Phys Lecture Laplace

7/31/2019 Phys Lecture Laplace

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Laplace and Poissons

Equations


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Derivation

We have the differential form of Gauss law

Using D = E and E = - in the above equation, we get

This is the Poissons Equation

. D

V

2D = V =-E V

2V

or


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20V

In free space , this equation becomes,

This equation is called the Laplace equation.

is called the Laplacian operator or simply Laplacian.

Note that the del operator is defined only in the

rectangular coordinates only, as

i j k

x y z

2


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The Laplacian in the three coordinate systems are

2 2 2

2

2 2 2

= ( )cartesian

x y z

2 2

2

2 2 2

1

1= ( )Cylindrical

z

2

2 2

2 2 2 2 2

1 1

1= sin ( )

sin sinr Spherical

r r r r r


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Applications

one dimensional solution of Laplace equation

2 2 2

2

2 2 2

= = 0 ( )V V V

V cartesian

x y z

--- (1)

V is a function of only one variable and is independent of theother two variables. Under this condition, in rectangular

coordinate system the Laplace equation reduces to


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ne dimensional solution of Laplace Equation in rectangular

coordinate system

Let V be a function of z only. Then in Rectangular coordinatesystem,the Laplaces Equation reduces to

2

2

20

VV

z

Since V is a function of z only, it is independentof x and y. Therefore,

Integrating both sides once we get

--- (2)


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Equation (3) represents a family of equi potential surfaceswith z taking up constant values

dVA

dz

Integrating both sides once again, we get

, A is an arbitrary constant --- (3)

V Az B , B is an arbitrary constant --- (4)

A and B are arbitrary constants to be evaluated under suitable

boundary conditions.


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Consider two such equi potential surfaces one at z = z1 and theother at z = z2. Let V= V1 at z = z1 and V = V2 at z = z2.

We immediately recognize that this is the case with a parallel

plate capacitor with a plate separation of z1 - z2 = d and apotential difference V1 - V2 .

Applying the above two conditions, called boundary conditions,we get,

1 1

2 2

V V Az B

V V Az B

--- (5)

--- (6)


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Solving equations (5) and (6) we getthe values for A and B as

2 1

2 1

1 2 2 1

2 1

V VA

z z

V z V z B

z z

Substituting the values of A and B in equation (4) we get,

2 1 1 2 2 1

2 1 2 1

2 1 1 2

2 1

( ) ( )

V V V z V z V zz z z z

V z z V z z

z z

--- ( 7)


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Further let, for simplicity, V1 = 0 and z1 = 0, V2 = Va

Then equation (7) reduces to

aV

V zd

We find that V is a linear function of z

V = V2 = Va

V = V1 = 0

d

z = z2

z = z1Fig 1 Parallel plate capacitor

--- (8)


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1 Given V, Determine E using the formula E = - V

2 Determine D using D = E

3 Find D on any one of the plates, D = DS= DS aS = DN aNon the chosen plate, and recognising that DN = S

4 Determine Q by surface integration of S over the surfacearea of the chosen plate using

5 Compute the capacitance using the formulaa

QC

V

S

S

Q dS


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Applying these five steps to the parallel plate capacitor we get,

0;

a

az

a

z

aS z S S N N N S

z

a a a

S S

zV V

d

VE V a

d

V

D E adV

D D a D a D a D d

V V VQ dS dS S

d d d

Therefore the capacitance of the parallel plate capacitor is

a

Q SC

V d


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We have, in cylindrical coordinates, the Laplace equation as

2 2

2

2 2 21 0

1V= ( )V V V Cylindricalz

One dimensional solution of Laplace Equation in cylindrical

coordinate system


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We consider that V is a function of only. In this casethe Laplace equation in cylindrical coordinates reduces to

0

V

Integrating this equation once we get,

10

V

V V AA o r

, , A an arbitrary constant

Case 1:


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Integrating once again, we get

V A l n B, B an arbitrary constant

From this equation, we observe that equipotential surfaces

are given by = constant and are cylinders. Example of theproblem is that of a coaxial capacitor or coaxial cable.

Let us create the boundary conditions by choosing V = Va at = a and V = 0 at = b, b > a.

Then we get from the above equation,

BbnlAVV

BanlAVV

b

a


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Solving these two equations for A and B we get

b

a

bVaVBdna

b

a

VVA abba

ln

lnln

ln

Substituting these the values of A and B in the generalExpression for V , we get

b

a

bVaV

b

a

VVV abba

ln

lnlnln

ln

Letting Vb = 0, we get


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a

b

b

VVa

ln

ln

ab

a

bVab

a

bVVVE aa ln

ln

ln

ln


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Therefore

dnaaDaDa

a

ba

VD

Then

Da

a

b

VED

a

ab

VE

NNNSa

S

Sa

a

ln

1

ln

1

ln

1

i.e.,

Hence

a

bV

aD

aSN

ln

1


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Therefore,

1

ln

12

ln

S S

S of one of S of one of the plates the plates with a with a

a

a

Q dS dS

V Sbaa

V a Lba

a

over a length L meters of the

coaxial cable

a

b

L

V

QC

a ln

2


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Therefore the capacitance per unit length of the line CL is

a

bCL

ln

2 F/m

Next we consider V as a function of only. In this case

The Laplace equation in cylindrical coordinate systemreduces to

2

2

2 2

10

V=

V

Case 2:


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i.e.,

2

2 2

1

0 d V

d

Excluding the value = 0 this equation becomes

2

20

d V

d

Integrating both sides we get

dV

Ad

Integrating once again, we get V A B

This is the general equation for V when V is a function of only.


