009 Lecture 2: May 20 th 2009 Physics for Scientists and Engineers II