PHY481: Electromagnetism

PHY481: Electromagnetism
Solving Laplace’s equation via “Separation of Variables”
1) Cartesian coordinates
2) Spherical coordinates
3) Cylindrical coordinates
Examples
Lecture 20
Carl Bromberg - Prof. of Physics
Cartesian coordinates

Determine nature of solutions on the rectangle (unbounded z)
– Grounded at x = –a/2 or +a/2 then Xk(x) = Akcos(kx) +Bksin(kx)
– Grounded at y = –a/2 or +a/2 then Yk(y) = Akcos(ky) +Bksin(ky)
– Other solution is then Uk(u) = Ckcosh(ku) + Dksinh(ku) (same k)
 Use periodic (sine or cosine) solution and boundary conditions to
determine legal values of k = nπx/a (same for both solutions)
 Generate general solution V(x,y) = Σ Xk(x)Yk(y)
 Use additional boundary conditions to determine coeficients
– Periodic B.C.: Vb(x0,y) = Σ Xk(x0)Yk(y) or Vb(x,y0) = Σ Xk(x)Yk(y0)
– Apply Fourier Integral to both sides: 1 a
Be sure to split Fourier integral (-a,-a/2,a/2,a)
# Vb (x, y0 )cos m! x a (or y version)
a
"a
– Use orthogonality on right side: a
cos ( n! x a ) cos ( m! x a ) = $ mn (or y version)
#
– Determine coef. Ak,Bk,Ck,Dk
"a
 Substitute into V(x,y) = Σ Xk(x)kY(y)
Lecture 20
Carl Bromberg - Prof. of Physics
1
From previous Exam 2
Boundary conditions V (x, ± a 2) = 0, x ! 0 V (0, y) = 2V y / a V (x, y) ! 0, as x ! "
0
Y Solution Y ( y) = Acos ( k! y a ) + Bsin ( k! y a )
Apply boundary condition on Y solution
y
V (0, y) = 2V0 y a , an odd function ! A = 0.
+a 2
Y (± a 2) = Bsin ( ±k " 2 ) = 0 ! k = 2n, an even integer.
Apply boundary condition on X solution (same k)
2n! x a
"2n! x a
V (0, y) =
X (x) = Ce
+ De
; V (x, y) # 0, as x # $, % C = 0
Most general solution consistent with boundary conditions
!
V (x, y) = " K n sin ( 2n# y a ) e
$2n# x a
n=1
Lecture 20
!
$
n=1
x
y
V =0
!a 2
n=1
a
Solution V (x, y) =
a
V (0, y) = " K n sin ( 2n# y a )
2V
1
K n = " V (0, y)sin ( 2n# y a ) dy = 4 20
a !a
a
#
2V0
!
Determine coefficients
2V0
V =0
( "1)n+1
n
a 2
"
0
n+1
2V0 $ ( !1) '
%
(
y sin ( 2n# y a ) dy =
# & n )
sin ( 2n! y a ) e
"2n! x a
Carl Bromberg - Prof. of Physics
2
Spherical coordinates: V(r, θ) = R(r)Θ(θ)
Laplace’s equation for potential in Spherical coordinates (no phi dep.)
1 # $ 2 #V '
1
# $
#V '
! V (r," ) = 2 & r
)( + 2
&% sin "
)(
%
#r
#"
r #r
r sin " #"
2
Separate solutions
Sep. of variables
V (r,! ) = R(r)"(! )
B
!
R(r) = Ar +
r
(
!+1
B! %
"
!
V (r,! ) = ) $ A! r + !+1 ' P! ( cos! )
r &
!=0 #
Boundary condition on a sphere
V (a,! ) = f (! )
1
# V (a,! )Pn (cos! )d (cos! )
"1
Lecture 20
!(" ) # F(cos" ) = P! (cos" )
Legendre Polynomials
P0 ( cos! ) = 1
P1 ( cos! ) = cos!
