Transcript Chapter 4 Energy and Potential

Chapter 4
Energy and Potential
4.1 Energy to move a point
charge through a Field
• Force on Q due to an electric field
FE
QE
• Differential work done by an external
source moving Q
dW
QE dL
• Work required to move a charge a finite
distance
4.2 Line Integral
• Work expression without using vectors
W

Q 

final
E L dL
initial
EL is the component of E in the dL direction
• Uniform electric field density
W
QE L BA
Example
y
E( x y )   x 
 
2
 .8 
A   .6 
 
1
Q  2
1
B   0 
 
1
Path:
2
x y
2
1
z
1
Calculate the work to cary the charge from point B to point A.
W
A
A
A
0
1
2
 0
 1
 2


Q
E( x y ) d x  Q
E( x y ) d y  Q  E( x y ) d z
0
1
2



B
B
B
Plug path in for x and y in E(x,y)
A


A1
 0

2

W  Q
E 0  1  x 0 d x  Q 
B
B
0

E
1

A2

1  y  0 1 d y  Q 

B
2
E( 0  0) d z
2
2
W  0.96
Example
y
E( x y )   x 
 
2
Q  2
 .8 
A   .6 
 
1
1
B   0 
 
1
Path:
3( x  1) z
y
1
(straight line)
Calculate the work to cary the charge from point B to point A.
A
W
A1
 0


Q
E( x y ) d x  Q 
0


B
B
0
1
A2

E( x y ) d y  Q 
1

B
E( x y ) d z
2
2
Plug path in for x and y in E(x,y)
A

 0

W  Q
E[ 0  3( x  1) ] d x  Q 
0


B

0
A1
B1

y
E
 1  0  1 d y  Q 

 3


A2
E( 0  0) d z W  0.96
B2
• Same amount of work with a different path
• Line integrals are path independent
2
4.3 Potential Difference
• Potential Difference
• Using radial distances from the point charge
4.3 Potential
• Measure potential difference between a
point and something which has zero
potential “ground”
V AB
VA  VB
Example – D4.4
 6x2 
 
E( x y  z)   6y 
 4 
 
a) Find Vmn
2 
M   6 
 
 1 
 3 
N   3 
 
2 
M0

VMN  
N
0
M1

6x d x  
N
2
1
M2

6y d y  
N
4 dz
2
b) Find Vm if V=0 at Q(4,-2,-35)
 4 
Q   2 



35


M0

VM  
Q
0
M1

6x d x  
Q
2
1
M2

6y d y  
Q
4 dz
VM  120
2
c) Find Vn if V=2 at P(1, 2, -4)
1 
P   2 
 
 4 
N0

VN  
P
0
N1

2
6x d x  
P
1
N2

6y d y  
P
4 dz  2
2
VN  19
VMN  139
4.4 Potential Field of a Point
Charge
• Let V=0 at infinity
• Equipotential surface:
– A surface composed of all points having the
same potential
Example – D4.5
 2 
P1   3 
 
 1 
9
Q  15 10
Q is located at the origin
 12
0  8.85 10
a) Find V1 if V=0 at (6,5,4)
6
P0   5 
 
4
V1 
 1  1 

P0 
4 0  P1

Q
V1  20.677
b) Find V1 if V=0 at infinity
V1 
Q
1
V1  36.047
4 0 P1
c) Find V1 if V=5 at (2,0,4)
2
P5   0 
 
4
V1 
 1  1 5

P5 
4 0  P1

Q
V1  10.888
Potential field of single point
charge
Q1
V ( r)
4
    0  r  r1
A
|r - r1|
Q1
Move A
from infinity
Potential due to two charges
V( r)
Q1
4   0 r  r1

Q2
4   0 r  r2
A
|r - r1|
Q1
|r - r2|
Q2
Move A
from infinity
Potential due to n point charges
Continue adding charges
V( r)
Q1
4   0 r  r1

Q2
4   0 r  r2
n
V( r)

m1
 .... 
Qm
4   0 r  r m
Qn
4   0 r  r n
Potential as point charges become
infinite
Volume of charge
Line of charge
Surface of charge
V( r)





