Lectures 3,4 (Ch. 22)Gauss’s law
E
1.Flux of2. Gauss’s law3. Applicationa)Finding of Fluxb)Finding of Chargec)Conductors propertiesd)Finding of E for symmetric charge distributions(sphere, cylinder, plane)
Flux of a uniform E
EAAEEACosAE
EonAorAonEprojecteither
should cross the surface to produce a flux
n
n
n
E
Compare to a flux of water
should cross the surface to produce a fluxv
A
A
A
?EAorEA 0 EAEAtotal
For the closed surface A
is out of the surface
General definition of flux of E
E
AdEd
E AdE
enclosedQnoconstN
NNif
NEAA
NE
outin
0
~~
d
Flux of E through the closed surface containing a point charge, q
2. Arbitrary surface with arbitrary q position
1. Sphere with q placed in the center
0
22
0
44
qr
r
qEAAdE
02
0
0
22
02
0
4
444
q
r
dAqAdEor
qR
R
qdA
R
q
22 4
)(
4
)()()(
R
RqdA
r
rqdArdArEAdE
AdE
Flux of E through the closed surface containing an ensemble of charges
q1 q2
qN
Superposition principle:
N
i
iN
ii
N
ii
N
ii
QqAdEAdEAdE
1 00111
N
iiEE
1
Charges outside the closed surface do not contribute to the flux through this surface
inAd
outAd For each solid angle flux in =-flux out or
densityechvolumeais
dVQ
AdElawsGauss volumeencl
surface
arg
:'00
0 outin NN
Gauss’s Law may be used 1) to find a flux 2) Q enclosed 3) E for some symmetric charge distributions.
Figure 22.15
Find the flux through the closed surfaces: A,B, C, D.
Figure 22.16
Find the flux through the indicated closed surfaces.
Find the flux through each side.
q
Find the charge inside
Q=?
E0=500N/CEL=400N/C
=10m
Conductors properties
1.E=0 within conductor in electrostatics(otherwise F=qE and hence the charges
would be moving, i.e. nonstatic)
2.q =0 in every point inside solid conductor. It follows from the Gauss’s law, indeed, since E=0
one has
q=0 0
0enclq
AdE
Shrinking a Gaussian surface to any point
inside a conductor one gets 0enclq
Conductors properties3. E near the surface
AEALaw
sGaussfromfollowsitEb
movingbewould
electronsotherwiseEEei
surfaceconductoratoEa
0
0
:
')
)
(..
,)
Is the surface charge density, i.e. a charge per unit area
E
Conductors properties
4.In the absence of external charge placed into the cavitya) inside a cavity E=0b) there is no charge on the inner surface of the conductor(“Scooping out theorem” in electrically neutral material)
-
-
++
Faraday’s cageInsert uncharged conductor into E
E=0, qinner=0
Is it safe to be in a carduring the thunderstorm?
☺
Conductors properties
5. In the presence of external charge q inside the cavitya)Charge –q is induced on the inner surface of the cavity. It follows from the Gauss’s law. Since E=0 inside a conductor one gets:
0
0enclq
AdE
Hence qqei
qqq
inner
innerencl
..
0
b) Charge qqc resides on the outer surface of a cavity.
It follows from the charge conservation law: qqqqqqqq cinnercoutercinnerouter
Faraday’s ice-pail experiment
Calculation of E using the Gauss’s lawThree types of the symmetry for a charge distribution
1. Spherical A=4 π r2, V=4 π r3/3
1. Cylindrical Aside=2 π rL, V= π r2L
2. Plane A=L1L2, V=AL3
Application of Gauss’s law• Look for the symmetry of charge distribution• Use the Gaussian surface of the same symmetry
• Then you’ll get A
qE
qEAAdE enclencl
00
lqorAqorVq
ondestributiechuniformafor
dlqordAqordVq
enclenclencl
enclenclencl
arg
here
volume
surfacelinear
charge density
1. r<R, E=0(conductor)
