topic_14 (electrostatic energy)

8/6/2019 Topic_14 (Electrostatic Energy)

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Electrostatic Energy

EE 141 Lecture NotesTopic 14

Professor K. E. OughstunSchool of Engineering

College of Engineering & Mathematical SciencesUniversity of Vermont

2009


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Motivation


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Potential Energy of a Static Charge Distribution

Whenever two charges qa and qb are brought within a distance Rab ofeach 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 theelectrostatic 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 rapidcharge accelerations occur, may be neglected.


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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 atinfinity 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

U2 = q2V(1)

2 = q2q1

40R12(1)

= q1V(2)1 = q1 q2

40R21(2)

where V(1)

2 denotes the electrostatic potential at P2 due to the

charge q1

at P1, and where V

(2)

1denotes the electrostatic potential at

P1 due to the charge q2 at P2.


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The work done in bringing a third charge q3 in from infinity to P3is then given by

U3 = q3V(1)

3 + q3V(2)

3 (3)

= q1V

(3)

1 + q2V

(3)

2 , (4)

and so on for the remaining charges q4, q5, . . . , qn, noting that

qkV(j)k = qjV

(k)j (5)

where V(j)k denotes the electrostatic potential at Pk due to the

charge qj at Pj.


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The total energy Ue = U1 + U2 + + Un can be written in twodifferent ways: First by adding Eqs. (1), (3), etc., giving

Ue = q2V(1)

2

+q3V(1)

3 + q3V(2)

3

+q4

V(1)

4+ q

4V

(2)

4+ q

4V

(3)

4+ + qnV

(1)n + qnV

(2)n + + qnV

(n1)N , (6)

or by adding Eqs. (2), (4), etc., giving

Ue = q1V(2)1

+q1V(3)

1 + q2V(3)

2

+q1V(4)

1 + q2V(4)

2 + q3V(4)

3

+

+ q1V

(4)

1 + q2V

(n)

2 +

+ qn1V

(n)

N . (7)


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Adding Eqs. (6) and (7) together and dividing by 2 then yields themore symmetric expression

Ue =1

2q1 V

(2)1 + V

(3)1 + V

(4)1 + + V

(n)1

+q2

V

(1)2 + V

(3)2 + V

(4)2 + + V

(n)2

+q3

V

(1)3 + V

(2)3 + V

(4)3 + + V

(n)3

+ + qn

V(1)n + V(2)n + V

(3)n + + V

(n1)n

.


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The potential energy of the assembled charge configuration is thengiven by

Ue =1

2

nk=1

qkVk (8)

whereqk = 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 theindividual charges; this is the energy that would be liberated if eachcharge was allowed to expand to an infinite volume. As aconsequence, Eq. (8) identically vanishes for a single point charge.


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For a continuous (macroscopic) volume charge distribution v(r), theexpression (8) for the electrostatic potential energy generalizes to

Ue =1

2

V

v(r)V(r)d3r (9)

For a continuous (macroscopic) surface charge distribution s(r), the

expression (8) for the electrostatic potential energy generalizes to

Ue =1

2

S

s(r)V(r)d2r (10)

For a continuous (macroscopic) line charge distribution (r), theexpression (8) for the electrostatic potential energy generalizes to

Ue =1

2

C

(r)V(r)d (11)

Notice that these expressions include the self-energies of the charges.


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Self-Energy of a Spherical Charge Distribution

For a uniform spherical charge distribution of charge Q and radius r0with charge density v = 3Q/(4r

30 ) for r r0, the absolute

electrostatic potential inside the sphere is given by (see Topic 7)

V(r) =Q

80r30

r20 r

2

+Q

40r0; r r0.

From Eq. (9), the self-energy of this spherical charge distribution is

Use =1

2

20

d

0

sin d

r00

v(r)V(r)r2dr

=3Q

2r30

r0

0

Q

80r30r20 r2 +

Q

40r0 r2dr

=3Q2

200r0 as r0 0 at fixed Q > 0.

On the other hand,

Use = v r20 0 as r0 0 at fixed v.

El E


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From Poissons equation

v(r) = 2

V(r) (12)at every point in the electrostatic field in a material with dielectricpermittivity . Substitution of this expression in Eq. (9) then gives

Ue =

2

V V(r

)2

V(r

)d

3

r, (13)

where V is any volume containing all of the charges in the system.From Greens first integral identity [see Eq. (40) of Topic 2]

V

2

+

d3

r =S nd

2

r

with (r) = (r) = V(r), one obtains

VV2V + (V)2d3r =

S

VV nd2r

.

El i E


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With this substitution, the expression (13) for the electrostatic energybecomes

Ue =

2

S

VV nd2r

V

(V)2d3r

(14)

Note that:

1 because V can be any volume that contains all of the charges inthe system configuration, the boundary surface S may then bechosen at an arbitrarily large distance from the chargedistribution;

2

because V(r) falls off at least as fast as 1/r as r, thenV(r) falls off at least as fast as 1/r2 as r, and because

the surface area ofS increases as r2 in that limit, then thesurface integral appearing in Eq. (14) decreases at least as fastas 1/r as r and can be made arbitrarily small by choosing

S sufficiently distant from the source charge distribution.

El i E


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Because V(r) = E(r), the electrostatic energy is then given by

Ue = 2

V

E2(r)d3r = 12

V

D(r) E(r)d3r (15)

where the volume V must now only be large enough to include allregions where the electrostatic field E(r) produced by the charge

distribution is nonzero. If this is not satisfied, then the electrostaticenergy is given by Eq. (13). Notice that this expression includes theself-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 electrostaticenergy density

ue(r) 1

2D(r) E(r) (J/m3) (16)

which is associated with the field energy at each point in the field.

topic_14 (electrostatic energy)

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