Transcript PHYS_3342_090611

The problem solving session will be Wednesdays from
12:30 – 2:30 (or until there is nobody left asking questions)
in FN 2.212
Torque on the electric dipole
r
r
p  qd
(electric dipole moment from “-” to “+”)
Electric field is uniform in space
Net Force is zero
 Torque is not zero
Net
  (qE )(d sin  )



  p E
(torque is a vector)
Stable and unstable equilibrium




p  E
p  E
Charge #2
Three point charges lie at the vertices of an
equilateral triangle as shown. Charges #2
and #3 make up an electric dipole.
The net electric torque that Charge #1 exerts
on the dipole is
+q
Charge #1
+q
y
–q
x
A. clockwise.
B. counterclockwise.
C. zero.
D. not enough information given to decide
Charge #3
Electric field of a dipole
E-field on the line connecting two charges
 1 1 
E  keq 2  2 
r2 r1 

-
A
+
E
r2
d
2 p ke
E 3
r

r1


E-field on the line perpendicular to the dipole’s axis
E  2E 2 sin

E2

E


2
 E2
d
r
E 2  ke
A
E1
-
q
r2
qd
E  ke 3
r
r
r
p
E  k e 3
r

when r>>d
+
d
General case – combination of the above two

 

3( p r )  p
E
r 3
5
r
r
Dipole’s Potential Energy
E-field does work on the dipole – changes its potential energy
Work done by the field (remember your mechanics class?)
dW   d   pE sin d
 
U p E
Dipole aligns itself to minimize its potential energy
in the external E-field.
Net force is not necessarily zero in the non-uniform
electric field – induced polarization and electrostatic
forces on the uncharged bodies
Chapter 22
Gauss’s Law
Charge and Electric Flux
Previously, we answered the question – how do we find
E-field at any point in space if we know charge distribution?
Now we will answer the opposite question – if we know E-field
distribution in space, what can we say about charge distribution?
Electric flux
Electric flux is associated with the flow of electric field through a surface
1
E~ 2
For an enclosed charge, there is a connection
r
between the amount of charge
2
and electric field flux.
S~r
E S  const
Calculating Electric Flux
Amount of fluid passing through
the rectangle of area A
dV
 A
dt
dV
  A cos 
dt
dV  
 A
dt
Flux of a Uniform Electric Field
 
 E  E A  EA cos 


A A n

n
- unit vector in the direction of normal to the surface
Flux of a Non-Uniform Electric Field


E   E d A
S
E – non-uniform and
A- not flat
Few examples on calculating the electric flux
Find electric flux
E  2 103[ N / C ]
Gauss’s Law
E 


 E d A
 qi
0
Applications of the Gauss’s Law
Remember – electric field lines must start and must end on charges!
If no charge is enclosed within Gaussian surface – flux is zero!
Electric flux is proportional to the algebraic number of lines leaving
the surface, outgoing lines have positive sign, incoming - negative
Examples of certain field configurations
Remember, Gauss’s law is equivalent to Coulomb’s law
However, you can employ it for certain symmetries to solve the reverse problem
– find charge configuration from known E-field distribution.
Field within the conductor – zero
(free charges screen the external field)
Any excess charge resides on the
surface


E d A0
S
Field of a charged conducting sphere
Field of a thin, uniformly charged conducting wire
Field outside the wire can only point
radially outward, and, therefore, may
only depend on the distance from the wire


Q
 E d A  0

E
2 r 0
- linear density of charge
Field of the uniformly charged sphere
Uniform charge within a sphere of radius r
r
q  Q 
a
3
'
E

r
3 0

Q - total charge
Q
- volume density of charge
V
Field of the infinitely large conducting plate
s
Q
A
s- uniform surface charge density
s
E
2 0
Charges on Conductors
Field within conductor
E=0
Experimental Testing of the Gauss’s Law