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Unformatted text preview: Lecture slides are posted each Thursday for that
week. Thursday, September 9, 2010 What is an electric flux through spheres of radii r1
and r2= 2r1 surrounding a charge of q=3 e? A : Φ1 = Φ2 = 0 B : Φ1 = Φ2 = 0
C : Φ1 = Φ2 /2 r2 3e
r1 D : Φ1 = 2Φ2
E: Φ1 = Φ2 /4
2
Thursday, September 9, 2010 What is a electric flux through spheres of radii r1
and r2= 2r1 surrounding a charge of q=3 e? A : Φ1 = Φ2 = 0 B : Φ1 = Φ2 = 0
C : Φ1 = Φ2 /2
D : Φ1 = 2Φ2
E: Φ1 = Φ2 /4 r2 3e
r1 equal: A 3
Thursday, September 9, 2010 iClicker What is the flux through the spherical surface
encompassing two charges q1= 3q, q2=2q
A q
5
0 B
C q
−
q 0 0 D
C 3q
2q q
3
0 0
4 Thursday, September 9, 2010 iClicker What is the flux through the spherical surface
encompassing two charges q1= 3q, q2=2q
A q
5
0 B
C q
−
q 0 0 D
C 3q q
3
0 2q qtot = 3q − 2q = q 0
5 Thursday, September 9, 2010 iClicker The total electric flux through two closed
surfaces is A: equal B: not equal 6
Thursday, September 9, 2010 iClicker The total flux of electric field through the closed
surface is A: > 0 B: =0 C: <0 +3q +3q +3q +3q
+3q q 7
Thursday, September 9, 2010 Gauss’s Law for Various Charge We have applied Gauss’s Law to a point charge and showed
that we get Coulomb’s Law Now let’s look at more complicated distributions of charge
and calculate the resulting electric field We will use a “charge density” to describe the distribution
of charge This charge density will be different depending on the
geometry August 12, 2010
Thursday, September 9, 2010 University Physics, Chapter 22 64 Cylindrical Symmetry (1) Let’s calculate the electric field from a conducting wire
with charge per unit length λ using Gauss’s Law q
E · dA =
0 We start by assuming a Gaussian surface in the form of a
right cylinder with radius r and length L placed around the
wire such that the wire is along the axis of the cylinder August 12, 2010
Thursday, September 9, 2010 University Physics, Chapter 22 65 Cylindrical Symmetry (2) From symmetry we can see that the electric field will
extend radially from the wire
E How?
• If we rotate the wire along its axis,
the electric field must look the same
• Cylindrical symmetry • If we imagine a very long wire,
the electric field cannot be different
anywhere along the length of the wire
• Translational symmetry Thus our assumption of a right cylinder as a Gaussian
surface is perfectly suited for the calculation of the
electric field using Gauss’ Law
August 12, 2010
Thursday, September 9, 2010 University Physics, Chapter 22 66 Cylindrical Symmetry (3) The electric flux through the ends of the cylinder is zero
because the electric field is always parallel to the ends The electric field is always perpendicular to the wall of the
cylinder so … and now solve for the electric field August 12, 2010
Thursday, September 9, 2010 University Physics, Chapter 22 67 Planar Symmetry (1) Assume that we have a thin, infinite nonconducting sheet
of positive charge
E
σ The charge density in this case is the
charge per unit area, σ From symmetry, we can see that the electric field will be
perpendicular to the surface of the sheet
August 12, 2010
Thursday, September 9, 2010 University Physics, Chapter 22 68 Planar Symmetry (2) To calculate the electric field using Gauss’ Law, we assume a
Gaussian surface in the form of a right cylinder with cross sectional area A
and height 2r, chosen to cut through the plane perpendicularly. Because the electric field is perpendicular to the plane
everywhere, the electric field will be parallel to the walls of the cylinder
and perpendicular to the ends of the cylinder. Using Gauss’ Law we get … so the electric field from an infinite
nonconducting sheet with charge density σ is August 12, 2010
Thursday, September 9, 2010 University Physics, Chapter 22 69 Spherical Symmetry (1) Now let’s calculate the
electric field from charge
distributed as a spherical
shell. Assume that we have a
spherical shell of charge q
with radius rS (gray). +
+ + + We will assume two different
spherical Gaussian surfaces
• r > rS (blue) i.e., outside +
+ +
+ • r < rS (red) i.e., inside
August 12, 2010
