lecture 9 electrostatics and poisson’s equation

8/2/2019 Lecture 9 Electrostatics and Poissons Equation

1/30

EECS 117

Lecture 9: Electrostatics and Poissons Equation

Prof. Niknejad

University of California, Berkeley

Universit of California Berkele EECS 117 Lecture 9 . 1/


2/30

Review of Divergence

If we try to measure the flux at a point, we are temptedto compute

SA dS is a small volume surrounding a

point

This flux will diminish, though, as we make the volumesmaller and smaller. We might suspect that normalizingby volume will solve the problem

In fact, the divergence of a vector is defined as follows

divA limV0

SA dS

V

We usually write

divA = A



3/30

Divergence in Rectangular Coor. (I)

x

y

z r

dxdy

dz

Consider the flux crossing a small differential volume atposition (x0, y0, z0). We can break up the flux SA dSinto several components=

x

x/2

+

x+

x/2

+

y

y/2

+

y+

y/2

+

z

z/2

+

z+

z/



4/30

Divergence in Rectangular Coor. (II)

Without loss of generality, consider the flux crossing thesurfaces in the xz plane, or the third and fourth terms inthe above calculation

The flux for these terms is simply

= Ay(x0, y0 y

2

, z0)xz + Ay(x0, y0 +y

2

, z0)xz

x

y

z

dx

dy

dzA

Ay



5/30

Divergence in Rectangular Coor. (III)

For a differential volume this becomes

SA dS = xzyAy

y

Thus the total flux is simply

SA dS = xyz Ax

x+ Ay

y+ Az

z

For rectangular coordinates, we have therefore proved

that

div A =

Axx

+Ayy

+Azz



6/30

Divergence in Rectangular Coor. (IV)

In terms of the operator

= x

x

+ y

y

+ z

z

Since the dot product of the operator and the vectorA gives

A = Ax

x+ A

y

y+ A

z

z

We can rewrite divA as simply A



7/30

Gauss Law (again)

Applying the definition of divergence to the electric fluxdensity D, we have

limV0

SD dSV

= qinsideV

= = D

Therefore we have the important result that at any given

point D = (x,y,z)

This is the analog to the equivalent statement that we

derived last lectureSD dS =

V

dV



8/30

Divergence Theorem

The Divergence Theorem is a direct proof of thisrelationship between the volume integral of thedivergence and the surface integral (applies to any

vector function A)V AdV =

SA dS

Here V is any bounded volume and S is the closedsurface on the boundary of the volume

Application of this theorem to the electric flux density Dimmediately gives us the differential form of Gauss lawfrom the integral form



9/30

Perfect Conductors (I)

Perfect conductors are idealized materials with zeroresistivity. Common metals such as gold, copper, aluminumare examples of good conductors.

A fuzzy picture of a perfect conduc-tor is a material with many mobilecharges that easily respond to an ex-ternal fields. These carriers behaveas if they are in vacuum and they canmove unimpeded through the metal.These mobile charges, though, can-

not leave the material due to a largepotential barrier (related to the workfunction of the material).

+ + + + +

+ + + + +

+ + + + +

+ + + + +

+ + + + +

+ + + + +



10/30

Perfect Conductors (II)

One may argue that the electric field is zero under staticequilibrium based on the following argument. Under staticequilibrium charges cannot move. So we might argue that if

an external field is applied, then all the charges in a perfectconductor would move to the boundary where they wouldremain due to the potential barrier. But this argumentignores the fact that the external field can be canceled byan internal field due to the rearrangement of charges (seefigure below).

+

+

+

+

+

+

+

+

+

+

+

+



11/30

Perfect Conductors (III)

Thus we have to invoke Gauss law to further prove thatin fact there can be no net charge in the body of aconductor. Because if net charge exists in the body,

Gauss law applied to a small sphere surrounding thecharge would require a field in the body, which wealready hypothesized to be zero. Therefore ourargument is consistent.

In conclusion, we are convinced that a perfect metalshould have zero electric field in the body and no netcharge in the body. Thus, we expect that if any netcharge is to be found in the material, it would have to beon the boundary.



12/30

Perfect Conductors (IV)

We could in fact define a perfect conductor as amaterial with zero electric field inside the material. Thisis an alternative way to define a perfect conductor

without making any assumptions about conductivity(which we have not yet really explored)

A perfect conductor is also an equipotential material

under static conditions, or the potential is everywhereconstant on the surface of a perfect conductor

This is easy to prove since if E 0 in the material, then

E d is likewise zero between any two points in thematerial body.



13/30

Tangential Boundary Conditions (I)

Its easy to show that the electric field must cross thesurface of a perfect conductor at a normal angle. Since

CE d 0

For any path C, choose a path that partially crosses into the

conductor. Since E = 0 inside the conductor, the onlycontribution to the integral are the side walls and Et, or thetangential component along the path.

In the limit, we can make thepath smaller and smaller un-til it is tangent to the surfaceand imperceptibly penetrates

the material so the side wallcontributions vanish.

