2.5 Electric charges and fields
ε
0
=
{\displaystyle \varepsilon _{0}=}
8.85×10−12 F /m = vacuum permittivity.
e = 1.602×10−19 C : negative (positive) charge for electrons (protons)
k
e
=
1
4
π
ε
0
=
{\displaystyle k_{e}={\tfrac {1}{4\pi \varepsilon _{0}}}=}
= 8.99×109 m/F
F
→
=
Q
E
→
{\displaystyle {\vec {F}}=Q{\vec {E}}}
where
E
→
=
1
4
π
ε
0
∑
i
=
1
N
q
i
r
P
i
2
r
^
P
i
{\displaystyle {\vec {E}}={\tfrac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\tfrac {q_{i}}{r_{Pi}^{2}}}{\hat {r}}_{Pi}}
E
→
=
∫
d
q
r
2
r
^
{\displaystyle {\vec {E}}=\int {{\tfrac {dq}{{r}^{2}}}{\hat {r}}}}
where
d
q
=
λ
d
ℓ
=
σ
d
a
=
ρ
d
V
{\displaystyle dq=\lambda d\ell =\sigma da=\rho dV}
E
=
σ
2
ε
0
{\displaystyle E={\tfrac {\sigma }{2\varepsilon _{0}}}}
= field above an infinite plane of charge.
Find E
E
→
(
r
→
)
=
1
4
π
ε
0
∑
i
=
1
N
R
^
i
Q
i
|
R
→
i
|
2
=
1
4
π
ε
0
∑
i
=
1
N
R
→
i
Q
i
|
R
→
i
|
3
{\displaystyle {\vec {E}}({\vec {r}})={\frac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\frac {{\widehat {\mathcal {R}}}_{i}Q_{i}}{|{\mathcal {\vec {R}}}_{i}|^{2}}}={\frac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\frac {{\vec {\mathcal {R}}}_{i}Q_{i}}{|{\mathcal {\vec {R}}}_{i}|^{3}}}}
is the electric field at the field point,
r
→
{\displaystyle {\vec {r}}}
, due to point charges at the source points,
r
→
i
{\displaystyle {\vec {r}}_{i}}
, and
R
→
i
=
r
→
−
r
→
i
,
{\displaystyle {\vec {\mathcal {R}}}_{i}={\vec {r}}-{\vec {r}}_{i},}
points from source points to the field point.
2.6 Gauss's law :
Φ
=
E
→
⋅
A
→
{\displaystyle \Phi ={\vec {E}}\cdot {\vec {A}}}
→
∫
E
→
⋅
d
A
→
=
∫
E
→
⋅
n
^
d
A
{\displaystyle \to \int {\vec {E}}\cdot d{\vec {A}}=\int {\vec {E}}\cdot {\hat {n}}\,dA}
= electric flux
q
e
n
c
l
o
s
e
d
=
ε
0
∮
E
→
⋅
d
A
→
{\displaystyle q_{enclosed}=\varepsilon _{0}\oint {\vec {E}}\cdot d{\vec {A}}}
d
Vol
=
d
x
d
y
d
z
=
r
2
d
r
d
A
{\displaystyle d\,{\text{Vol}}=dxdydz=r^{2}drdA}
where
d
A
=
r
2
d
ϕ
d
θ
{\displaystyle dA=r^{2}d\phi d\theta }
A
sphere
=
r
2
∫
0
π
sin
θ
d
θ
∫
0
2
π
d
ϕ
=
4
π
r
2
{\displaystyle A_{\text{sphere}}=r^{2}\int _{0}^{\pi }\sin \theta d\theta \int _{0}^{2\pi }d\phi =4\pi r^{2}}
Surface Integrals
Calculating
∫
f
d
A
{\displaystyle \int fdA}
and
∫
f
d
V
{\displaystyle \int fdV}
with angular symmetry
Cyndrical:
d
A
=
2
π
r
d
z
;
d
V
=
d
A
d
r
{\displaystyle dA=2\pi r\,dz;\,dV=dA\,dr}
. Spherical:
∫
d
A
=
4
π
r
2
,
d
V
=
4
π
r
2
d
r
{\displaystyle \int dA=4\pi r^{2},\;dV=4\pi r^{2}\,dr}
More Gauss Law
Calculating
∫
f
d
A
{\displaystyle \int fdA}
and
∫
f
d
V
{\displaystyle \int fdV}
with angular symmetry
Cyndrical:
d
A
=
2
π
r
d
z
;
d
V
=
d
A
d
r
{\displaystyle dA=2\pi r\,dz;\,dV=dA\,dr}
. Spherical:
