Energy in Electric Field 1

Energy in the Electric Field

© Frits F.M. de Mul

Energy in Electric Field 2

Energy in the Electric Field

Available: Electric Field E produced by a distributed charge density ρ [C/m3]

E

E

volume V surface A

ρ (x,y,z)

Question: How much energy did it cost to build up the electric field E by positioning the charge density ρ ???

Energy in Electric Field 3

Expressions for the Energy

E

E Following expressions will be derived:

(ρf = density of free charges ;

V = potential function)

. A.2

1 ∫∫∫ ρ=V

f dvVW

B.21 dvW

V

ED∫∫∫ •=volume v surface A

ρ (x,y,z)

Energy in Electric Field 4

Energy = f (ρf ,V ) (1)

Q

PE

Assume : charge Q in O produces E-field in P ; potential in P : VP(Q)

How much energy W is needed to bring charge q from infinity to P ?

q

q

q

W = q.VP(Q)

Suppose N charges Qi (i=1..N) in O :

Each charge produces its own V in P

)(;. ..1

∑=

==Ni

iPPP QVVVqW

Energy in Electric Field 5

Energy = f (ρf ,V ) (2)

Q1

PQ3

Q2

Qi

Qj-1 Suppose j-1 charges Qi (i=1..j-1), not necessarily all in O , but in Pi

q

Call q=Qj , to be placed at P=Pj

Energy needed to place j th charge Qj at Pj in field of Q1 ... Qj-1 :

)( 1..1

∑−=

=ji

iPjj QVQWj

Total energy needed to position all N charges Qj at Pj (j=1..N) , with preceding Qi (i=1..j-1) present :

∑ ∑= −=

=Nj ji

iPj QVQWj

..1 1..1

)(

Energy in Electric Field 6

Energy = f (ρf ,V ) (3)

Q1

PN QN

Q2

Qi

QN-1 Total energy needed to position all N charges Qj at Pj (j=1..N) , with preceding Qi (i=1..j-1) present :

∑ ∑= −=

=Nj ji

iPj QVQWj

..1 1..1

)( ∑ ∑= ≠=

=Nj jiNi

iPj QVQj

..1 ;..121 )(

Summation i,j=1..N ; factor 1/2 to avoid “double-count”

Summation is over all charges, each in field of all other charges.

Energy in Electric Field 7

∑ ∑= ≠=

=Nj jiNi

iPj QVQWj

..1 ;..121 )(

Suppose all charges are distributed as charge density ρ [C/m3] :

Energy = f (ρf ,V ) (4)

Q1

PN QN

Q2

Qi

QN-1

“Summation over all charges, each in field of all other charges” now means:1. Divide v into volume elements dv , with charge ρ.dv

2. Calculate potential from all other charges at that spot.

3. Integrate over volume v : . 21 ∫∫∫ ρ=

V

dvVW

ρ(x,y,z)

Energy in Electric Field 8

Energy = f (ρf ,V ) (5)

E

E

volume v surface A

ρ (x,y,z)

. 21 ∫∫∫ ρ=

V

dvVW

If dielectric material present:

V originates from all charges (free and bound) ;

ρ originates from free charges only (being “transportable”) :

. 21 ∫∫∫ ρ=

Vf dvVW

Energy in Electric Field 9

Energy = f (D,E) (1)

E

EEnergy to build charge distribution = energy to destroy it.

How ?

Transport all charges to infinity and record energy.

How ?

Place conducting sphere with radius = 0 at O ; “Blow up” till radius = infinity;

Sweep all charges to infinity.volume v surface A

ρ (x,y,z)

Energy in Electric Field 10

Energy = f (D,E) (2)

Or

Suppose: now radius = r

cross section “Blowing up” the sphere:

dr

Those charges originally inside the sphere, now lie on the surface of the sphere, and produce same (average) E-field as if they were inside (Gauss).

E

E

E

E

EQuestions:

1. how much energy is involved in increasing radius from r to r+dr ??2. integrate answer from r=0 to infinity.

Energy in Electric Field 11

Energy = f (D,E) (3)

Or

Question 1:

How much energy is involved in increasing radius from r to r+dr ??

cross section

Consider surface element dA

Free charge in dA: dQf = σf .dA

Work to shift dA from r to r+dr:dF

dW = dF.dr

E

= dQf .E.dr

dW = σf .dA.E.dr = E.σf . dv

dr

Energy in Electric Field 12

Energy = f (D,E) (4)

Or

cross section

E

dr

dA

Work to shift dA from r to r+dr:

dW = σf .dA.E.dr = E.σf . dv

Question: what are E and σf ??

DD

Gauss pill box:

σf .dA = D.dA

What E acts on dA? see next slide

Energy in Electric Field 13

Energy = f (D,E) (5)

Or

cross section

Ddr

dA

Etot

Etot = total field, just outside conducting sphere: Etot = σ /ε0 .

Eself

Eself = 1/2 σ /ε0 .

Eact

Electric field acting on charge in pill box : (apparently due to all “other” charges)

Eact = Etot - Eself = 1/2 Etot

Energy in Electric Field 14

Energy = f (D,E) (6)

Or

cross section

D

E

dr

dA

Work to replace dA from r to r+dr:

dW = σf .dA.E.dr = E.σf . dv

Gauss pill box: σf .dA = D.dA

Acting Eact = 1/2 Etot (= 1/2 E )

dW = 1/2 D.E.dA.dr

dW = 1/2 D.E.dv

Energy in Electric Field 15

Energy = f (D,E) (6)

Or

cross section

D

E

dr

dA

dW = 1/2 D.E.dA.dr

dW = 1/2 D.E.dv

“Blow up” the sphere: (all charges to infinity)

21 dvW

V

ED∫∫∫ •=

Energy in Electric Field 16

Conclusions

E

E

volume v surface A

ρ (x,y,z)

Following expressions are derived:

(ρf = density of free charges ; V = potential function)

. A.2

1 ∫∫∫ ρ=V

f dvVW

B.21 dvW

V

ED∫∫∫ •=

Energy in Electric Field 17

Example: Parallel-plate capacitor

V++++++++++

--------------

ED

+Q

- Q

d

A . A.

2

1 ∫∫∫ ρ=V

f dvVW

221

21

21 CVVQdAVW f

Af ==σ= ∫∫

B.21 dvW

V

ED∫∫∫ •=Gauss: D = σf E = V/d

221

21

21

21

21 .. CVVQdVAdA

d

Vdv

d

VW fff

volf ==σ=σ=σ= ∫∫∫

the end