Download P - Oregon State University

Tuesday, 28 June 2016
Overview
• Continuous Charge distributions
•
•
•
•
Finish up line of charge
Ring of charge
Disk of Charge
Sphere of charge
• Applications:
• Parallel plate capacitor
Announcements
• Homework (due Wednesday)
• Will take about a week to grade.
• Good idea to make a hard or electronic copy of what you hand in.
If 8 nC of charge are placed on the square
loop of wire, the linear charge density will be
46%
37%
A. 800 nC/m.
B. 400 nC/m.
C. 200 nC/m.
D. 8 nC/m.
E. 2 nC/m.
9%
4%
800 nC/m. 400 nC/m. 200 nC/m.
4%
8 nC/m.
2 nC/m.
QuickCheck 26.6
If 8 nC of charge are
placed on the square loop
of wire, the linear charge
density will be
A. 800 nC/m.
B. 400 nC/m.
C. 200 nC/m.
D. 8 nC/m.
E. 2 nC/m.
Slide 26-45
Continuous Charge Distributions
The linear charge
density of an object of
length L and charge Q
is defined as
Linear charge density,
which has units of
C/m, is the amount of
charge per meter of
length.
Slide 26-43
Q

L
Y-axis
Line of charge
x=0, y=L/2
𝑑𝑄 = 𝜆𝑑𝑦
X-axis
L
P=(xP,yP)
x=0, y=-L/2
*no symmetry here
Q

L
Line of charge: calculating…
Y-axis
1) Find components of E-field due to dQ:
𝑑𝑞
• Find 𝐸𝑑𝑞 : 𝐸𝑑𝑞 = 𝐾 2
𝑟
2
- Find 𝑟𝑑𝑞 : 𝑟 = 𝑥𝑃 − 0 2 + 𝑦𝑃 − 𝑦
• Find components: 𝐸𝑑𝑞𝑋 𝑎𝑛𝑑 𝐸𝑑𝑞 𝑦 :
x=0, y=L/2
dq=λdy
𝐸𝑑𝑞 𝑥 = 𝐸𝑑𝑞 cos 𝜃
(x=0,y)
L
𝐸𝑑𝑞 𝑦 = −𝐸𝑑𝑞 sin 𝜃
X-axis
2
cos 𝜃 =
sin 𝜃
𝑥𝑝
𝑟
𝑦−𝑦𝑃
=
𝑟
𝐿/2
P (xP,yP)
θ
2) Determine limits of integration: ‫׬‬−𝐿/2 … . 𝑑𝑦
3) Integrate components:
𝐿/2
𝐸𝑥 = න
𝐸𝑑𝑞𝑥
−𝐿/2
𝐿/2
x=0, y=-L/2
𝐸𝑥 = න
𝐾
𝜆𝑑𝑦
2
2
−𝐿/2 𝑥𝑝
+ 𝑦𝑝 − 𝑦
𝐿/2
𝜆𝑑𝑦
=‫׬‬−𝐿/2 𝐾 2
2
(𝑥𝑝 + 𝑦𝑝 −𝑦 ) 3/2
𝑥𝑝
𝑥𝑃2 + 𝑦𝑃 − 𝑦
2
Integrating…
xP dy
E x K 
2
2 3/ 2
( xP  ( y  y P ) )
L / 2
L/2
ds
s

2
2
3
/
2
 (s  a ) a 2 s 2  a 2
Let 𝑠 = 𝑦 − 𝑦𝑃 → 𝑑𝑠 = 𝑑𝑦
See Appendix A-3, or you can verify this by doing
integration by parts.
Integrating
L / 2 yP
ds
E x  KxP  
2
2 3/ 2
( xP  s )
 L / 2 y P
ds
s
 ( s 2  a 2 )3 / 2  a 2 s 2  a 2
Adding together
L / 2 yP
ds
E x  KxP  
2
2 3/ 2
( xP  s )
 L / 2 y P
L / 2 yP
E x  KxP 
s
x
2
P
x s
2
P
2
 L / 2 y P
Adding together
K 
E x
xP 

L / 2  yP
x  ( L / 2  yP )
2
P
2



2
2 
xP  ( L / 2  y P ) 
 L / 2  yP
Can you write an equation for the y-component?
(Check this yourself.)
K 
E y
xP 

1
x  ( L / 2  yP )
2
P
2



2
2 
xP  ( L / 2  y P ) 
1
Problem-Solving Strategy: The
electric field of a continuous
distribution of charge
Check:

