Transcript
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Unit 15 Electrostatics and Gausss Law
15.1 Electric charge
15.2 Coulombs law
15.3 Shell theorems for electrostatics
15.4 Electric field
15.5 Electric field lines
15.6 Shielding and charging by induction
15.7 Electric flux
15.8 Gausss law
15.1 Electric charge
There are two kinds of charges, namely, positive (+) charge and negative () charge.
Like charges repel
Unlike charges attract
Objects with zero net change are said to be electrically neutral. Electric charges are generated after rubbing between materials.
Example
After rubbing plastic rod (or amber rod) with fur, the plastic rod (or
amber rod) becomes negatively charged and the fur is positively
charged.
+ +
+
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After rubbing glass rod with silk, the glass rod becomes positively charged and the silk is
negatively charged.
A familiar example of an electrically neutral object is the atom.
Atoms have a small, dense nucleus with a positive charge surrounded
by a cloud of negatively charged electrons. All electrons have exactly
the same electric charge. This charge is very small, and is defined to
have a magnitude, e = 1.60 1019 C. S.I. unit of charge is coulomb,
C. Clearly, the charge on an electron, which is negative, is e. This is
one of the defining, or intrinsic, properties of the electron. Another
intrinsic property of the electron is its mass, me:
me= 9.11 1031 kg
In contrast, the charge on a proton one of the main constituents of nuclei is exactly +e. As
a result, since atoms have equal numbers of electrons and protons, their net charge is
precisely zero. The mass of the proton is
mp= 1.673 1027 kg.
Note that this is about 2000 times larger than the mass of the electron. The other main
constituent of the nucleus is the neutron, which, as its name implies, has zero charge. Its mass
is slightly larger than that of the proton:
mn= 1.675 1027
kg.
Example
How is it that rubbing a piece of amber with fur gives the amber a
charge?
Answer:
Rubbing the fur across the amber simply results in a transfer of
charge from the fur to the amber with the total amount of chargeremaining unchanged. Before charging, the fur and the amber are
both neutral. During the rubbing process some electrons are
transferred from the fur to the amber, giving the amber a net
negative charge, and leaving the fur with a net positive charge. At
no time during this process is charge ever created or destroyed.
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This, in fact, is an example of one of the fundamental conservation laws of physics:
Conservation of electric charge.
When charge is transferred from one object to another it is generally due to the movement of
electrons. In a typical solid, the nuclei of the atoms are fixed in position. The outer electrons
of these atoms, however, are often weakly bound and fairly easily separated. The atom that
loses an electron is a positive ion, and the atom that receives an extra electron becomes a
negative ions. This is charging by separation.
Example
Find the amount of positive electric charge in one mole of helium atoms.
Answer:
Note that the nucleus of a helium atom consists of two protons and two neutrons. The totalpositive charge in a mole is
CCeNA51923 1093.1)1060.1)(2)(1002.6()2( == .
15.1.1 PolarizationWe know that charges of opposite sign attract, but it is
also possible for a charged rod to attract small objects
that have zero net charge. The mechanism responsible
for this attraction is called polarization. When a charged
rod is far from a neutral object the atoms in the object
are undistorted. As the rod is brought closer, however,
the atoms distort, producing an excess of one type of
charge on the surface of the object (in this case a negative charge). This induced charge is
referred to as a polarization charge. Since the sign of the polarization charge is the opposite
of the sign of the charge on the rod, there is an attractive force between the rod and the
object.
15.1.2 Conductor and insulator
Conductors: materials that allow electric charges to move more or less freely, e.g. metals
Insulators: materials in which charges are not free to move, e.g. nonmetallic substances, say,
amber.
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On a microscopic level, the difference between conductors and
insulators is that the atoms in conductors allow one or more of their
outermost electrons to become detached. These detached electrons,
often referred to as conduction electrons, can move freely
throughout the conductor. The right figures show the charging of a
conductor by touching it with charged rod.
15.2 Coulombs law
Electric force Coulombs law
2
21
r
qqkFe = , where k(electrostatic constant > 0) is a constant.
The electrostatic constant 229
0
/1099.8
4
1CmNk ==
, where 0 is called permittivity
constant of free space, and 012 2 2885 10= . /C N m .
Gravitational force Newtons gravitational law
2
21
r
mmGFg = , where
2211/1067.6 kgmNG = .
The negative sign is inserted to represent an attractive force.
Remarks:
1. Fundamental laws cannot be derived!
etc.
law,Newtons
law,Coulombs
#
are concluded according to results in experiments and have survived in
every experimental test.
