Download An old rule of thumb

2D case:
W
conservative


 [U (r2 )  U (r1 )]
U ( x, y )
U ( x, y )
Fx  
; Fy  
x
y
2 or 3D cases:
If
or

dU
F  
dr
U ( x, y, z )
U ( x, y, z )
U ( x, y, z )
Fx  
; Fy  
; Fz  
x
y
z
then

U ( r2 )
 


dU 
W   F  dr      dr    dU  U (r2 )  U (r1 )
L

dr
U ( r1 )
W
con


 [U (r2 )  U (r1 )]
Several dimensions: U(x,y,z)
U ( x, y, z )
U ( x, y, z )
U ( x, y, z )
Fx  
; Fy  
; Fz  
x
y
z
Partial derivative is taken assuming all other arguments fixed
Compact notation using vector del, or nabla:

     
F  U ,   i 
j k
x
y
z

dU
Another notation: F   
dr
Geometric meaning of the gradient
U :
Direction of the steepest ascent;
Magnitude
U : the slope in that direction

F  U : Direction of the steepest descent

Magnitude F : the slope in that direction
http://reynolds.asu.edu/topo_gallery/topo_gallery.htm
1)The electric potential V in a region of space is given by
V ( x, y)  A( x  3 y )
2
2
where A is a constant. Derive an expression for the
electric field at any point in this region.
2)The electric potential V in a region of space is given by
c
V (r ) 
3r 3
where c is a constant. The source of the field is at the
origin. Derive an expression for the electric field at any
point in this region.
Exercise 5 p. 52
An electron moves from one point to another where the
second point has a larger value of the electric potential by
5 volts. If the initial velocity was zero, how fast will the
electron be going at the second point?
Problem 3 p. 45
Electric potential V is a scalar!
An
old
rule
ofofthumb:
you
have
totostudy
2-3
hours
aaweek
An
old
rule
thumb:
you
have
study
2-3
hours
week
An
old
rule
of
thumb:
you
have
to
study
2-3
hours
a
week
outside
the
class
per
each
credit
hour
An
old
rule
of
thumb:
you
have
to
study
2-3
hours
a
week
outside
the
class
per
each
credit
hour
An
old
rule
of
thumb:
you
have
to
study
2-3
hours
a
week
outside
the
class
per
each
credit
hour
An
old
rule
of
thumb:
you
have
to
study
2-3
hours
a
week
outside
the
class
per
each
credit
hour
An
old
rule
of
thumb:
you
have
to
study
2-3
hours
a
week
outside
the
class
per
each
credit
hour
An
old
rule
of
thumb:
you
have
to
study
2-3
hours
a
week
outside
the
class
per
each
credit
hour
An
old
rule
of
thumb:
you
have
to
study
2-3
hours
a
week
outside
the
class
per
each
credit
hour
An
old
rule
of
thumb:
you
have
to
study
2-3
hours
a
wee
outside
therule
class
per each
credit
hour
An
old
of
thumb:
you
have
to
study
2-3
hours
a
we
outside
the
class
per
each
credit
hour
An
old
rule
of
thumb:
you
have
to
study
2-3
hours
a
w
outside
the
class
per
each
credit
hour
An
old
rule
of
thumb:
you
have
to
study
2-3
hours
a
outside
the
class
per
each
credit
hour
outside
the
class
per
each
credit
outside the class per each credithour
hour
Outline
•
•
•
•
•
Area vector
Vector flux
More problems
Solid angle
Proof of Gauss’s Law
Electric field lines
These are fictitious lines we sketch which point in
the direction of the electric field.

1) The direction of E at any point is tangent to the
line of force at that point.
2) The density of lines of force in any region is
proportional to the magnitude of E in that region
Lines never cross.
Density is the number of lines going through an area (N)
divided by the size of the area
At R1
R1 R
2
At R2
N
density 
2
4R1
N
density 
2
4R2
N
At any r density 
4r 2
1
q
For a charge q located at the origin E 
40 r 2
density  E
It is important that the force is proportional to 1
r2
Gauss’s Law
The total flux of electric field out of any
closed surface is equal to the charge
contained inside the surface divided by  0 .
  Qenclosed
 E  dS 
S
0
 
E  dS
What is
water flow?
or flux of any vector, e.g. velocity of a

Consider a flow with a velocity vector v .
Let S be a small area perpendicular to v .
v
v
S
a)
b)

S  Sn
S

n
Area
 vector
Flux:
 
  vS cos  v  S
a) The volume of water flowing through S per unit time is
vS

b) Now S is tilted with respect to v . The volume of water flowing
through S per unit time is vS cos


 is the angle between velocity vector v and unit vector n
normal to the surface S.

Flux of electric field E
S
S

The flux of E
 
d  E  dS
 
 s   E  dS
S
 
   E  dS
Have a great day!
Hw: All Chapter 3 problems
and exercises
Reading: Chapter 4