Thursday, September 9, 2010 University Physics, Chapter 22 72 Spherical Symmetry (2) Let’s start with the Gaussian surface outside the sphere of charge,
r > rS (blue) We know from symmetry arguments that the electric field will be radial
outside the charged sphere
• If we rotate the sphere, the electric field cannot change
• Spherical symmetry Thus we can apply Gauss’ Law and get . … so the electric field is August 12, 2010
Thursday, September 9, 2010 University Physics, Chapter 22 73 Spherical Symmetry (3) Let’s take the Gaussian surface inside the sphere
of charge, r < rS (red) We know that the enclosed charge is zero so By symmetry, we find that the electric field is zero
everywhere inside spherical shell
of charge Thus we obtain two results
• The electric field outside a
spherical shell of charge is the same
as that of a point charge
• The electric field inside a spherical shell of charge is zero
August 12, 2010
Thursday, September 9, 2010 University Physics, Chapter 22 74 Spherical Symmetry: Uniform Distribution Next, let’s calculate the electric field from charge
distributed uniformly throughout a sphere. Assume that we have a solid sphere of charge Q with radius
R with constant charge density per unit volume ρ We will assume two different spherical Gaussian surfaces
• r2 > R (outside) • r1 < R (inside) R Charge density inside ρ (C/m^3) August 12, 2010
Thursday, September 9, 2010 University Physics, Chapter 22 75 Spherical Symmetry: Uniform
Consider a Gaussian surface with radius r2 > R
Again by spherical symmetry we know that the electric field
will be radial and perpendicular to the Gaussian surface.
Gauss’ Law gives us total charge area
R Solving for E we find same as a point
charge!
August 12, 2010
Thursday, September 9, 2010 University Physics, Chapter 22 78 Spherical Symmetry: Uniform Let’s start with a Gaussian surface with r1 < R From spherical symmetry we know that the electric field
will be radial and perpendicular to the Gaussian surface. Gauss’s Law gives us volume . area R Solving for E we find August 12, 2010
Thursday, September 9, 2010 University Physics, Chapter 22 76 Spherical Symmetry: Uniform R In terms of the total charge Q … August 12, 2010
Thursday, September 9, 2010 University Physics, Chapter 22 77 Efield for charged sphere with
constant density of charges E/E(R)
1.0 0.8 0.6 0.4 0.2 0.5 1.0 1.5 2.0 2.5 3.0 r/R 21
Thursday, September 9, 2010 Gauss theorem in differential form
q
E · dA =
0 q q q= ρdV dV = r2 dΩdr
dA = r2 dn 1
div E =
ρ
4π0 12
E · dA = r E · dn = ρr dΩdr
0
1
Divergence of electric field equals
E · dn = dΩdr
0
local charge density
2 div E = ∂x Ex + ∂y Ey + ∂z Ez
22
Thursday, September 9, 2010 Electric Field from a Ring of Charge
We can’t solve it by Gauss’s Law Use integration …
Total charge Q is distributed
homogeneously over a ring
of radius R. Find Efield on
the axis at distance z. August 12, 2010
Thursday, September 9, 2010 University Physics, Chapter 22 79 dφ Take section
Its charge is dφ
dφ
dq =
Q
2π It produces Efield Edq dq
dφ
Q
=k 2
=k
2
R +z
2π R2 + z 2 From symmetry: only Ez is important Ez,dq dφ
Q
z
√
= Edq cos θ = k
2 + z2
2π R
R2 + z 2
24 Thursday, September 9, 2010 dφ
Q
z
√
Ez,dq = Edq cos θ = k
2π R 2 + z 2 R 2 + z 2
Qz
Integrate over dφ E = k
2 + z 2 ) 3 /2
(R
E
0.3 ∝ 1/z 0.2 2 0.1 z/R
0.5 1.0 1.5 2.0 2.5 3.0 25
Thursday, September 9, 2010 Shielding An interesting application of Gauss’ Law: The electric field inside a charged conductor is zero
• movie from Wikipedia. August 12, 2010
Thursday, September 9, 2010 University Physics, Chapter 22 60 Shielding An interesting application of Gauss’ Law: The electric field inside a charged conductor is zero
• movie from Wikipedia. August 12, 2010
Thursday, September 9, 2010 University Physics, Chapter 22 60 Pointed Surfaces
E field is always perpendicular to the surface of a
charged conductor
Now consider a sharp point. The field lines are
much closer together, i.e., the field is much
stronger and looks much more like the field of a
point charge. August 12, 2010
Thursday, September 9, 2010 University Physics, Chapter 22 80 Find electric field between two plates of area A
opposite charges q, distance d apart.