P2 ( cos! ) = ( 3cos ! " 1) 2
2
P3 ( cos! ) = (5cos ! " 3cos! ) 2
3
Orthogonality
1
# Pm (cos! ) Pn (cos! ) d (cos! ) =
"1
Carl Bromberg - Prof. of Physics
2$ mn
2n+1
3
Familiar problem in spherical coordinates
1
Orthogonality of Legendre polynomials:
# Pm (cos! ) Pn (cos! ) d (cos! ) =
"1
Consider a grounded conducting sphere radius, a, in a
constant external field, E0 in z direction
E field
Boundary condition V = 0 on sphere
#
(
V (a,! ) = 0 = $ A! a + B! a
!=0
!
" ( !+1)
2$ mn
2n+1
)P (cos! )
!
Multiply by Pn ( cos! ) and integrate
1
B! ' 1
$
!
&% A! a + !+1 )( # P! ( cos! ) Pn ( cos! )d ( cos! )
# V (a,! )Pn (cos! )d (cos! ) = 0 = +
a
!=0
"1
"1
Boundary condition
n
! ( n+1)
2n+1
V (r ! ",# ) = $ E0 z = $ E0 r cos#
0 = An a + Bn a
2 ( 2n + 1) Bn = ! An a
*
cos! = P1 (cos! )
!!=1
(
A1 = ! E0
)
uniform field
(
B! %
"
!
V (r,! ) = ) $ A! r + !+1 ' P! ( cos! )
r &
!=0 #
Lecture 20
dipole
dipole moment
3
= ! E0 r cos"
+
E0 a cos"
r
2
Carl Bromberg - Prof. of Physics
3
E0 a =
p
4!" 0
4
Another spherical coordinate problem
Disk, charge density σ, radius R. Determine potential on z-axis, then at all angles.
!
V (r,0) = V (z,0) =
4"# 0
2"
%
0
R
d$ %
0
(z
&d &
2
+&
)
2 12
=
! ( 2
2 12
)(z + R ) ' z *+
2# 0
(
Use an expansion of it to
B! %
"
!
V
(r,
!
)
=
A
r
+
P! ( cos! )
) $# !
!+1 '
find potential everywhere.
&
r
!=0
2
2 12
(z + R )
(
= z 1+ R
2
z
)
2 12
(
2
2
4
4
)
= z 1 + R 2z ! R 8z + ...
2
! $ 2
!R
2 12
&
V (z,0) =
(z
+
R
)
#
z
=
%
'
2" 0
2" 0
2
$1
&
R
( # 3 + …)
% 2z 8z
'
Match coefficients
2
"R
All A! = 0 V (r,! ) =
2# 0
2
1
R
B0 = ; B2 = !
2
8
Lecture 20
2
f (! ) = 1 + f "(0)! + f ""(0) ! 2
12
f = (1 + ! )
f " = (1 + ! )
; f (0) = 1
#1 2
f "" = #(1 + ! )
*$ B0 '
$ B2 '
+& 1 ) P0 ( cos! ) + & 3 ) P2 ( cos! ) .
%r (
,% r (
/
2
"# R
V (r,! ) =
4#$ 0
12
(1 + ! )
2 12
(1 + x )
/ 2 ; f "(0) = 1 2
#1 2
/ 4 ; f ""(0) = #1 4
2
= 1+ ! 2 " ! 8
2
4
= 1+ x 2 ! x 8 +…
,.% 1 ( % R 2 ( % 3cos! + 1(
0.
