Download Topic 50 Notes 50 Applications and and interpretation of Stokes theorem

Topic 50 Notes
Jeremy Orloff
50
Applications and and interpretation of Stokes
theorem
Read section V15 in the supplementary notes.
Divergence and Stokes’
Z Z Theorems Zin
Z Zdel notationZ Z Z
Divergence Theorem:
F · n dS =
divF dA =
∇ · F dA.
S
D
D
I
ZZ
ZZ
Stokes’ Theorem:
F · dr =
curlF · n dS =
∇ × F · n dS.
C
Maxwell’s Equations
S = closed surface
C1 = closed curve
E = electric field
Q = charge inside S
ρ
= charge density inside S
S
S
D
S1
B
=
=
=
interior of S
surface capping C1
magnetic field
J
=
current density
k and k1 are constants depending on units.
Integral
Z Z form of Maxwell’s equations
1’)
E · n dS = 4πkQ
Z ZS
2’)
B · n dS = 0
I S
ZZ
d
3’)
E · dr = −
B · n dS
dt
IC1
Z Z S1
ZZ
k1 d
E · n dS
4’)
B · dr = k1
J · n dS +
k dt S1
C1
S1
Gauss-Coulomb Law.
Gauss’ law of magnetism.
Faraday’s law.
Ampere’s law.
Differential form of Maxwell’s equations (equivalent to the integral form).
1) ∇ · E = 4πkρ
Gauss-Coulomb Law.
2) ∇ · B = 0
Gauss’ law of magnetism.
∂B
3) ∇ × E = −
Faraday’s law.
∂t
k1 ∂E
4) ∇ × B = k1 J +
Ampere’s law.
k ∂t
In words these statements say the following:
1,1’) Flux of E through S = 4πk×charge enclosed.
2,2’) There is no magnetic monopole. (No net magnetic charge enclosed.)
1
50 APPLICATIONS AND AND INTERPRETATION OF STOKES THEOREM 2
3,3’) A changing magnetic field induces an electric field.
4,4’) Magnetic fields are induced by either a current or a changing electric field.
We discussed Gauss’ law in topic 46 with respect to gravitation. Here’s a quick recap
for electricity.
hx, y, zi
For a charge q at the origin the electric field is Eq = k
(here ρ is distance
ρ3
from the origin, not charge density).
By direct computation the flux of Eq through a sphere centered on the origin is 4πkq.
Everywhere but the origin ∇ · Eq = 0, so the extended divergence theorem gives flux
(
4πkq if S goes around q
of Eq through S =
0
otherwise.
In general, if q is placed anywhere else the same formula holds.
Now suppose q1 , q2 , . . . , qn are charges with fields E1 , E2 , . . . , En . Let E = E1 +
. . . + En be the net electric field. The net flux through S is the sum of the flux
contributed by each field Ej . That is, the flux of E through S = 4πk · ×(total charge
inside S).
For a continuous charge distribution we replace sums by integrals and find the same
thing. This proves (1’).
Before showing (1) and (1’) are equivalent we state a very reasonable theorem.
RRR
Theorem: Assume h(x, y, z) is a continuous function and
h dV = 0 for every
D
volume D then h(x, y, z) = 0.
Proof: We argue by contradiction. Suppose for some point (x0 , y0 , z0 ) we have
h(x0 , y0 , z0 ) > 0. Since h is continuous we can put aRRR
small ball D around (x0 , y0 , z0 )
such that h(x, y, z) > 0 throughout D. This implies
h dV > 0, which contradicts
D
our assumption. Therefore it can’t be positive. Likewise it can’t be negative. QED
RRR
RRR
Corollary: Assume f (x, y, z) and g(x, y, z) are continuous and
f dV =
g dV
D
D
for every volulme D then f (x, y, z) = g(x, y, z).
Proof: Let h = f − g and apply the theorem.
Now we show (1) and (1’) are equivalent.
ZZ
ZZZ
The divergence theorm implies
E · n dS =
∇ · E dV .
S
D
ZZZ
The definition of density says: 4πkQ = 4πk
ρ dV .
D
ZZ
ZZZ
ZZZ
Thus,
E · n dS = 4πkQ ⇔
∇ · E dV = 4πk
ρ dV .
S
D
D
Since this is true for every D the above corollary says it is equivalent to ∇ · E = 4πkρ.
Showing (2), (3) and (4) are equivalent to (2’), (3’) and (4’) respectively is similar.
QED
50 APPLICATIONS AND AND INTERPRETATION OF STOKES THEOREM 3