exam 03 – ANDERSON, ZACH – Due: Apr 3 2008, 11:00 pm
1
E & M  Basic Physical Concepts
Electric force and electric Feld
Electric force between 2 point charges:

F

=
k

q
1

q
2

r
2
k
=8
.
987551787
×
10
9
Nm
2
/C
2
?
0
=
1
4
π k
.
854187817
×
10

12
C
2
/N m
2
q
p
=

q
e
=1
.
60217733 (49)
×
10

19
C
m
p
.
672623 (10)
×
10

27
kg
m
e
=9
.
1093897 (54)
×
10

31
kg
Electric Feld:
A
E
=
>
F
q
Point charge:

E

=
k

Q

r
2
,
A
E
=
A
E
1
+
A
E
2
+
···
±ield patterns:
point charge, dipole,
±
plates, rod,
spheres, cylinders,
...
Charge distributions:
Linear charge density:
λ
=
Δ
Q
Δ
x
Area charge density:
σ
A
=
Δ
Q
Δ
A
Surface charge density:
σ
surf
=
Δ
Q
surf
Δ
A
Volume charge density:
ρ
=
Δ
Q
Δ
V
Electric ²ux and Gauss’ law
±lux:
ΔΦ =
E
Δ
A
⊥
=
A
E
·
ˆ
n
Δ
A
Gauss law:
Outgoing Flux from S,
Φ
S
=
Q
enclosed
<
0
Steps:
to obtain electric ±eld
–Inspect
A
E
pattern and construct
S
–Find Φ
s
=
±
surface
A
E
·
d
A
A
=
Q
encl
<
0
, solve for
A
E
Spherical:
Φ
s
=4
π r
2
E
Cylindrical:
Φ
s
=2
π r @ E
Pill box:
Φ
s
=
E
Δ
A
, 1 side;
= 2
E
Δ
A
, 2 sides
Conductor:
A
E
in
= 0,
E
±
surf
= 0,
E
⊥
surf
=
σ
surf
<
0
Potential
Potential energy:
Δ
U
=
q
Δ
V
1 eV
≈
1
.
6
×
10

19
J
Positive charge moves from high
V
to low
V
Point charge:
V
=
kQ
r
V
=
V
1
+
V
2
=
Energy of a chargepair:
U
=
kq
1
q
2
r
12
Potential di³erence:

Δ
V

=

E
Δ
s
±

,
Δ
V
=

A
E
·
Δ
As
,
V
B

V
A
=

²
B
A
A
E
·
dAs
E
=

dV
dr
,
E
x
=

Δ
V
Δ
x
³
³
³
fix y,z
=

∂V
∂x
, etc.
Capacitances
Q
=
CV
Series:
V
=
Q
C
eq
=
Q
C
1
+
Q
C
2
+
Q
C
3
+
,
Q
=
Q
i
Parallel:
Q
=
C
eq
V
=
C
1
V
+
C
2
V
+
,
V
=
V
i
Parallel platecapacitor:
C
=
Q
V
=
Q
Ed
=
<
0
A
d
Energy:
U
=
²
Q
0
V dq
=
1
2
Q
2
C
,
u
=
1
2
?
0
E
2
Dielectrics:
C
=
κC
0
,
U
κ
=
1
2
κ
Q
2
C
0
,
u
κ
=
1
2
?
0
κ E
2
κ
Spherical capacitor:
V
=
Q
4
π<
0
r
1

Q
4
0
r
2
Potential energy:
U
=

A
p
·
A
E
Current and resistance
Current:
I
=
dQ
dt
=
nq v
d
A
Ohm’s law:
V
=
IR
,
E
=
ρJ
E
=
V
=
,
J
=
I
A
,
R
=
ρ=
A
Power:
P
=
IV
=
V
2
R
=
I
2
R
Thermal coe´cient of
ρ
:
α
=
Δ
ρ
ρ
0
Δ
T
Motion of free electrons in an ideal conductor:
aτ
=
v
d
→
qE
m
τ
=
J
nq
→
ρ
=
m
2
τ
Direct current circuits
V
=
Series:
V
=
eq
=
1
+
2
+
3
+
,
I
=
I
i
Parallel:
I
=
V
R
eq
=
V
R
1
+
V
R
2
+
V
R
3
+
,
V
=
V
i
Steps:
in application of Kirchho²’s Rules
–Label currents:
i
1
,i
2
3
,...
–Node equations:
∑
i
in
=
∑
i
out
–Loop equations:
“
∑
(
²E
)+
∑
(
∓
iR
)=0”
–Natural:
“+” for looparrow entering

terminal
“

” for looparrowparallel to current ³ow
RC circuit:
if
dy
dt
+
1
RC
y
= 0,
y
=
y
0
exp(

t
)
Charging:
E 
V
c

Ri
= 0,
1
c
dq
dt
+
R
di
dt
=
i
c
+
R
dt
=0
Discharge:
0=
V
c

=
q
c
+
R
dt
,
i
c
+
R
dt
Magnetic Feld and magnetic force
μ
0
π
×
10

7
Tm
/
A
Wire:
B
=
μ
0
i
2
Axis of loop:
B
=
μ
0
a
2
i
2(
a
2
+
x
2
)
3
/
2
Magnetic force:
A
F
M
=
i
A
@
×
A
B
→
qAv
×
A
B
Loopmagnet ID:
Aτ
=
i
A
A
×
A
B
,
Aμ
=
iA
ˆ
n
Circular motion:
F
=
mv
2
r
=
q v B
,
T
=
1
f
=
2
v
Lorentz force:
A
F
=
q
A
E
+
×
A
B
Hall e³ect:
V
H
=
F
M
d
q
,
U
=

·
A
B
Sources of
A
B
and magnetism of matter
BiotSavart Law:
Δ
A
B
=
μ
0
4
π
i
Δ
>
=
×
ˆ
r
r
2
,
B
=
μ
0
4
π
q>v
×
ˆ
r
r
2
Δ
B
=
μ
0
4
π
i
Δ
y
r
2
sin
θ
,
sin
θ
=
a
r
,
Δ
y
=
r
2
Δ
θ
a
Ampere’s law:
M
=
±
L
A
B
·
dAs
=
μ
0
I
encircled
Steps:
to obtain magnetic ±eld
–Inspect
A
B
pattern and construct loop
L
–Find
M
and
I
encl
, and solve for
A
B
.