-'& *) + ' 3 * '&
*) + …1
& 4r )
2
./ r
.2
Carl Bromberg - Prof. of Physics
5
From previous Exam 2: Ring of charge
Potential on z axis
3
#( x $ )d x $
Q ( 2
Q
2 %1 2
2
2 %1 2
)
(1 + a z )
=
& x % x$ = 4!" z + a
4!" 0 z
0
Expansion in powers of a/z let ! = " 2 = a 2 z 2
1
V (z) =
4!" 0
f (! ) = (1 + ! )
"1 2
; f #(! ) = " 12 (1 + ! )
"3 2
; f ##(! ) = + 43 (1 + ! )
"5 2
2
4
'
Q $
a
a
V (z) =
& f (0) + f #(0) 2 + f ##(0) 4 + …)
(
4!" 0 z %
z
2z
2
4
2
4
'
'
Q $
a
3a
Q $1 a
3a
=
& 1 * 2 + 4 + …) =
& * 3 + 5 + …)
%
(
(
4!" 0 z
4!" 0 % z 2z
2z
8z
8z
Potential in spherical coordinates with boundary condition at large r
"
V(r,# ) = & B! r
V(r ! ",# ) = 0 $ A! = 0
!=0
% ( !+1)
P! ( cos# )
Matching coefficients for solutions on the z-axis
"
V(z,0) = # B! z
!=0
Lecture 20
! ( !+1)
2
4
)
Q &1 a
3a
=
+ 3 + 5 +… =
( ! 3 + 5 + …+
*
z
4$% 0 ' z 2z
z
z
8z
B0
B2
B4
Carl Bromberg - Prof. of Physics
6
Ring of charge, continued
Matching coefficients for solutions on the z-axis
2
4
)
Q &1 a
3a
V(z,0) = # B! z
=
+ 3 + 5 +… =
( ! 3 + 5 + …+
' z 2z
*
z
4
$%
z
z
8z
!=0
0
No odd ! solutions
"
! ( !+1)
B0
B2
B4
2
Q
Q a
Q 3a
B! = 0 for odd !, B0 =
; B2 =
; B4 =
4!" 0
4!" 0 2
4!" 0 8
4
Insert coefficients into general solution
2
4
(
Q %1 a
3a
V ( r,! ) =
' $ 3 P2 ( cos! ) + 5 P4 ( cos! ) + …*
)
4"# 0 & r 2r
8r
Lecture 20
Carl Bromberg - Prof. of Physics
7
From previous Exam 2
Surface charge density ! (" ) = ! 0 cos " . Find Potential inside and out
! ( x" ) d x"
V (z) = $
x # x"
3
=
2 2&
%0R
4&' 0
% R
= 0
2' 0
1
"
!1
$
0
3
2
2
for z < ! R
2
2
2
=
1
2 2
3R z
1
2 2
3R z
1
# ! R 2 ! 2Rxz + z 2 ( R 2 + Rxz + z 2 ) %
$
& !1
#$ ! ( R ! z )2 [ R 2 + Rz + z 2 ] + ( R + z )2 [ R 2 ! Rz + z 2 ]%&
= {! ( R ! z )[ R + Rz + z
] + ( R + z )[ R2 ! Rz + z 2 ]} 3R2 z 2 = 2z 3R2
2
2
2
2
2 2
2
= {! ( z ! R )[ R + Rz + z ] + ( R + z )[ R ! Rz + z ]} 3R z = 2R 3z
2
2
2
2
2 2
2
= {( z ! R )[ R + Rz + z ] ! ( R + z )[ R ! Rz + z ]} 3R z = !2R 3z
z < R : V (z) = [! 0 3" 0 ] z ;
Lecture 20
2
= z + R ! 2Rz cos#
R + z # 2Rzx
=
for z > R
2
2
2
x ! x " = ( z ! Rcos# ) + R sin #
xdx
R + z ! 2Rzx
for ! R < z < R
2
R + z # 2Rz cos)
0
2 1
$
= (# 0 cos$ ) R sin $ d$ d%
cos) sin ) d)
d( $
#1
2
2
&
xdx
2
! ( x " ) d x " = # ($ ) d x "
2
2
3
'2
z > R : V (z) = #$! 0 R 3" 0 %& z ;
3
!2
z < ! R : V (z) = ! $%" 0 R 3# 0 &' z
Carl Bromberg - Prof. of Physics
8
Cylindrical coordinates
Laplace’s equation in cylindrical coordinates
Separation of variables
2
1 # $ #V ' 1 # V
! V (r," ) =
&% r
)( + 2 2 = 0
r #r
#r
r #*
2
V (r,! ) = R(r)"(! )
r dependence φ dependence
2
2
r 2
r # $ #R ' 1 # *
! V (r," ) =
=0
&% r )( +
2
V
R #r #r
* #"
Constants sum to zero
n
2
!n
Separate solutions
! n (" ) = Cn cos n" + Dn sin n"
n
Rn (r) = An r + Bn r
2
!n
R0 (r) = Aln r + B
Solution with series to insure boundary conditions can be satisfied
#
(
V (r,! ) = Aln r + B + $ An r + Bn r
n=1
Lecture 20
n
"n
)(C cos n! + D sin n! )
n
n
Carl Bromberg - Prof. of Physics
9
Cylinder problem
Half cylinders, radius R. V = +V
Find potential inside and out.
