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LAPLACE’S EQUATION, POISSON’S EQUATION AND UNIQUENESS

THEOREM

CHAPTER 6

6.1 LAPLACE’S AND POISSON’S EQUATIONS

6.2 UNIQUENESS THEOREM

6.3 SOLUTION OF LAPLACE’S EQUATION IN ONE VARIABLE

6. 4 SOLUTION FOR POISSON’S EQUATION

6.0 LAPLACE’S AND POISSON’S EQUATIONS AND UNIQUENESS THEOREM

- In realistic electrostatic problems, one seldom knows the charge distribution – thus all the solution methods introduced up to this point have a limited use.

- These solution methods will not require the knowledge of the distribution of charge.

6.1 LAPLACE’S AND POISSON’S EQUATIONS

To derive Laplace’s and Poisson’s equations , we start with Gauss’s law in point form :

vED

VE Use gradient concept :

v

v

V

V

2Operator :

Hence :

(1)

(2)

(3)

(4)

(5) => Poisson’s equation

is called Poisson’s equation applies to a homogeneous media.

22 / mVV v

0v When the free charge density

=> Laplace’s equation(6)22 / 0 mVV

2

2

2

2

2

22

z

V

y

V

x

VV

In rectangular coordinate :

6.2 UNIQUENESS THEOREM

Uniqueness theorem states that for a V solution of a particular electrostatic problem to be unique, it must satisfy two criterion :

(i) Laplace’s equation

(ii) Potential on the boundaries

Example : In a problem containing two infinite and parallel conductors, one conductor in z = 0 plane at V = 0 Volt and the other in the z = d plane at V = V0 Volt, we will see later that the V field solution between the conductors is V = V0z / d Volt.

This solution will satisfy Laplace’s equation and the known boundary potentials at z = 0 and z = d.

Now, the V field solution V = V0(z + 1) / d will satisfy Laplace’s equation but will not give the known boundary potentials and thus is not a solution of our particular electrostatic problem.

Thus, V = V0z / d Volt is the only solution (UNIQUE SOLUTION) of our particular problem.

6.3 SOLUTION OF LAPLACE’S EQUATION IN ONE VARIABLE

Ex.6.1: Two infinite and parallel conducting planes are separated d meter, with one of the conductor in the z = 0 plane at V = 0 Volt and the other in the z = d plane at V = V0 Volt. Assume and between the conductors.

02 0v

Find : (a) V in the range 0 < z < d ; (b) between the conductors ;

(c) between the conductors ; (d) Dn on the conductors ; (e) on the conductors ; (f) capacitance per square meter.

E

D s

Solution :

0vSince and the problem is in rectangular form, thus

02

2

2

2

2

22

z

V

y

V

x

VV (1)

(a)

0

0

2

2

2

2

22

dz

V

dz

d

dz

Vd

z

VV

We note that V will be a function of z only V = V(z) ; thus :

BAzV

Adz

dV

Integrating twice :

where A and B are constants and must be evaluated using given potential values at the boundaries :

00

BVz

dVA

VAdVdz

/0

0

(2)

(3)

(4)

(5)

(6)

(7)

)(0 Vzd

VV

Substitute (6) and (7) into general equation (5) :

dz 0

)/(ˆˆ

ˆˆˆ

0 mVd

Vz

z

Vz

z

Vz

y

Vy

x

VxVE

(b)

)/(2

ˆ 200 mCd

VzED

(c)

)/(2

)ˆ(2

ˆˆ

2

ˆ2

ˆˆ

200

00

00

000

mCd

V

zd

VznD

d

V

zd

VznD

dzs

zs

(d) Surface charge :

0

/

V

ds

VQC

s

ab

)/(/2

/

2

0

00

0

2

mFdV

dV

VmC s

(e) Capacitance :

z = 0

z = d

V = 0 V

V = V0 V

Ex.6.2: Two infinite length, concentric and conducting cylinders of radii a and b are located on the z axis. If the region between cylinders are charged free and , V = V0 (V) at a, V = 0 (V) at b and b > a. Find the capacitance per meter length.

