Electrostatic energy of a charge
distribution.
N.B.: in vacuum – not in a material !!! And, with the
boundary condition that Φ(∞) = 0
This one we have already seen! For a system of N (point) charges
qi q j
1
U 
2 i , j 1, N rij
Which, for a continuous charge distribution becomes:
U
1
 (r)(r)dV
2 V
The integral can be computed over the whole space, but is only nonzero in places where there is charge (ρ≠0). So, the formula is
saying that all the electrostatic energy is concentrated in the
region where there are charges, if not directly on the charges
themselves. But… we can operate on the equations:
U


1
1
E

(
r
)

(
r
)
d
V


(
r
)
dV 

2 V
2 space
4
1
8
1
8
 [  E]    E]dV 
space
 E n dS 
Surface
1
8
2
E
 dV 
space
1
8
1
8
   (E)dV 
space
1
8
 [E  E] dV 
space
2
E
 dV
space
Advanced EM - Master
in Physics 2011-2012
1
In the previous page, the key passage was to apply Gauss’s theorem to
convert from a volume integral to a surface integral. Since the volume
was the whole space, the surface enclosing it is a spherical surface of
radius r with r∞. Then, this surface integral is zero because as r
grows the surface grows as r² but the integrand (ΦE·n) decreases
as 1/
r³.
Where then is the total energy distributed? Only where the
charges are?: Or over the whole space, where E≠0?
The two formulas are equivalent. And this leaves us in doubt on where
this energy is. But… we are only in the field of electrostatics. We will
have to check which formula applies in the general case, i.e. moving
charges, and see if it is one of these two or an entirely different one.
In fact, we have used in the derivation the formula:
E   
But we know that this formula is only valid in electrostatics: the general
formula is (anticipating a little):
E    
1 A
c t
Magnetostatic Energy
(i.e., from a distribution of constant currents)
This situation is similar to one just dealt with in electrostatics. And, we
also get similar formulas for the energy distribution, despite the large
differences, i.e. E is the gradient of a scalar, while B is the curl of a
Vector potential.
Advanced EM - Master
in Physics 2011-2012
2
The two different formulas for the magnetostatic energy are:
U Mag
1

2
U Mag 
 j A dV
space
1
8
2
B
 dV
space
Again, both integrals are done over the whole space. But the first
one is really done only over a limited region, where are contained the
currents which generate the fields. The charges and the currents
are, of course, all the currents and all the charges.
The charges present under the form of matter have also to be taken
into account.
One last remark: in the last form of the energy distribution that we
have derived, the fields are sufficient to account for all energy: the
charges and the currents are not present in the formulas. This may
be a simple consequence of the relations between charges or
currents and fields. But it may also be a more fundamental fact.
In any case, the fields, which originally were defined as a simple
mathematical artifice to define the forces experimented by a charge
as the value of the charge times a field which just lies there,
E  F / q1
The fields, again, seem to have more and more physical
substance.
Advanced EM - Master
in Physics 2011-2012
3
The electrostatic stress tensor
The electric field exerts a force on charges: in fact, it is from
this force that the electric field is defined.
Well, how can we calculate the force that a given electric field
exerts on a given charge distribution? Keeping in mind, of course,
that the two, the charge distribution and the field can not be
random – they have to be compatible with each other. Well, given
the field E, the force it exerts on an infinitesimal charge ρ dV is:
dF  E(r )  (r ) dV
 E(r)  (r) dV
F
volume
Where the volume of integration is any volume where charges are
contained – well, possibly a volume contained by only one closed
surface.
Here too we can, like we did before, take away the charges – i.e.
the ρ - from the formula by replacing it with div(E)/4π
F
1
4
 E   E dV
space
And here too we have an integrand as three functions (the
components of E) times a derivative of another (E) again,
which we can treat as we did before as a derivative of the
product of two functions minus the product of the derivatives
of the first three functions times the other one (E).
Advanced EM - Master
in Physics 2011-2012
4
Given the fact that the derivatives are in fact a vector operator, we
have to be a bit careful in treating them as simple derivatives. This
is done by using the standard “Vector Formulas”, as are given, p.ex.,
in the cover pages of Jackson. Here we shall use the following one:
   C  C      C
There is however a complication: the scalar quantity in this formula (ψ)
is, in the equation we want to transform, the vector E. That is no
problem: E = Exi+Eyj+Ezk. The vector formula above will be applied
three times taking as scalar functions the three components of E.
F
F

1
4
 [i E (  E)  j E (  E)  k E (  E)]dV
x
y
z
volume
1
 [i Ex (  E)  j Ey (  E)  k Ez (  E)]dV 
4 volume
1
{i [  ( Ex E)  E  Ex ]dV  j [  ( E y E)  E  E y ]dV  K  [  ( Ez E)  E  Ez ]dV
4 V
V
V
F


1
4
1
4
 [i E E  n  j E E  n  k E E  n] dS 
x
y
z
Surface
 [i(E  E )]  j(E  E
x
y
)  K (E  E z )] dV 
Volume
1
[  E(E  n) dS   (E  )E dV
V
4 S
Advanced EM - Master
in Physics 2011-2012
5
F


