Study the properties and laws of electric field, magnetic field and electromagnetic field that they...
-
date post
22-Dec-2015 -
Category
Documents
-
view
220 -
download
0
Transcript of Study the properties and laws of electric field, magnetic field and electromagnetic field that they...
Study the properties and laws of electric field, magnetic field and electromagnetic field that they are stimulated by charges and currents.
Volume 2Volume 2
Electromagnetism Electromagnetism
Chapter 8
Electrostatic Field in Vacuum
stimulated by static charges with respect to observer.
Inertial reference frame
§8-1 Coulomb’s Law 库仑定律
§8-2 The Electric Field 电场 电场强度
§8-4 Gauss’ Law 高斯定理
§8-5 Electric Potential 电势
§8-3 Electric Field Line and Flux 电力线 电通量
§8-6 Equipotential Surface and Potential Gradient 等势面 电势梯度§8-7 The Electric Force Exerted on a Moving Particle 运动带电粒子所受电场力
§8-1 Coulomb’s Law
1. Two kinds of electric charges
positive charge
Like charges repel each other
Negative charge
Unlike charges attract each other
2. quantization of charge
Electron is the smallest negative charge in nature.
experiments show :
e=1.60217733×10-19 C(Coulomb)
q= Ne
Proton is the smallest positive charge in nature.
The magnitude of electric charge possessed by a body is not continuous.
integer
Electrification by rubbing
3. Conservation of charge
Positive charges and negative
charges have same magnitude.
A B A B
A B
B
A
Electrification by induction :
inducible charges have same magnitude.
The conservation of charge : in any interaction the net algebraic amount of electric charge remains constant.
1q
2q21r
21F
12F
21321
2121 r
r
qqkF
21221
2121 r̂
r
qqkF
2112 FF
or
Coulomb’s Law :
4. Coulomb’s Law
Point charge :The size of charged bodies
<< their distance
In SI :
04
1
k
0=8.8510-12 C2/Nm2
----permittivity of vacuum (真空介电系数)
f1
f2
fn
f
nffff
21
5. Superposition principle of electrostatic forces
Assume there are many point
charges in space:q0 、 q1 、q2 、 q3 … qn ,the resultant force
acting on q0 :
---vector addition
q1
q2
qn
q0
§8-2 The Electric Field
Viewpoint of action-at-a distance:
Viewpoint of field:
1.Viewpoints of the interaction about electric charges
charge charge
charge field charge
Field is a kind of matter.
The behavior of electric field as a kind of matter:
force : E-field exerts a force on the charges in
it.
work : E-field does work on charges during
the charges move in it.
induction and polarization : in the field,
conductor and dielectric produce induction
and polarization.
2. Electric field Test charge:
small size--point charge small charge magnitude—no influence for ori
ginal field.
Test results:
D0qC
0q
A0q
B0q
same q0 is put on different points in space,
0q
P 02q
P
F
203q
P
F
3 Put different test charges on same point,
the electric forces that the test charges suffer change.
the direction and the magnitude of the force that q0
suffers is different at different points---E-field is different at different points
the ratio
000 3
3
2
2
q
F
q
F
q
F
=constant vector at same point.
the electric field is defined :
0q
FE
SI unit :牛顿 / 库仑 (N/C) or 伏特 / 米 (V/m)
nFFFF
21
3. the superposition principle of electric field
There are q1、 q2 、 q3 … qn in space,
q0 is put on the point P, the force acting on q0 :
00
2
0
1
0 q
F
q
F
q
F
q
FE n
At point P, the E-field is set up by q1、 q2 、 q3 … qn :
n
iin EEEEE
121
P
r
q
.point charge:
Put qoon point P , using Coulomb Law , qo suffe
rs
r
r
r
qqF
20
04
1
r
r
r
q
q
FE
200 4
1
--the E-field of
a point charge
4. The distributions of electric field about several different charged bodies
The distribution is spherical symmetry
1
12
1
1
01 4
1
r
r
r
qE
There areq1 , q2 ,… , qn in space
Each charge set up its field at point P :
.the point charge system
n
n
n
nn r
r
r
qE
2
04
1
q1
q2
q3
r1
r3
r2
P
the total field at P :
nEEEE
21
n
i i
i
i
i
r
r
r
q
12
04
.A continuously distributed charged body
rr
rdqEd
2
041
r
r
r
dqEdE
q
204
1
At point P, element ch
arge dq produces :
The total field at P produced by entire charged
body:
Pdq
r
q
kdEjdEidEE zyx
According to the distribution of charge, dq is written as follow:
for Cartesian coordinate system :
dv
ds
dl
dq
line distribution
area distribution
volume distribution
xr
dqdEx 3
04
1
y
r
dqdE y 3
04
1
5. Examples of calculating E-field
Steps : divide charged body into many small charge e
lements . write out produced by dq at point PEd
jdEidEEd yx
rr
rdqEd
2
041
?dq
set up a coordinate system, write components of , such as Ed
total E-field
xx dEE yy dEE
jEiEE yx
calculate the components of , such as E
[Example] Calculate the E-field at point P produced by a charged line.
