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ma F = Coulomb's Law = + - ψ ψ ψ 2 2 hf E = Muhammad Nadeem School of Electrical Engineering &Computer Sciences 2 mc E = dt dB E / - = × ma F = Physics 20th Century 21st Century t i V x = + - ψ ψ ψ h h 2 2 hf E = o ε ρ / = E mv P = P r L × = h P x . 0 = B dt dE j B / o o o ε μ μ - = × P h = λ h G k o ε o μ R c

### Transcript of Coulombs Law

maF =

Coulomb's Law

∂=+∂− ψψψ22hfE =

School of Electrical Engineering &Computer Sciences

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2mcE =

dtdBE /−=×∇

maF =Physics

20th Century 21st Century

tiV

x ∂∂=+

∂∂− ψψψ

hh2

22hfE =

oερ /=•∇ E

mvP =

PrL ×= h≈∆∆ Px.

0=•∇ B

dtdEjB /ooo

εµµ −=×∇

P

h=λ

h

G

k

oµ R

c

� Electromagnetism is one of the fundamental forces in nature, and the dominant force in a vast range of natural and technological phenomena

�The electromagnetic force is solely responsible for the

ElectromagnetismElectromagnetismElectromagnetismElectromagnetism

�The electromagnetic force is solely responsible for thestructure of matter, organic, or inorganic

� Physics, chemistry, biology, materials science

�The operation of most technological devices is based onelectromagnetic forces. From lights, motors, and batteries, to communication and broadcasting systems, as well as microelectronic devices.

Electromagnetism

Electricity

Magnetism

Optics

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Electric ChargesElectric ChargesElectric ChargesElectric Charges

� All ordinary matter contains both positive and negative charge.

� You do not usually notice the charge because most matter charge because most matter contains the exact same number of positive and negative charges.

� An object is electrically neutral when it has equal amounts of both types of charge.

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SILK

The Transfer of Charge

Electric Charges

Glass Rod

Some materials attract electronsmore than others.

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The Transfer of Charge

SILK-+

Electric Charges

Glass Rod

•As the glass rod is rubbed against silk, electrons are pulled off the glass onto the silk.

•Glass and silk are insulators: charges stuck on them stay put.

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The Transfer of Charge

SILK

--+

+

Electric Charges

Glass Rod

-+

•Usually matter is charge neutral, because the number of electrons and protons are equal. But here the silk has anexcess of electrons and the rod a deficit.

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The Transfer of Charge

SILK

-

+++++

----

Electric Charges

Glass Rod

+

Electric charge is always conserved. That is, when one object is rubbed against another, charge is not created in the process. The electrified state is due to a transfer of charge from one object to the other. One object gains some amount of negative charge while the other gains an equal amount of positive charge.

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� Attractive electric forces are responsible for the behavior of a wide variety of commercial products.

� For example, the plastic in many contact lenses, etafilcon, is made up of molecules that electrically attract the protein

Like charges repel, unlike charges attract.

made up of molecules that electrically attract the protein molecules in human tears. These protein molecules are absorbed and held by the plastic so that the lens ends up being primarily composed of the wearer’s tears. Because of this, the wearer’s eye does not treat the lens as a foreign object, and it can be worn comfortably.

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History

600 BC Greeks first discover attractiveproperties of amber when rubbed.

1600 AD Electric bodies repel as well as attract 1735 AD du Fay: Two distinct types of electricity

Electric Charges

1735 AD du Fay: Two distinct types of electricity1750 AD Franklin: Positive and Negative Charge1770 AD Coulomb: “Inverse Square Law”1890 AD J.J. Thompson: Quantization of

electric charge - “Electron”

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Summary of things we know:

� There is a property of matter called electric charge. (In the SI system its units are Coulombs.)

� Charges can be negative (like electrons) or positive (like protons).

Electric Charges

protons).� In matter, the positive charges are stuck in place in the

nuclei and electrons revolve around it. Matter is negatively charged when extra electrons are added, and positively charged when electrons are removed.