q
A:E=
Ad0
q
B:E=
2A0
q
C:E=
A0
qd
D:E=
A0
q
E:E=
dA0
Thursday, September 9, 2010 iClicker Find electric field between two plates of area A
opposite charges q q
A:E=
Ad0
q
B:E=
2A0
q
C:E=
A0
qd
D:E=
A0
q
E:E=
dA0
Thursday, September 9, 2010 From single plane:
Total Eﬁeld q
E=
A 0 Electric Potential Energy Electrostatic force is conservative force • For a conservative force, the work is pathindependent
• Work over closed path is 0 B A E E
• circulation of electrostatic field is zero E · dl = 0 August 12, 2010
Thursday, September 9, 2010 NB: there is inductive Eﬁeld, for which
work over closed path is not 0 (circulation
not 0)
2 Electric Potential Energy Electrostatic force is conservative force • For a conservative force, the work is pathindependent
• Work over closed path is 0 B
dl
A E E
• circulation of electrostatic field is zero E · dl = 0 August 12, 2010
Thursday, September 9, 2010 NB: there is inductive Eﬁeld, for which
work over closed path is not 0 (circulation
not 0)
2 Is this a possible configuration of static Efield? Ex(y) Thursday, September 9, 2010 Is this a possible configuration of static Efield? Ex(y) E.dl>0, large E.dl=0 E.dl< 0, small Thursday, September 9, 2010 NO, since
E.dl=0 E · dl = 0 Is this a possible configuration of static Efield? Ex(y) E.dl>0, large E.dl=0 E.dl< 0, small NO, since
E.dl=0 E · dl = 0 This is a possible conﬁguration of nonelectrostatic ﬁelds Thursday, September 9, 2010 Line of Electrostatic field cannot be closed
on themselves. E NB: there is inductive Eﬁeld, for which
work over closed path is not 0 (circulation
not 0, field lines do close on themselves) Thursday, September 9, 2010 E · dl = 0 Line of Electrostatic field cannot be closed
on themselves. E NB: there is inductive Eﬁeld, for which
work over closed path is not 0 (circulation
not 0, field lines do close on themselves) Thursday, September 9, 2010 E · dl = 0 Electric potential b
aq
E Thursday, September 9, 2010 Charge q moves from
point a at x=0 to point b
at x=L in constant Efield E. How does it’s
energy change? Electric potential y
L x
b aq
E Thursday, September 9, 2010 Charge q moves from
point a at x=0 to point b
at x=L in constant Efield E. How does it’s
energy change? Dificult way:
y L
m 2
2
E=
vx + vy
2
vy = const = vy,0 x
b a
E Thursday, September 9, 2010 vx = vx,0 + at = vx,0 + F
qE
t = vx,0 +
t
m
m x = vx,0 t + at2 /2 = L
2
−mvx,0 + m2 vx,0 + 2qELm
t=
qE
2qEL
2
vx =
+ vx,0
m
m 2
2
E=
v + vx,0 + qEL
2 y,0
∆E = E − E0 = qEL ∆EK = qEL Eﬁeld did work on a particle When moved along the ﬁeld, energy of a positively
charged particle increased, energy of a negatively
charged particle decreased Thursday, September 9, 2010 Electric potential Electric potential = ability of Efield to do
potential energy
work. U = mgh Recall gravity: Φ
V g = gh kinetic energy Gravity (in constant grav. field):
• potential V= g h
• potential energy U = m V Thursday, September 9, 2010 Electric potential
potential energy ΦE
V Electric potential kinetic energy
y
L a
E Thursday, September 9, 2010 x Energy conservation:
change in kinetic energy of a particle =  change in
potential energy of a particle change in kinetic energy of a particle = work done
by Efield Thursday, September 9, 2010 Electric potential Φ
∆EK = qEL ∆ E k = − ∆E p
∆Ep = q ∆Φ
V ∆Φ = −E ∆x
V Electric potential = change in energy of unit
