1
5. The electric field from a uniformly charged conducting sphere with total charge Q and radius R,
kQ
e
for r > R
r2
Er =
(17)
0
for r < R
Coulomb law
F = ke
ke =
Q1 Q2
, r̂
r2
1
= 8.98 × 109 Nm2 /C2
4πo
(1)
(2)
Z
o = 8.85 × 10−12 C 2 /(N m2 )
B
∆V = VB − VA = −
(3)
E · dx
(18)
A
For a constant electric field Ez in the z direction
E=
F
qo
(4)
Q
r̂
r2
(5)
∆V = −E · d
= −Ez d cos(θ)
= −Ez z
(19)
(20)
(21)
you
WAB
= UB − UA = q(VB − VA )
(22)
∆U = q ∆V
(23)
ke Q
r
(24)
For a point charge
E = ke
For field lines
or
# of lines
E∝
Area
(6)
For a point charge
qE
a=
m
(7)
V (r) =
Gauss Law
W you = U = ke
q1 q2
q2 q3
q1 q3
+ ke
+ ke
r12
r23
r13
(25)
Z
ΦE =
E · dA
(8)
= E·A
for constant field
= EA cos(θ)
for constant field
ΦE = 4πke Qnet
Qnet
=
o
(9)
(10)
(11)
(12)
Ex = −
1. Uniformly Charged slab.
E=
σ
2o
2. Two charged slabs
σ
Inbetween the plates
E = o
0
Outside the plates
(13)
(14)
3. The electric field from a long line of charge
Er =
2ke λ
r
(15)
4. The electric field from a uniformly charged insulating sphere with total charge Q and radius R,
ke Q
for r > R
r2
(16)
Er = ke Q
r
for r < R
2
R R
∂V
∂x
V =
Ez = −
∂V (z)
∂z
(26)
Er = −
∂V (r)
∂r
(27)
Ey = −
X
∂V
∂y
∆q
ke
=
r
Ez = −
Z
ke
∂V
∂z
dq
r
(28)
(29)
1. Uniformly charged insulating sphere with charge Q
and radius R
ke Q
for r > R
r
(30)
V (r) = ke Q
2
3
−
r
/R
for
r<R
2R
2. Conducting Sphere with charge Q and radius R
ke Q
for r > R
(31)
V (r) = kerQ
for r < R
R
3. The potential of a charged disc of radius a
p
x2 + a2 − x
V = 2πke σ
(32)
2
4. For a ring of radius a
V = ke √
Q
x2 + a2
(33)
V = IR
(48)
P = I 2R
(49)
Capacitance
C∆V = Q
(34)
J=
1. For two plates of area A and separation d the capacitance is
C = o
A
d
I
= nqvd
A
J = σE
(35)
ρ=
2. For a coaxial cable of length L with inner radius a
and outer radius b the Capacitance is
L
C=
2ke ln(b/a)
1
σ
R=ρ
(36)
`
A
(50)
(51)
(52)
(53)
Series
with ke = 1/(4πo )
Req = R1 + R2 . . .
(54)
1
1
1
=
+
...
Req
R1
R2
(55)
Parallel:
Resistors in k
Ceq = C1 + C2 + C3 + . . .
(37)
Series:
1
1
1
1
=
+
+
+ ...
Ceq
C1
C2
C3
(38)
Kirchoff Laws
1. For each wire indicate a current with an arrow
2
U=
Q
1
C(∆V )2 =
2
2C
σind =
1−
1
κ
(41)
σo
(42)
C = κCo
(43)
p ≡ 2aq
(44)
where we have written (−I2 ) and (−I3 ) because
these currents are drawn exiting rather than entering the vertex.
3. For every closed loop, draw a circle and indicate
the loop direction. The sum of the potential drops
going around the loop is zero.
X
∆V = 0
(57)
(a) If the current is moving with loop direction
(Fig. 5) the voltage drop across the resistor is
(∆V )R = −IR
τ =p×E
(45)
U = −p · E
(46)
Currents and Circuits
I=
(56)
(40)
Eo
κ
2. The sum of the currents entering a vertex is zero.
Thus in Fig 4. the sum of the vertex
I1 + (−I2 ) + (−I3 ) = 0
U
1
= o E 2
Vol
2
E=
(39)
dQ
Charge passing through surface A
=
dt
∆t
(47)
(58)
If the current and loop direction are opposite
get +IR
3
(b) If the loop direction is the with the battery
(Fig. 3) the voltage change is
(∆V )E = +E
(59)
t
I(t) = −Io e− RC
(65)
with Io = Q/RC.
Forces and Magnetic Fields
If the loop and battery are opposite −E
(c) For each capacitor if the loop direction is in
the same as the current direction (Fig. 1) the
voltage drop is
(∆V )C = −
q
C
(60)
where q is the charge on the capacitor and C is
the capacitance. If current and loop direction
are opposite get +q/C.
Capacitor Charging
F = qv × B
r=
mv
qB
dF = IdL × B
(66)
(67)
(68)
For a uniform magnetic field
t
q(t) = Q (1 − e− RC )
E − t
I(t) =
e RC
R
(61)
(62)
with
• Q = CE .
F = ILCD × B
(69)
where LCD is the line connecting C to D. A corrlary is
that a closed loop in a uniform magnetic field experiences
no net force (it does experience a torque though).
= IA
(70)
τ =µ×B
(71)
U = −µ · B = −µB cos(θ)
(72)
• Io = E /R.
τ = RC
(63)
Capacitor Discharging
t
− RC
q(t) = Qe
(64)
4