Transcript Chapter 21: Electric Charge and Electric Field

Chapter 22: Electric Potential
Electric Potential Energy
 Review
of work, potential and kinetic energy
• Consider a force acts on a particle moving from point a to point b.
The work done by the force WAB is given by:
WAB  
B
A
 
B
F  d    F cos d
A

d  : an infinitesimal displacement along theparticle's path


 : theangle between F and d  at each pointalong thepath
• If the force is conservative, namely when the work done by the force
depends only on the initial and final position of the particle but not on
the path taken along the particle’s path, the work done by the force F
can always be expressed in terms of a potential energy U.
Electric Potential Energy
 Review
of work, potential and kinetic energy
• In case of a conservative force, the work done by the force can be
expressed in terms of a potential energy U:
WAB  U A  U B  (U B  U A )  U AB  U BA
U i : thepotentialenergyat pointi
• The change in kinetic energy K of a particle during any displacement is
equal to the total work done on the particle:
WAB  K  K B  K A
• If the force is conservative, then
WAB  K  K B  K A  U AB  (U B  U A )
 K A U A  KB  U B
Electric Potential Energy
 Electric
potential energy in a uniform field
• Consider a pair of charged parallel metal plates that generate a uniform
downward electric field E and a test charge q0 >0
A
+ + + ++ + + ++ +
q0
d

E
- - - - - - - - - B
F  q0 E ; it does not depend the test chargelocation
WAB  Fd  q0 Ed  0
the force is in the same direction as the
net displacement of the test charge
conservative
force
• In general a force is a vector:
Fg
m
g

F  ( Fx , Fy , Fz ) ; Fy  q0 E, Fx  Fz  0
Note that this force is similar to the force due to gravity:

Fg  ( Fg , x , Fg , y , Fg , z ) ; Fg , y  mg, Fg , x  Fg , z  0
Electric Potential Energy
 Electric
potential energy in a uniform field (cont’d)
• In analogy to the gravitational force, a potential can be defined as:
U  q0 Ey (c. f . U g  mgy)
• When the test charge moves from height ya to height yb , the work done
on the charge by the field is given by:
WAB  U AB  (U B  U A )  (q0 EyB  q0 EyA )  q0 E ( y A  yB )
If y A  yB , U AB  0 and thepotentialdecreases.
• U increases (decreases) if the test charge moves in the direction
opposite to (the same direction as) the electric force
UAB <0

E
+
+
UAB>0
A


F  q0 E
B

E
+
+
U
B
 A
F  q0 E

E
AB


F  q0 E
-
UAB<0
>0
A
B

E
-
B


F  q0 E
A
Electric Potential Energy
 Electric
potential energy of two point charges
b
rb

E
• The force on the test charge at a distance r
qq0  0 repulsive
1 qq0
Fr 
qq0  0 attractive
4 0 r 2
• The work done on the test charge
rB
qq0
qq0 1 1
dr

(  )
2
rA 4
4 0 rA rB
0 r
WAB   Fr dr  
rA
q0
r
ra
+
a
q

E
rB
1
Electric Potential Energy
 Electric
potential energy of two point charges (cont’d)
• In more general situation
tangent to the path

E

F

dr


d
B
WAB  
rB
rA
r
  rB
rB 1
qq0
F  d   F cos d  
dr
2
rA
rA 4
0 r
dr
qq 0  1 1 
    U A  U B

