Transcript alternate - BYU Physics and Astronomy

iClicker Quiz
(1) I have completed at least 50% of the reading and studyguide assignments associated with the lecture, as indicated on
the course schedule.
a) True
b) False
iClicker Quiz
I have a copy of the text "Physics for Scientists and Engineers
Vol 2" by Serway, and I have purchased the current course
packet and the Formulas & Chapter Summaries booklet from
the bookstore.
a) True
b) False
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I registered my quiz transmitter via the course website (not on
the iclicker.com website). I realize that until I do so, my quiz
scores will not be recorded.
a) True
b) False
iClicker Quiz
I have downloaded and studied my personal Physics 220
homework data sheet and the numeric response guide, and
now understand how to correctly submit online homework
responses.
a) True
b) False
iClicker Quiz
A cat slides down a rubber pole and falls into metal pail A, which rests on a
wooden shelf. The impact breaks the shelf, causing metal pails B and C,
which were in contact, to separate and fall to the floor. What is the final
charge on pail C?
a) Positive
b) Negative
c) Zero
Coulomb’s torsion balance
E31
E32
At what position x will the Coulomb forces
acting on the charged gray bead be balanced?
F=qE
F
E
+
F
Depending on the sign of the charge, the force experienced by a
charged particle in an electric field can either point with or
against the field.
Fields due to individual point charges
Fields due to multiple charges
iClicker Quiz
The field magnitude is greatest
smallest at
at point:
point:A,
A, B,
B, C?
C?
Electric monopoles
We sometimes refer to the total
charge Q of an extended charged
object as its “monopole moment”.
When you back far enough away,
any charged object, regardless of
its shape, looks like a simple point
charge.
In the expression for the electric
field of a monopole, the vector r
originates from the “center of
charge”.
 kQrˆ
E 2
r
Electric dipoles
r
Equal and opposite charges of
magnitude Q = |Q+| = |Q|.
r
p
r 
r+ points from the negative
charge to the positive charge.
We now ask “What is the electric field produced by a dipole?”
Define dipole moment: p  Qr
Like r+, the dipole moment vector p always points from the
negative charge to the positive charge.
E  E  E 
kQr k (Q)r

3
3
r
r
 k
3(p  rˆ )rˆ  p
r3
Limit : r   0 and Q   such that p  Qr  is constant.
Moments of a distribution (extra details for purists)
(If you tell anyone that I mentioned this, I’ll deny it!)
Monopole moment (scalar)
q    (r )dV
dQ

r
Dipole moment (vector)
p   r (r ) dV
Quadrupole moment (polar rank-2 tensor)
~
~
Q   (3 r  r  r 2 I )  (r ) dV
Choose origin at the
center of charge
(like center of mass)
The electric field of a charge distribution can be expressed as a series
expansion involving successively higher moments of the distribution.
~
~
qrˆ
3(p  rˆ )rˆ  p
5(rˆ  Q  rˆ )  Q  rˆ
Ek 2 k
k
 ...
3
4
r
r
r
Off-axis
dipole field
kQr k (Q)r kQ(r  r )
E  E  E  3 

r
r3
r3

 

Q(r  r )  a ˆi  y ˆj   a ˆi  y ˆj  Q(2a ˆi )  p
For r  a , r  y
kp
kp
E 3  3
r
y
r
E
p
r
p
Serway 23.6
On-axis dipole field
d = 2a
+Q
p
E
-Q
p
E
x
E  E  E 
kQr k (Q)r

3
3
r
r
 1
 1
1 
1 
ˆ
ˆ




 kQi  2  2   kQi 
2 
2
r 
 ( x  a) ( x  a) 
 r
kQˆi
2
2
 2 1  a / x   1  a / x 
x
kQˆi   2a   2a  
 2  1    1   
x 
x  
x 
  4a  2k (2aQˆi ) 2k p
ˆ
 3
 kQi  3  
3
x
x
 x 


Binomial approximation
(1  s) p  1  ps if
s  1
Try using the general dipole formula
(only for the purists among you)
3(p  rˆ )rˆ  p
Ek
r3
Perpendicular case
On-axis case
p   p ˆi and rˆ  ˆj
3( pˆi  ˆj)ˆj  ( pˆi )
Ek
y3
0ˆj  pˆi  kp
k
 3
3
y
y
p   p ˆi and rˆ  ˆi
3( pˆi  ˆi )ˆi  pˆi
Ek
x3
 2kpˆi 2kp

 3
3
x
x
See interactive applet at http://lectureonline.cl.msu.edu/~mmp/applist/applets.htm
Calculating the electric force/field due to a continuous charge distribution

r


qdQ r
dF  k
q
r3
dQ

r
dQ


dQ r
dE  k 3
r
qQ
F   dF  ke
L
dx
L / 2 (r  x)2
L/2
L/2
qQ  1 
 ke


L  r  x   L / 2
Q
dQ   dx  dx
L
ke qdQ
qQ dx
dF 
k
2
(r  x)
L (r  x) 2
 ke
qQ  1
1 

L  r  L / 2 r  L / 2 
L
L
(r  )  (r  )
qQ
2
2
 ke
L (r  L )( r  L )
2
2
qQ
 ke 2
r  L2 / 4

Find E at the center of a hemispherical shell of uniform charge density.
The electricity vs. gravity analogy
Coulomb’s Electric Force Law
Q

r
q
 kQrˆ
E 2
r

 kqQrˆ
F  qE  2
r



 qE
F  ma  a 
m
Newton’s Gravitational Force Law
M

r
m

GMrˆ
g 2
r


GmMrˆ
F  mg  
r2


 
F  ma  a  g
E-field perpendicular to conducting surfaces
E-field perpendicular to conducting surfaces
E = /0 = 4πk
Electric field strength at
the surface of a charged
conductor is proportional
to local surface charge
density.



F  qE  ma

 qE
a
m
vx  vx 0  axt  vx 0
 qE 
v y  v y 0  a yt  
t { y0  0, v y 0  0}
 m
 qE  2
y  y0  v y 0 t  a y t  
t
 2m 
1
2
2
a  constant
v  v0  at
x  x0  v0t  12 at 2
v  v  2a ( x  x0 )
2
2
0
F  qE
WE    F  ds   q  E  ds
Uniform field :
 E  ds  E   ds  E  s  E x
K  WE  qEx