2. Objects are considered as point particle or point charge, if rdd
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4. The magnitude of the force of interaction between two point changes is directly
proportional to the product of the charges and inversely proportional to the square of
the distance between them.
F F1 2= Obey Newtons third law
5. Unit: International System of Units or Metric System (SI)Charge q: measured in Coulomb orC.
Remark:
Electrostatic constant is related to the speed of light c:
00
1
=c ,
where smc /10998.28= , and 0 = 4 10
-7N/A
2, the permeability of free space.
Example
Compare the electric and gravitational forces between a proton and an electron in a hydrogen
atom.
Answer:
Taking the distance between the two particles to be the radius of hydrogen, mr 111029.5= ,
we find that the electric force has a magnitude
Nm
CCCmN
r
qqkF
pe
e
8
211
1919229
21022.8
)1029.5(
)1060.1)(1060.1()/1099.8(
=
== .
Similarly, the magnitude of the gravitational force between the electron and the proton is
N
m
kgkgkgmN
r
mmGF
pe
g
47
211
27312211
21063.3
)1029.5(
)10673.1)(1011.9()/1067.6(
=
== .
Hence, we obtain the ratio of the two forces
000,000,000,000,000,000,000,000,000,000,000,000,260,21026.21063.3
1022.8 3947
8
==
=
N
N
F
F
g
e
+ +1q 2q
1FK
2FK
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Example
We study the classical model for Hydrogen atom. The electron undergoes a circular motion
with a radius a0 called Bohr radius. Find the speed of electron.
The constants are given as below.
a m A011
529 10 0529= =. .D
me= 9.11 1031
kg
Answer:
The force acting on the electron is obtained in the last example, where ege FFFF += .
But,0
2
a
vmF e= (the centripetal force). Hence, v
Fa
me
2 0=
vFa
mm s
e
= =
=
0
8 11
31
68 22 10 529 10
911 10218 10
. .
.. / .
15.2.1 Superposition of Coulombs force
The force exerted on charge 1 by charge 2:
122
12
21
0
12
4
1r
r
qqF
=
G
where12
r : unit vector along12
rG
.
q1 and q2 same sign repulsion
opposite sign attraction
Similarly, The force exerted on charge 2 by charge 1:
212
21
21
0
21
4
1r
r
qqF
=
G
where 21r : unit vector along 21rG
.
The total force acting on charge q due to coulombs forces F1, F2and F3.
F F F F = + +1 2 3 (Principle of superposition)
The direction of forces shown in the right figure, representing that the
charge q is of opposite charge ofq1, q2and q3. q
1q 3q
2q
1FK
2FK
3FK
1rK
2rK
1q
2q
Point charge
2112 rrrKKK
=
O x
y
Proton
electron
0a
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Example
Three charges, each equal to +2.90 C, are placed at three corners of a square 0.500 m on a
side. Find the magnitude and direction of the net force on charge number 3.
Answer:
The magnitude of force exerted on charge 3 by charge 1:
Nm
CCmN
r
qkF 151.0
)500.0(2
)1090.2()/1099.8(
)2(2
26229
2
2
31 =
==
The magnitude of force exerted on charge 3 by charge 2:
Nm
CCmN
r
qkF 302.0
)500.0(
)1090.2()/1099.8(
2
26229
2
2
32 =
==
The components of 31FG
and 32FG
:
NNFF ox 107.0)707.0)(151.0(0.45cos31,31 ===
NNFF oy 107.0)707.0)(151.0(0.45sin31,31 ===
NNFF ox 151.0)1)(302.0(0cos32,32 ===
NNFF oy 0)0)(151.0(0sin32,32 ===
The components of the resultant force:
NNNFFF xxx 409.0302.0107.0,32,31 =+=+=
NNNFFF yyy 107.00107.0,32,31 =+=+=
The resultant force acting on charge 3:
NFFF yx 423.022 =+=
31FG
32FG
FG
1 x
y
2
r2
3
r=0.500 m
r=0.500 m
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The direction of the resultant force on charge 3:
o
x
y
F
F7.14tan
,3
,31 =
= .
15.3 Shell theorems for electrostaticsTheorem 1:
A uniform spherical shell of charge behaves, for external points, as if all its charge were
concentrated at its center.
Fq q
r=
1
4 0
1
2
Theorem 2:
A uniform spherical shell of charge exerts no force on a charged particle placed inside the
shell.
F= 0
Remarks: The theorems are similar to the gravitational case.
