大学物理期末复习 (2)

Electric potential（simply the Potential）
WP “０” unit：V (volt) 1V 1 J/C U E Fra Baidu bibliotekdl r Q0
The potential at a point equals the work required to
bring a unit positive charge from this point to the zero point of electric potential .
Calculate Potential. Two Methods: Applying element analysis method
U
dq 4 0 r
Q
Applying the definition of potential
U
"0"
r
E dl
Example 1
dl dq ds dV
Q
q0dq r dF 40 r 3
F
q0dq r 40 r 3
dq
Q
r
q0 + P
(4). Coulomb Law in the Dielectric
f12
4 0 r r
q1q2
r 3 12
2. Electric Field Intensity
×
×
√
4 . Calculating the Electric Field
Problem-solving strategy: Analysis the distribution of charges. Solution 1. Applying the superposition principle of the field intensity.
Unit N/C 、V/m
(2) The principle of superposition
n E E1 E 2 E n E i i 1
(3). Electric Field Lines
Electric field lines are not real. Field lines are not material objects. They are used only as a pictorial representation to provide a qualitative description of the field.
The relationship between the electric potential energy and Electric potential

In general, the electric potential is a function of all three spatial coordinates. If V is given in terms of rectangular coordinates, the electric field components Ex ,Ey, and Ez can be found from V(x,y,z) as the partial derivatives
Example 7
r
E
R2
R 1

5. Electric potential energy , Electric potential
Electric potential energy of q0 at a point
“0”
wa＝wa wb Aab q0

a
E dl
Example 2
Example 3
Example 4
There are two concentric charged spherical shells of radius R1 and R2 . Charge quantities distribute uniformly.
r R1 , R2 r ,
E E E
R1， Q1
R1 r R 2 ,
R2， Q2
Example 5
Example 6
The charge line density of an infinite uniform charged cylindrical surface of radius R is ,find the electric field intensity.
(1) Definition
Magnitude: the electric field
F E q0
force on unit positive charge
Direction : direct the direction of
force exerted on a positive test charge .
× ×
If the electric flux of a Gauss surface equals zero, then field intensity at every point on the Gauss surface is zero. Gauss theorem is tenable only to the electrostatic field whose distribution is symmetrical in space.
General Revision of College Physics
Chapter 1. Electrostatic Field
1. Coulomb Law in the Vacuum
(1) Coulomb force
q1q2 f12 r 3 12 4 0 r
(2) The principle of superposition
In electrostatic field E0, electric field lines begin and end on charges. In Induced electric field Ev, the electric field lines form closed loops, with no beginning and no end.
Solution 2. Applying Gauss’s law to symmetric charge distribution.
The three symmetries:
Spherical symmetry
Cylindrical symmetry Plane symmetry
Problem-solving strategy:
Example 2
Example 3.
Find the electric field intensity and the electric potential.
Q
q
B
A R1 O
R2
R3
6. Equipotential volumes and surfaces
7. SUMMARY Find electric potential
(b) in the +x direction
(c) in the -x direction.
In a certain region of space, the electric field is zero. From this we can conclude that the electric potential in this region is (a) zero (c) positive (b) constant (d) negative.
1、superposition principle
Find electric field intensity
Qi r E （separated） 3 i 4 0 ri
1、superposition principle
U
i
4 0 ri
Qi
（separated）
U
If the charges are countable, the resultant field is the vector sum of the fields due to the individual charges. when confronted with problems that involve a continuous distribution of charge, Element Analysis Method
Analysis the symmetry of the field intensity distribution .
Select appropriate gaussian surface.
Select appropriate coordinates, apply Gauss’s law.
Example 1
The total electrical force on given charge is the vector sum of the electrical forces caused by the other charges, calculated as if each acted alone.
(3) Find Electrostatic Force Force Due to the System of Point Charges. Force Due to Continuous Charge Distributions
Field lines are never cross. The field at any point has a unique direction.
3. GAUSS’S
In vacuum,
LAW
e E dS
S

1
0
Q
i
In dielectric,
D dS Q0
U Ex x
For example, if
U Ey y

U Ez z
V 3x2 y y 2 yz
,then
Ex 6 xy
In a certain region of space, the electric potential is zero everywhere along the x axis. From this we can conclude that the x component of the electric field in this region is (a) zero
1 s Estatic dS 0 qi
Evortex dS 0
s
The electric field lines are denser in the place where the field intensity is stronger, and the electric field lines are sparser in the place where the field intensity is weaker.
S
Electric flux of the Gauss surface is related with charges in Gauss surface and is not related with charges out of Gauss surface.
√
The field intensity at a point on the Gauss surface is related with charges in the Gauss surface and is not related with charges out of the Gauss surface. If the electric flux of a Gauss surface equals zero, there must be not charge in the Gauss surface.