1
4
1
4
 [i E E  n  j E E  n  k E E  n] dS 
x
y
z
Surface
 [i(E  E )]  j(E  E
x
y
)  K (E  E z )] dV 
Volume
1
[  E(E  n) dS   (E  )E dV ]
V
4 S
The first term (of the last formula) is easy to understand.
The second term is less easy: we have used the definition of
the quantity
( A  )B  Ax
B
B
B
 Ay
 Az
x
y
z
Now, we can use the Vector Formula:
(A  B)  (A  )B  (B  )A  A  (  B)  B  (  A)
to calculate
(E  E)  2(E  )E  2E  (  E)
(E  E)  2(E  )E
The last formula is understood by remembering that the curl of the
electric field is zero in electrostatics. We use it by replacing (in the
last expression for F)
1
2
(E  E) for (E  )E .
Advanced EM - Master
in Physics 2011-2012
6
F
We obtain then:
and are ready to use
another Vector Formula:
1
1
[  E(E  n) dS   ( E 2 ) dV ]
4 S
2V
  dV   n dS
V
in order to eventually
obtain:
S
F
1
1 2
[
E
(
E

n
)

E n] dS
4 S
2
If we write separately the three components of this vector equation,
we see that it can be written in the form:
F   TE n dS
S
Where the newly introduced quantity
TE is the
Electrostatic stress tensor
 2 E2
E x 
2

1 
TE 
Ex E y
4 

 Ex Ez

Ex E y
E2
Ey 
2
2
E y Ez
Advanced EM - Master
in Physics 2011-2012

Ex Ez 

E y Ez 

E2 
2
Ez  
2 
7
The last formula shows that the force exerted on all the charges
contained in a volume V is equal to the integral –on the surface S
which encloses the volume V- of the product of the Electrostatic
stress tensor with the unitary vector normal to the surface.
Remember at this point that the product of a tensor with a
vector is a vector, a vector whose direction depends on n but not
in a simple way: it is not necessarily parallel, nor orthogonal to n.
What has been obtained now is very similar to what we discussed
a few pages ago, about the energy of a system of charges within
their fields: it is distributed over the whole space and can be
accounted for using only the fields. Here we show that the force
exerted over a system of (static) charges and their field can be
accounted for with only the knowledge of the field on the
enclosing surface of that system.
What said so far for electrostatics is also valid for
magnetostatics. The magnetic stress tensor is defined
 2 B2
B x 
2

1 
TM 
Bx B y

4

 Bx Bz

Bx B y
B2
By 
2
2
B y Bz
In presence of charges and currents, there
will be both electrical and magnetic forces,
and the total force will be determined by the
Maxwell stress tensor T

Bx Bz 

B y Bz 

B2 
2
Bz  
2
T  TE  TM
For a physical introduction to tensors in Physics, see
Feynman II, ch.31.
Advanced EM - Master
in Physics 2011-2012
8
Magnetostatics
Let us start from the Maxwell equations. In the case of statics, electrical
and magnetic fields are independent, so only eqs 3 and 4 are relevant.
Β  0
4
Β 
J
c
The B vector field has zero divergence: therefore it is the curl of
another vector field, A, so far not yet determined, such that
B  A
If we have a field, let’s call it A0, that satisfies this condition, then
also A1 = A0 + grad Φ, with Φ any finite function of space
coordinates, satisfies it. This indetermination is called ”the gauge
invariance”. Usually the gauge is fixed by determining the div (A).
In the case of statics, the Coulomb gauge is normally chosen:
A  0
From previous courses it is known that the general (again…)
solution ot Magnetostatic problems is
1 j(r )  r
B(r1 )   2 3 12 dV
c V r12
Biot-Savart’s law, bears
some resemblance to the
electrostatics law:
r   
V
Advanced EM - Master
in Physics 2011-2012
 r'
r  r'
dV '
9
The two laws compared at the end of previous slide are also somewhat
different: one deals with the potential, the other with the field. One is
scalar, the other a vector equation, not amenable to three simple ones
because the various components are intermixed.
So, if the standard solution for the fields is of difficult handling, let’s
try and see if the situation is simpler if –like we did for electrostatics,
we try to calculate the potential ,i.e. A first.
Here is how the calculation proceeds, but before the Vector Formula
which will be used in the demonstration is shown.
  (  C)  (  C)   2 A
B   A
  (  A) 
because of the3rd Maxwell Eq.
4
j
c
(  A)   2 A 
(  A)  0
4
2A  
j
c
Vector Formula
4th Maxwell Equation (ME)
4
j
c
Vector formulaapplied to A
in theCoulombgauge!
t hisis theequat ion!!
The last equation is really analogue to the electrostatic one.
It consists of three scalar equations, one for each component
of the vector potential which is computed by integrating the
same component of the current, with exactly the same
algorithm as for the scalar potential! Let us write them all
down.
Advanced EM - Master
in Physics 2011-2012
10
 2   4
4
 2 Ax  
Jx
c
4
2
 Ay  
Jy
c
4
2
 Az  
Jz
c
These four equations are the same, apart from small changes in the
constants. And, as an obvious consequence, they admit the same
solutions, so that we can apply the “general solution” to determine
the vector potential A.
1 j(r2)
A(r1 )  
dV2
c V r1  r2
A striking similarity between electricity and magnetism! It can
not, however, be carried on too far, since in magnetism we do
not have the same boundary conditions as in electrostatics:
known potential on some surface.
Advanced EM - Master
in Physics 2011-2012
11
Cylindrical wire of infinite length run by a current
z
We already know that this system generates a
magnetic field turning around the wire with
modulus B=2I/cr.
.
.
x
The wire has radius a, then
J=(0,0,I/a2π) ; then, A=(0,0,Az)
k
θ
y
Ax=0
Ay=0
Az = -(π/c)jr2
j
.
inside the wire r≤a
= -(π/c)ja2(1+2ln[r/a]) outside
the wire
i.e., the same type of solutions
than for the electric potential of a
cylinder of uniform charge density.
If we now compute the curl of A
we find in cylindrical coordinates
(r,θ,k)
2 Ir

2
ca
2I


cr
B
for r  a ( is theunit vector along the axis)
for r  a
Advanced EM - Master
in Physics 2011-2012
12