15.4 Electric field
Gravitational field (a vector field)
rr
MG
m
rr
mMG
m
Fg e
e
2
2
=
==
GG
Gravitational field: Gravitational force per unit mass
Electric fieldE
G
rr
q
q
FE e
4
12
00 =
GG
Electric field: Electrostatic force per unit charge
SI unit of electric field: Newton/Coulomb orN/C
Total charge qon spherical shell
1q
Earth
m
Test body
A test chargewith positivecharge
0qeFK
Charged
particle q
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Remark:
1. Why do we need to introduce the concept of Electric field?
Introducing the field as an intermediary between the charges, we can represent the interaction
as:
Our problem of determining the interaction between the charges is therefore reduced to two
separate problems: (1) determine, by measuring or calculation, the electric field established
by the first charge at every point in space, and (2) calculate the force that the field exerts on
the second charge placed at a particular point in space.
2. Principle of superposition in electric field:
The resultant electric field Eat a point is given by E E E E= + +1 2 3 , where E1,E2, and E3
are the electric fields experienced at that point due to charge 1, 2 and 3 respectively.
15.4.1 Discrete and continuous charge distribution
a) Discrete case
E E E E= + + + 1 2 3
b) Continuous case
i) When charge is uniformly distributed along a line.
Linear charge density
q
lcharge per unit length ( =
dq
dl)
ii) Charge on a surface (uniformly distributed)
Surface charge density
q
S charge per unit area ( =
dq
dS)
iii) On a volume
Volume charge density
q
Vcharge per unit volume =
dq
dV
l
q Total
charge chargefield
Surface area S
Total charge q
Volume V
Total charge q
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Example (Challenging)
Find the electric field at a point P, which is at the top of the center of a charged ring. The
total charge on the ring is q.
Answer:
Due to the symmetry of the ring, the electric field along z direction can be calculated as
follows.
= cos'EdE , where cos =z
rand
220
20
'
4
1
4
1
Rz
ds
r
dqdE
+==
.
The linear charge density of the ring
= =q
R
dq
ds2, whereR is the radius of ring.
N
Ez
z Rds
R
=+
1
4 02 2 3 2
2
( ) /
=+
1
4
2
0
2 2 3 2
z R
z R
( )
( ) /
Plug in the expression, 2 R q= , hence, the electric field at any point P, a perpendicular
distancezfrom the plane and center of ring,2/322
0 )(4 Rz
zqE
+=
.
Remark:
Whenz >> R, that is the distance is much larger than the dimension of the ring,
1 1
0
12 2 3 2 2 3 2 3( ) ( )/ /z R z z+
+
=
Eq
z
4 02
This is as if the case where 0R , the ring seems to be a point charge!
'EdG
P
EG
'EdG
Total charge qds
z
'EdG
P
r
R
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Example (Challenging)
Find the electric field at a point P, which is at the top of the center of a uniformly charged
disk. The total charge on the disk is q.
Answer:
The surface charge density
=q
R 2
The differential area of the ring
rdrdA 2=
)2( rdrdAdq ==
The electric field experienced at a perpendicular distance
z from the center of disk due to the differential ring.
Remember that we have an expression for the ring in the
previous section.
2/322
0
2/322
0 )(
2
4)(4 rz
drrz
rz
zdqdE
+=
+=
+==R
rz
drrzdEE
0 2/322
0 )(
2
4
Now, we let 2ry = and we have rdrdy 2= . The integral +R
rz
drr
0 2/322 )(
2becomes
+2
0 2/32 )(
R
yz
dy, which gives
2
0
2
1
2
0
))(2(4
R
yzz
E
+=
. Hence, we have
)1(2 220 Rz
zE
+=
.
Remark:
WhenR >>z(infinite sheet of charge), we have E=
2 0.
Whenz>>R, we have zz R R
z
Rz2 2 2
2
2
21
1
12+
=
+
That is ER
z=
2
0
24. By using
=
q
R 2, we have
2
04 z
qE
= (Result of point charge!)
dA
Rq
EG
dr
r
r
Disk of Radius R
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15.5 Electric field lines
Rules for drawing electric field lines
Electric field lines:
(i) Point in the direction of electric field vectorEat every point;
(ii) Start at positive (+) charges or at infinity;(iii) End at negative () charges or at infinity;(iv) Are more dense where Ehas a greater magnitude. In particular, the number of lines
entering or leaving a charge is proportional to the magnitude of the charge.
ExampleWhich of the following statements is correct: Electric field lines (a) can or (b) cannot
intersect?
Answer:
By definition, electric field lines are always tangent to the electric field. Since the electric
force, and hence the electric field, can point in only one direction at any given location, it
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follows that field lines cannot intersect. If they did, the field at the intersection point would
have two conflicting directions.
