physics for scientists and engineers ii, summer semester 2009 lecture 5: may 29 th 2009 physics for...

Physics for Scientists and Engineers II , Summer Semester 2009

Lecture 5: May 29th 2009

Physics for Scientists and Engineers II


Electric Field Calculation from Electric Potential

operator.gradient thecalled is

ˆˆˆ :notation vector using or,

,,,,,, :generalIn

:charge)point a of field (e.g.,symmetry spherical with fieldsFor

:0 If

Vz

ky

jx

iVE

z

zyxVE

y

zyxVE

x

zyxVE

dr

dVE

drEsdEdV

dx

dVEdxEdVEE

dzEdyEdxEsdEdV

zyx

r

r

xxzy

zyx


Example: Electric Dipole (similar to Ex. 25.4)

x

y

q -q

a a

P

zero!t isn'it that know We:Answer

zero?it Is P?point at field electric for themean that doesWhat

0

:y) 0, xes(coordinat Ppoint at Potential Electric

2222

ya

q

ya

qk

r

qkV e

i i

iep

qq EEE qE

qE



x

y

q -q

a a

P

P.point at with x variesV how know toneed We

Ppoint at Ppoint at E calculate toneed We

:direction- xin the points Pat field electric that theknow We

x

dx

dV

qq EEE qE

qE



x

y

q -q

x+ax-a

P

5)Example23. (c.f., 2

,0

:0)for xdirection -in x field electric theng(calculati 0 xof case special For the

,

23

2223

2223

22

23

2223

222222

ya

aqk

ya

a

ya

aqkyxE

yax

ax

yax

axqk

x

VE

yax

q

yax

qkyxV

eex

exe



0,0

:0)for xdirection -yin field electric theng(calculati 0 xof case special For the

,

23

2223

22

23

2223

222222

ya

y

ya

yqkyxE

yax

y

yax

yqk

y

VE

yax

q

yax

qkyxV

ey

eye



y

3x

222

22

22

222

322

32

23

2223

222222

1E

:away)(far aFor x

4

11

0,

:0)yfor direction -in x field electric theng(calculati 0y of case special For the

,

xqak

xa

axqk

axax

axaxqk

xaaxqk

ax

ax

ax

axqkyxE

yax

ax

yax

axqk

x

VE

yax

q

yax

qkyxV

e

e

eeex

exe


Electric Potential Due to Continuous Charge Distributions

B

A

e

e

sdE-ΔV

V

E

r

dqkV

r

dqkdV

using calculated becan then law, sGauss' using nscalculatio

fromknown ison distributi charge a todue if ely,Alternativ

P

r

dq


Example

P(x=a,y=b)

y

xO

x

L

dx

b

a

dq=dxr

L

e

eee

bax

dxkbaV

bax

dxk

bax

dqk

r

dqkdV

022

2222

),(


Example

2222

2222

)()(

22

22

022

ln)(ln),(

ln)(ln),(

1

1),(

)(1

)(1 :onSubstituti

),(

22

22

22

22

yxxyxLxLL

QkyxV

baabaLaLL

QkbaV

k

dkk

axkax

dk

axkkbaV

axk

ax

bax

ax

dx

dkbaxaxk

bax

dxkbaV

e

e

baLaL

baa

e

baLaL

baa

e

L

e


Example

02

L x:case Special

.............),(

11),(2222

x

y

ex

E

y

yxVE

yxyxLL

Qk

x

yxVE

VE


Example: Chapter 25, Problem 38

R

O

2R 2R

Find the electric potential at O. Linear charge density =

Three contributions to V: left straight piece, curved piece, and right straight piece.

3ln2

3lnln

3lnln

333

333

etotal

ee

circlehalf

e

circlehalf

e

circlehalf

ecurved

e

R

Re

R

R

e

R

R

eright

e

R

Re

R

R

e

R

R

eleft

kV

kRR

kds

R

kds

R

k

R

dqkV

kxkx

dxk

x

dqkV

kxkx

dxk

x

dqkV


Example: …and another one

R

O

R

Find the electric potential at O. Charge +Q is evenly distributed.

