Download Recall: The Electric Field

Recall: The Electric Field
§ A charge creates an
electric field around itself
and the other charge feels
that field.
+
+
Test charge q
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Review - Point Charge
The electric field created by a
point charge, as a function of
position r,
The force exerted by an electric
field on a point charge q located at
position r
… direction: tangent to the field line
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Electric field points away from the positive charge
E
+q
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Electric field points toward the negative charge
E
-q
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Continuous Distribution of charges
§ Two methods:
• Method #1: Divide the distribution into infinitesimal elements dE and
integrate to get the full electric field
• Method #2: If there is some special symmetry of the distribution, use
Gauss’ Law to derive the field from
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Calculating E-field from Coulomb’s law
§ From a point charge
§ From continuos distribution of charges
Integrate over volume
E=k
Z
kdq
dE = 2 r̂
r
dq
r̂
2
r
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E-field from a uniformly charged rod
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E-field from a uniformly charged rod
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E-field from a uniformly charged rod
§ Simmetry:
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Electric Field from a Ring of Charge
Total charge Q is distributed
homogeneously over a ring of
radius R. Find E-field on the
axis at distance z.
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d
Take section
Its charge is
d
d
dq =
Q
2⇡
It produces E-field
Edq
dq
d
Q
=k 2
=
k
R + z2
2⇡ R2 + z 2
From symmetry: only Ez is important
Ez,dq
d
Q
z
p
= Edq cos ✓ = k
2⇡ R2 + z 2 R2 + z 2
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d
Q
z
p
Ez,dq = Edq cos ✓ = k
2⇡ R2 + z 2 R2 + z 2
Qz
Integrate over d
E=k
3/2
2
2
(R
+
z
)
E
0.3
/ 1/z
0.2
2
0.1
z/R
0.5
1.0
1.5
2.0
2.5
3.0
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Field on the axis from a charged disk
from previous:
Z
Qx
Ex = k 2
(a + x2 )3/2
dQx
Ex = k 2
(a + x2 )3/2
dQ = 2 ⇥ada
a=R
z
Ex = k
2 ⇥a 2
da
=
2 )3/2
(a
+
z
0
!
1
2 k⇥ 1 p
1 + (R/x)2
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Field on the axis from a charged plate
Ex = 2 k⇥ 1
R!1
p
1
1 + (R/x)2
!
Ex = 2 k⇥
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Gauss’ Law
§ Two methods fro calculating E-field from continuous
distribution of charges:
• Method #1: Divide the distribution into infinitesimal elements dE and
integrate to get the full electric field
• Method #2: If there is some special symmetry of the distribution, use
Gauss’ Law to derive the field from
Gauss’ Law
The flux of electric field
through a closed surface is
proportional to the charge
enclosed by the surface.
First we need to
define the
concept of flux
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Water Flux
§ Let’s imagine that we put a ring with area A perpendicular to a stream of water flowing with velocity v
§ The product of area times velocity, Av, gives the volume of water passing through the ring per unit time
• The units are m3/s
§ If we tilt the ring at an angle θ, then the
projected area is Acosθ, and the volume of water
per unit time flowing through the ring is Av cosθ.
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Electric Flux (1)
§ We call the amount of water flowing through the ring the “flux of
water”
!
§ We can make an analogy with electric field lines from a constant
electric field and flowing water
!
!
!
§ We call the density of electric field lines through an area A the
electric flux
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Electric Flux (2)
§ Consider a constant electric
field E passing through a
given area A
§ The angle θ is the angle
between the electric field
vector and the area vector
§ The density of electric field
vectors passing through a
given area A is called the
electric flux
N.B.: Nothing is actually flowing!
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Surfaces and Normal Vectors
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Electric Flux for Closed Surface
§ Assume an electric field and a closed surface, rather than
the open surface associated with our ring analogy
!
§ In this closed-surface case, the total electric flux through
the surface is given by an integral over the closed surface
!
!
§ The differential area vectors
always point out of the closed
surface
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Gauss’s Law
§ We can now formulate Gauss’s Law as
§ Flux through a closed area equals the enclosed charge
§ q is the total charge inside a closed surface
• We call this surface a Gaussian surface
• This surface could be our imaginary box
• This surface could be any closed surface
§ Usually we chose this closed surface to have symmetries
related to the problem we are trying to study
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Gauss’s Law (5)
§ Gauss’s Law (named for German mathematician and
scientist Johann Carl Friedrich Gauss, 1777-1855)
states
!
• (q = net charge enclosed by Gaussian surface S).
§ If we add the definition of the electric flux we get
another expression
for
Gauss’
Law
Z
!
!
q
E · dA =
✏0
§ Gauss’s Law: the electric field flux through S is
proportional to the net charge enclosed by S
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Gauss’s Law and Coulomb’s Law Are
Equivalent
§ Let’s derive Coulomb’s Law from Gauss’s Law
§ We start with a point charge q
§ We construct a spherical surface with radius r surrounding
this charge
• This is our “Gaussian surface”
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Gauss’ Law and Coulomb’s Law (2)
§ The electric field from a point charge is radial, and thus is
perpendicular to the Gaussian surface everywhere
!
!
!
!
Z
q
E · dA =
✏0
§ The electric field has the same magnitude anywhere on the
surface, directed radially, along the local normal
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Gauss’ Law and Coulomb’s Law (3)
§ Now we are left with a simple integral
over a spherical surface
§ So for Gauss’ law related to a point charge
we get
!
!
§ Which gives
!
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End of lecture
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