2. r>R, 0
24
qrE
2r
kqE
q
Dielectrcic sphere
03
0
03
32
3
0
32
34
4
3/4
3
44:.1
r
R
QrE
R
QrrE
R
Q
rrERr
Q
2. r>R, 0
24
qrE
2r
kqE
Q
L
Q
L is the total length of the line
sheetA
Q
Q
Conducting spherical layer
a
b
Q
q
2
2
)(.3
0.2
.1
r
qQkEbr
conductorEbrar
kqEar
q-q Q+q
1. Find a surface charge density on inner and outer sides.2.Find E(r).
2
2
4
4
b
qQa
q
b
a
Dielectric spherical layer
q
Q a
b
)(
)((
3/)(4
3/)(44.3
)(.2;.1
332
33
2
33
0
332
22
abr
arkQ
r
kqE
ab
Q
arqrE
r
qQkEbr
r
kqEar
Earnshaw’s Theorem, 1842 • A collection of point charges cannot be maintained
in a stable stationary equilibrium configuration solely by the electrostatic interaction of the charges.
• Suppose a charge q is at stable equilibrium in p. P. • It means if it is moved away from this point in any
direction there should be a restoring force (opposite to the direction of a displacement.)
• It means that there is a flux of E through the nearby surface surrounding p. P.
• According to Gauss’s theorem p. P should be occupied with a charge.
• But we supposed that the charge was shifted from this point.
P
J.J. Thomson’s “plum pudding” model of atom
Sir J.J.Tomson (1856 -1940) Nobel prize for discovery of e,1906
Such atom would be unstable!
Long conducting cylinder2a
1.r<a E=0 (conductor)2.r> a the same as for theline of charge
aL
aL
L
Q 22
Connection between the linear and surface charge density
r
kE
2
Long dielectric cylinder2a
1.r>a the same result as for conducting cylinder:
20
2
2
0
2
22
2
.2
a
rk
La
QrE
La
Q
lrrlE
ar
r
kE
2
Conducting cylindrical shell with a line of charge
a b
1 2
1 21
NB: linear charge density on an inner surface of a shell ison an outer surface
1.r<a the same result as for a line in the center:
r
kE
)(2 21
2.a<r<b E=0 (conductor)
3. r>b
1
21
r
kE 12
Dielectric cylindrical shell with a line of charge
a b
1 2
1.r<a and 3.r>b the same results as for conducting cylindrical shell
2.a<r<b
)(
)(22
)()(
)(2
22
2221
222
222
0
22
0
1
abr
ark
r
kE
ababL
Q
larlrlE
A conducting slab
x
z
ya
-a
L1
L2
L2 >>a,L1>>a, r<< L2 ,r<<L1
1.|x|<a E=0 (conductor)
2.|x|>a
2EA=2σA/ε0→E= σ/ε0
σ =Q/2S, S= L1 L2
E=Q/2ε0S
A dielectric slab
x
z
ya
-a
L1
L2
L2 >>a,L1>>a, r<< L2 ,r<<L1
1.|x|>a2EA=σ*A/ε0→E= σ*/2ε0
NB: σ* =Q/S, S= L1 L2
E=Q/2ε0S the same as for conductor
2.|x|<a 2EA=ρA2x/ε0
ρ=Q/S2a= σ*/2a
E=σ*x/2aε0
0
Conclusions1. Inside of charged bodiesConductors E=0Dielectrics E ~ rIndependent on the symmetry of charge distribution
2.Outside of charged bodiesE is the same for conductors and dielectrics.It depends on the symmetry of charge distribution:Sphere E ~ 1/r2
Cylinder E ~1/r (r<<L)Plane E=const (r<< L1 and L2)
3. For finite size bodies at r→∞ E=kQtotal/r2
4. E jumps on the surface of the charged conductor
E changes continuously on the surface of dielectric