12ln

2ln2lnln2

22

R

QkV

R

QkV

R

Qkkxk

x

dxk

x

dqkV

etotal

eQ

ee

R

Re

R

R

e

R

R

eQ

R

+Q

-Q


Electric Potential Due to a Charged Conductor

Charged conductor in electrostatic equilibrium (no net charge movement)

A

B

surface). ialequipotent

an is (surface surface theno everywhere same theis

conductor a of surface at the potential electric The

0

:surface thealongpath a Choosing

path) theofnt (independe

AB

B

A

AB

VV

sdE

sdEVV

C

conductor. theof surface on the potential electric

theequalsconductor a inside potential electric The

0

0 :conductor theinsidepath a Choosing

path) theofnt (independe

AC

B

A

AC

VV

E

sdEVVThe electric potential is constantthroughout a charged conductor inelectrostatic equilibrium.


Electric Potential and Field of a Charged Spherical Conductor

+

+

+

+

++++

+

+

++

V

R

Qke

r

Qke

2r

QkeE


Two Connected Charged Spheres (far apart, so electric field of one sphere does not significantly the affect charge distribution on the other sphere)

r1

r2

metal wire

q1

q2

1

22

1

22

2

1

22

1

221

2

1

22

222

1

11

2

1

2

1

2

2

1

1

44

:sphereeach on density charge Surface

:conductors t the throughousame theis potential electric The

r

r

r

r

r

r

qr

rq

r

q

r

q

r

r

q

q

r

qk

r

qkV ee


Two Connected Charged Spheres (far apart)

1

2

2

22

21

1

2

22

21

1

2

12

2

222

1

11

01

2

2

1

1

22

1

22

2

1

22

1

221

2

1

2

1

2

1

2

2

1

1

EE

:elyAlternativ

)E :(remember :surface At the

r

r

r

r

r

r

q

r

r

q

E

E

r

qk

r

qk

r

r

E

E

r

r

r

r

r

r

qr

rq

r

r

q

q

r

qk

r

qkV

ee

ee


Electric Field in the Cavity within a Conductor

Charged conductor in electrostatic equilibrium (no net charge movement)

A

B

cavity. theinside are charges no as

long asregion free-field a is wallsconductingby surroundedcavity A

cavity e within theverywhere0

path) theofnt (independe 0

E

sdEVVVVB

A

ABBA


Chapter 26: Capacitance and Dielectrics

VQ

increases. Q difference potential theincreasing versa, viceOr,

linearly increases Vcharge ofamount theincreasingWhen

conductors twoebetween th exists V difference potentialA

+Q -Q

Assume you have two charged conductors having equal but opposite amounts of charge on them:


Definition of “Capacitance”

Faraday Michael ofhonor in (farad) F1V

C1 :ecapacitanc of Units

.conductors twoebetween th difference potential electricΔV and

conductoreither on charge of magnitude Q where

,:eCapacitanc

V

QC


The Plate Capacitor

+-

+Q-Q

d

Area = A

A battery has a potential difference V (“voltage”)between it’s two terminals.

Assume: Before the wires are connected, Q=0 on the plates.1) There will be an electric field within the wire going from the left plate to the negative terminal and 2) There will be an electric field within the wire going from the positive terminal to the right plate.Electrons will move opposite to the field lines (from the negative terminal to the left plate and from the right plate to the positive terminal)The left plate gets charged negatively and the right plate gets charged positively as electrons leave it.

The increasing charges on the plates create an increasing additionalelectric field in the wires, opposite to that produced by the battery terminals.Once enough charge is on the plates, the electric field in the wires is zero.The capacitor is now “fully charged”.The higher the voltage of the battery, the more charge can accumulate on the capacitor.

physics for scientists and engineers ii, summer semester 2009 lecture 5: may 29 th 2009 physics for...

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summer semester

electric potential slide

y slide

r q q slide

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charge q

dx r slide

y qq aa p slide