4 0  rA rB 
A
U
qq0
4 0 r
1
Natural and consistent definition of the
electric potential
Electric Potential Energy
 Electric
potential energy of two point charges (cont’d)
• Definition of the electric potential energy
qq0
U
4 0 r
1
• Reference point of the electric potential energy
Potential energy is always defined relative to a reference point
where U=0. When r goes to infinity, U goes to zero. Therefore
r=  is the reference point. This means U represents the work
to move the test charge from an initial distance r to infinity.
0
If q and q0 have the same sign, this work is POSITIVE ; otherwise
it is NEGATIVE.
U
U
0
qq0>0
qq0<0
Electric Potential Energy
 Electric
potential energy with several point charges
• A test charge placed in electric field by several particles

q0  q1 q2
q
q
   ...  0 i i
U
4 0  r1 r2
ri
 4 0
ri : thedistancebetween the chargei and the test charge
• Electric potential energy to assemble particles in a configuration
U
1
4 0

i j
qi q j
rij
rij : thedist ancebetween the chargei and j
Electric Potential
 Example
: A system of point charges
q1=-e
q2=+e
-
+
x=a
x=0
Work done to take q3 from x=2a to x=infinity
q3=+e
+
x=2a
q3  q1 q2   e   e  e   e 2
   
W U 



4 0  r13 r23  4 0  2a a  8 0 a
Work done to take q1,q2 and q3 to infinity
qi q j
1
1  q1q2 q1q3 q2 q3 


U




4 0 i  j rij
4 0  r12
r13
r23 

1  (e)(e) (e)(e) (e)(e) 
e




4 0  a
2a
a  8 0 a
Electric Potential Energy
 Two
interpretations of electric potential energy
• Work done by the electric field on a charged particle moving in the field
Work done by the electric force when the particle moves from A to B
electric
AB
W

 U A  U B ; displacement drAB
• Work needed by an external force to move a charged particle
slowly from the initial to the final position against the electric force
Work done by the external force when the particle moves from B to A




Fext   Felectric ; displacement drBA  drAB
WBext
A  U A U B
Electric Potential
 Electric
potential or potential
• Electric potential V is potential energy per unit charge
U
V
or U  q0V
q0
1 V = 1 volt = 1 J/C = 1 joule/coulomb
WAB  U A U B  q0 (VA VB )  q0VAB
potential of A with respect to B
work done by the electric force
when a unit charge moves from
A to B
work needed to move a unit
charge slowly from b to a
against the electric force
Electric Potential
 Electric
potential or potential (cont’d)
• Electric potential due to a single point charge
U
1 q
V 
q0 4 0 r
• Electric potential due to a collection of point charges
U
1
V 
q0 4 0
qi
i r
i
• Electric potential due to a continuous distribution of charge
U
1
dq
V 
q0 4 0  r
Electric Potential
 From
E to V
• Sometimes it is easier to calculate the potential from the known
electric field
WAB  
B
A
 
 
B
F  d    q0 E  d 
A
VA  VB  
B
A
 
B
E  d    E cos d
A
VA  VB  
A
B
 
E  d
The unit of electric field can be expressed as:
1 V/m = 1 volt/meter = 1 N/C = 1 newton / coulomb
Electric Potential

 
V f  Vi    E  ds
f
Example :
i

V f Vi    E  dr  kq 
R
q
0 Vi  k
R


V 
k
kq
R
1
4 0
Replace R with r
V
1
q
4 0 r
E
kq
r2

1
1
dr

kq
r2
rR
Electric Potential
Example:
q1
+
m, q0
A B
q2
-
=0

Energyconservation  K A  U A  K B  U B
1
2
K  m , U  qV
2
1
2
0  q0VA  m  q0VB
2
2q0 (VA  VB )

m
Electric Potential
 Unit:
electron volt (useful in atomic & nuclear physics)
• Consider a particle with charge q moves from a point where the potential
is VA to a point where it is VB , the change in the potential energy U is:
U B  U A  q (VB  VA )  qVBA  qVAB
• If the charge q equals the magnitude e of the electron charge
1.602 x 10-19 C and the potential difference VAB= 1 V, the change
in energy is:
U A  U B  (1.6021019 C)(1V)  1.6021019 J
 1 eV
meV, keV, MeV, GeV, TeV,…
Calculating Electric Potential

