elect magnet 2
TRANSCRIPT
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2 2
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1.1.
1.1. 1.1.
1.2. . 1.2. .
1.3. 1.3.
1.4. 1.4. 1.5. 1.5.
1.6. 1.6.
1.7. 1.7.
1.8. 1.8.
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1.5. 1.5.
? ?
DDqq
PP : :
!!
2
q
E k r r
D
D
ii2
i i
q E k r
r
D
i
2q 0i i
q E k lim r
rD
D
2
dqk r
r
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, , [[//33
]]
, , [[//22]]
, , [[//]]
EE EExx, E, Eyy, E, Ezz (3(3 ))
Q
V Q
S
Q
l
dQ,
dV
2
dV E k ,
r
dQ,
dS
dQ
dl
2
dS E k ,
r
2
dlE k
r
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20.7 20.7 The electric field dueThe electric field due
to a charged rod:to a charged rod: AArod of lengthrod of length llhas ahas auniform positive chargeuniform positive chargeper unit lengthper unit length and aand atotal chargetotal charge QQ..
Calculate the electricCalculate the electricfield at a pointfield at a point PPalongalongthe axis of the rod, athe axis of the rod, adistancedistance ddfrom one endfrom one end(Fig.).(Fig.).
..ll
QQ .. d
P .x
y
Pd l
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Solution:Solution: For thisFor this
calculation, the rod iscalculation, the rod is
taken to be along thetaken to be along the xxaxis.axis.
:: xx
..
The ratio ofDq, thecharge on the segment toDx, the length of the
segment, is equal to theratio of the total charge tothe total length of therod.
Dq Dx
.
q Q
x l
D
D
x
y
P
d l
xDx Dq=Dx
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d>>ld>>l ll--
DDq=q=DDxx
PP x
y
Pd l
xDx Dq=Dx
2 2
q x E k k
x x
D DD
DE
l d
2
d
dxE k
x
2 2E k l / d kQ / d
l d
2d
dxE k x
l d
d
1 1 1k k
x d l d
lk
d( l d )
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20.8 20.8 The electric field of a uniform ring of chargeThe electric field of a uniform ring of charge. A ring. A ring
of radiusof radius aa has a uniform positive charge per unithas a uniform positive charge per unitlength, with a total chargelength, with a total charge QQ. Calculate the electric. Calculate the electricfield along the axis of the ring at a pointfield along the axis of the ring at a point PP lying alying adistancedistance xx from the center of the ring (Fig.).from the center of the ring (Fig.).
..
aa
QQ..
xx PP . .
aP
x+
+
+
+
+
+
+
+
+
+
+
+
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DDqq
: :
xx
. . : :
(x>>a)(x>>a)
(x=0)(x=0)
Dq
a P
x+
+
+
+
+
+
+
+
+
+
+
+
DE
2
q
E k r
D
D
x E E cosD D ED ED
r
DEx
ED
Dq
122 2r ( x a ) , cos x / r
x E E cosD D 2q x
kr r
D
3
22 2
kxq
( x a )D
32
x 2 2
kxE q
( x a )D
3
22 2
kxQ
( x a )
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20.9 20.9 . . RR
. . xx . .
rr drdr 22pprdrrdr
dq=2dq=2pprdrrdr
20.8 20.8-- ((aa rr))
r=0r=0-- r=Rr=R
x 32 2 2
kxdE ( 2 rdr )
( x r )p
Q=pR2 P x 0,0, RR :: x 0 E 2 k / 2p
R
xPr
dr
dq
32
R
x 2 20
2rdr E kx
( x r )p
1
22 2
x... 2 k 1
( x R )p
32
R2 2 2
0kx ( x r ) d( r )p
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1.6. 1.6.
1. 1. EE
2. 2. N~EN~E
, ,
. .
+ -
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, ,
, ,
2 2
N 1 E
4 r rp
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, ,
((..))--
+ 2q q
!
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1.7. 1.7.
EE qq qqEE
IIII FF=q=qEE=m=maa
(( ))
qEam
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(x,y)
tt
t=x/vt=x/v00 y~xy~x22
YY EE
vv00ii
vvx0x0=v=v00, v, vy0y0=0=0
eEa j
m
x 0 y
eEv v const, v at t
m
2 2
0
1 1 eE
x v t , y at t 2 2 m
+ + + + + +
_ _ _ _ _ _
(0,0)
v
v0i
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5,35,31010--1111
FFee/F/Fgg3310103939..
2 19 29 8
e 2 11 2
e ( 1,6 10 ) F k 9 10 8,2 10
r ( 5,3 10 )
31 27 e p 11 47
g 2 11 2
m m ( 9,11 10 )( 1,67 10 ) F G ( 6 ,7 10 ) 3,6 10
r ( 5,3 10 )
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20.10 20.10
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1.8. 1.8.
(())
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. .
, ,
, ,
CC
AA
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