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普通物理學甲下 (202 101A2)
General Physics(A)(2)
台大物理 吳俊輝
Electricity II
Flux (used in Gauss Law)


The concept of flux – fluid flow across a surface.
(通量: 單位時間通過的量)
Intuitively, it is the component of the fluid velocity that is
normal to the area of interest, i.e. the flow across the area:
flux   (v cos ) A  v  A

Gaussian surface



A Gaussian surface is a closed
surface which we use to calculate
the flux of E field.
Electric flux: the electric field has
been given the “properties” of a
fluid-flow field. This allows us to
determine the properties of electric
field more easily.
The total net electric flux across a
Gaussian surface is

 E  dA
dA方向為由Gaussian surface內部朝外
Gauss’s Law

E

dA

q
enc

where qenc is the sum
of all charges enclosed
by the Gaussian surface.
Examples: choosing a Gaussian surface

Gauss’ Law can be applied to
any closed surface:
  E  dA  qenc

e.g.:

 E  dA 
qenc


q1  q2  q3

高斯定律的用處：求電場，及
研究電荷系統的性質…
1. Spherical Symmetry  Coulomb’s Law


From symmetry, we know that the
electric field is the same at a given
distance r from the center of the
charge.
By Gauss Law, we have

 E  dA  q
enc
左式 
 E  dA  
EdA
sphere

E
dA  E (4 r 2 )
sphere
故
E

q
4 r 2
The electric field points away from
the charge center as we know from
the area integral.
2. Charge on an isolated conductor
將電荷置於絕緣導體上時…
 電荷會分佈在導體的外表面!!!

By the Gauss Law:   E  dA  qenc
(a) electric field must be zero inside
a conductor, otherwise the charges
would flow. Therefore, excess
charges must be on the conductor
exterior.
(b) electric field inside a conductor
must be zero, therefore, no charge
is on the inner surface!
3. Planar geometry and Gauss’ Law



Excessive + charge on a large planar
conductor.
Let’s use Gauss’ Law:
Again, the symmetry dictates that E must
point perpendicular away from the plane
(  is charge per unit area).
E  dA  EA
 E  dA  EA   A

Therefore,

E

Again, the field points away from the plane
from the area integration.
4. Cylindrical symmetry

A very long plastic wire with
uniform charge per unit length l
E  dA  E (2 rdh)
 E  dA  E 
dA  E (2 rh) 
cylinder
qenc

lh
E (2 rh) 

l
E
2 r

So the electric field with 1/r away
from a long wire.
Electrical work (WE), energy (U), and potential (V)

電荷在電場中移動時，電場會對其作功，
造成其位能的改變：
f
U  U f  U i  WE    dWE
i
f
f
   F  dsE  q  E  ds
i

i
Electrical potential: V 
U
q
f
U
V 
   E  ds
i
q

The unit of V: joule/coulomb (J/C) = volt (V).
Equipotential Surface


Surfaces of constant V (electric potential)
Equipotential Surface  E
Electron volt (eV)

能量的單位：joule (J), electron volt (eV).

1 eV = energy needed to move a particle with
charge e across a potential difference of 1 V.

So
1eV  (1.60 1019 C )(1J / C )  1.60 1019 J
Some practical knowledge about
electricity
 Power
P=IV
 Current I=dq/dt :
1. AC: Alternating Current
2. DC: Direct Current
 Ohm’s Law V=IR
Medical device - defibrillator

Energy is stored in a capacitor to stop fibrillation
of heart attack patients (100kW of pulsed energy
can be applied to a patient)
+
+
+
-
-
-
+
-
+
-
+ +
-
-
Apply to patient
Exercise 1 (習題1)
 Ch
21: 25, 60, 83
 Ch 22: 22, 41, 65, 87
 Ch 23: 21, 29, 47, 81
 Deadline:
3/6, 10:20AM