� Like charges repel, unlike charges attract.� Charges travel in conductors, not in insulators� Force of attraction or repulsion ~ 1 / r2

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Charge is Quantized

� q = multiple of an elementary charge e:� e = 1.6 x 10-19 Coulombs

Electric Charges

Neq =

Charge Mass Diameterelectron - e 1 0proton +e 1836 ~10-15mneutron 0 1839 ~10-15m

Protons and neutrons are made up of quarks, whose charge is quantized in multiples of e/3. Quarks can’t be isolated.

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� Coulomb’s showed that the electric force between two stationary charged particles is

� inversely proportional to the square of the separation r between the particles and directed along the line joining them;

� proportional to the product of the charges q1 and q2 on the two particles;

� is attractive if the charges are of opposite sign and repulsive if the

Coulomb’s Coulomb’s Coulomb’s Coulomb’s LawLawLawLaw

is attractive if the charges are of opposite sign and repulsive if the charges have the same sign.

� the magnitude of the electric force between two point charges:

221

r

qqkF =

229109 −×= CNmK

oπε4/1=K

212121085.8 −−−×= mNCo

Vector form of Coulomb’s LawVector form of Coulomb’s LawVector form of Coulomb’s LawVector form of Coulomb’s Law

Q2

21F

r

+

+

12r̂

21F

+

12F

- Q2

r̂21r̂

Q1

21r̂12F

+

Force on 1 due to 2 Force on 2 due to 1

12

12221

12 r̂r

qkqF = 212

2121 r̂

r

qkqF =

`2112 ˆˆ rr −=

+

Q1

12r̂r

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SuperpositionSuperpositionSuperpositionSuperposition

-

Force on charge is vector sum of forces from all charges.

1F

14F 12F2q

+

+

-

14

13F1q

3q

4q

1413121 FFFFvvvv

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� Always attractive

CoulombCoulombCoulombCoulomb vs. vs. vs. vs. NewtonNewtonNewtonNewton

Attractive or repulsive

12212

21

012 ˆ

4

1r

rF

QQ

πε= 122

12

2112 r̂

rF

mmG−=

� Always attractive� 1/r2

� very weak � important on very large

scales, planets, the Universe

� Attractive or repulsive

� 1/r2

� very strong

� only relevant relatively local scales

2

0

2

4Gm

e −>>πε

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Gravitational and Electric Forces in the Hydrogen Atom

+e -eM

m

r12m = 9.1 10-31 kgM = 1.7 10-27 kgr12 = 5.3 10

-11 m

Gravitational force Electric ForceGravitational force Electric Force

Fg = 3.6 10-47 N

rF G

Mm

rrg =

122

\$rF

Qq

rre =

14 0 12

2πε\$

Fe = 3.6 10-8N

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Consider three point charges located at the corners of a right triangle as shown in Figure, where q1=q3=5µC, q2=-2µC and a=0.1m. Find the resultant force exerted on q3 .

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Na

qqkF 11

)2( 2

3131 ==

The force F31 is repulsive and makes an

Consider three point charges located at the corners of a right triangle as shown in Figure, where q1=q3=5µC, q2=-2µC and a=0.1m. Find the resultant force exerted on q3 .

31F

32F

The force F31 is repulsive and makes an angle of 45° with the x axis. Therefore,

9.7

45cos3131

== FF x

9.7

45sin3131

=

= FF y

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Net force will be

Na

qqkF x 9

)( 232

32 == Along –ve x-axis

032 =yF

ˆˆ +−=r

ji

ji

jFiFFF yxx

ˆ9.7ˆ1.1

ˆ9.7ˆ)99.7(

ˆˆ)( 3132313

+−=

+−=

+−=r

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Two small beads having positive charges 3q and q are fixed atthe opposite ends of a horizontal, insulating rod, extendingfrom the origin to the point x =d. As shown in Figure, a thirdsmall charged bead is free to slide on the rod. At what positionsmall charged bead is free to slide on the rod. At what positionis the third bead in equilibrium? Can it be in stable equilibrium?

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333 qqkqqk=

At equilibrium, net force on q3 is zero,

3231 FF =

Two small beads having positive charges 3q and q are fixed at the opposite ends of a horizontal, insulating rod, extending from the origin to the point x =d. As shown in Figure, a third small charged bead is free to slide on the rod. At what position is the third bead in equilibrium? Can it be in stable equilibrium?