charge, when moved distance ∆x along electric
field Only difference is important > Electric potential difference = change in energy
of unit charge, when moved distance  ∆x along
electric field
Thursday, September 9, 2010 Potential is useful when we do not care about
direction of velocity, only energy
v
v0
a
E Thursday, September 9, 2010 b Electric Potential Energy (2) Like gravitational or mechanical potential energy, we must
define a reference point from which to define the
electric potential energy Typically, the electric potential energy is defined to be
zero when all charges are infinitely far apart We can then write a simpler definition of the electric
potential taking the initial potential energy to be zero, The negative sign on the work:
• If E does positive work then U < 0 • If E does negative work then U > 0
August 12, 2010
Thursday, September 9, 2010 3 Constant Electric Field Let’s look at the electric potential energy when we move a charge
q by a distance d in a constant electric field The definition of work is W =F·d For a constant electric field the
force is … so the work done by the electric field on the charge is W = eE · d = qEd cos θ
August 12, 2010
Thursday, September 9, 2010 University Physics, Chapter 23 4 Constant Electric Field Let’s look at the electric potential energy when we move a charge
q by a distance d in a constant electric field The definition of work is W =F·d For a constant electric field the
force is … so the work done by the electric field on the charge is W = eE · d = qEd cos θ
August 12, 2010
Thursday, September 9, 2010 University Physics, Chapter 23 4 Constant Electric Field  Special Cases If the displacement is in the same
direction as the electric field
• A positive charge loses potential energy
when it moves in the direction of the
electric field. If the displacement is in the
direction opposite to the electric
field
• A positive charge gains potential energy
when it moves in the direction opposite
to the electric field.
August 12, 2010
Thursday, September 9, 2010 University Physics, Chapter 23 5 Definition of the Electric Potential The electric potential energy of a charged particle in an electric
field depends not only on the electric field but on the charge of
the particle We want to define a quantity to probe the electric field that is
independent of the charge of the probe We define the electric potential as “potential energy per unit charge of a
test particle” Unlike the electric field, which is a vector, the electric
potential is a scalar
• The electric potential has a value everywhere in space but has no direction
• Units: Volt, symbol V, 1V = 1J/C
August 12, 2010
Thursday, September 9, 2010 6 Electric Potential V The electric potential, V, is defined as the electric
potential energy, U, per unit charge, V=U/q Actually, U = q V , potential energy is charge times
potential The electric potential is a characteristic of the electric
field, regardless of whether a charged object has been
placed in that field
(because U ∝ q) The electric potential is a scalar The electric potential is defined everywhere in space as a
value, but has no direction The electric potential is defined up to an arbitrary constant
August 12, 2010
Thursday, September 9, 2010 7 Electric potential ~ “mountain” in energy space.