#
(
V (r,! ) = Aln r + B + $ An r + Bn r
n=1
n
"n
0
on right half, and V = -V
0
on left half
)(C cos n! + D sin n! )
n
n
Boundary conditions at r = 0 and r = ∞
"
VInt (R,! ) = VExt (R,! )
VInt (r,! ) = # An r cos n!
n
n
An R = Bn R
n=1
"
VExt (r,! ) = # Bn r
$n
cos n!
n=1
"
VInt (r,! ) = # cn
n=1
"
VExt (r,! ) = # cn
n=1
r
R
R
r
cos n!
Lecture 20
; Bn = cn R
n
#% V 1st & 4rd quadrants
f (! ) = $ 0
nd
th
&%"V0 2 & 3 quadrants
cos n!
sin ( (2 j + 1) ! 2 ) = ("1)
cn
Be careful to break Fourier
integral into pieces !!
n
n
= cn
R
Use Orthogonality
n
n
An =
!n
j
cj =
4V0 (!1)
j
" (2 j + 1)
Orthogonality
n
2"
$ cos n! cos m! = "# nm
0
Complete solution
( #1) j r 2 j+1
VInt (r,! ) =
cos(2 j + 1)!
%
" j=0 2 j + 1 R 2 j+1
4V0
$
( #1) j R 2 j+1
VExt (r,! ) =
cos(2 j + 1)!
%
2 j+1
" j=0 2 j + 1 r
4V0
$
Carl Bromberg - Prof. of Physics
10
From previous Exam 2 (Method of Images)
Two charges above grounded plane.
Determine potential and force on charge q.
P
z
q
Image
-d
–q
Potential above the plane
1
1
$
+
&
( x + d )2 + ( z # d )2
q & ( x # d )2 + ( z # d )2
V (x,0, z) =
%
4!" 0 &
1
1
#
#
&
( x # d )2 + ( z + d )2
( x + d )2 + ( z + d )2
'
d
q
+d
d
x
–q
(
&
&
)
&
&
*
2
$ î
k̂
î + k̂ '
#
#
% 2
(
2
2
& 4d
4d
8d 2 )
2
$*
q
2- *
2- '
%, 1 #
=
î
#
1
+
/. ,+
/. k̂ (
2 &+
)
4
4
16!" 0 d
q
F=
4!" 0
Lecture 20
Carl Bromberg - Prof. of Physics
11
Capacitor problem from previous Exam 2
Cylindrical capacitor d = b ! a << a For Va=V, determine C, Q, and U.
i) Means spacing very small
ii) Means cylinder areas are nearly the same
iii) Field between the cylinders must be ~ the
field between parallel plates with the same area.
Initial state
2
2
1Q
Q (b ! a)
2!" 0 aL
! 0 A 2"! 0 aL
U=
=
Q=
V
C=
=
2 C
4"# 0 aL
b# a
d
b# a
Force to pull inner cylinder out changing the overlap. dL = L! " L (negative)
dL
dU
dL
W = dU = !U
=!
L
U
L
Without initial assumptions
V (r) =
C=
V0 ln ( b r )
ln ( b a )
E(r) =
dU
U
=!
dL
L
V0
ln ( b a ) r
E(a) =
2
dU U Q ( b ! a )
F=!
= =
2
dL L
4"# 0 aL
V0
ln ( b a ) a
=
!
Q
=
" 0 2#" 0 aL
Q 2!" 0 L 2!" 0 aL because
ln ( b a ) = ln [( a + d ) a ] = ln [1 + a / d ] ! ( b " a ) a
=
#
V ln ( b a )
b$ a
Lecture 20
Carl Bromberg - Prof. of Physics
12