Lecture 13 Method of Images Steady Currents

8/2/2019 Lecture 13 Method of Images Steady Currents

1/21

EECS 117

Lecture 13: Method of Images / Steady Currents

Prof. Niknejad

University of California, Berkeley

University of California, Berkeley EECS 217 Lecture 13 p. 1/


2/21

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



3/21

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



4/21

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



5/21

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.



6/21

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



7/21

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



8/21

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?



9/21

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!



10/21

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



11/21

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/


12/21

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).



13/21

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



14/21

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.



15/21

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



16/21

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


17/21

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


C t C ti it


18/21

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


E Ch


19/21

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


E Ch ( t)


20/21

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


R l ti Ti f G d C d t


21/21

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.


Lecture 13 Method of Images Steady Currents

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