Positive charges what to roll along the steepest
decent. ∆Φ = − E ∆x →
V 1.0 y
0 2 E = −∇Φ
V
2 E
0.5 0.0 2 Thursday, September 9, 2010 0
x 2 Negative charges what to roll along the steepest
decent of minus potential ∆Φ = − E ∆x →
V
E = −∇Φ
V
2 0
x 2 0.0
5.0 E 2 y
0 2 0.1 Thursday, September 9, 2010 Math Reminder  Partial Derivatives Given a function V(x,y,z), the partial derivatives are they act on x, y, and z independently Example: V(x,y,z) = 2xy2 + z3 Meaning: partial derivatives
give the slope along the
respective direction August 12, 2010
Thursday, September 9, 2010 33 Calculating the Field from the Potential (2) We can calculate any component of the electric field by taking the
partial derivative of the potential along the direction of that
component We can write the components of the electric field in terms of
partial derivatives of the potential as In terms of graphical representations of the electric potential, we
can get an approximate value for the electric field by measuring the
gradient of the potential perpendicular to an equipotential line August 12, 2010
Thursday, September 9, 2010 34 iClicker
An electric potential is
in volts.
Which of the following expressions describes the
associated electric field, in units of volts per meter?
a)
b)
c)
d)
e) Thursday, September 9, 2010 reminder: E = −∇V iClicker
An electric potential is
in volts.
Which of the following expressions describes the
associated electric field, in units of volts per meter?
reminder: a)
b)
c)
d)
e) E = −∇V Ex = −∂x V = 10x
Ey = − ∂ y V = 1
Ez = − ∂ z V = 1 Thursday, September 9, 2010 The Volt The commonly encountered unit joules/coulomb is
called the volt, abbreviated V, after the Italian physicist
Alessandro Volta (17451827) With this definition of the volt, we can express the units of
the electric field as For the remainder of our studies, we will use the unit V/m
for the electric field August 12, 2010
Thursday, September 9, 2010 10 Example: Energy Gain of a Proton (1) A proton is placed between two parallel
conducting plates in a vacuum as shown. +  Distance between the plates is d, Efield
inside is E. The proton is released from rest
close to the positive plate.
Question: What is the kinetic energy of the
proton when it reaches the negative plate? August 12, 2010
Thursday, September 9, 2010 11 Example: Energy Gain of a Proton (1) A proton is placed between two parallel
conducting plates in a vacuum as shown. +  Distance between the plates is d, Efield
inside is E. The proton is released from rest
close to the positive plate.
Question: What is the kinetic energy of the
proton when it reaches the negative plate? August 12, 2010
Thursday, September 9, 2010 11 Two ways to solve the problem
E d ∆Φ = EL
Vd Dynamical
v = at, x = at2 /2, E = mv 2 /2
qE
a=
m
qE t2
x=
m2
qE T 2
L=
d m2
2d
Lm
T=
qE
d
qE 2Lm
v=
m
qE E = mv 2 /2 = qEL
d Thursday, September 9, 2010 Energetic E = q ∆Φ
V Two ways to solve the problem
+ E  d ∆Φ = EL
Vd Dynamical
v = at, x = at2 /2, E = mv 2 /2
qE
a=
m
qE t2
x=
m2
qE T 2
L=
d m2
2d
Lm
T=
qE
d
qE 2Lm
v=
m
qE E = mv 2 /2 = qEL
d Thursday, September 9, 2010 Energetic E = q ∆Φ
V Example: Energy Gain of a Proton (2)
initial final Conservation of energy
ΔK = − ΔU = + 450 eV Because the acceleration of a charged particle across a potential
difference is often used in nuclear and high energy physics, the
energy unit electronvolt (eV) is common An eV is the energy gained by a charge e that accelerates across an
electric potential of 1 volt August 12, 2010
Thursday, September 9, 2010 12 ...
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This note was uploaded on 09/09/2011 for the course PHYS 241 taught by Professor Wei during the Fall '08 term at Purdue UniversityWest Lafayette.
 Fall '08
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