P
d
l
q
1
2
L 、 q 、 d 、 1 、 2 are known
Solution . divide q dq
. Any dq produces at PEd
204
1
r
dqdE
P
d
l
q
1
2dqr
EdThe magnitude of : Ed
The direction of shows in Fig.
Ed
x
y
o
. set up Cartesian coordinate, jdEidEEd yx
ydE
xdE
. calculate Ex 、 Ey cosdEdE x sindEdE y
P
d
l
q
1
2dqr
Ed
cosdEdEE xx
cos
4cos
4
12
02
0 r
dx
r
dq
)( dxdq
sin
dr In the figure : dctgx
dd
dx2sin
x
y
o
ydE
xdEP
d
l
q
1
2dqr
Ed
xE 2
1
cos4 0
dd
)sin(sin4 12
0
d
Same as yE 2
1
sin4 0
dd
)cos(cos4 21
0
d
jEiEE yx
. the total E
( 1 ) If P locates on the mid-perpendicular plane of the line, i.e.
021 180
21 sinsin 21 coscos
0xE
22
0 24
l
dd
lE y
Discussion
( 2 ) If P is very close to the line ld --the length of the charged line tends to infinity
021 1800 、
0xE
dEE y
02
E-field distribution of the infinite line with uniform charge
( 3 ) If P is far away from the line ld 0
21 90
0xE
44
22
0
ldd
qEE y
2
04 d
q
The charged line can be regarded as a point charge.
Question : If P locates on the elongating line of the charged line shown as in figure,
2
04
1
r
dqEE x
How do we calculate Ex 、 Ey ?
laa
qE
1
4 0
l aP
q
la
a r
dr2
04
1
Caution !(1) If the charged body is not a point charge, we
can not use the formula
(2) If the directions of for different are not same, we can not integrate directly.
dqEd
Ed
r
r
r
qE
2
04
1
directly,
use only for point charge.
integrating its components
[Example] Find the E-field of an uniform charged ring on its axis. ( q 、 R 、 x are known)
x x
r
P
q R·
Solution Divide q dq
204
1
r
dqdE
direction Ed
EdidEEd
//
//dE
dE
0 Ed
x x
r
P
q R·
dq
cos// dEdEE
r
x
r
dq
204
1
dqr
x3
04
23
220 )(4 Rx
qx
Direction: along x axis
Discussion Discussion ( 1 ) at x=0 , E=0.
When x , 01
2
xE
E has extreme values on x axis.
let 0dx
dEWe get Rx
2
2
( 2 ) when x>>R , 323
22 xRx
204 x
qE
can be regarded as a point charge.
E
x0
R2
2
R2
2
R
P xx0
d
rEd
[Example] thin
round plate with
uniform charge
area density ,
radius R. find its
field on the axis.
21220
12 xR
xE
( 1 ) when x « R ,
discussiondiscussion
02122
xR
x
02
EThe E-field set up by uniform sheet charge of a infinite plane.
( 2 ) when x » R ,
2
1
2
2
21221
x
R
xR
x
2
2
2
11
x
R
20
2
2
0 42
1
2 x
q
x
RE
as a point charge
AB
AE
BE
§8-3 Electric field line and flux
1. E-field line ( line )E
E
dS
dS --area element perpendicular to line.E
de – the number of line crossing .dSE
dS
dE e
the tangential direction of line at any point gives the direction of at that point.
E
E
the density of line gives the magnitude of
E
E
line originate on positive charges and terminate on negative charges (or go on infinity). They never originate or terminate on a no-charge point in finite space.
E
2. Electric flux eE or
The properties of line.E
Two lines never intersect at a point.E
– the number of line crossing any area.E
ESe
cosESESe
SE
S
S
S
n
The plane S is at right angle to the uniform E-field.
The plane S is at any angle with E
:the unit vector at the normal direction of the plane.
n
nSS
n
E
dS
S
EdSd e SdE
S ee d
The total E-flux crossing S :
Take any dS on S :
An arbitrary surface S is placed in a no-uniform E-field.
S
SdE
Ede < 0
Se sdE
If S is a closed surface :
de >0
Stipulation : the direction of is outward.n
n
0 Se sdE
0 Se sdE
? e relate to the charge in the surface.
If there is no any charge in the closed surface,
the number of line entering it equals the number of line going out it.