Example: A charged conducting sphere
+
+
R +
+ +
+
+
+
E
E
E0
0
1
q
4 0 R 2
1 q
E
4 0 r 2
r
V
V
R  r :V 
R  r :V 
1
q
4 0 R
1
q
V
4 0 r
0
Using Gauss’s law we calculated the electric
field.
Now we use this result to calculate the potential
and we take V=0 at infinity.
r
R  r :V 
1
q
4 0 r
the same as the potential
due to a point charge
1
q
4 0 R
1
q
4 0 R
inside of the conductor
E is zero. So the potential
stays constant and is
the same as at the surface
Equipotential Surface
 Equipotential
surface
• An equipotential surface is a 3-d surface on which the electric potential V
is the same at every point
• No point can be at two different potentials, so equipotential surfaces for
different potentials can never touch or intersect
• Because potential energy does not change as a test charge moves over an
equipotential surface, the electric field can do no work
• E is perpendicular to the surface at every point
• Field lines and equipotential surfaces are always mutually perpendicular
Equipotential Surface
 Examples
of equipotential surface
Equipotential Surface
 Equipotentials
and conductors
• E = 0 everywhere inside a conductor
- At any point just inside the conductor the component of E tangent to
the surface is zero
- The tangential
component of E is also zero just outside the surface

E
vacuum
E
E//
conductor

E 0
If it were not, a charge could move around a
rectangular path partly inside and partly outside
and return to its starting point with a net amount
of work done on it.
• When all charges are at rest, the electric field just outside a conductor must
be perpendicular to the surface at every point
• When all charges are at rest, the surface of a conductor is always an
equipotential surface
Equipotential Surface
 Equipotentials
and conductors (cont’d)
• Consider a conductor with a cavity without any charge inside the cavity
- The conducting cavity surface is an equipotential surface A
- Take point P in the cavity at a different potential and it is on a
different equipotential surface B
Guassian surface
- The field goes from surface B to A or A to B
- Draw a Gaussian surface which surrounds the surface B inside
cavity
B
A
P
equipotential
surface through P
- The net flux that goes through this Gaussian surface is not zero
because the electric field is perpendicular to the surface
conductor
- Gauss’s law says this flux is zero as there is no charge inside
surface of cavity
- Then the surfaces A and B are at the same potential
• In an electrostatic situation, if a conductor contains a cavity and if no
charge is present inside the cavity, there can be no net charge anywhere
on the surface of the cavity
Equipotential Surface
 Electrostatic
shielding
Potential Gradient
 Potential
gradient
• Potential difference and electric field
B
A
VA  VB   dV   dV
A
B
• Potential difference and electric field
VA  VB  
B
A
B
B
A
A
  dV  
 
E  d
 
E  d

E  E xiˆ  E y ˆj  E z kˆ

d   dxiˆ  dyˆj  dzkˆ
 
 dV  E  d   Ex dx  E y dy  Ez dz
Potential Gradient
 Potential
gradient (cont’d)
f
f ( x  x)  f ( x)
 lim
...

x

0
x
x
• E from V
V
Ex  
x
V
Ey  
y
V
Ez  
z

 V ˆ V ˆ V
E  
i
j
y
z
 x

ˆ
k 

• Gradient of a function f

ˆ  ˆ  ˆ  
f   i  j  k  f
y
z 
 x


E  V
If E is radial with respect to
a point or an axis
V
Er  
r
Potential Gradient
 Potential
gradient (cont’d)
Exercises

Exercise 1
Exercises

Exercise 1 (cont’d)
Exercises

Exercise 1 (cont’d)
Exercises

Exercise 2
Exercises

Exercise 3
Exercises

Exercise 4
Exercises

Exercise 4 (cont’d)
Exercises

Exercise 4 (cont’d)
Exercises

Exercise 4 (cont’d)
Exercises

Exercise 5: An infinite line charge + a conducting cylinder
Q
-Q
Outer metal braid
r
r

E is in radial direction
1 l
Er 
2 0 r
Va  Vb  
b
a

  b
l
E  d    Er dr 
a
2 0
r
l
ln b
2 0 ra
Signal wire
line charge density l

b
a
dr
r