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o  Today’s Lecture: –  Class news. –  Electric flux. –  Gauss’s Law –  Applica<ons of Gauss’s Law 2
Class News o 
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The first quiz will be on Wednesday Feb 8th during the class period. The format will be 15 mul<ple-­‐choice ques<ons and two long ques<ons. I will post an example of a quiz from a previous year later today. You may bring one handwriOen 8½ inch by 11 inch (double-­‐sided) sheet of equa<ons for the quiz. The material to be covered is from chapters 22, 23 and 24 and material presented in lecture up to today. The quiz will be held in the usual class room – Wiley 125. Be sure that you understand and follow the instruc<ons in the syllabus about calculators, cell phones, etc. Bring your student id with you. I will be away from campus Thursday and Friday, so I willnot hold ofice hours then. Physics-1001-Spring-2017
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E-­‐field from a charged plate. Area A, total charge Q
radius R
⎛
E x = 2π kσ ⎜ 1−
⎝
x
x +R
2
⎞
2 ⎟
⎠
Q
Where σ = 2
πR
is the
charge per unit area.
da
a
r
dEx
θ
x
dE⊥
R
We have shown that in the limit of large x E x = k
Q
x2
σ
2ε 0
So the electric field from an infinite plane of charge is
the same for all values of x.
In the limit of R → ∞
E x = 2π kσ =
The Electric Flux o  We define the physical quan<ty called the electric flux (ΦE). o  The magnitude of the electric flux through a surface with area A in a uniform electric field of magnitude E is defined as ΦE ≡ EA cos θ © 2015 Pearson Education,
Inc.
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Electric Flux © 2015 Pearson Education,
Inc.
Electric Flux For simple cases where the surface is flat and the field
uniform we have ΦE=|E|.|A|cosθ.
For more complex situations:
! !
Φ E = ∑ E j .δ A j
! !
! !
and Φ E = lim ∑ E j .δ A j = ∫ E.dA
δ A→0
!
!
Note δ A has the magnitude of δ A and the direction
of the normal to the surface.
We can write it as n̂.δ A.
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Electric Flux from a Charge For a sphere the electric
field at the surface is:
!
q
E=k 2
r
While the surface area at
radius r is:
+
A = 4π r 2
As the field is always normal to the
surface then cosθ is always 1.
! ! ! ! q
and Φ E = "
E.d
∫S A = E . A = ε 0
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Gauss’s Law It turns out that this statement is true for all surfaces and is in
fact a general result that we can use to calculate the electric
fields where we know the charge distribution.
Mathematically:
! ! q
E.d
"∫S A = ε 0 is one form of Gauss’s Law.
In words: For any closed surface the flux of the electric field
through the surface is equal to the charge contained within that
surface divided by the permittivity of free space.
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Gauss’s Law Since the charge is a scalar quantity and the result holds no
matter where it is in the enclosing surface, we can generalize
this to larger values of q.
! !
2q
"∫ E.dA = ε
For two charges:
S
0
And more generally for any charge distribution ρ(r)
contained inside the surface then:
! ! 1
E.d
"∫S A = ε 0
!
"∫ ρ (r )dV
V
where V is the volume inside the surface and the left
integral is over the volume.
Check with a simple example: 10
Uniform E-field through a
cylindrical Gaussian surface.
1)  By symmetry let the axis of
the cylinder be parallel to the
field.
2)  Let the radius of the cylinder
be r.
! !
ΦE = "
E
∫ ⋅ dA =
∫
! !
E ⋅ dA +
back flat
surface
∫
! !
! !
E ⋅ dA + ∫ E ⋅ dA.
curved
surface
front flat
surface
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! !
ΦE = "
E
∫ ⋅ dA
S
! !
= ∫ E ⋅ dA +
back flat
surface
∫
! !
! !
E ⋅ dA + ∫ E ⋅ dA.
curved
surface
front flat
surface
= E.A cos(π )+ 0 + E.A cos(0)
= −1+ 0 + 1 = 0
So the integral of the flux through the surface is 0 and by
! !
Gauss’s Law E ⋅ dA = q we get q = 0 as expected.
"∫
S
ε0
Clicker Ques<on o  A cylindrical piece of insula<ng material is placed in an external electric field, as shown. The net electric flux passing through the surface of the cylinder is 1. 
2. 
3. 
posi<ve. nega<ve. zero. 6
Clicker Ques<on o  A cylindrical piece of insula<ng material is placed in an external electric field, as shown. The net electric flux passing through the surface of the cylinder is 1. 
2. 
3. 
posi<ve. nega<ve. zero. © 2015 Pearson Education,
Inc.
Gauss’s Law and Coulomb’s Law o  Remember Coulomb’s law can always be used to calculate the electric field from a source charge situa<on, typically through the use of integra<on. o  However, Gauss’s law is much easier to apply mathema<cally for symmetric source charge distribu<ons: –  This is because the integra<ons can be trivial since the electric field is constant over the Gaussian surface and can be factored outside the integrand. © 2015 Pearson Education,
Inc.
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Using Gauss’s Law Gauss’s law allows you to calculate the electric field for charge distribu<ons that exhibit spherical, cylindrical, or planar symmetry without having to carry out any integra<ons. © 2015 Pearson Education,
Inc.
1) Find the Symmetry: © 2015 Pearson Education,
Inc.
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2) Make a drawing Sketch the charge distribu<on and the electric field Drawing a number of field lines, remembering that the field lines start on posi<vely charged objects and end on nega<vely charged ones. © 2015 Pearson Education,
Inc.
3) Draw the Gaussian Surface Draw a Gaussian surface such that the electric field is either parallel or perpendicular (and constant) to each face of the surface. If the charge distribu<on divides space into dis<nct regions, draw a Gaussian surface in each region where you wish to calculate the electric field. © 2015 Pearson Education,
Inc.
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4) Solving for E o  For each Gaussian surface determine the charge qenc enclosed by the surface. o  For each Gaussian surface calculate the electric flux ΦE through the surface. Express the electric flux in terms of the unknown electric field E. o  Use Gauss’s law to relate qenc and ΦE and solve for E. © 2015 Pearson Education,
Inc.
Some examples: • 
• 
• 
• 
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Electric field inside a ball charge.
At the surface of a conductor.
From a long line of charge.
From a large charged plate.
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Clicker Question
A thin spherical metal shell of radius R carries a uniformly
distributed charge q. It is surrounded by a concentric thin spherical
metal shell of radius 2R that carries a uniformly distributed charge
–q.
For which of the following regions enclosed by a concentric
spherical Gaussian surface of radius r is the electric field non-zero?
1. r < R
2. R < r < 2R
3. r > 2R
© 2015 Pearson Education,
Inc.
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Clicker Question
A thin spherical metal shell of radius R carries a uniformly
distributed charge q. It is surrounded by a concentric thin spherical
metal shell of radius 2R that carries a uniformly distributed charge
–q.
For which of the following regions enclosed by a concentric
spherical Gaussian surface of radius r is the electric field zero?
Answer all that apply.
1. r < R
2. R < r < 2R
3. r > 2R
© 2015 Pearson Education,
Inc.
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