E
E
If there are charges in the closed surface,
the number of line entering it does not equal the number of line going out it.
E
E
1. The E-flux crossing a sphere surface with a point charge q in its center.
q
S
sdESe
22
0
44
rr
q
dsES
SdsE
0q
§8-4 Gauss’s Law of electrostatic field
q
S
Discussion Discussion
e does not relate with r .
S’ If S’is an arbitrary
closed surface surro
unding q,
then
sdES
e
0q
sdES
2. If the charge is outside the closed surface S,
+ qS
e = 0
3. e crossing any closed surface S with point charge system.
Inside S :
iqq 1 1iq
Outside S:
nii qq 1
Then e
iq
S
S
sdE
000
1
iqq
内S
ie q0
1
q
4. e crossing any closed surface S with any charged body
S
qe dq
0
1
Inside S
Summary above results Gauss’LawIn any electrostatic field, the electric flux crossing any closed surface equals the algebraic sum of the charges enclosed by the surface divided by 0
内S
iSe qSdE0
1
The closed surface S –Gaussian surface
or q
S
e dqdSE0
1
S 内
Notes:
① is the algebraic sum of the charges enclosed by the Gaussian surface
内S
iq
i.e., the e crossing S depends on the charges enclosed by S only, and has nothing to do with the charges outside S.
the total at any point of S is concerned with all the charges (inside and outside S).
E
0 S
e dSE
indicates that the electrostatic field is a 有源场
If the distribution of charges and its E-field has some symmetry, Gauss’s Law can be used to calculate the E-field.
Common steps: Analyze the symmetry of charges and its E-field. Choose a suitable enclosed surface as Gaussian s
urface S. Calculate e crossing S.
Calculate the algebraic sum of charges inside S. Use Gauss’s Law to calculate E 。
4. Applications of Gauss’s Law
[Example 1] Calculate the E-field distribution of a infinite line with uniform charges. (Assume the linear density of charges is )
Solution : Analyze the symmetry
--axial symmetry
λ
E
Choose a suitable Gaussian surface
--cylinder surface
hr
P
PE1S
2S
3S
Se SdE
Calculate e crossing S :
321 SSS
SdESdESdE
rhE 2 Calculate the algebraic sum of charges inside S
hqi Use Gauss’s Law :
ii
s
e qSdE0
1
0
2 h
rhE
rE
02
E-field distribution of the infinite line with uniform charge
Similar problem: the charge distribution on a infinite cylinder surface with radius R. The charges per meter of length of the cylinder isλ.
axial symmetry
Gaussian surface – cylinder surface
( 1 ) r R>
S
SdE
SdE
下 SdE
上 SdE
侧
i
iq : l
rlE 2
use i
i
s
e qSdE0
1
rlE 20
l
G-surface
rE
02
e
0 E
( 2 )r R
0E
( r > R )
Distribution of E-field of the cylinder charges
r02
)( Rr<
[Example 2] Calculate the E-field distribution of a infinite plane with uniform charges. (Assume the area density of charges is σ)
Solution:Analyze the distribution character of E-field
P
E-field :area symmetry
σ
SE
E
σ
E
.P
Make a cylinder surface through point P as Gaussian surface.
G-surface
02
E
S
SdE
SdE
左 SdE
右 SdE
侧0 SESE SE2
Use i
i
s
E qSdE0
1
SE20
S
Sqi
i :
e
[Example 3] Calculate the E-field distribution of a sphere surface with uniform charge q.
(1) r > R
spherical symmetry
Make a sphere as Gaussian surface through P.
R
+++++ +
+++
+ ++++
+ +
qrP
G-surface
e
24 rE
S
SdE
S
EdS 00cos
qqi
i
204
1
r
qE
use i
i
s
e qSdE0
1
0
24
qrE
(2) r < R
G-surface
E-field—spherical symmetry
Gaussian surface--sphere
P.r
E
+ +
++++ +
+
+
+ ++++
+ +
R
q
S
e SdE
S
EdS 00cos
24 rE 0
iiq
0 E
use i
i
s
e qSdE0
1
04 2 rE
204
1
r
q
The distribution of E-field for charged spherical surface:
E0 ( r>R )
( ) r < R
12r
r
E
0 R
R
q( 1 ) <r R
r
G-sueface
Similar question : uniform charged spherical body
E
S
SdE
24 rE e
3
3
4rqi
33
rR
q3
3 3
4
3
4r
R
q
24 rE 33
0
rR
q
r
R
qE
304
1
( 2 )r R>
G-surface
R
r
E
.P
S
SdE
24 rE 0
q
204
1
r
qE
The distribution of E-field with charged spherical body:
rR
q3
04
1
E
( r > R )204
1
r
q
)( Rr <
rR
E
o
204
q
R