∫
d
A
=
4
π
r
2
,
d
V
=
4
π
r
2
d
r
{\displaystyle \int dA=4\pi r^{2},\;dV=4\pi r^{2}\,dr}
2.7 Electric potential The alpha-particle is made up of two protons and two neutrons.
Δ
V
A
B
=
V
A
−
V
B
=
−
∫
A
B
E
→
⋅
d
ℓ
→
{\displaystyle \Delta V_{AB}=V_{A}-V_{B}=-\int _{A}^{B}{\vec {E}}\cdot d{\vec {\ell }}}
= electric potential
E
→
=
−
∂
V
∂
x
i
^
−
∂
V
∂
y
j
^
−
∂
V
∂
z
k
^
=
−
∇
→
V
{\displaystyle {\vec {E}}=-{\tfrac {\partial V}{\partial x}}{\hat {i}}-{\tfrac {\partial V}{\partial y}}{\hat {j}}-{\tfrac {\partial V}{\partial z}}{\hat {k}}=-{\vec {\nabla }}V}
q
Δ
V
{\displaystyle q\Delta V}
= change in potential energy (or simply
U
=
q
V
{\displaystyle U=qV}
)
P
o
w
e
r
=
Δ
U
Δ
t
=
Δ
q
Δ
t
V
=
I
V
=
e
Δ
N
Δ
t
{\displaystyle Power={\tfrac {\Delta U}{\Delta t}}={\tfrac {\Delta q}{\Delta t}}V=IV=e{\tfrac {\Delta N}{\Delta t}}}
Electron (proton) mass = 9.11×10−31 kg (1.67× 10−27 kg). Elementary charge = e = 1.602×10−19 C.
K
=
1
2
m
v
2
{\displaystyle K={\tfrac {1}{2}}mv^{2}}
=kinetic energy . 1 eV = 1.602×10−19 J
V
(
r
)
=
k
q
r
{\displaystyle V(r)=k{\tfrac {q}{r}}}
near isolated point charge
Many charges:
V
P
=
k
∑
1
N
q
i
r
i
→
k
∫
d
q
r
{\displaystyle V_{P}=k\sum _{1}^{N}{\frac {q_{i}}{r_{i}}}\to k\int {\frac {dq}{r}}}
.
2.8 Capacitance
Q
=
C
V
{\displaystyle Q=CV}
defines capacitance .
C
=
ε
0
A
d
{\displaystyle C=\varepsilon _{0}{\tfrac {A}{d}}}
where A is area and d<<A1/2 is gap length of parallel plate capacitor
Series
:
1
C
S
=
∑
1
C
i
.
{\displaystyle {\text{Series}}:\;{\tfrac {1}{C_{S}}}=\sum {\tfrac {1}{C_{i}}}.}
Parallel:
C
P
=
∑
C
i
.
{\displaystyle {\text{ Parallel:}}\;C_{P}=\sum C_{i}.}
u
=
1
2
Q
V
=
1
2
C
V
2
=
1
2
C
Q
2
{\displaystyle u={\tfrac {1}{2}}QV={\tfrac {1}{2}}CV^{2}={\tfrac {1}{2C}}Q^{2}}
= stored energy
u
E
=
1
2
ε
0
E
2
{\displaystyle u_{E}={\tfrac {1}{2}}\varepsilon _{0}E^{2}}
= energy density
2.9 Current and resistors
Electric current: 1 Amp (A) = 1 Coulomb (C) per second (s)
Current=
I
=
d
Q
/
d
t
=
n
q
v
d
A
{\displaystyle I=dQ/dt=nqv_{d}A}
, where
(
n
,
q
,
v
d
,
A
)
{\displaystyle (n,q,v_{d},A)}
= (density, charge, speed, Area)
I
=
∫
J
→
⋅
d
A
→
{\displaystyle I=\int {\vec {J}}\cdot d{\vec {A}}}
where
J
→
=
n
q
v
→
d
{\displaystyle {\vec {J}}=nq{\vec {v}}_{d}}
=current density .