C

E = 0

E

t

n



14/30

Tangential Boundary Conditions (II)

CE d = Et 0

We are thus led to conclude that Et 0 at the surface

Another argument is that since E = and

furthermore since the surface is an equipotential, thenclearly Et is zero since there can be no change inpotential along the surface


N l B d C di i (I)


15/30

Normal Boundary Conditions (I)

Consider a pill-box hugging the surface of a perfectconductor as shown below

E = 0

E

t

n


N l B d C di i (II)


16/30

Normal Boundary Conditions (II)

The electric flux density is computed for this surface.Like before, well make the volume smaller and smalleruntil the sidewall contribution goes to zero. Since the

bottom is still in the conductor and the field is zero, thisterm will not contribute to the integral either. The onlyremaining contribution is the normal component of thetop surface

D dS = DndS = Qinside = dSs

We have thus shown that Dn = s


E l S h i l Sh ll


17/30

Example: Spherical Shell

+Q

Neutral Conductor

Consider a charge placed at the center of a spherical

shell made of perfect conducting materials

Lets find the surface charge density on the inner andouter surface of the conductor


E l ( t)


18/30

Example (cont)

Lets apply Gauss law for the dashed sphere lyinginside the conductor. Since E 0 on this surface, thecharge inside bus likewise be zero Qinside = 0 which

implies that there exists a uniform charge density (bysymmetry) of inner = Q/Si where Si = 4a

2

If we now consider a larger sphere of radius r > b, since

the sphere is neutral, the net charge is just the isolatedcharge Q. Thus

Dr4r2 = Q + insideS1

Q

+outsideS2 = Q

Thus outside = Q/S2 =Q

4b2


E l Fi ld Sk t h


19/30

Example: Field Sketch

The fields have been sketched in the figure below. Radialsymmetry is preserved and an induced negative andpositive charge density appear at the shell surfaces

+

___ _ _

__

_ __

__

_

__

_

_

_

_

_

__

_

_

___

_____

____

__

__

___

_

_

_

_

__

____

____

++++++++

++

+++++++++++++++

+++

++

+++

++++++++++++++

+++

++

++

+++

++++++++++++++++

++++

++++++


E l El t i Fi ld S


20/30

Example: Electric Field Summary

We have found the following relations to hold

Dr =Q

4r2 r < a

Dr = 0 a r bDr =

Q4r2 r > b

A plot is shown below

a b

1

r2

Er

(r)

r


E ample: Potential


21/30

Example: Potential

Lets compute the potential from the fields. Take thereference point at to be zero and integrate along aradial path. For r > b

=

r

Erdr =Q

4r

Likewise for a r b, since Er is zero for the pathinside the conducting sphere, the potential remains

constant at Q40b until we exit the conductor

Finally, once outside the conductor for r < a wecontinue integrating

= Q4b ra

Q40r2dr


Example: Potential (cont)


22/30

Example: Potential (cont)

Thus we have

=Q

40 1

b

+1

r

1

aA plot of the potential is shown below. Note thatpotential on the conductor is non-zero and its value

depends on the amount of charge at the center

a b r

r(r)


G d d Sh ll


23/30

Grounded Shell

What if we now ground the shell? That means that = 0 on the surface of the conductor. The work done inmoving a point charge from infinity to the surface of the

shell is thus zero. Choose a radial path

=

r

Erdr = 0

Since the function Er monotonic, Er = 0 everywhereoutside the sphere! A grounded spherical shell acts like

a good shield.If Er(b) = 0, then s = 0 at the outer surface as well.


Charged Spherical Shell


24/30

Charged Spherical Shell

To find the charge on the inner sphere, we can use thesame argument as before unchanged soinside = Q/S1

But now the material is not neutral! Where did thecharge go?

Imagine starting with the ungrounded case where the

positive charge is induced on the outer surface. Thenground the sphere and we see that the charge flows outof the sphere into ground thus charging the material

negative


Charged Shell Field Sketch


25/30

Charged Shell Field Sketch

The fields have been sketched in the figure below. Radialsymmetry is preserved and an induced negative chargedensity appears at inner shell surfaces. The field outside

the shell is zero.

+

___ _ _

__ _

__

__

__

_

_

_

_

_

__

__

_

___

_____

____

__

_

____

_

_

_

__

_

______

__


Poissons Equation (aka Fish Eq )


26/30

Poisson s Equation (aka Fish Eq.)

We already have an equation for calculating E and directly from a charge distribution

If we know E, then we can find by simply taking the

divergence: = E. If we know is it possible topredict ? Sure:

= E =

The operator A is a new operator and we call it the

Laplacian 2

Thus we have Poissons Equation 2 =

In a charge free region we have 2 = 0. This is known

as Laplaces equation


Laplacian in Rect Coordinates


27/30

Laplacian in Rect. Coordinates

Lets untangle the beast operator

=

x

x +

y

y +

z

z

x

x +

y

y +

z

zOK, all cross products involving mutual vectors die (e.g.x y 0) so we have

2 =

2

x2+

2

y2+

2

z2


Example: Parallel Plate Structure


28/30

Example: Parallel Plate Structure

+ + + + + + + + + + + + + + + + + + + + + + + +

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _EV1

s,t

s,b

Consider a battery connected to a grounded parallelplate

Suppose the plates are made of perfect conductors. Ifthe plates are large, then we expect to be a functionof z alone (ignore the edge effects). Thus the operator

2

degenerates and inside the plates we have 0

z

d

2

z2= 0


Example (cont)


29/30

Example (cont)

Thus = az + b. The boundary conditions on the topand bottom plate determine a and b uniquely. Since(z = 0) = 0 we have b = 0. Also since (z = d) = V1,

a = V1/d

(z) =V1d

z

And the field is easily computed

E = = z

z

V1d

z

= z

V1d


Example (cont)


30/30

Example (cont)

The field is constant and points in the z direction.Since the charge density on the inner plate surface isequal to the normal component of the field there

s,b = n E = V1d

s,t = + V1d


lecture 9 electrostatics and poisson’s equation

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