Coulombs Law
35
ma F = Coulomb's Law ∂ = + ∂ - ψ ψ ψ 2 2 hf E = Muhammad Nadeem School of Electrical Engineering &Computer Sciences [email protected] 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
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Transcript of Coulombs Law
maF =
Coulomb's Law
∂=+∂− ψψψ22hfE =
Muhammad Nadeem
School of Electrical Engineering &Computer Sciences
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
oε
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.
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.
SILK
The Transfer of Charge
Electric Charges
Glass Rod
Some materials attract electronsmore than others.
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.
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.
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.
� 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.
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”
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
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.
� 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
r̂
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
SuperpositionSuperpositionSuperpositionSuperposition
-
Force on charge is vector sum of forces from all charges.
1F
14F 12F2q
+
+
-
14
13F1q
3q
4q
1413121 FFFFvvvv
� Always attractive
CoulombCoulombCoulombCoulomb vs. vs. vs. vs. NewtonNewtonNewtonNewton
Attractive or repulsive
12212
21
012 ˆ
4
1r
rF
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
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
$rF
rre =
14 0 12
2πε$
Fe = 3.6 10-8N
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 .
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
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
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?
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.
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 −=
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.
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)
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
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?
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
2/Qq [email protected]
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
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
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
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