1q 2q3q0=x

23

23

)()(

3

xd

qqk

x

qqk

−=

dx 7.0= dx 28.2=

1 230=x

22)(3 xxd =−

0362 22 =+− ddxx

System can be in stable equilibrium ifq3 is [email protected]

x

Three charged particles lie on a straight line and are separatedby a distance d as shown. Charges q1 and q2 are held fixedwhile charge q3 is free to move. If the charge q3 is found to bein equilibrium under the action of electric forces, find q1 inin equilibrium under the action of electric forces, find q1 interms of q2.

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qkqqkq

At equilibrium, net force on q3 is zero,

3231 FFrr

−=

Three charged particles lie on a straight line and are separated by a distance d as shown. Charges q1 and q2 are held fixed while charge q3 is free to move. If the charge q3 is found to be in equilibrium under the action of electric forces, find q1 in terms of q2.

1q 2q 3q

232

231

)()2( d

qkq

d

qkq −=1q 2q 3q

1421 qq −=

21 4qq −=

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Two identical small charged spheres, each having a charge q andmass of 0.03 kg, hang in equilibrium as shown in Figure. Thelength of each string is 0.15 m, and the angle θ is 5°. Find themagnitude of the charge on each sphere.magnitude of the charge on each sphere.

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Repulsive electric force between charges is

Two identical small charged spheres, each having a charge q and mass of 0.03 kg, hang in equilibrium as shown in Figure. The length of each string is 0.15 m, and the angle θ is 5°. Find the magnitude of the charge on each sphere.

22 )2/( akqFe =

013.0)5sin()15.0(

sin

=== θLa

(I)

013.0)5sin()15.0( ==

eFT =θsin

mgT =θcos

System is in equilibrium, equating forces acting on one charge

025.0)5tan()8.9)(03.0(tan === θmgFe

(II)

(III)

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From (I)

K

Faq e

24=

2

T T

F

C

q

8

9

2

104.4

109

)026.0()013.0(4

−×=×

=FE FE

FG=mg FG=mg

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A certain charge Q is to be divided into two parts, Q-q and q. What is the relation of Q to q if the two parts, placed a given distance apart, are to have a maximum coulomb’s repulsion?distance apart, are to have a maximum coulomb’s repulsion?

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The repulsive force between two charges Q-q and q , if they are distance r apart, will be

A certain charge Q is to be divided into two parts, Q-q and q. What is the relation of Q to q if the two parts, placed a given distance apart, are to have a maximum coulomb’s repulsion?

2

))((

r

qqQkF

−=We want to maximize this force with respect to variation in q. it can We want to maximize this force with respect to variation in q. it can be done by setting

0=dq

dF

0)2(

2=−

r

qQk

Three point charges q, -2q, and q are located along the x axisand a fourth charge Q is placed on y-axis, as shown in Figure. Ify>>a, show that the net electric force experience by Q isy>>a, show that the net electric force experience by Q is

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Force due to q1

Three point charges q, -2q, and q are located along the x axis and a fourth charge Q is placed on y-axis, as shown in Figure. If y>>a, show that the net electric force experience by Q is

JkQq

ikQq

F ˆsinˆcos221 θθ +=

r

Qjy

kQqaF ˆ3

4

2

−=

Jr

ir

F ˆsinˆcos221 θθ +=

1q 2q 3qForce due to q2

Jy

kQqF ˆ2

22 −=r

Force due to q3

Jr

kQqi

r

kQqF ˆsinˆcos

223 θθ +−=r

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Net force on Q will be

jFiFF yxˆˆ +=

r

where 0321 =++= xxxx FFFF

321

sin2

sinkQqkQqkQq

FFFF yyyy

+−=

++=

θθ

r r

θ θ

22

222

2sin

2

sinsin

y

kQq

r

kQq

ryr

−=

+−=

θ

θθ

From fig.

22 yar += 22sin

ya

y

r

y

+==θ

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22/322

22222

2

)(

2

22

y

kQq

ya

kQqy

y

kQq

ya

y

ya

kQqFy

−+

=

−++

=

ayif >>

[ ]22/3223 )/1(2 −−− −+= yyayykQqFy

ayif >>

[ ]2222 )2/31(2 −− −−= yyaykQqFy

4

23

y

kQqaFy −=

jy

kQqajFiFF yx

ˆ3ˆˆ4

2