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From this equation, we observe that equipotential surfacesare given by = constant and are planes To visualize this, choose twosuch equipotential surfaces, V = Va at = and V = 0 at =0.

Insulating gap

V = Vaat =

V = 0at =0


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For the chosen boundary condition, we get

andaVA

Thus the general expression for V becomes

0 0 0

aV V A B

V A B B

aV

V


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Once again we follow the 5 step procedure to determine

the capacitance of the system1 1

aVV

E V

Note that E is a function of and not of . But the vector field EIs a function of . Now,

a

VD E

a

NNSSS

VaDaDD


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S

a

N

V

D

The surface integration on S gives Q:

S

a

S

a

S

S dSVdsVdSQ

and we get immediately the value for the capacitance ofthe corner reflector system as

Sa

S

a

a

dSV

dSV

V

QC


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Next we consider the Laplace Equation in spherical coordinates:

2

2 2

2 2 2 2 2

1 1 1sin 0

sin sin

V V VV r

r r r r

In this system we consider that V is a function of r only.Then the Laplace equation reduces to

01 2

2

2

r

V

rrV

Again we exclude r = 0 from our solutions. Multiplying bothSides by r2 we get

02

r

Vr

--- ( a)

--- ( b)

One dimensional solution of Laplace Equation in spherical

coordinate system

Case 1:


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Integrating once again, we get

2 Vr A

r

2

V Aor

r r

Ao r V B

r

where A and B are arbitrary constants to be evaluated. This

equation represents a family of equi potential surfaces forr = constant.

Let us choose two such equipotential surfaces at r = a andR = b, b > a , such that at r = a, V = Va and at r = b, V = Vb

--- ( c)

--- ( d)


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a

b

AV B

a

A

V Bb

Solving these two equations we get

--- ( e)

--- ( f)

1 1

a bV VA

a b

and

1 1

1 1

a bV Vb aB

b a

We immediately recognise that this is the example of concentric

spheres or Spherical capacitor


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Substituting the values of A and B in equation (d), we get,

1 1

1

1 1 1 1

a ba b

V VV V b a

Vr

a b b a

1 1

1

1 1 1 1

b aa b

V VV V a b

o r Vr

a b a b

Let Vb = 0 Then equation (g) becomes

--- (g)


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1 1

1 1a

r b

V V

a b

--- (h)

2

1 1

1

1 1 1 1

a

a

Vr bE V V r

r

a b a b

2

1 1

aV

D E rr

a b


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2

1 1

aS S S N N

VD D D a D a r

ra b

21 1

a

N S

VD

a

a b

Therefore the charge Q on the capacitor plate is

2 21 1 1 1

a a

S

S S S

V VQ dS dS dS

a a

a b a b


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Therefore we get the expression for the spherical capacitor as

2

2

4 4

1 1 1 1

a

a a

Q V aC

V a V

a b a b

4

1 1C

a b

For an isolated sphere , i.e., as b we get

4C a


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Finally let us consider V as a function of only . In this caseThe Laplaces equation reduces to

2

2

10

V= sinsin

V

r

We exclude r = 0 and = n/2, n = 0, 1, 2, 3, . . .

Multiplying both sides of the above equation by r2 sin, we get

0

sinV

Case 2:


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Integrating the equation with respect to we get,

or

sin V A

sin

V AA is an arbitrary constant

Integrating once again, we get,

2 ln(tan / )V A B

This equation represents a family of equipotential surfaces forconstant . Let us consider two such equipotential surfaces at =/2 and = and let V = 0 at =/2 and V = Va at =0.

--- (i)


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The equipotential surfaces are cones as shown in figure below.Such a system is called a conical antenna

Insulating gap

Ground plane

V = Va at =

V = 0 at = 0

Applying these two boundary conditions to the equation (i),Solving for A and B and substituting these values in (i),we get,

2

2

ln(tan / )

ln(tan / )aV V


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We have 2

2

ln(tan / )

ln(tan / )aV V

We use E = -

V to find the field strength, as1

2

sin (tan / )

aVV

E V a ar r ln

Next we determine D using D = 0 E as

2

sin (tan / )

aV

D E ar ln


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2

sin (tan / )

aS

V

r ln

The total charge Q on the conical surface is therefore,

2

0 2

sinsin (tan / )

aS

S o

VQ ds r d dr

r ln

0

2

2

(tan / ) a

VQ dr

ln


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This equation leads to an infinite value of charge and capacitance.

Therefore we have to consider a cone of finite size.

Our expression for Q is approximate, since, theoretically, the

potential surface = extends from r = 0 to r = . But our physicalconical surface extends from r = 0 to say, r = r1 . The approximatecapacitance is

12

2

(cot / ) a

rC

ln

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Phys Lecture Laplace