MIT OpenCourseWare 6.641 Electromagnetic Fields, Forces ...

MIT OpenCourseWare 6.641 Electromagnetic Fields, Forces, and Motion, Spring 2005

Please use the following citation format: Markus Zahn, 6.641 Electromagnetic Fields, Forces, and Motion, Spring 2005. (Massachusetts Institute of Technology: MIT OpenCourseWare). (accessed MM DD, YYYY). License: Creative Commons Attribution-Noncommercial-Share Alike.

Note: Please use the actual date you accessed this material in your citation.

For more information about citing these materials or our Terms of Use, visit:

6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn

Lecture 10: Solutions to Laplace's Equation In Cartesian Coordinates

I. Poisson's Equation

? E = 0 E = -

i E =

i

( ) =

2

= -

(Poisson's

Equation)

II. Particular and Homogeneous Solutions

2P

=

-

( ) r' dV '

( ) Poisson's Equation

P

r

=

V' 4 r - r'

2h = 0

Laplace's Equation

2

(P

+

h ) = -

= P + h must satisfy boundary conditions

III. Uniqueness of Solutions

Try 2 solutions a and b

2a 2b

=

-

=

-

2 ( a - b ) = 0

d = a - b 2d = 0 d = 0 i d d = d 2 d + d i d = d 2

i d d dV = d d i da = d 2 dV = 0

V

S

V

on S, d = 0 or d i da = 0

d = 0 a = b on S

d i da = 0

a = b n n

on SE

na = Enb on S

6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn

Lecture 10 Page 1 of 8

A problem is uniquely posed when the potential or the normal derivative of the potential (normal component of electric field) is specified on the surface surrounding the volume.

IV. Boundary Conditions

1. Gauss' Continuity Condition

Courtesy of Krieger Publishing. Used with permission.

( ) D i da = sfdS D2n - D1n dS = sfdS

S

S

( ) n

i

D2

-

D1

=

sf

2. Continuity of Tangential E

Courtesy of Krieger Publishing. Used with permission.

6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn

Lecture 10 Page 2 of 8

 ( ) E i ds = E1t - E2t dl = 0 E1t - E2t = 0

C

( ) n ? E1 - E2 = 0

Equivalent to 1 = 2 along boundary

V. Solutions to Laplace's Equation in Cartesian Coordinates, (x, y )

2 (x, y) =

2 + 2 = 0 x2 y2

1. Try product solution: (x, y ) = (x ) Y (y )

Y

(y )

dY2 (x )

dx2

+

X

(x )

d2 (y )

dy2

=

0

Multiply through by 1 :

XY

1 d2 X dx2

=

- 1 d2 Y Y dy2

=

- k

2

only a function of x

k=separation constant only a function of y

d2 = - k 2 dx2

;

d2 Y d y 2

=

k 2 Y

2. Zero Separation Constant Solutions: k=0

d2 dx2

=

0

=

a1 x

+

b 1

d2 Y dy2

=

0

Y

=

c1y

+

d 1

(x, y ) = Y = a2 + b2x + c2y + d2xy

3. Non-Zero Separation Constant Solutions: k0

d2 + k 2 = 0 dx2

X = A1 sin kx + A2 cos kx

6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn

Lecture 10 Page 3 of 8

d2 Y dy2

-

k 2 Y

=

0

Y = B1 eky + B2 e-ky = C1 sinh ky + D1 cosh ky

Courtesy of Krieger Publishing. Used with permission.

(x, y ) = (x ) Y (y ) = D1 sin kxeky + D2 sin kxe-ky + D3 cos kxeky + D4 cos kxe-ky

= E1 sin kx sinh ky + E2 sin kx cosh ky + E3 cos kx sinh ky + E4 cos kx cosh ky

6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn

Lecture 10 Page 4 of 8

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download