lecture 13 method of images steady currents
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8/2/2019 Lecture 13 Method of Images Steady Currents
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EECS 117
Lecture 13: Method of Images / Steady Currents
Prof. Niknejad
University of California, Berkeley
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Point Charge Near Ground Plane
Consider a point charge q sitting a distance d above aconductive ground plane. Wed like to find the electricfield E everywhere
Clearly E = 0 for z < 0 due to the conductor. Also, since = 0 everywhere except for point (0, 0, d). Thus we cansay that satisfies Laplaces equation for all points
z > 0 except (0, 0, d) where the point charge lives
d
q
z > 0
E = 0
(0, 0, d)
E = zEz
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Method of Images
Consider now a problem we have met before. If wehave two charges of equal and opposite magnitude atz = d, then the solution is trivial. How is this problem
related?Notice that for this new problem the electric field isnormal to the xy plane at z = 0
E(z = 0) =q(xx + yy + dz)
4(x2 + y2 + d2)3/2+
q(xx + yy dz)
4(x2 + y2 + d2)3/2+
Combining the terms we have a z-directed vector
E(z = 0) =q2dz
4(x2
+ y2
+ d2
)3/2
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The Image Charge
Now consider placing a conductive plane at z = 0. Dothe fields need to change? The boundary condition isalready satisfied by this field configuration. In other
words, this problem satisfies the same boundaryconditions as our original problem
This negative charge is the image charge that actually
solves our original problem!Recall that the solution of Poissons equation isuniquely determined by the boundary conditions. Since
this image problem satisfies the problem for z > 0 andalso the boundary conditions, its the solution!
Of course its only the solution for points z > 0 since we
know that E = 0 for z < 0
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E Field Plots
+
_
+
The vector field plot for a dipole distribution is shown on theleft. The upper half field plot is also the solution to themonopole charge placed near a ground plane.
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Surface Charge Density
For a conductive boundary, the normal component of Dis equal to the surface charge density
s(x,y, 0) = z E(z = 0) = 2qd4(x2 + y2 + d2)3/2
As expected, there is a negative charge density induced
on the surface of the plane
s(x,y, 0) =2qd
4(x2
+ y2
+ d2
)3
/2
-4 -2 0 2 4 6
s(x, 0, 0) =2qd
4(x2 + d2)3/2
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Forces on Dielectrics
Due to polarization, a dielectric will feel a force when inthe presence of a field
We can use the principle of virtual work to calculatehow much work it takes to slightly displace a dielectricin a field. This work must go into the energy of the field
Example: Consider the force on a dielectric insulator
placed part-way into a parallel plate capacitor. Letsassume that the plates of the capacitor are charged to avoltage difference of V0
++++++++++++++++++++++++
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
0F
d1
t
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Energy Calculation
The energy stored in the capacitor volume is (ignoringfringing fields)
E = 12
C1V20 + 12C0V20
=1
2
wd1
tV20 +
1
2
w( d1)0t
V20
Now displace the dielectric slightly to the right
E
=1
2
w(d1 + d)
t V2
0 +1
2
w( d1 d)0t V
2
0
The energy of the system has been increased due to
the additional amount of material polarization. Does thisenergy come from us pushing or the battery?
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Energy Change and Force
The energy difference between the final and initial stateis simply
E = E E = wdV2
0
2t( 0)
If the motion is incremental, we can say that E = dF
The force is therefore given by
F =2
2t
V20 ( 0) =1
2
wtE20( 0)
Its not clear at this point if the force is positive ornegative!
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Electrons in Metals
Electrons are the charge carriers in most metals. Theyare free to move unimpeded in the crystal at lowtemperatures. At high temperature the random vibration
(phonons) of the positive ion lattice can scatterelectrons. Likewise, the presence of impurities can alsoscatter electrons.
At thermal equilibrium with no external fields, therefore,the electrons are in motion with random direction andspeed. The net velocity of the ensemble of electrons iszero.
The thermal velocity is quite high ( 12
mv2 = kT2
) with
vth 107cm/s
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Drift Velocity
When a carrier in a conductor is accelerated by anexternal field, it initially travels in the direction of thefield. But after scattering with an impurity or phonon,
the electrons new trajectory is random andindependent of the field
Lets suppose that the average time between collisions
is tc. Then in this period the momentum of electrons isincreased by qE due to the external field, mvd/tc = qE.
