Transcript I-5
I-5 Special Electrostatic Fields
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Main Topics
•
•
•
•
Electric Charge and Field in Conductors.
The Field of the Electric Dipole.
Behavior of E. D. in External Electric Field.
Examples of Some Important Fields.
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A Charged Solid Conductor I
• Conductors contain free charge carriers of one or
both polarities. Charging them means to introduce
in them some excess charges of one polarity.
• A special case are metals :
• every atom which joins metal structure, often crystallic,
keeps some of its electrons in its vicinity but the
valence electrons, which are bounded by the weakest
forces, are shared by the whole structure and they are
the free charge carriers. They can move within the
crystal when electric (or other) force is acting on them.
• It is relatively easy to add some excess free electrons to
metal and also to take some out of it.
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A Charged Solid Conductor II
• Adding electrons means charging the metal
negatively.
• Taking some electrons out means charging it
positively.
• For our purposes we can consider the ‘holes’
left after missing electrons as positive free
charge carriers each with charge +1e.
• So effectively the charged metal contains
excess charges either negative or positive,
which are free to move.
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A Charged Solid Conductor III
• Excess charges repel themselves and since
they are free to move as far as to the
surface, in equilibrium, they must end on a
surface.
• In equilibrium there must be no forces
acting on the charges, so the electric field
inside is zero and also the whole solid
conductor must be an equipotential region.
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A Hollow Conductive Shell I
• In equilibrium again:
• the charges must remain on the outer surface.
• the field inside is zero and the whole body is an
equipotential region.
• The above means the validity of the Gauss’
law.
• To proof that let’s return to the Gauss’ law.
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The Gauss’ Law Revisited I
• Let us have a positive point charge Q and a
spherical Gaussian surface of radius r
centered on it. Let us suppose radial field:
kQ
E (r ) p
r
• The field lines are everywhere parallel to
the outer normals, so the total flux is:
e E (r ) A 0 Qr
1
2 p
• But if p2 the flux would depend on r !
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The Gauss’ Law Revisited II
• The validity of the Gauss’ law p = 2.
• By using a concept of the solid angle it can be
shown that the same is valid if the charge Q is
anywhere within the volume surrounded by the
spherical surface.
• By using the same concept it can be shown that
the same is actually valid for any closed surface.
• It is roughly because from any point within some
volume we see any closed surface confining it
under the solid angle of 4.
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A Hollow Conductive Shell II
• Let first the shell be spherical. Then the charge
density on its surface is constant.
• From symmetry, in the center the intensities from
all the elementary surfaces that make the whole
surface always compensate themselves and E 0
• For any other point within the sphere they
compensate themselves and E 0 only if p = 2.
• Again, using the concept of solid angle, it can be
shown, the same is valid for any closed surface.
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A Hollow Conductive Shell III
• Conclusion: The existence of a zero electric
field within a charged conductive shell is
equivalent to the validity of the Gauss’ law.
• This is the principle of:
• experimental proof of the Gauss’ law with a
very high precision: p – 2 = 2.7 3.1 10-16.
• of shielding and grounding (Faraday’s cage).
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Electric Field Near Any
Conducting Surface
• Let us take a small cylinder and submerge it into
the conductor so its axis is perpendicular to the
surface.
• The electric field
• within the conductor is zero
• outside is perpendicular to the surface
• A non-zero flux is only through the outer cup
E
0
• Beware the edges! is not generally constant!
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The Electric Dipole I
• Materials can produce non-zero electric fields in
their vicinity even when the total charge in them is
compensated.
• But they must contain so called electric multipoles
in which the centers of gravity of positive and
negative charges are not in the same point.
• The fields produced are not centrosymmetric and
decrease generally faster than the field of the
single point charge.
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The Electric Dipole II
• The simplest multipole is the electric dipole.
• It is the combination of two charges of the same
absolute value but different sign +Q and –Q.
• They are separated by vector l , starting in –Q.
• We define the dipole moment as :
p Ql
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The Electric Dipole II
• Electric dipoles (multipoles) are important
because they are responsible for all the
electrical behavior of neutral matter.
• The components of material (molecules,
domains) can be polar or their dipole
moment can be induced.
• Interactions of dipoles are the basis of some
types of (weaker) atomic bonds.
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Behavior of the Electric Dipole in
External Electric Fields
• In uniform electric fields the dipoles are
subjected to a torque which is trying to turn
their dipole moments in the direction of the
field lines
• In non-uniform electric fields the dipoles
are also dragged.
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Some Examples
• The field of homogeneously charged sphere
• Parallel uniformly charged planes
• Electrostatic xerox copier
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Homework
• Now, you should be able to solve all the
problems due Monday!
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Things to read
• The first five lectures cover :
Chapters 21, 22, 23 !
• Advance reading
Chapter 24 - 1, 2, 3
• Who is really interested should try to see
the physicist “Bible”:
“The Feynman Lectures on Physics”
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The Solid Angle I
• Let us have a spherical surface of radius r.
From its center we see an element of the
surface da under a solid angle d :
da
d 2
r
We see the whole spherical surface under :
4r
2 4
r
2
The Solid Angle II
If there is a point charge Q in the center the
elementary flux through da is:
da cos
d e E da E da cos kQ
2
r
Since the last fraction is d, the total flux is:
e kQ d kQ4
Q
0
^
Intensities near more curved
surfaces are stronger!
• Let’s have a large and a small conductive spheres
R, r connected by a long conductor and let’s
charge them. Charge is distributed between them
to Q, q so that the system is equipotential:
2
2
Q q a r
Q r R
R
; 2
2
R r A R
S R r
r
^
Potential of Electric Dipole I
• Let us have
and
a
a charge –Q at the origin
+Q in dl. What is the potential in r ? We use
the superposition principle and the gradient:
( r ) ( r ) ( r dl )
kQ kQ
kQ
( dl ) grad (
)
r
r
r
How to calculate grad(1/r)?
• r is the distance from the origin :
1
2
2
2 12
r x y z (x y z )
r
2
2
2
• e.g. the first components of the gradient is :
2 12
3
( x y y )
x
2
2
2
1
2
( 2 )( x y y ) 2 x 3
x
r
2
2
Potential of Electric Dipole II
• The first two terms cancel:
kQdl r kp r
(r )
3
3
r
r
• The potential has axial symmetry with the
dipole in the axis and axial anti-symmetry
perpendicular to it. It decreases with 1/r2!
^
Electric Dipole - The Torque
• Let us have a uniform field with intensity E
Forces on both charges contribute
simultaneously to the torque:
l
T 2 QE sin
2
• The general relation is a cross product:
T p E
^
Electric Dipole - The Drag
• Let us have
a non-uniform field with
intensity E and a dipole parallel to a field
line (-Q in the origin).
F QE (0) QE (dl )
dE
QE (0) QE (0) Qdl
dx
• Generally:
F gradE p
^
The vector or cross product I
Let c a b
Definition (components)
ci ijk a j bk
The magnitude c
c a b sin
Is the surface of a parallelepiped made by a , b.
The vector or cross product II
The vector c is perpendicular
to the plane
made by the vectors a and b and they have to
form a right-turning system.
ux
uy
uz
c ax
ay
az
bx
by
bz
ijk = {1 (even permutation), -1 (odd), 0 (eq.)}
^