L / 2  yP
 L / 2  yP
K 
E x


2
2
xP  x 2  ( L / 2  y )2
x

(

L
/
2

y
)
P
P
P
 P
Oregon State University PH 213, Class #4




QuickCheck 26.8
At the dot, the y-component of the electric field due to the
shaded region of charge is
dfghgf
A. A.A
B. B
C. B.C
D. D
E. C.E
67%
D.
E.
4%
6%
A
B
14%
10%
C
Slide 26-52
D
E
QuickCheck 26.8
At the dot, the y-component of the electric field due to the
shaded region of charge is
A.
B.
C.
D.
E.
Slide 26-53
The rod has a continuous and
uniform net charge distribution.
Which of the following actions will
increase the electric field strength
at the position of the dot?
1. Lengthen the rod
without moving its
center or changing its
charge
2. Widen the rod without
moving its center or
changing its charge
3. Shorten the rod without
moving its center or
changing its charge
4. None of the above.
49%
26%
15%
Lengthen the
rod without
moving its
center or
changing its
charge
Oregon State University PH 213, Class #5
10%
Widen the
rod without
moving its
center or
changing its
charge
Shorten the
rod without
moving its
center or
changing its
charge
None of the
above.
The rod has a continuous and
uniform net charge distribution.
Which of the following actions will
increase the electric field strength
at the position of the dot?
1. Lengthen the rod without moving its center or
changing its charge
2. Widen the rod (into and out of the page) without
moving its center or changing its charge
3. Shorten the rod without moving its center or
changing its charge
4. None of the above.
Electric field of ring
Easy on axis!
s-axis
θ
R
P (x=0,y=0,zP)
z-axis
Electric field of ring
Easy on axis!
s-axis
θ
R
P (x=0,y=0,z=zP)
2
𝑟 =
𝑧𝑃2
+𝑅
z-axis
dq
EK 2
r
2
zP
Ez  E
r
Cylindrical Coordinate system: r=(r,θ,z)
Electric field of ring
z P Rd
Ez  K  2
( z P  R 2 )3 / 2
Easy on axis!
(s=R,θ,z=0)
R
P (zP,s=0,θ)
r  zP  R
2
2
dq
EK 2
r
2
zP
Ez  E
r
Electric field of ring
z P R 2
Ez  K 2
( z P  R 2 )3 / 2
Easy on axis!
(s=R,θ,z=0)
R
P (zP,s=0,θ)
r  zP  R
2
2
dq
EK 2
r
2
zP
Ez  E
r
Electric field of ring
KQzP
Ez  2
( z P  R 2 )3 / 2
Easy on axis!
(s=R,θ,z=0)
R
P (zP,s=0,θ)
r  zP  R
2
2
dq
EK 2
r
2
zP
Ez  E
r
A Ring of Charge
 Consider the on-axis
electric field of a positively
charged ring of radius R.
 Define the z-axis to be the
axis of the ring.
 The electric field on the
z-axis points away from
the center of the ring,
increasing in strength until
reaching a maximum
when |z| ≈ R, then
decreasing:
Slide 26-55
Which of the following statements is not true
for the electric field of a semi-circle:
A. We expect a 1/(distance squared) dependence
for the field
B. We expect the electric field in the y-direction to
be zero
C. The charge density will allow us to write:
dQ = (λ)d(Θ)
42%
31%
21%
7%
D. We need to take vector components before
integrating
We expect a
1/(distance
squared)
dependence
for the field
We expect
The charge
density will
the electric
allow us to
field in the ydQ
direction to write:
= (λ)d(Θ)
be zero
We need to
take vector
components
before
integrating
Electric field of semi-circle
Which of the following statements is not true:
1.We expect a 1/(distance squared) dependence for the field
2.We expect the electric field in the y-direction to be zero
3.The charge density will allow us to write:
dQ = (λ)d(Θ)
4.We need to take vector components before integrating
Let’s look at planes…
• Surface charge:
Q

A
Q   ( x, y)dxdy
Oregon State University PH 213, Class #5
Continuous Charge Distributions
The surface charge
density of a twodimensional distribution
of charge across a
surface of area A is
defined as:
Surface charge
density, with units
C/m2, is the amount of
charge per square
meter.
Slide 26-46
What is the differential amount of area dA, of a ring
of width dR at a distance R from the center?
78%
1.
2.
3.
4.
2πR
πR2
(2πR) dR
(πR2 )dR
11%
11%
0%
2πR
πR2
(2πR) dR
(πR2 )dR
What is the differential amount of area dA, of a ring
of width dR at a distance R from the center?
1. 2πR
2. πR2
3. (2πR) dR
4. (πR2 )dR
A Disk of Charge
Electric field of Disk
(s=R,θ,z=0)
R
P (zP,s=0,θ)
Let’s look at limits for a charged disk:
What about a square plane of charge?
Square – Try this one for yourself
z-axis
Square – Try this one for yourself
y-axis
L
P (xP,y=0,z=0)
L
x-axis
Q

L
Recall: Line of charges
x=0, y=L/2
KQ 
1
E
r  r 2  ( L / 2) 2

L
r
x=0, y=-L/2
P




z-axis
Square – Try this one for yourself
y-axis
P (xP,y=0,z=0)
   dy
x-axis
Square – Try this one for yourself
L/2
E

L / 2


2
2 
y  xP 
KL
E  KLxP
L/2

L / 2


2
2
2 
y  xP  ( L / 2) 

1

y 2  xP2 

1
xP
y x
2
2
P
dy

dy
y 2  xP2  ( L / 2) 2 
1
Square – Try this one for yourself
E  KLxP
L/2

L / 2

1

y 2  xP2 


dy
y 2  xP2  ( L / 2) 2 
1
wL
E  4 K tan (
)
2 xP
1
1
2
w
1
2
L
L x
2
2
P
Square – Try this one for yourself
Now what if the square was very large? L>>xP
E  4K tan ()
1

E  2K 
2 0