c2 2 gauss law
TRANSCRIPT
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Gauss’ Law – electric flux
• Consider an imaginary sphere of radius R centered on charge Q at origin:
• FLUX OF ELECTRIC FIELD LINES ?
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Gauss’ Law – total electric flux
• FLUX OF ELECTRIC FIELD LINES (through surface S):
• ΦE = “measure” of “number of E-field “lines”passing through surface S, (SI Units: Volt-meters).
• TOTAL ELECTRIC FLUX (ΦETOT )
associated with any closed surface S, is a measure of the (total) charge enclosed by surface S.
• Charge outside of surface S will contribute nothing to total electric flux ΦE (since E-field lines pass through one portion of the surface S and out another – no net flux!)
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Gauss’ Law - calculation
• Consider point charge Q at origin. • Calculate ΦE passing through a sphere of radius r:
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Gauss’ Law in integral form
• Gauss’ Law in integral form:
0
S) surfaceby enclosed charge (electric
S surface closedgh flux throu Electric
ε=
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Gauss’ Law – discrete charges
• If ∃ (= there exists) lots of discrete charges qi , all enclosed by Gaussian surface S’
• by principle of superposition
• Then
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Gauss’ Law – volume charge
• If ∃ volume charge density ρ(r) , then
• Using the DIVERGENCE THEOREM:
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Gauss’ Law in Differential Form
• This relation holds for any volume v ⇒ the integrands of ∫v ( ) dτ' must be equal
• So, Gauss’ Law in Differential Form:
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The DIVERGENCE OF E(r) – 1/3
• Calculate from Coulomb’ law
Extend over all space !!
NOT a constant !!
• r : Field point P• r’: source point S
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The DIVERGENCE OF E(r) – 2/3
• Recall that
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The DIVERGENCE OF E(r) -3/3
Thus
or
Now, we have the Gauss’s law:
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Gauss’ Law in Integral Form - revisited
• By the differential form of Gauss’ law:
• We have,
• Apply the Divergence Theorem
• Thus: Gauss’ Law in Integral Form
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GAUSS’ LAW AND SYMMETRY
• Use of symmetry can be extremely powerful in terms of simplifying seemingly complicated problems.
• Examples of use of Geometrical Symmetries and Gauss’ Lawa) Charged sphere
– concentric Gaussian sphere、spherical coordinatesb) Charged cylinder
– coaxial Gaussian cylinder、cylindrical coordinatesc) Charged box / Charged plane
– use rectangular box、rectangular coordinatesd) Charged ellipse
– concentric Gaussian ellipse、 elliptical coordinates
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APPLICATIONS OF GAUSS’ LAW - Example 2.2 -1/3
• Find / determine the electric field intensity E(r) outside a uniformly charged solid sphere of radius R and total charge q
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APPLICATIONS OF GAUSS’ LAW - Example 2.2 -2/3
• Gauss’ law
(by symmetry of sphere)
(for Gaussian sphere)
• By symmetry, the magnitude of E is constant ∀ any fixed r !!
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APPLICATIONS OF GAUSS’ LAW - Example 2.2 -3/3
or
The electric field (for r > R) for charged sphere is equivalent to that of a point charge q located at the origin!!!
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Example 2.3
• Consider a long cylinder of length L and radius S that carries a volume charge density ρ that is proportional to the distance from the axis s of the cylinder, i.e.
a) Determine the electric field E(r) inside this long cylinder.
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Example 2.3 (conti.)
• Gauss’ law :
• Enclosed charge :
• By cylindrical Symmetry
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Example 2.3 (conti.)
• from cylindrical symmetry
= constant on cylindrical Gaussian surface
• What are the vector area element?
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Example 2.3 (conti.)
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Example 2.3 (conti.)
Note:• On LHS and RHS endcaps E(r) is not constant, because
r is changing there• However, note that
=> Gaussian endcap terms do not contribute!!!
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Example 2.3 (conti.)
• Putting this all together now:
where
or
=>
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Example 2.3 (conti.)
b) Find ELECTRIC FIELD E(r) outside of this long cylinder
• use Coaxial Gaussian cylinder of length l (<< L) and radius s (> S):
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Example 2.3 (conti.)
• Enclosed charge (for s > S):
• symmetry of long cylinder
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Example 2.3 (conti.)
• vector area element
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Example 2.3 (conti.)
• Now• Then
∴ Electric field outside charged rod (s = r > S) :
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Example 2.3 (conti.)
• Inside (s < S): • Outside (s > S):
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Example 2.4
• An infinite plane carries uniform charge σ. Find the electric field.
• Use Gaussian Pillbox centered on ∞-plane:
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Example 2.4 (conti.)
• from the symmetry associated with ∞-plane
• six sides and six outward unit normal vectors:
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Example 2.4 (conti.)
• Then
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Example 2.4 (conti.)
• since
• The integrals is separated into two regions:
• Then
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Example 2.4 (conti.)
So, non-zero contributions are from bottom and top surfaces
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Example 2.4 (conti.)
• Thus, we have:
• These integrals are not over z, and E(z) = constant for z = zowe can pull E(z) outside integral,
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Example 2.4 (conti.)
• Now, what is Qencl ?
• Note:
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Curl of E(r)
• Consider point charge at origin:
• By spherical symmetry (rotational invariance)1. E(r) is radial.2. thus static E-field has no curl.
• Calculation : see next page
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Curl of E(r)
• Let’s calculate:
• In spherical coordinates:
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Curl of E(r)
• thus
• So, around a closed contour C
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Curl of E(r)
• Use Stokes’ Theorem
• Since must be true
for arbitrary closed surface S,
this can only be true for all ∀ closed surfaces S IFF (if and only if):
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Curl of E(r) - discrete charges
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Curl of E(r) - discrete charges
• It can be shown that
FOR ANY STATIC CHARGE DISTRIBUTIONSTATIC = NO TIME DEPENDENCE / VARIATION
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Helmholtz theorem
• Vector field A(r) is fully specified if both its divergenceand its curl are known.
• Corollary:Any differentiable vector function A(r) that goes to zero faster than 1 /r as r → ∞ can be expressedas the gradient of a scalar plus the curl of a vector:
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Helmholtz theorem for electrostatics
• For the case of electrostatics:
• Thus
• i.e.
with
~ valid for localized charge distributions~ NOT vaild for infinite-expanse charge distributions
( Electrostatic Potential )