Transcript CHAPTER 23

23.1 THE ELECTRIC FIELD
•
The electric field strength E:
E=F/qt
(23.1)
 Kq Q
F  2t rˆ
r
 KQ
E  2 rˆ
r
(23.2)
Any charged particle does not interact directly with other
charged particle; rather, it responds to whatever field it
encounters .
E is the property of a point in space and depends
only on the source of the field Q
F=q E SI unit N/C (23.3)
E is the resultant field strength due to all the net charges present.
F=mg
(g: N/kg)
Example 23.1: On a clear day there is an electric field of approximately
100N/C directed vertically down at the earth’s surface. Compare the electrical
and gravitational forces on an electron.
FE  eE
 (1.6  10 19 C )(100 N / C )
 1.6  10 17 N
Fg  mg
 (9.1  10 31 Kg )(9.8 N / Kg )
 8.9  10 30 N
Fg
 5.6  10 13
FE
The gravitatio nal force on particle such as the electron
and proton mat be safely ignored in problems involvoing
electric fields.
Problem-Solving Guide for the Electric Field
1. First draw the field vectors at the given location. (Their direction can be
found by imagining that there is a positive charge at that point.)
2.Find the (scalar) magnitude of the field strength due to each charge. The
signs of the charges must be ignored. This step may be ensured by writing
the magnitude in the form E=k｜Q ｜ /r2
(23.5)
3. Place the origin at the point at which E is being calculated. The choice of
coordinate axes determines the signs of the components of the field strength
E
Example 23.2: A point charge Q1=20μC is at (-d,0) while Q2=-10μC is at
(+d,0). Find the resultant field strength at a point with coordinates (x,y) Take
d=1.0m and x=y=2m.
r1  ( x  d ) 2  y 2  13  3.6m,
r2  ( x  d ) 2  y 2  5  2.2m
k Q1 (9.0  109 N  m 2 / C 2 )(2  10 5 C )
E1  2 
 1.4  10 4 N / C
2
r1
13m
k Q2 (9.0  109 N  m 2 / C 2 )(10 5 C )
E2  2 
 1.8  10 4 N / C
2
r2
5m
  
E  E1  E2
Ex  E1x  E2 x  E1 cos 1  E2 cos  2 , E y  E1 y  E2 y  E1 sin 1  E2 sin  2
sin1  y / r1 , sin 2  y / r2 , cos 1  ( x  d ) / r1 , cos  2  ( x  d ) / r2
3
1.0
 (1.8  10 4 N / C )
 3.5  103 N / C
3.6
2.2
2
2
E y  (1.4  10 4 N / C )
 (1.8  10 4 N / C )
 8.6  103 N / C
3.6
2.2

E  3.5  103 iˆ  8.6  103 ˆj N / C
Ex  (1.4  10 4 N / C )
The properties of lines of force
1.
2.
3.
4.
5.
Electrostatics field line always start on positive charges and end on negative charges
The number of lines that originate from, or terminate on, a charge is proportional to the
magnitude of the charge.
The direction of the field at a point is along the tangent to the line of force, as shown in
fig.
The field strength is proportional to the density of the lines, that is , the number of lines
per unit area, intercepted by a surface normal to the field.
Line of force never cross.
Example 23.3: Sketch the field lines for two point charges 2Q and –Q.
(a) Symmetry
(b) Near field
(c) Far field
(d) Null point
(e) Number of lines
EXAMPLE 23.4: A proton travels a distance of 4 cm parallel to a
uniform electric field E=103iN/c, as shown in fig. 23.15 If its
initial velocity is 105 m/s, find its final velocity.
eE (1.6  10 19 C )(103 N / C )
10
2
a


9
.
6

10
m
/
s
m
1.67  10 27 Kg
v 2f  vi2  2ax
 (105 m / s) 2  2(9.6  1010 m / s 2 )(4  10 2 m)
 1.77  1010 m2 / s 2
v f  1.3  105 m / s
EXAMPLE 23.6: A thin insulating rod of length L carries a uniformly
distributed charge Q, find the field strength at a point along its axis at s
distance a from one end.
dq / Q  dx / L
dq  dx,   Q/L
k(dx )
dE 
x2
aL
dx
 1
E  k  2   k - 
 x a
a x
1
1
kQ
 k ( ) 
a aL
a(a  L)
a L
note : a  L , E 
kQ
a2
EXAMPLE 23.7: What is the field strength at a distance R from an infinite
line of charge with linear charge density λC/m?
kdl
(i )
2
r
r  R sec and l  R tan
dE 
dl  R sec 2d .
k d
(ii)
R
dE x  dEcos .
dE 
Integrating from - 1 to 2 :
k  2
k
k
2


cos

d


sin


(sin  2  sin 1 )



1


1
R
R
R
for an infiniteline,1   2 ; therefore
E
E
2k
R
(23.9)
EXAMPLE 23.8: nonconductoring disk of radius a has a uniform surface
charge density σC/m2. What is the field strength at a distance y form the
center along the center axis?
kdq y 2
, r  x 2  y 2 , dq  dA   (2xdx)
2
r r
a
a
2xdx
dx2
E  ky  2
 ky  2
2 3/2
(x

y
)
(x  y2 )
0
0
dEy  dE cos 
a


-2
 ky  2
2 1/2 
 (x  y )  0


y
 2k 1  2
2 1/ 2 
 (a  y ) 
for large valuesof y, we use (1  z)n  1  nz
2 1 / 2
(a  y )
2
a 2 1/ 2
 (1  2 )
y

1 a2
 1  ( 2 )  ...
2 y
On substituting thisibto (i) and using Q  a 2 ,
we find E  kQ/y2
EXAMPLE 23.9: Find the field due to the following: (a) an infinite sheet of
charge with surface density +σ; (b) two parallel infinite sheets with charge
densities +σ and - σ
•
We may use the result in Ex 23.8 for the electric
field due to a disk. In the limit as a→∞, the second
term in Eq.(i) of Example 23.8 vanished and we left with
E=2πkσ, or
•
(a) E=σ/2ε0
(23.10)
•
(B)
(23.11)
E=σ/ε0
Field Due to a Dipole
kQ
E  E  2
r  a2
E y  ( E   E  ) cos 
2kQ
a
- 2
(r  a 2 ) (r 2  a 2 )1 / 2
- k2aQ
 2
(r  a 2 ) 3/2
Torque in a Uniform Field
      r F , where r  (d sin  / 2)
d
  2(qE)( sin  )  pE sin  (23.15)
2
  
  P  E (23.16)
Potential Energy




Wext    d   d
Wext  
2

pE sind  pE( cos 2  cos1 )
1
WEXT  U  U 2  U 1 , U 1  0 at 1   / 2
 
U   pE cos   p  E
(23.17)
π/2
Interaction between Dipoles
• (Axis)
2kp1
E1  3
x
dE1
F2  p 2
dx
p 2  E1
dE1 / dx  1 / x 4
1
F 7
x
Discussion
• RQ : ELECTROSTATICS II
• Physlet physics Problem: 22.1,22.4,
Exercises of chapter 23
• Questions:
• Exercises:25,26,35,40,44
• Problem:2,4,7,11,18,19,20