Download Ch 1.13 Gauss*s Law and Poisson*s Equation

講者： 許永昌 老師
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Contents
 Gauss’s Law
 Poisson’s Equation
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Gauss’s Law
 closedEds=q/e0,
q1
 Prove:
1. Prove that E1=0 except at the origin.
2. Therefore we get SE1ds-S’E1ds=0 based on
.
3. Prove that S’E1ds=q1/e0 for arbitrary small sphere.

Etot=E1+E2+…, where Ei is contributed from qi, we get
closedEtotds=qinside/e0. (這裡用到的條件是E與q呈線性關係)
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Poisson’s Equation
 2j= - r/e0.
 Since E=-j and E=r/e0, we get 2j= - r/e0.
 One of the solution is
j r  
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4e 0
r  r '
 r - r ' d '.
 Laplace Equation:
 2j=0.

It is homogeneous.
 If 2j1= - r/e0 and 2j2= 0, we get
 2(j1(r) +aj2(r))= - r (r) /e0.
 You can use j2(r) to match the boundary conditions.
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Summary
 Gauss’s Law: closedEds=qinside/e0.
 It means that the net electric flux of a closed surface is the
same as the total charge inside this closed region.
 D=r: or you can say the charge is the divergence source of
electric field.
 Possion’s Equation:
 2j= - r/e0.
 It tells us the relationship between electric potential j and
charge distribution r.
 With the boundary conditions, we possibly can get an unique
j from a provided r.
 We will discuss it briefly in Ch 1.14e.
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Homework
 1.13.3e (1.14.3)
 1.13.4e (1.14.4)
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nouns

(Physics): closedEds=q/e0
 Be careful, when we consider the EM field in a medium,
we should use closedDds=q.


(Physics): 2j= - r/e0.
(Physics): 2j=0.
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