Transcript Ch. 26 slides

Chapter 26
Electric Field
Phys 133
Electric Field
New long range interaction
KqA qB
FA on B = 2 rˆ
rAB
A creates field in space
changes the environment
EA =
FA on B
qB
KqA
= 2 rˆ
rAB
B interacts with field
æ KqA ö
F on B = qB EA = qB ç 2 rˆ÷
è rAB ø
Phys 133
Procedure
•Place test charge
•Find force
•remove test charge
Phys 133
Simulation: Electric Field
Phys 133
Understanding Fields (in general)
Vector field worksheet (field at tail, relative lengths)
a) v ( x , y ) = 2i + 1 j
b) v ( x , y ) = xi + 2 yj
y
c)
v ( x , y ) = 2 yi + xj
y
y
x
x
x
x
y
vx
vy
x
y
vx
vy
x
y
vx
vy
1
2
0
4
1
3
5
2
2
2
2
2
1
1
1
1
1
2
3
4
1
2
2
1
1
2
3
4
2
4
4
2
1
2
2
4
1
2
3
1
2
4
6
2
1
2
2
4
Phys 133
Vector field worksheet (cont.)
Now here are some for practice.
d) v ( x , y ) = - 2i + 1 j
y
e)
v ( x , y ) = - yi + ( x + y ) j
y
x
x
y
1
2
3
4
1
2
2
1
vx
vy
x
x
y
1
2
3
4
1
2
2
1
Phys 133
vx
vy
Vector field worksheet (ans)
x
y
vx
vy
x
y
vx
vy
1
2
3
4
1
2
2
1
-2
-2
-2
-2
1
1
1
1
1
2
3
4
1
2
2
1
-1
-2
-2
-1
2
4
5
5
Phys 133
Electric Field: point charge
Positive point charge
E due to +Q
Negative point charge
Q
= +K 2 r
r
E due to -Q
Q
= -K 2 r
r
Radial from source, depends on distance
Phys 133
Electric Field (multiple charges)
Net force at some point is sum of
the forces due to all objects around
åF
on q
= F1 on q + F2 on q +… (superposition)
Defined
F1 on q
E1 =
q
Therefore,
Fon q F1 on q F2 on q
å q = q + q +…
or
E tot = E1 + E 2 +… (superposition)
Phys 133
Do workbook 26.5ac, 7adf
Phys 133
Electric field of a dipole along the y-axis…
Phys 133
Electric field of a dipole along the y-axis…
E net = å E i = E + + E -
E net = å E i = E + + E -
E+
i
i
(E ) = (E ) + (E )
q E total
q
r+
E-
q
E + = +K
net x
r-
E - = +K
r+ = r- = r = d 2 + y 2
q
E+ = E+ = K 2
r
q
r-2
Phys 133
- x
Kq
Kq
= 2 cosq + 2 cos q
r
r
Kq
= 2 2 cosq
r
Kq æ d ö
Kqd
=2 2
=2 3
è
ø
r r
r
2Kqd
=
3
2
2 2
(d + y )
q
q
r+2
+ x
(E ) = (E ) + (E )
total y
+ y
- y
=0
Electric field of a dipole along the y-axis
E net,x =
[
Kp
d2 + y
where p = 2dq
]
2 3/2
E net,y = 0 (symmetry)
What is the electric field of a dipole for y >> d?
E net,x =
Kp
( ) ùû
y 3 é1 +
ë
Kp é
= 3 1+
y ë
Kp æ
= 3 è1y
2 3/2
d
y
()
d
y
3
2
2 -3 / 2
ù
û
()
d
y
2
Kp
ö
+… ø » 3
y
Phys 133
Vector Field and Field Lines
Two ways to describe the same thing
Dipole with +/-
Vector field
Field lines
Phys 133
More Vector Field and Field Lines
Dipole with +/+
Vector field
Field lines
Phys 133
Electric Field lines
Rank the electric field
strength in order from largest
to smallest.
A: E1 < E2 < E3 = E4
B: E3 = E4 < E2 < E1
C: E2 = E3 < E4 < E1
D: E1 < E4 < E2 = E3
Phys 133
Electric Field of a “blob”
Q
Q
Q
l= , h= , r=
L
A
V
Phys 133
Charge Density
Charge Q is spread uniformly on a rectangle of sides a and b.
a) What is the surface charge density?
b) The original rectangle, #1, is then broken into two smaller
rectangles, #2 and #3. Compare the surface charge densities,
1, 2, 3. and the charges Q1, Q2, Q3.
Phys 133
Charge Density ans
Charge Q is spread uniformly on a rectangle of sides a and b.
a) What is the surface charge density?
Q1 Q
h1 = =
A1 ab
b) The original rectangle, #1, is then broken into two smaller
rectangles, #2 and #3. Compare the surface charge densities,
1, 2, 3.
Charge is uniformly spread over rectangle.
Piece #2 has a 1/3 the area, so 1/3 the charge.