03

Solution : Use Laplace’s equation in cylindrical coordinate :

and V = f(r) only :

011

2

2

2

2

22

z

VV

rr

Vr

rrV

BrAVr

A

r

V

Ar

Vr

r

Vr

r

r

Vr

rrV

ln

0

012

and V = f(r) only :

(1)

BrAV ln

BbAV

BaAVV

br

ar

ln0

ln0

Boundary condition :

babV

Bba

VA

/ln

ln;

/ln00

Solving for A and B :

ab

rbVV

/ln

/ln0

Substitute A and B in (1) :

(1)

bra ;

rabr

VED

rabr

V

r

VrVE

ˆ/ln

ˆ/ln

ˆ

0

0

ab

rbVV

/ln

/ln0

abb

VrD

aba

VrD

brs

ars

/lnˆ

/lnˆ

0

0

Surface charge densities:

abV

b

ab

Va

brsbr

arsar

/ln

22

/ln

22

0

0

Line charge densities :

oab V

d

V

QC

Capacitance per unit length:

)/(/ln

2/

0

mFabV

mC

6/ and 0 0

VV 100 6/ EV and

Ex.6.3: Two infinite conductors form a wedge located at

is as shown in the figure below. If this region is characterized by charged free. Find . Assume V = 0 V at

and at .

z

x = 0

= /6

V = 100V

Solution : V = f ( ) in cylindrical coordinate :

01

2

2

2

2 dVd

rV

BAV

Ad

dV

d

Vd

0

2

2

/600

)6/(100

0

6/

0

A

AV

BV

Boundary condition :

Hence :

ˆ600

ˆ1

r

d

dV

rVE

600V

6/0 for region :

BAV

A

d

dV

Ad

dV

d

dV

d

d

d

dV

d

d

rV

2/tanlnsin

sin

0sin

0sinsin

12

2

= /10

= /6

V = 50 V

xy

z

6/ and 10/ E

Ex.6.4: Two infinite concentric conducting cone located at

10/ . The potential V = 0 V at

6/ and V = 50 V at . Find V and between the two conductors.

Solution : V = f ( ) in spherical coordinate :

2/tanlnsin

d

Using :

BAV 2/tanln BAV

BAV

12/tanln50

20/tanln0

6/

10/

Boundary condition :

Solving for A and B :

20/tan12/tan

ln

20/tanln50 ;

20/tan12/tan

ln

50

BA

1584.0

2/tanln1.95

20/tan

2/tanln

20/tan

12/tanln

50

V

ˆ

sin1.95

ˆ1

r

ddVr

VE

6/10/ Hence at region :

and

6. 4 SOLUTION FOR POISSON’S EQUATION

0v When the free charge density

Ex.6.5: Two infinite and parallel conducting planes are separated d meter, with one of the conductor in the x = 0 plane at V = 0 Volt and the other in the x = d plane at V = V0 Volt. Assume and between the conductors.

04 0v

Find : (a) V in the range 0 < x < d ; (b) between the conductors E

Solution :

BAxx

V

Axdx

dVdx

Vd

V v

2

20

0

02

2

2

V = f(x) :

2

2

0

00

2

00

0

d

d

VA

Add

VV

BV

dx

x

Boundary condition :

BAxx

V 2

20

dx 0In region :

xd

Vxd

xV 00

2

xxd

d

V

xdx

dVE

ˆ2

ˆ

00

;

xrv 1 and 0 Ex.6.6: Repeat Ex.6.5 with

BxAV

x

A

dx

dV

Adx

dVx

Vxdx

d

Exdx

d

E

D v

)1ln(

1

1

01

01

0

0

Solution :

)1ln(

)1ln(

0

0

0

0

d

VA

dAVV

BV

dx

x

Boundary condition :

xdx

Vx

dx

dVE

d

xVV

ˆ)1ln()1(

ˆ

)1ln(

)1ln(

0

0

dx 0In region :BxAV )1ln(