15.6 Shielding and charging by induction
In a perfect conductor there are enormous numbers f electrons
completely free to move about. This simple fact has some rather
interesting consequences. Consider, for example, a solid metal sphere
attached to an insulating base as shown in figure. Suppose a positive
charge Q is placed on the sphere. The question is: How does this charge
distribute itself on the sphere when it is in equilibrium? In particular,
does the charge spread itself uniformly throughout the volume of the
sphere, or does it concentrate on the surface?
The answer is that the charge concentrates on the surface. Why should
this be the case? First, assume the opposite that the charge is spread
uniformly throughout the spheres volume. If this were the case, a charge at location A would
experience an outward force due to the spherical distribution of charge between it and the
center of the sphere. Since charges are free to move, the charge at A would respond to this
force by moving toward the surface. Clearly, then, a uniform distribution of charge within the
spheres volume is not in equilibrium. In fact, the argument that a charge at pointA will move
toward the surface can be applied to any charge within the sphere. The preceding result holds
no matter what the shape of the conductor. In general,
excess charge placed on a conductor, whether positive or negative, moves to the exterior
surface of the conductor.
15.6.1 ShieldingWhen electric charges are in equilibrium, the electric field within a conductor is zero; E= 0.
A straightforward extension of this idea explains the phenomenon of shielding, in which a
conductor shields its interior from external electric fields.
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We also noted that the electric field lines contact conductor surfaces at right angles. If an
electric field contacted a conducting surface at an angle other than 90o, the result would be a
component of force parallel to the surface. This would result in a movement of electrons and
hence would not correspond to equilibrium.
15.6.2 Charging by induction
One way to charge an object is to touch it with a charged rod; but since
electric forces can act at a distance, it is also possible to charge an object
without making direct physical contact. This type of charging is referred
to as charging by induction.
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15.7 Electric flux
The electric flux is defined as , which is the product of the electric field and the area of
surface AE= .
Example
Consider a surface with area A and a uniform electric field penetrating the surface
perpendicularly. The electric flux is given by EA= .
Example
Consider a surface with area A and a uniform electric field penetrating the surface with an
angle with the normal of surface. The electric flux is given by AE= . Or in scalar
form: cos EA .
e
Surface areaA
EG
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Remarks:
1. For a plane surface, the area vectorA is defined as
A = Ae , where e is the normal vector.
2. For a curved surface, we have d dA eA = .When the electric fieldEis not uniform, i.e.,E=E(x,y,z), or if the surface is not a
plane, then =i
iAE or in the integral form AE d .
3. If the surface through which the flux is calculated is closed, the sign of the flux is asfollows:
The flux is positive for field lines that leave the enclosed volume of the surface. The flux is negative for field lines that enter the enclosed volume of the surface.
15.8 Gausss law
In order to understand Gausss law, we first look at the following. Consider a point charge q
and a spherical surface of radius r and centered on the charge. The electric field on the
surface of the sphere has the constant magnitude
2r
qkE= .
Since the electric field is everywhere perpendicular to the
spherical surface, it follows that the electric flux is simply E
times the areaA = 4r2
of the sphere:
( ) kqrr
qkEA 44 2
2=
==
Plug in k= 1/(40), we obtain the Gausss law
0
q= .
Thus we find the very simple result that the electric flux through a sphere that encloses a
charge q is the charge divided by the permittivity of free space, 0. This is a very nice result!!
As the electric field of many symmetrical system can be found readily with Gausss law.
Gausss law states that the flux of the electric field over the Gaussian surface equals to the
net charge enclosed by that surface.
Normal of the surface
e
AreaA
EG
+q
Gaussian surface with area 4r2
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Remarks:
1. Symmetrical situations arise in all fields ofphysics and, when possible, it makes sense
to cast the laws ofphysics in forms that take full advantage of this fact.
2. Gausss law is a new formulation of Coulombs law that can take advantage of
symmetry.
3. The integral form of Gausss law: = qdAE0 , where 0 is the permittivity
constant.
Example
Consider the surface S shown in the figure. Is the electric flux through this surface (a)
negative, (b) positive, or (c) zero?
Answer:
Since the surface S encloses no charge, the net electric flux
through it must be zero, by Gausss law. That a charge +q is
nearby is irrelevant, because it is outside the volume enclosed
by the surface.
We can explain why the flux vanishes in another way. Notice
that the flux on portions ofSnear the charge is negative, since
field lines enter the enclosed volume there. On the other hand,
the flux is positive on the outer portions ofSwhere field lines
exit the volume. The combination of these positive and negative contributions is a net flux of
zero. That is, the answer is (c).
