Name:
Physics 322 Final
May 11, 2023
Phys 322 Exam 2 formula sheet
Electric field of a distribution of charges:
~ r) =
E(~
Potential:
V (~b)
Z
V (~a) =
~b
~a
Legendre Polynomials:
Z
1
4⇡✏0
⇢(~r0 )
r
2
r̂
d⌧ 0
1
V (~r) =
4⇡✏0
~ · d~l
E
Z
1
P0 (x) = 1, P1 (x) = x, P2 (x) = (3x2
2
Z
Boundary conditions:
1
1
d cos ✓Pl (cos ✓)Pl0 (cos ✓) =
?
Eabove
?
Ebelow
=
✏0
~ ||
, E
above
2
2l + 1
⇢(~r0 )
r
d⌧ 0
1)
ll0
~ ||
E
below = 0
Multipole expansion:
V (~r) =
✓
◆
Z
1
1 X 1
1
Q p~ · ~r
0n
0
0 )d⌧ 0 =
~
r
P
(cos
↵
)⇢(
r
+
+
·
·
·
n
4⇡✏0
rn+1
4⇡✏0 r
r3
n=0
Bound charges:
~ · P~ ,
r
⇢b =
Electric displacement:
Linear media
~ = ✏0 E
~ + P~ ,
D
P~ =
= P~ · n̂
b
~ ·D
~ = ⇢f
r
~
e ✏0 E,
✏r =
✏
=1+
✏0
e
boundary conditions
?
Dabove
?
Dbelow
=
f,
~ //
D
above
~ //
~//
D
below = Pabove
Energy of a continuous charge distribution in matter
Z
1
~ ·E
~
W =
d⌧ D
2
current
I~ = ~v
~ = ~v
K
Magnetic force
~ =
F~ = q~v ⇥ B
Biot-Savart law
~ = µ0
B
4⇡
Z
8
Z
J~ = ⇢~v
~ dl
I~ ⇥ B
dI~0 ⇥
r
2
r̂
//
P~below
Name:
Physics 322 Final
May 11, 2023
Boundary conditions:
~ ||
B
above
?
?
Babove
= Bbelow
Vector potential
~ = µ0
A
4⇡
~ =r⇥A
~
B
~ above
@A
@n
~ above = A
~ below
A
Magnetic moment
m
~ =I
Vector potential of dipole field
H field
Z ~ ~0
J(r )
r
d⌧ 0
~ below
@A
=
@n
~
µ0 K
d~a
~ ⇥ r̂
~ dip = µ0 m
A
4⇡ r2
Torque on magnetic dipole
Bound currents
Z
~ ||
B
below = µ0 K
~ =m
~
N
~ ⇥B
~
J~b = r ⇥ M
~b = M
~ ⇥ n̂
K
~
~ = B
H
µ0
~ = J~f
r⇥H
~
M
Magnetic susceptibility of linear media
~ =
M
Ohm’s law
~
mH
~ = µH
~
B
µ = µ0 (1 +
~
J~ = E
Motional EMF
d
dt
E=
Inductance
= LI
Energy of magnetic fields
1
W =
2µ0
Maxwell’s equations
Z
B 2 d⌧
⇢
✏0
~
r·B = 0
~
@B
~ =
r⇥E
@t
~ =
r·E
~
~ = µ0 J~ + µ0 ✏0 @ E
r⇥B
@t
9
m)