The average velocity of carriers increases proportional
to the field Evd =
qtcm
E = E
The constant is the mobility and the drift velocity vdis so named because most of the motion is still randomas vd vth University of California, Berkeley EECS 217 Lecture 13 p. 11/
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Electron Drift
v =
1
N
i
vi = 0 v =
1
Ni
vi xvd
In the left figure, electrons are moving with random velocity(net motion is zero). On the right figure, electrons aremoving with random velocity plus a drift component to theright (drift has been exaggerated).
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Definition of Current
Current is defined as the amount of charge crossing aparticular cross sectional area per unit time. Consider adensity of N carriers carrying a charge q and moving
with drift velocity vd. How many cross a perpendicularsurface A?
In a time t, carriers move a distance of vdt. So any
carriers in the volume Avdt will cross the surface if theyare all moving towards it. The current is therefore
I =
AvdtNq
t = vdN qA
The vector current density J is therefore
J = N qvd
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Conductivity
Since we found that the velocity is proportional to thefield, it follows that J is also proportional
J = N qvd = N qE
The constant = qN is known as the conductivity ofthe material. It varies over several orders of magnitude
for conductors and dielectricsThis is in fact a statement of Ohms law V = RI inmicroscopic form. Often we work with the resistivity of
the material = 1
Since is very easy to measure, we can indirectlycalculate tc for a material. For a good conductor
107
S/m.
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Ohms Law
We assert that the ratio between the voltage andcurrent in a conductor is always constant. We see thisstems from J = E
I =
AJ dS =
A
E dS
V =CE d
The ratio is called the resistance
R =V
I=
CE dA E dS
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Resistance
The integrals over E depend only on the geometry ofthe problem and scale with voltage
Equivalently, if we increase V, the flux lines for E
remain the same but of a higher magnitude. Thus thecurrent J is the same and only I increases by the samefactor that V increases.
Example: Block of conductor. The electric field will beshown to be everywhere constant and parallel to theconductor
V = E0
I = AE0
J = E0 =I
A
R = VI = AUniversity of California, Berkeley EECS 217 Lecture 13 p. 16/
Ch C ti
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Charge Conservation
One of the fundamental facts of nature is chargeconservation. This has been experimentally observedand verified in countless experiments.
Therefore if we look at a volume bounded by surface S,if there is net current leaving this volume, it must beaccompanied by a charge pile-up in the region of
opposite polarity
SJ dS =
t
V
dV
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C t C ti it
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Current Continuity
We can transform this simple observation by employingthe Divergence Theorem
SJ dS =
V JdV
For any volume we have found that
V JdV
t
V
dV =
V
J +
t
dV = 0
If this is true for any volume V, it must be true that
J +
t
= 0
University of California, Berkeley EECS 217 Lecture 13 p. 18/
E Ch
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Excess Charge
Since J = E, we have
E +
t
= 0
If the conductivity is constant, we have
E +
t = 0
It is experimentally observed that Gauss law is satisfied
in all reference frames of constant velocity so E =
+
t= 0
University of California, Berkeley EECS 217 Lecture 13 p. 19/
E Ch ( t)
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Excess Charge (cont)
If we assume that is a function of time only, theequation
+
d
dt= 0
Let = / and solve the above equation
= 0e
t
Thus the charge density in a conductor decaysexponentially to zero with a rate determined by the time
constant = /, or the relaxation time of the material
Earlier we deduced that conductors have zero volumecharge density under electrostatic conditions. Now we
can estimate the time it takes to reach this equilibrium
University of California, Berkeley EECS 217 Lecture 13 p. 20/
R l ti Ti f G d C d t
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Relaxation Time for Good Conductors
In fact, a good conductor like copper has 108S/m
Its hard to measure the dielectric constant of a goodconductor since the effect is masked by the conductiveresponse. For 0, we can estimate the relaxationtime
10 1012
108s 1019s
This is an extremely short period of time!
How do electrons move so fast ? Note this result is
independent of the size of the conductor.
Answer: They dont, only a slight displacement in thecharge distribution can balance the net charge and thus
transfer it to the border region.
University of California, Berkeley EECS 217 Lecture 13 p. 21/
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