Gauss’ Law (Review)
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Gauss’ law – form of
Coulomb’s law
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qenc is the total charge enclosed
by a Gaussian surface
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Flux is proportional to # of E
field lines passing through a
Gaussian surface
ε0Φ = qenc
r r
Φ = ∫ E • dA
Conductors (Example)
A ball of charge -50e lies at the center of a
hollow spherical metal shell that has a net
charge of -100e. What is the charge on a)
the shell’s inner surface and b) its outer
surface?
Conductors
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Pick a cylindrical Gaussian
surface embedded in the
conductor
Sum the flux through surface
Inside conductor E = 0 so
Φ=0
Along walls of the cylinder
outside the conductor E is ⊥
to A so Φ = 0
Outer endcap Φ = EA
Conductors
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Using Gauss’ law and Φ = EA
ε0Φ = ε0EA= qenc
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If σ is charge per unit area, then
qenc = σA
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So E for a conducting surface is
σ
E=
ε0
Conductors
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E just outside a conductor is proportional to
surface charge density at that location
σ
E=
ε0
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If – charge on conductor, E toward conductor
If + charge on conductor, E directed away
Gauss’ Law
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Infinitely long insulating rod
with linear charge density λ
Pick Gaussian surface of
cylinder coaxial with rod
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What does E look like?
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Φ = 0 for the endcaps
Φ = EA for cylinder
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Gauss’ Law
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Substituting in Gauss’ law gives
ε0Φ = ε0EA= qenc
A = 2πrh qenc = λh
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E for a line of charge is
λ
E =
2 πε 0 r
Gauss’ Law
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Apply Gauss’ law to a uniformly charged
spherical shell S2
r r
ε0Φ = qenc = ε0 ∫ E • dA
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E radiates out || to A so
r
r
qenc
∫ E • dA = ∫ EdA =
ε0
A = 4πr 2
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Substitute to find E
1
q
E =
,r ≥ R
2
4πε 0 r
Gauss’ Law
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E outside of a charged
spherical shell is same as
E of point charge at
center of shell.
Charge inside S1 is zero,
so by Gauss’ law E=0
inside shell, r < R.
If a charge is placed
inside there will be no
force on it.
q
E = k 2
r
Gauss’ Law
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Non-conducting sheet
of charge σ
r r
ε 0 ∫ E • dA = qenc
ε 0 ( EA + EA) = σA
σ
E=
2ε 0
Gauss’ Law
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Conducting sheet of charge
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charge spreads
over two surfaces
z σ1 is charge on one
surface,
z σ1 = σ/2
σ1
E =
ε0
Gauss’ Law
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Positive and negative
charged conducting
plates put together
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Excess charges moves
to inner faces
New total surface
density, σ, is equal to
2σ1
E=
2σ 1
ε0
σ
=
ε0
Gauss’ Law (Checkpoint)
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Two large, parallel, non-conducting sheets with
identical + charge and a sphere of uniform +
charge. Rank magnitude of net E field for 4
points (greatest first).
Gauss’ Law (Checkpoint)
• E due to sheets
E =0
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E due to point charge
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Magnitude depends on distance r from point
charge
3 and 4 tie, then 2, then 1
q
E=k 2
r
Gauss’ Law
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Non-conducting solid
sphere of radius R and
total (uniform) charge q
Gaussian sphere outside
sphere
q
E=k 2, r≥R
r
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Same as shell
Gauss’ Law
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Use series of Gaussian
spheres for inside
q′
E =k 2
r
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Full charge enclosed within
R is uniform so q’ within r
is proportional to q
q′
q
=
3
3
4
4
3 πr
3 πR
Gauss’ Law
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Enclosed charge at r is
3
r
q′ = q 3
R
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E field inside sphere
kqr
E = 3 ,r ≤ R
R