3 ELECTROSTATICS

df r = cos θdf =
df r = 2 ∫
π /2
0
l
Thus
f r = 2∫
π /2
4πε 0 r
0
df r
qρ l qρ l π /2 (sin θ ) 0 = = 2πε 0 r 2πε 0 r 500 × 10−9 × 100 × 10−9 = −9 2π × (10 / 36π ) × 5 = 1.8 × 10 N
0
πε 0
=
2 ρR
πε 0
3.４ Electric Flux and 3.４ Electric Flux Density
Electric Flux – the number of lines of force due to a charge is equal to the magnitude of the charge in coulombs. Electric Flux Density – the flux per unit area.
4
Ex. An infinite, uniformly charge line with a charge density of 100nC/m extends along the z axis. Find the force experienenced by a charge of 500nC located at (3,4,0). Solution We begin our computing the force in
D = ε0E
16
C/m2
Electric flux
17
3.４ Electric Flux and 3.４ Electric Flux Density
3.4.1 Definition Electric Flux
ψ = ∫ D ⋅ ds
s
C
3.4.2 Gauss’s Law
∫ D ⋅ ds = Q
6
−4
3.３ Electric Field Intensity 3.３
F ( x, y , z ) qt
V/m （N/C） ）
E ( x , y , z ) = lim
qt → 0
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3.３ Electric Field Intensity 3.３
A point charge
F q Ep(r) = = a 2 r qt 4πε0r
ρl aρ E ( ρ , φ , z ) = E ρ a ρ + E φ aφ + E z a z = 2πε0ρ
12
(rectangular coordinate system)
(cylindrical coordinate system)
Ex. The charge is uniformly distributed in the shape of a ring of radius b. Determine the electric field intensity at any point on the axis of the ring. Solution dq dq
s
∇ ⋅ D = ρv
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3.５ The Electric Potential 3.５
(1) Potential difference
∫
b
a
E ⋅ dl = ∫ Edl cos α
a rb ra
b
= ∫ E dr =∫
rb ra
3 ELECTROSTATICS
3.１ 3.１ 3.２ 3.２ 3.３ 3.３ 3.４ 3.４ 3.５ 3.５ 3.６ 3.６ 3.７ 3.７ 3.８ 3.８ 3.９ 3.９ 3.10 3.11 13.12 Introduction Coulomb’ Coulomb’s Law Electric Field Intensity Electric Flux and Electric Flux Density The Electric Potential Electric Dipole Materials in an Electric Field Energy Stored in an Electric Field Boundary Conditions Capacitor and Capacitance Poisson’ Poisson’s and Laplace’s Equations Laplace’ Mathod of Images
F/m
Ex. A charge of 0.7mC is located at P(2,3,6) and another charge of 4.9uC is situated at Q(0,0,0). Calculate the force exerted by one charge on the other. Solution The distance vector from the 4.9uC charge to the 0.7mC charge is
3.２ Coulomb’s Law 3.２ Coulomb’
q1q2 a12 F21 = ⋅ 2 4πε 0 R q1q2 a21 F12 = ⋅ 2 4πε 0 R F21 = − F12
The permittivity of free space
N
N
ε 0 = 8.85 × 10−12
3
10−9 = 36π
3.１ Introduction 3.１
In this chapter, we study static electric fields (elelctrostatics), due to charges at rest. A rest. charge can either be concentrated at a point or distributed in some fashion. In any case, the fashion. charge is assumed to be constant in time. time. Some aspects of electrostatic fields discussed in this chapter may appear to be a repetition of what you have already studied in physics. physics. Some repetition is necessary, not only to maintain a continuous link from one section to another, but also to motivate the learning process. process. We are convinced that a discerning 2 student will find this repetition helpful. helpful.
qρl dθ cosθ df r = cosθdf = 4πε0r
5
qρ l a 0 dl cylindrical coordinates. We get df = 2 4πε 0 a rdθ , If l = r tanθ , dl = 2 cos θ r a= , cos θ
Ex. An infinite, uniformly charge line with a charge density of 100nC/m extends along the z axis. Find the force experienenced by a charge of 500nC located at (3,4,0). Solution qρ dθ cos θ
L2
Ex. A finite line extends from x=-L1 to x=L2 and carries a line charge density. Determine the electric field intensity at P.
ρl
x
1
2
1
Байду номын сангаас
ρl ay E p ( y ) = E y a y + E x ax = 2πε 0 y
F r − r' q Ep (r) = = ⋅ 2 qt 4πε0 r − r' r − r' q (r − r ' ) q = a = 3 2 R 4πε 0 R 4πε 0 r − r '
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3.３ Electric Field Intensity 3.３
n point charges
E (r ) =
0
dEr = 2 ∫
dE r
⑴
Er = 2 ∫
π /2
0
dEr = 2∫
π /2
0
dEr
ρl ρl π /2 (sin θ ) 0 = = 2πε 0 r 2πε 0 r
⑵ ⑶
E = 2∫
π /2
0
cosφdE =
2 ρs
πε 0
ρ s = ρdR
dE = 2 ρdr
πε 0
R
E=∫
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2 ρdr
dE =
4πε0R
2
=
4πε0 (b + z
2
2
)
,
z dq dE z = dE = 2 ⋅ 2 b + z 4πε b2 + z 2
b
z
E z = ∫ dE =
l
13
2ε 0 (b + z
2
ρ lbz
2
)
Ex. Obtain the electric field intensity at a point on the axis. Solution dq = ρl dl