Class 14

Boundary Value Problems

Boundary Value Problems

• So far the electric field has been obtained using Coulomb’s law or Gauss law where the charge distribution is known throughout the region or by using where the potential distribution is known. In practical problems the charge or potential is known only at some boundaries and it is desired to know the field or potential throughout the region. Such problems are tackled using Poisson or Laplace equation.

VE −∇=

Poisson and Laplace Equations

VED ρε =∇=∇

..

VE −∇=

VV ρε =∇−∇ ).(

EquationPoissonV V

ερ−=∇2

EquationLaplaceV

regionfreeechaFor

0

arg2 =∇

Poisson and Laplace Equation

• The Laplace equations in all the 3 coordinate systems are as given below

in Cartesian coordinates

in cylindrical coordinates

In spherical coordinates

2

2

2

2

2

22

z

A

y

A

x

AA

∂∂+

∂∂+

∂∂=∇

2

2

2

2

22 1

)(1

z

AAAA

∂∂+

∂∂+

∂∂

∂∂=∇

φρρρ

ρρ

2

2

2222

22

sin

1)(sin

sin

1)(

1

φθθθ

θθ ∂∂+

∂∂

∂∂+

∂∂

∂∂=∇ A

r

A

rr

Arrr

A

General Procedure for solving Laplace or Poisson Equation:

• Solve Laplace or Poisson equations for V by (a) direct substitution for single variable or (b) by method of separation of variables for more than one variable. The solution at this point is not unique because of the integration constants

• Apply the boundary conditions to determine the integration constants giving a unique solution for V.

• Having found V, find and .• If desired find the charge Q induced on a conductor

surface using and where Dn is the component of D normal to the conductor. If necessary the capacitance between two conductors can be found using C=Q/V.

VE −∇=

ED

ε=

∫= dSQ sρ ns D=ρ

Practice Example 6.1

• In a one dimensional device, the charge density is given by . If at x=0 and V=0 at x=a, find V and E.

ax

v0ρρ = 0=E

Solution 6.1

ερ

ερ

a

x

x

V v 02

2

−=−=∂∂

1

20

2C

a

x

x

V +−=∂∂

ερ

21

30

6CxC

a

xV ++−=

ερ

Solution 6.1

• Substituting V=0 at x=a we get

21

30

60 CaC

a

a ++−=ε

ρ

1

20

2C

a

xa

x

VVE x −=

∂∂−=−∇=

ερ

0;00 1 =∴== CxatE

( )3303

02 66

xaa

Va

aC −=∴=

ερ

ερ

xaa

xE

ερ2

20=

Practice Exercise 6.3

• Two conducting plates of size 1 x 5m are inclined at 450 to each other with a gap of width 4mm separating them as shown in the figure. Determine an approximate value of the charge per plate, if the plates are maintained at a potential difference of 50V. Assume the medium between them has .5.1=rε

Solution 6.3

• The potential varies only with respect to .

excludedisV

V 0;01

2

2

22 ==

∂∂=∇ ρ

φρ

212

2

;0 CCVV +==

∂∂ φ

φ000 2 =⇒== CVatφ

ππφ 200

504 1 =⇒== CVat

φπ200=∴V

φπρφρa

VVE

2001 −=∂∂−=−∇=

Solution 6.3

The gap between the plates is 4mm. Straight lines are extended from the near ends to meet at O as shown in the figure. The distance from O to the tip of the horizontal plate is found as follows.

sr aaED ρρπ

ερπ

εεε φφ =−=

−××== 0

00

3002005.1

mmll

2265.55.22sin

2;2

5.22sin ===

∫ ∫∫ ∫ ∫ =−===5

0

1

00523.0

05

0

1

00523.0

2.22300

arg nCdzd

dzddSeCh ss ρρ

περρρ