Download VOLTAGE

VOLTAGE
Now that we know that the electrostatic field is conservative we can define a potential
energy for it. We recall the definition of potential energy:
PEf – PEi = Work I do to move object from initial to final position at constant speed

This requires there to be no net force on the object. Hence if the field exerts a force F , I must



exert F M = - F . Thus the work I do in moving the object a distance d r is:

 

dW  FM  dr   F  dr
Thus
 
dPE   F  dr
PE F  PEi 
f 

 F  dr
i
We now apply this to the electric field:


F  qE
 
dPE  qE  dr
f 

PE F  PEi  q  E  dr
i
It is convenient to define a new concept, voltage, as the potential energy/charge:
 
dV  E  dr
VF  Vi 
f 

 E  dr
i
Note that
dV 
 
V
V
V
dx 
dy 
dz   E  dr  E x d x  E y dy  E z dz
x
y
z
Hence
Ex  
V
x
Ey  
V
y
Ez  
V
z

Recalling the definition of  we get


E  V
This is analogous to altitude lines on a topographic map. The change in altitude gives the change
in energy/pound.
As with potential energy, only change in energy is defined. Thus we can choose the zero
of voltage to be wherever we wish. For point charges a convenient choice is when the two
charges are infinitely far apart (same thing we did for gravity). Then the voltage at distance r
from a charge Q is:
r
V(r)  V() 

 


kQ
r
2
 V(r) 
dr 
kQ
r
r


kQ
r
kQ
r
ENERGY STORED IN ELECTRIC FIELDS
We now consider the energy stored in an assembly of charges. There are two things to
consider: creation of the charges at infinite separation, and moving them to their final locations.
We will consider only the second. The first – the “self energy” of the charges - is an unsolved
problem, basically because of the 1/r2 nature of the force which leads to infinities. Thus we will
calculate the work needed to assemble the already existing charges to their final locations.
To do this we note that since electrostatics is conservative we have the potential energy
discussed above. Hence we can calculate the work from:
dPE = dW = (dq)V
Now bring the first charge from infinity to its location. This is free since V = 0 (no charges yet
present). Now bring the second to a distance r12 from the first. If we take the zero of potential
energy to be at infinite separation we have:
V
kq1
r12
Hence
W2 
kq1q 2
r12
where W is the work required.
Now bring up the third charge. This time there is potential energy due to both charges 1
and 2. Because of superposition they add:
V
kq1 kq 2

r13 r23
and
W3 
kq1q3 kq 2q3

r13
r23
Continuing in this fashion we find the total work done – and thus the potential energy of the
assembly – to be:
N
N 1
kqi q j
i2
j i
rij
W  PE  

1
where rij is the separation between charges i and j. Writing this out we have:
W
kq 2q1 kq 2q3 kq 2q 4


 ...
r12
r13
r14

kq3q1 kq3q 2 kq3q 4


 ...
r13
r23
r34

kq N q1 kq N q 2
kq q
kq q
kq q

 ...  N k  N k 1  ...  N N 1
r1N
r2N
rkN
rk 1,N
rN,N 1
Now consider the sum
N
kqi q j
i, j
rij

i j
1

kq1q 2 kq1q3
kq q kq q

 ...  2 1  2 3  ...
r12
r13
r12
r23

kq3q1 kq3q 2 kq3q 4
kq q kq q
kq q


 ...  N 1  N 2 ...  N N 1
r13
r23
r34
rN1
rN2
rN,N 1
Note that this is exactly the same as the sum above except that each term appears twice. Thus:
W  PE 
 N kq j  1
1 N kqi q j 1
  qi  

   qi Vi  PE
2 i, j
rij
2 i  j i rij  2 i
i j
1

1
where Vi is the voltage at charge i due to all the other charges.
We now consider what happens if we have a continuous charge distribution. Then the
sum becomes an integral, but over what? The charge in an infinitesimal volume, d3r is:

q    r  d3r



where ρ( r ) is the charge density at position r . The work required to add a charge dq at r to an
existing charge Δq at r is:

dW = dq*V(q, r )

where V(q,r) is the voltage at r due to all the other charges:

  k  r  d3r '
V  q, r     
rr'

Hence the energy in the tiny volume d3r is:



dW   d  r V  , r  d3r
 0
Then the total energy in the volume is:


W   d3r  dV  , r 
vol
 0
But since V~ρ we have:
dV d

 dV  Vd
V

Then
  V   dV  V  2V
W 
1  3 V
1      3
  r ' V r ' d r '  PE
d r    V  


2 
2
0
vol
We can put this in another useful form by using the first Maxwell equation
 

  E  4k 
g0
Then
PE 
1   3
 g   E Vd r '
0
2
But


 
 
   VE   V  E  V    E 
 
 

 V    E     VE  E V


Since E  V we get
 


V    E      VE   E 2
 PE 
 

1   
1
g0    VE  E 2 d3r '  g0   VE  dA   E 2d3r '
2
2

vol
 S

But at large distances
V~
Hence
1
r
 1
E~
r2

1
1
VEA ~ r 2   0
r
r3
We can interpret this as meaning there is an energy density ε0E2/2 associated with an electric
field.
APPLICATIONS
As an example of these ideas consider a charge distribution given by:

Q 2
r
 r  
R5
1 r  R 


0 r  R 
To find the energy of the distribution we first need to find the voltage. This we can do with
Gauss’s Law because of the spherical symmetry. We know that the field will be that of a point
charge at the origin equal to the total charge inside radius r. Hence we need that charge. It is
given by:

q     r ' d vol





r
Q



5
2






R r  0  0  0
Q
R
5
Q
R5
r
2
2
4Q







r 2 r 2 sin  drdd
r 4 sin  drd
r  0  0
r
4



r 0
r



R 5 r 0
r dr   cos  

0
4Q  r 
r dr 
 
5 R
4
5
Since the electric field of a point charge is just:
 kq
k4Q rˆ
E
rˆ 
5 r2
r2
And
 r 5
 
 R 
 1


r  R

r  R 
r
V(r) 
 
 
E  dr

r
R

4
kQ
1  3 


   1
dr 
 
 r dr


5

 5    r 2
R
 

R
4kQ  1 r 4  R 4 

 

5  R
4R 5 
4kQ  5
r 4  4kQ 
r4 

5  4 
 

5R  4 4R 4  20R 
R 
Thus
1
1 Q
PE   Vd3r 
2
2 R5
R







2



r  0  0  0
r2
4kQ 
r4 
 5  4  r 2 sin  drdd
20R 
R 
R
R
82 kQ 2   4
1  8  82 kQ 2  R 5 1 5  82 kQ2 8 Q 2 4 1
5  r dr 

 R 

 r dr  
5


20R 9 45R g0

20R 6  0
R4 0
20R 6  5 9 
Next we do it using our result for the energy stored in an electric field. This time we
have:
5

 k 4Q  r 
E  rˆ
 R 
5 
r2
 1
Then

r  R

r  R 
g
PE  0  E 2d Vol
2
R
2

2
10
g0  4Q  2  
 r  sin  drdd 




 
k






2  5 
 r 0  0  0  R 

r2











2



r  R  0  0
sin  dddr
r2

R

10
g0 162Q2 2    r  dr  dr 
k 4    


2 
2
R
2
25
 r
r


R
0

g 323Q2 2  1 R 9 1 
k 
 0
 
25
 R10 9 R 
g 323Q2 k 2 10 2Q 2  10 4Q2
 0


25R
9 g0 225R 45g0 R
as before!!