E
→
=
ρ
J
→
{\displaystyle {\vec {E}}=\rho {\vec {J}}}
= electric field where
ρ
{\displaystyle \rho }
= resistivity
ρ
=
ρ
0
[
1
+
α
(
T
−
T
0
)
]
{\displaystyle \rho =\rho _{0}\left[1+\alpha (T-T_{0})\right]}
, and
R
=
R
0
[
1
+
α
Δ
T
]
{\displaystyle R=R_{0}\left[1+\alpha \Delta T\right]}
,
where
R
=
ρ
L
A
{\displaystyle R=\rho {\tfrac {L}{A}}}
is resistance
V
=
I
R
{\displaystyle V=IR}
and Power=
P
=
I
V
=
I
2
R
=
V
2
/
R
{\displaystyle P=IV=I^{2}R=V^{2}/R}
2.10 Direct current circuits
2.13 Electromagnetic induction
2.14 Inductance
2.15 Alternating current circuits
AC voltage and current
v
=
V
0
sin
(
ω
t
−
ϕ
)
{\displaystyle v=V_{0}\sin(\omega t-\phi )}
if
i
=
I
0
sin
ω
t
.
{\displaystyle i=I_{0}\sin \omega t.}
RMS values
I
r
m
s
=
I
0
2
{\displaystyle I_{rms}={\tfrac {I_{0}}{\sqrt {2}}}}
and
V
r
m
s
=
V
0
2
{\displaystyle V_{rms}={\tfrac {V_{0}}{\sqrt {2}}}}
Impedance
V
0
=
I
0
X
{\displaystyle V_{0}=I_{0}X}
Resistor
V
0
=
I
0
X
R
,
ϕ
=
0
,
{\displaystyle V_{0}=I_{0}X_{R},\;\phi =0,}
where
X
R
=
R
{\displaystyle X_{R}=R}
Capacitor
V
0
=
I
0
X
C
,
ϕ
=
−
π
2
,
{\displaystyle V_{0}=I_{0}X_{C},\;\phi =-{\tfrac {\pi }{2}},}
where
X
C
=
1
ω
C
{\displaystyle X_{C}={\tfrac {1}{\omega C}}}
Inductor
V
0
=
I
0
X
L
,
ϕ
=
+
π
2
,
{\displaystyle V_{0}=I_{0}X_{L},\;\phi =+{\tfrac {\pi }{2}},}
where
X
L
=
ω
L
{\displaystyle X_{L}=\omega L}
RLC series circuit
V
0
=
I
0
Z
{\displaystyle V_{0}=I_{0}Z}
where
Z
=
R
2
+
(
X
L
−
X
C
)
2
{\displaystyle Z={\sqrt {R^{2}+\left(X_{L}-X_{C}\right)^{2}}}}
and
ϕ
=
tan
−
1
X
L
−
X
C
R
{\displaystyle \phi =\tan ^{-1}{\frac {X_{L}-X_{C}}{R}}}
Resonant angular frequency
ω
0
=
1
L
C
{\displaystyle \omega _{0}={\sqrt {\tfrac {1}{LC}}}}
Quality factor
Q
=
ω
0
Δ
ω
=
ω
0
L
R
{\displaystyle Q={\tfrac {\omega _{0}}{\Delta \omega }}={\tfrac {\omega _{0}L}{R}}}
Average power
P
a
v
e
=
1
2
I
0
V
0
cos
ϕ
=
I
r
m
s
V
r
m
s
cos
ϕ
{\displaystyle P_{ave}={\frac {1}{2}}I_{0}V_{0}\cos \phi =I_{rms}V_{rms}\cos \phi }
Transformer voltages and currents
V
S
V
P
=
N
S
N
P
=
I
P
I
S
{\displaystyle {\tfrac {V_{S}}{V_{P}}}={\tfrac {N_{S}}{N_{P}}}={\tfrac {I_{P}}{I_{S}}}}
3.2 Geometric optics and image formation
1
S
1
+
1
S
2
=
1
f
{\displaystyle {\frac {1}{S_{1}}}+{\frac {1}{S_{2}}}={\frac {1}{f}}}
relates the focal length f of the lens, the image distance S1 , and the object distance S2 . The figure depicts the situation for which (S1 , S2 , f) are all positive: (1)The lens is converging (convex ); (2) The real image is to the right of the lens; and (3) the object is to the left of the lens. If the lens is diverging (concave ), then f < 0. If the image is to the left of the lens (virtual image ), then S2 < 0 .