midterm 04 – LIN, KEVIN – Due: May 2 2008, 10:00 am
1
Electric force and electric Feld
Electric force between 2 point charges:

F

=
k

q
1

q
2

r
2
k
= 8
.
987551787
×
10
9
N m
2
/C
2
ǫ
0
=
1
4
π k
= 8
.
854187817
×
10
−
12
C
2
/N m
2
q
p
=
−
q
e
= 1
.
60217733 (49)
×
10
−
19
C
m
p
= 1
.
672623 (10)
×
10
−
27
kg
m
e
= 9
.
1093897 (54)
×
10
−
31
kg
Electric Feld:
v
E
=
v
F
q
Point charge:

E

=
k

Q

r
2
,
v
E
=
v
E
1
+
v
E
2
+
···
±ield patterns:
point charge, dipole,
b
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
⊥
=
v
E
·
ˆ
n
Δ
A
Gauss law:
Outgoing Flux from S,
Φ
S
=
Q
enclosed
ǫ
0
Steps:
to obtain electric ±eld
–Inspect
v
E
pattern and construct
S
–Find Φ
s
=
c
surface
v
E
·
d
v
A
=
Q
encl
ǫ
0
, solve for
v
E
Spherical:
Φ
s
= 4
π r
2
E
Cylindrical:
Φ
s
= 2
π r ℓE
Pill box:
Φ
s
=
E
Δ
A
, 1 side;
= 2
E
Δ
A
, 2 sides
Conductor:
v
E
in
= 0,
E
b
surf
= 0,
E
⊥
surf
=
σ
surf
ǫ
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
=
k Q
r
V
=
V
1
+
V
2
=
...
Energy of a chargepair:
U
=
k q
1
q
2
r
12
Potential di³erence:

Δ
V

=

E
Δ
s
b

,
Δ
V
=
−
v
E
·
Δ
vs
,
V
B
−
V
A
=
−
i
B
A
v
E
·
dvs
E
=
−
dV
dr
,
E
x
=
−
Δ
V
Δ
x
v
v
v
fix y,z
=
−
∂V
∂x
, etc.
Capacitances
Q
=
C V
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
E d
=
ǫ
0
A
d
Energy:
U
=
i
Q
0
V dq
=
1
2
Q
2
C
,
u
=
1
2
ǫ
0
E
2
Dielectrics:
C
=
κC
0
,
U
κ
=
1
2
κ
Q
2
C
0
,
u
κ
=
1
2
ǫ
0
κE
2
κ
Spherical capacitor:
V
=
Q
4
π ǫ
0
r
1
−
Q
4
π ǫ
0
r
2
Potential energy:
U
=
−
v
p
·
v
E
Current and resistance
Current:
I
=
dQ
dt
=
nq ±
d
A
Ohm’s law:
V
=
I R
,
E
=
ρJ
E
=
V
ℓ
,
J
=
I
A
,
R
=
ρℓ
A
Power:
P
=
I V
=
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
=
±
d
→
q E
m
τ
=
J
nq
→
ρ
=
m
nq
2
τ
Direct current circuits
V
=
I R
Series:
V
=
I R
eq
=
I R
1
+
I R
2
+
I R
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
,i
3
,...
–Node equations:
∑
i
in
=
∑
i
out
–Loop equations:
“
∑
(
±E
) +
∑
(
∓
iR
)=0”
–Natural:
“+” for looparrow entering
−
terminal
“
−
” for looparrowparallel to current ³ow
RC circuit:
if
dy
dt
+
1
R C
y
= 0,
y
=
y
0
exp(
−
t
R C
)
Charging:
E −
V
c
−
R i
= 0,
1
c
dq
dt
+
R
di
dt
=
i
c
+
R
di
dt
= 0
Discharge:
0 =
V
c
−
R i
=
q
c
+
R
dq
dt
,
i
c
+
R
di
dt
= 0
Magnetic Feld and magnetic force
μ
0
= 4
π
×
10
−
7
T m
/
A
Wire: