Electrostatic Energy
EE 141 Lecture Notes
Topic 14
Professor K. E. Oughstun
School of Engineering
College of Engineering & Mathematical Sciences
University of Vermont
2014
Motivation
Potential Energy of a Static Charge Distribution
Whenever two charges qa and qb are brought within a distance Rab of
each other, work is expended against the Coulomb force [see Topic 1,
Eq. (3)] in consummating the process. Once the charges are in place,
the persistence of the Coulomb force makes the energy stored in the
electrostatic field potentially available whenever demanded.
If it is assumed that the charges are moved slowly enough into place
(i.e. reversibly), then their kinetic energies may be neglected and any
loss due to electromagnetic radiation effects, significant if rapid
charge accelerations occur, may be neglected.
Potential Energy of a Static Charge Distribution
Consider the energy stored in a fixed configuration of n charges,
given by the work required to assemble the charge configuration.
• Assume that all n charges q1 , q2 , . . . , qn are initially located at
infinity in their zero potential state.
• Upon bringing just q1 from infinity to its final position P1 , no work
is expended because no other charges are present.
• The work done in bringing q2 from infinity to P2 is given by
(1)
U2 = q2 V2
(2)
= q1 V1
(1)
q1
4πǫ0 R12
q2
= q1
4πǫ0 R21
= q2
(1)
(2)
where V2 denotes the electrostatic potential at P2 due to the
(2)
charge q1 at P1 , and where V1 denotes the electrostatic potential at
P1 due to the charge q2 at P2 .
Potential Energy of a Static Charge Distribution
• The work done in bringing a third charge q3 in from infinity to P3
is then given by
(1)
(2)
(3)
(3)
(3)
(4)
U3 = q3 V3 + q3 V3
= q1 V1 + q2 V2 ,
and so on for the remaining charges q4 , q5 , . . . , qn , noting that
(j)
(k)
qk Vk = qj Vj
(j)
where Vk denotes the electrostatic potential at Pk due to the
charge qj at Pj .
(5)
Potential Energy of a Static Charge Distribution
• The total energy Ue = U1 + U2 + · · · + Un can be written in two
different ways: First by adding Eqs. (1), (3), etc., giving
Ue
(1)
= q2 V2
(2)
(1)
+q3 V3 + q3 V3
(1)
(2)
(3)
+q4 V4 + q4 V4 + q4 V4
+ · · · + qn Vn(1) + qn Vn(2) + · · · + qn Vn(n−1) , (6)
or by adding Eqs. (2), (4), etc., giving
Ue
(2)
= q1 V1
(3)
(3)
+q1 V1 + q2 V2
(4)
(4)
(4)
+q1 V1 + q2 V2 + q3 V3
(n)
+ · · · + q1 V1
(n)
+ q2 V2
(n)
+ · · · + qn−1 Vn−1 . (7)
Potential Energy of a Static Charge Distribution
Adding Eqs. (6) and (7) together and dividing by 2 then yields the
more symmetric expression
h
i
1
(2)
(3)
(4)
(n)
Ue =
q1 V1 + V1 + V1 + · · · + V1
2
h
i
(1)
(3)
(4)
(n)
+q2 V2 + V2 + V2 + · · · + V2
h
i
(1)
(2)
(4)
(n)
+q3 V3 + V3 + V3 + · · · + V3
i
h
(1)
(2)
(3)
(n−1)
.
+ · · · + qn Vn + Vn + Vn + · · · + Vn
Potential Energy of a Static Charge Distribution
The potential energy of the assembled charge configuration is then
given by
n
1X
Ue =
(8)
qk Vk
2
k=1
where
qk = charge of the k th particle located at Pk ,
Vk = absolute potential at Pk due to all of the charges except qk .
Notice that this expression does not include the self-energy of the
individual charges; this is the energy that would be liberated if each
charge was allowed to expand to an infinite volume. As a
consequence, Eq. (8) identically vanishes for a single point charge.
Potential Energy of a Static Charge Distribution
For a continuous (macroscopic) volume charge distribution ̺v (r), the
expression (8) for the electrostatic potential energy generalizes to
ZZZ
1
(9)
̺v (r)V (r)d 3r
Ue =
2
V
For a continuous (macroscopic) surface charge distribution ̺s (r), the
expression (8) for the electrostatic potential energy generalizes to
ZZ
1
̺s (r)V (r)d 2r
Ue =
(10)
2
S
For a continuous (macroscopic) line charge distribution ̺ℓ (r), the
expression (8) for the electrostatic potential energy generalizes to
Z
1
Ue =
(11)
̺ℓ (r)V (r)d ℓ
2 C
Notice that these expressions include the self-energies of the charges.
Self-Energy of a Spherical Charge Distribution
For a uniform spherical charge distribution of charge Q and radius r0
with charge density ̺v = 3Q/(4πr03) for r ≤ r0 , the absolute
electrostatic potential inside the sphere is given by (see Topic 7)
Q
Q
V (r ) =
r02 − r 2 +
; r ≤ r0 .
3
8πǫ0 r0
4πǫ0 r0
From Eq. (9), the self-energy of this spherical charge distribution is
Z
Z π
Z r0
1 2π
dφ
sin θd θ
̺v (r )V (r )r 2 dr
Use =
2 0
0
Z r0 0
3Q
Q
Q
2
2
=
r 2 dr
r0 − r +
3
3
2r0 0 8πǫ0 r0
4πǫ0 r0
2
3Q
=
→ ∞ as r0 → 0 at fixed Q 6= 0.
20πǫ0 r0
Alternatively,
̺v
Use = r02 → 0 as r0 → 0 at fixed ̺v .
ǫ
Electrostatic Energy
From Poisson’s equation
̺v (r) = −ǫ∇2 V (r)
(12)
at every point in the electrostatic field in a material with dielectric
permittivity ǫ. Substitution of this expression in Eq. (9) then gives
ZZZ
ǫ
Ue = −
V (r)∇2 V (r)d 3 r ,
(13)
2
V
where V is any volume containing all of the charges in the system.
From Green’s first integral identity [see Eq. (40) of Topic 2]
ZZZ
I
3
2
φ∇ ψ + ∇φ · ∇ψ d r =
φ∇ψ · n̂d 2 r
V
S
with φ(r) = ψ(r) = V (r), one obtains
I
ZZZ
3
2
2
V ∇ V + (∇V ) d r =
V ∇V · n̂d 2 r
V
.
S
Electrostatic Energy
With this substitution, the expression (13) for the electrostatic energy
becomes
I
ZZZ
ǫ
2
2 3
(14)
V ∇V · n̂d r −
(∇V ) d r
Ue = −
2 S
V
Note that:
1
because V can be any volume that contains all of the charges in
the system configuration, the boundary surface S may then be
chosen at an arbitrarily large distance from the charge
distribution;
2
because V (r) falls off at least as fast as 1/r as r → ∞, then
∇V (r) falls off at least as fast as 1/r 2 as r → ∞, and because
the surface area of S increases as r 2 in that limit, then the
surface integral appearing in Eq. (14) decreases at least as fast
as 1/r as r → ∞ and can be made arbitrarily small by choosing
S sufficiently distant from the source charge distribution.
Electrostatic Energy
Because ∇V (r) = −E(r), the electrostatic energy is then given by
ZZZ
ZZZ
1
ǫ
2
3
E (r)d r =
D(r) · E(r)d 3 r
(15)
Ue =
2
2
V
V
where the volume V must now only be large enough to include all
regions where the electrostatic field E(r) produced by the charge
distribution is nonzero. If this is not satisfied, then the electrostatic
energy is given by Eq. (13). Notice that this expression includes the
self-energies of all the charges in the system and is positive-definite,
whereas the expression given in Eq. (8) can be negative.
The integrand appearing in Eq. (15) is defined as the electrostatic
energy density
1
ue (r) ≡ D(r) · E(r)
2
(J/m3 )
(16)
which is associated with the field energy at each point in the field.
ElectrostaticEnergy-Thermodynamic Interpretation
The electrostatic energy Ue of a system of charges is derived from
the work done on the system in bringing it to its charged state [see
Eqs. (1)–(8)]. The question then arises as to what thermodynamic
property of the system this work represents.
For a reversible process, the First Law of Thermodynamics
(conservation of energy) states that
dWi = TdS + dWm ,
(17)
where dWi represents the change in internal energy of the system, dS
is the entropy and T is the absolute temperature, so that TdS is the
heat added to the system during the process, and where dWm is the
mechanical work done on the system.
ElectrostaticEnergy-Thermodynamic Interpretation
The work increment dWm may be identified with the change in
internal energy dWi of the system only for an adiabatic process for
which dS = 0.
Because the temperature T will, in general, change during an
adiabatic process, and because the dielectric permittivity is a function
of the temperature, only isothermal processes may be considered in a
thermodynamic interpretation of the electrostatic energy.
The quantity of interest here is the Helmholtz free energy
F = Wi − TS.
(18)
ElectrostaticEnergy-Thermodynamic Interpretation
Differentiation of Eq. (18) and combining the result with Eq. (17)
then gives
dF = dWi − TdS − SdT
= dWm − SdT .
(19)
Hence, for an isothermal process (dT = 0)
dF = dWm ,
(20)
and the electrostatic energy is seen to be part of the free energy of
the system. This energy represents the maximum work that can be
extracted from the electrostatic field at a later time.