15.8.1 Gaussian surface
Gaussian surface is a closed surface, e.g. sphere, cube, cylinder, etc. And, Gausss law tells
how the fields at the Gaussian surface are related to the charges contained within that surface.
Charges
Gaussian surface
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Example
Calculate the electric field at a external point due to a point charge.
Answer:
We form a spherical Gaussian surface, centered at the charge, to enclose the charge q. The
electric field,E, is uniform on the spherical Gaussian surface.
From Gausss law == qdAE00
As AE d// andEis uniform on the surface, we have
= qdAE0 or qrE =)4( 20
2
04 r
qE
= ,
which gives the Coulombs law.
Remark 1: Gausss law equivalent
Coulombs law.
Remark 2:
Suppose we have a spherical Gaussian surface, and if the electric field vectors are of uniform
magnitude and point radially outwards as shown. One can conclude that a net positive charge
must lie within the surface and that it must have spherical symmetry.
Example
Use Gausss law to investigate the space under a uniform electric field.
?
Charge inside?
Spherical Gaussian surface
E
EG
+q
Gaussian surface with area 4r2
dAG
b
EG
dAG
adA
G
c EG
EG
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Answer:
We form a cylindrical Gaussian surface, which is immersed in a uniform field.
The electric flux:
++==cba
dddd AEAEAEAE , where E A = d EdAcos.
(i) Surface a = 180 EAda
= AE
(ii) Surface b = 90 0=b
dAE
(iii) Surface c = 0 EAdc
= AE
00 =++= EAEA
No charge is enclosed in the Gaussian surface.
Example
Find the electric field at a point very near to the surface of a charged conductor.
Answer:
Suppose charges q are on the right surface, the surface charge density is given by =q
A, or
in the language of calculus, ( )r
=
dq
dA .
Now, we form the Gaussian surface as shown in figure. Near the conductor, the surface is
flat, so theE-field is //to e of the surface.
Apply the Gausss law, we have
qd = AE0
++=cba
dAE
i) For surface a, E= 0 (inside the conductor)
ii) For surface c, eE , hence we haveE A = d 0 .
iii) For surface b, AEAE =b
d .
So,
0
0
EA q Eq
A= = or E=
0.
EGa
c
Gaussian
surfacee
b
Charges q on surface
Conductor
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Example
Find the electric field radiated from an infinite long charged
plastic rod with linear charge density , where q
h( q: total
charge enclosed by Gaussian surface)Answer:
From cylindrical symmetry,Eis along radial direction.
By Gausss law: qd = AE0 .
Surfaces a and b do not contribute to the integral
0 2E rh q h = =
So Er
=
2 0
Example
Find the electric field radiated from an infinite plane sheet of
charge, with surface charge density q
A
Answer:
Gausss law qd = AE0
EAdba
2=+= AE
So qEA =02 00 22
==
A
qE
Example
Find the electric field inside and outside a spherical shell of charge q.
Answer:Due to spherical symmetry, the direction ofEis along radial
direction andEis uniform on the sphere.
For Gaussian surface S2 : qd = AE0
r1
r2
S1
S2
Spherical shell of
Charge q
Spherical Gaussian surfaces
Gaussian surface
+
++
++
+
+
+
+
+
rEG
h
a
c
b
e
e
EG
Gaussian surface at top
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or2
0
2
204
14
r
qEqEr
== , which gives the first shell theorem.
For Gaussian surface S1 : qd = AE0 , where q = 0 .
0042
10 == EEr ,
which gives the second shell theorem.
Example
Find the electric fieldE(r), inside and outside a uniformly
charged insulating sphere, with total charge Q and radiusR.
Answer:
In order to findE
with different r, we apply the Gausss law.
The volume charge density is given by
= =Q
V
Q
R4
3
3
.
The charge enclosed by the Gaussian surface S1 is
Q VQ
R
r Qr
Rencl = = =
'
4
3
4
33
33
3.
From the spherical symmetryEhas the same value on the surface S1.
enclQd = AE0 or 33
2
0 )4(R
QrrE =
That is, rR
QE
304
1
= , forr < R (Er).
At the surface of sphere, r = R, hence, we obtain EQ
R=
1
4 02
.
When r > R, the Gaussian surface S2
encloses all the charges.
Q Qencl =
02
4E r Q =
Hence, we obtain2
04
1
r
QE
= , which is
the result of coulomb s law.
rR
S1S2
EG
Radial direction
Insulating
sphere withcharge inside
E
r
rR
QE
3
04
1
=
2
041
RQE
=
204
1
r
QE
=
R
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