Q2
Q3
Q
h2 =
=
=
A2 (a 3)b ab
h1 = h2 = h3
Q3
2Q 3
Q
h3 =
=
=
A3 (2a 3)b ab
Phys 133
Electric Field (continuous distribution)
q
E po int = K 2 r√
r
th
r1
r23 ri
(i) bit of charge
E123
P
Ei E
23
E1
r123
Enet = å Ei
i
Phys 133
Strategy
 Draw a picture, pick a coordinate system.
 Identify, P, the place in space you want the field.
 Pick a generic (no special points) chunk of charge DQ for
which you know the field.
 Determine the components of (Ex, Ey, …) at P due to chunk
expressed in terms of variables (x, y, …r, q…).
 Express the charge DQ in terms of charge density and
infinitesimal variable(s) (dz, dq…)
 Express all quantities (angles, distances, etc.) in terms of
coordinates.
 Add up contributions, sum becomes definite integral with limits
corresponding to charge.
 Evaluate integral and simplify result as much as possible.
Phys 133
Problem
A thin rod of length L and charge Q. Find components of the
electric field vector along the dotted line.
Phys 133
Express
Draw
Identify,
Determine
picture,
all
P,the
quantities
the
components
place
pickDQ
ain
(angles,
coordinate
space
of you
(E
distances,
want
E
the
etc.)
atfield.
Pin
Pick aageneric
the
charge
chunk
of
in
charge
terms
DQ.
of
charge
density
x,system.
y, …)
due infinitesimal
terms
toof
chunk
coordinates.
expressed
in terms
(x,
and
variable(s)
(dz, of
dqvariables
…)
y, …r, q…).
Problem ans
cosq i =
x
=
ri
yi
sin q i = =
ri
x
x 2 + y i2
yi
x 2 + y i2
y
ith
yi
ri
qi
x
P
(E )
i y
(E )
i x
qi
Ei
Phys 133
x
Problem ans x
Add up contributions, sum becomes
definite integral with limits corresponding
to charge.
( )
E net, x = å E i
i
x
L
Q
dy
= K xò 2
L 0 x + y2
[
]
3/2
é
ù
Qê
L
ú
=K
L ê x x 2 + y 2 1/ 2 ú
Phys 133 û
ë
[
]
Problem ans y
Add up contributions, sum becomes
definite integral with limits corresponding
to charge.
( )
E net, y = å E i
i
y
L
E net,y
Q
y dy
= -K ò 2
L 0 x + y2 3/2
é
Q ê1
1
= -K
2
2
L ê x xPhys
+
L
133
ë
[
]
[
ù
ú
1/ 2
ú
û
]
Problem ans complete
é
ù
Qê
L
ú
Enet, x = K
L ê x é x 2 + L2 ù1/2 ú
û û
ë ë
E net, y
Phys 133
é
Q ê1
1
= -K
L ê x x 2 + L2
ë
[
ù
ú
1/ 2
ú
û
]
Problem 26.49
A plastic rod with linear charge
density  is bent into a quarter
circle. Find the electric field at the
origin.
a) Write expressions for x and y
components of field due to a small
piece at angle q.
b) Write integrals for the
components of total field.
c) Evaluate, find
E net
Phys 133
Problem 26.49 ans
qi
Ei
ri
( )
E net, x = å E i
i
p
2
Þ òK
ldq
0
=K
R
l
R
i
x
cos q
x
(i)th bit of charge
p
2
Þ òK
ldq
0
(1)
( )
E net, x = å E i
=K
R
l
R
sin q
(1)
Phys 133
ri = R
Field of quarter circle of charge
Phys 133
Other charge geometries
•
•
•
•
Line
Hoop
Disk (from hoops)
Plane (from disk)
Phys 133
Field of line (infinite)
Points directly away, decrease with distance
2l
E line = K
r
Phys 133
Field of hoop, along axis
(E )
ring z
Phys 133
=
1
zQ
4 pe 0 ( z 2 + R 2) 3 2
Field of disk, along axis
( E disk ) z
Q
h=
A
Phys 133
é
ù
h ê
z
ú
=
12e0 ê ( z 2 + R 2 )1 2 ú
ë
û
Field of plane (“infinite”)
h
Eplane =
2e 0
perpendicular to
the plane
Phys 133
Plane from lots of lines (perpendicular to page)
Straighten up, no distance dependence
Phys 133
Parallel-plate Capacitor
•Net charge is zero
•Charge lost from +Q side ends up on -Q side
Phys 133
Find the electric field inside capacitor
h
E inside =
e0
Phys 133
E outside = 0
Phys 133
Charges in an Electric Field
Electric Field
(due to charges; we can calculate)
Charge in Electric Field
(experiences a force)
Fon q = qE
Phys 133
Do Workbook 26.28 & 31
Phys 133
Problem 26.52
A proton traveling at a speed of 1.0x106 m/s
enters the gap between the plates of a 2.0-cm
wide parallel plate capacitor. The plates have a
surface charge density of +/- 1.0x10-6 C/m2.
How far is the proton deflected when reaching
far end of capacitor?
(Ans: 2.2 mm)
e=1.6 x 10-19C; mp=1.67 x 10-27kg; o=8.85 x10-12C2/N-m2
Phys 133
Dipole in an Electric Field
Phys 133
Phys 133