ELECTRIC FORCES ON CHARGES
Lorentz Force Law:
f = q (E + v × μo H) Newtons
⇒
a = f/m = qE/m ≈ eV/mL [m s-2]
Kinematics*:
t
v = ∫ a(t)dt
0
= v o + zˆat
cathode
E⊥
anode, phosphors
-
-V
heated
filament
f = qE = ma
deflection plates
cathode ray tube (CRT)
+
z
ˆ o t + at 2 /2
z = zo + z•v
L
Electron charge = -e
= -1.6x10-19 [C]
Electron kinetic energy wk:
wk = fs = (eV/L)L = eV [Joules]
Electron volt = energy of 1 electron moving 1 volt = e Joules
* For E in -z direction
L5-1
ELECTRIC FORCES
ON CHARGED CONDUCTORS
Force on free charges:
z
f = qE = −eE
Force:
Surface charge:
[N]
[N m-2 ] 0
Electric pressure: Pe = ρs <E>
ρs = ­ εoEo
-2
[C m ]
Eo
Average E:
< E> =
[V m-1]
2
1
2
-2
Electric pressure:
=
­
ε
E
[N
m
]
e
o o
2
P
unlike charges attract
E(z)
⎯Eo
Deepest electrons
experience zero force
Eo
Attractive pressure
⎯E
Repulsive forces:
Like charges repel,
-e
+ρs
E=0
+ρs
⎯E
L5-2
ENERGY METHOD
FOR FINDING FORCES
Force, work, and energy:
f
dw
dw = f ds ⇒ f =
[N]
ds
C = εοA/s
2
2
1
Q
1
Q
s
1
w = 2 CV2 = 2 C = 2 ε A [J]
o
Q ≠ f(s) if C is open circuit
1 Q2
dw
f=
=
2 εo A
ds
f
(εoEA)2
= ( 1 εoE2 )A
=
2εo A
2
= -PeA [N]
1
2
⇒ Pe = 2 εoE [N m-2] = [J m-3]
is the force externally
applied to the upper plate
ds
C
+V
f
s
+Q
0
⎯E
-Q
The static pressure Pe of the
electric field on the upper
plate is the same with a
battery attached:
1
Pe = ρs <E> = ρs Eo
2
Electric fields always pull on conductors → attractive force
L5-3
ENERGY METHOD WITH A BATTERY
Incremental work dw:
ds
2
w = CV
2
dw = f ds = -V dQ + (V2/2) dC
To battery To capacitor
dQ = VdC ⇒ dw = -(V2/2)dC
C
+Q
⎯E
-Q
f
+V
s dQ
0
Externally applied force f
Force and pressure:
2
dw = − V 2 dC* = V 2 εo A = εoE A = −P A
f=
e
ds
2 ds
2 s2
2
Pe = -εo E2/2 (as before) Q.E.D.
*C = εA/s
L5-4
LATERAL FORCES – ENERGY METHOD
Energy derivative:
W
dw
f = − dD (externally applied)
2
A’ = Ws
+Q
C
A’
E
2
w=Q = Q s
2C 2εo WD
Fringing
s field
D
2
Q
s = (εoEWD) s = ( 1 ε E2 )Ws = P A ' [N]
=
e
f
2 o
2εo WD2
2εo WD2
2
f
-Q
Energy derivative:
2
Q
L
A’ = Ws
d( )
+Q
W
dw = − 2C
=
−
+V C
f
dD
dD
f
ε >> εο s
ε (L-D)W
E
C = CD + Co = εDW + o
s
s
D
-Q
0
W
= [D(ε - εo ) + εoL]
s
(ε − εo )E2
=
A' [N] = ΔPe A '; A' = Ws (E& pushes, E⊥ pulls)
f
2
L5-5
ROTARY ELECTROSTATIC MOTORS
Energy derivative:
2
Q
d( )
dw
Torque:
T[Nm] = −
= − 2C
dθ
dθ
2
εo A
R
, A=2 θ
=
s
2
2
εoE2 R
C
Q
s
Therefore : T =
=
A'
[Nm]
2 2
2
2
2εoR θ
v
+
stator
θ
R
rotor
A’ = 2Rs
+
T
≅ pressure × gap-area A' × R
2
Motor power:
Peak power:
v
n=4
P = Tω
[W]
Average power: Pavg = P/2
(duty cycle = ½)
Segmentation advantage:
θ
+
-
rotor
stator
T [Nm] = -dw/dθ ∝ A’ ∝ nRs (n = # segments)
L5-6
ELECTROSTATIC SENSORS
Cantilevered microphone:
Rs
V+
+
Vs
-
out
R
s
f
plate area A
C, V
Cantilevered capacitor C at V volts
s → s + δ ⇒ CV2/2 → C’V’2/2 initially, then:
Voltage decays to Vo = VsR/(R + Rs) ⇒
Voltage pulse to amplifier, Δw ≈ wδ/s > Eb > ~10-20 [J]
Minimum detectable δ ≈ sEb/w [m] (hears brownian motion)
L5-7
MIT OpenCourseWare
http://ocw.mit.edu
6.013 Electromagnetics and Applications
Spring 2009
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