V( r)





V( r)






 v r prime

4   0 r  r prime

 L r prime

4   0 r  r prime

 S r prime
d v prime
d L prime

4   0 r  r prime
d S prime
Example
Find V on the z
axis for a uniform
line charge L in
the form of a ring
V( r)






 L r prime

4   0 r  r prime
d L prime
Conservative field
No work is done (energy is conserved) around a
closed path
KVL is an application of this
4.6
Potential gradient Relationship between
potential and electric field intensity
V=-



Ed dL
Two characteristics of relationship:
1. The magnitude of the electric field intensity is given
by the maximum value of the rate of change of potential
with distance
2. This maximum value is obtained when the direction
of E is opposite to the direction in which the potential is
increasing the most rapidly
Gradient
• The gradient of a scalar is a vector
• The gradient shows the maximum space rate of change
of a scalar quantity and the direction in which the
maximum occurs
• The operation on V by which -E is obtained
E = - grad V = - V
Gradients in different coordinate
systems
The following equations are found on page 104 and
inside the back cover of the text:
gradV
gradV
gradV
V
x
V

V
r
a x 
a  
a r 
V
y
a y 
V
z
a z
1 V
V
 a  
a z
 
z
1 V
1
V
 a  
 a 
r 
r sin   
Cartesian
Cylindrical
Spherical
Example 4.3
Given the potential field, V = 2x2y - 5z, and a point P(-4,
3, 6), find the following: potential V, electric field intensity
E
potential
VP = 2(-4)2(3) - 5(6) = 66 V
electric field intensity - use gradient operation
E = -4xyax - 2x2ay + 5az
EP = 48ax - 32ay + 5az
Dipole
The name given to two point charges of equal magnitude
and opposite sign, separated by a distance which is small
compared to the distance to the point P, at which we want
to know the electric and potential fields
Potential
To approximate the
potential of a dipole,
assume R1 and R2 are
parallel since the point
P is very distant
V
V
Q
4   0
 
1
 R1
Q d  cos   
2
4   0 r



R2 
1
Dipole moment
The dipole moment is assigned the symbol p and is
equal to the product of charge and separation
p = Q*d
The dipole moment expression simplifies the
potential field equation
Example
An electric dipole located at the origin in free space
has a moment p = 3*ax - 2*ay + az nC*m. Find V at
the points (2, 3, 4) and (2.5, 30°, 40°).
 3 10 9 


p   2 10 9 



9 
 1 10 
 12
0  8.854 10
2
P   3 
 
4
V 
p
4  0  P
 2.5 



 30

Pspherical  
180 

 
40



 180 

2

P
P
Transform this
into rectangular
coordinates
 2.5 sin  30    cos  40   

 


 180   180  





  sin  40  
Prectangular   2.5 sin  30
 


 180   180  


 



2.5 cos  30


 180 

V 
p

4  0 Prectangular
V  0.23

2

Prectangular
Prectangular
 0.958 
Prectangular   0.803 


 2.165 
V  1.973
Potential energy
Bringing a positive charge from infinity into the field
of another positive charge requires work. The work is
done by the external source that moves the charge into
position. If the source released its hold on the charge,
the charge would accelerate, turning its potential
energy into kinetic energy.
The potential energy of a system is found by finding
the work done by an external source in positioning the
charge.
Empty universe
Positioning the first charge, Q1, requires no work (no field present)
Positioning more charges does take work
Total positioning work = potential energy of field = WE =
Q2V2,1 + Q3V3,1 + Q3V3,2 + Q4V4,1 + Q4V4,2 + Q4V4,3 + ...
Manipulate this expression to get
WE = 0.5(Q1V 1 + Q2V2 + Q3V3 + …)
Where is energy stored?
The location of potential energy cannot be precisely
pinned down in terms of physical location - in the
molecules of the pencil, the gravitational field, etc?
So where is the energy in a capacitor stored?
Electromagnetic theory makes it easy to believe that
the energy is stored in the field itself