University Electromagnetism: Energy in the electric field (field and displacement vs. density and potential)

1. Energy in the Electric Field © Frits F.M. de Mul

2. Energy in the Electric Field Available: Electric Field E produced by a distributed charge density  [C/m 3 ]  Question: How much energy did it cost to build up the electric field E by positioning the charge density    E E volume V surface A  ( x,y,z )

3. Expressions for the Energy E E Following expressions will be derived: (  f = density of free charges ; V = potential function) volume v surface A  ( x,y,z )

4. Energy = f (  f ,V ) (1) How much energy W is needed to bring charge q from infinity to P ? W = q.V P ( Q ) Suppose N charges Q i ( i= 1.. N ) in O : Each charge produces its own V in P Q P E Assume : charge Q in O produces E -field in P ; potential in P : V P ( Q ) q q q

5. Energy = f (  f ,V ) (2) Energy needed to place j th charge Q j at P j in field of Q 1 ... Q j-1 : Total energy needed to position all N charges Q j at P j ( j= 1.. N ) , with preceding Q i ( i= 1.. j- 1) present : Q 1 P Q 3 Q 2 Q i Q j-1 Suppose j-1 charges Q i ( i= 1.. j-1 ), not necessarily all in O , but in P i q Call q=Q j , to be placed at P=P j

6. Energy = f (  f ,V ) (3) Q 1 P N Q N Q 2 Q i Q N-1 Total energy needed to position all N charges Q j at P j ( j= 1.. N ) , with preceding Q i ( i= 1.. j- 1) present : Summation i,j= 1.. N ; factor 1/2 to avoid “double-count” Summation is over all charges, each in field of all other charges.

7. Energy = f (  f ,V ) (4) Q 1 P N Q N Q 2 Q i Q N-1 “Summation over all charges, each in field of all other charges” now means: 1. Divide v into volume elements dv , with charge  .dv 2. Calculate potential from all other charges at that spot. 3. Integrate over volume v :  ( x,y,z ) Suppose all charges are distributed as charge density  [C/m 3 ] :

8. Energy = f (  f ,V ) (5) If dielectric material present: V originates from all charges (free and bound) ;  originates from free charges only (being “transportable”) : E E volume v surface A  ( x,y,z )

9. Energy = f ( D,E ) (1) E E Energy 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  ( x,y,z )

10. Energy = f ( D,E ) (2) Questions: 1. how much energy is involved in increasing radius from r to r+dr ?? 2. integrate answer from r= 0 to infinity. O r 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 E

11. Energy = f ( D,E ) (3) O r Question 1: How much energy is involved in increasing radius from r to r+dr ?? cross section Free charge in dA : dQ f =  f .dA  Work to shift dA from r to r+dr : dW =  f .dA. E.dr = E.  f . dv dr Consider surface element dA dF dW = dF.dr E = dQ f . E.dr

12. Energy = f ( D,E ) (4) O r cross section E dr dA = E.  f . dv Question: what are E and  f ?? What E acts on dA ? see next slide Work to shift dA from r to r+dr : dW =  f .dA. E.dr D D Gauss pill box:  f .dA = D.dA

13. Energy = f ( D,E ) (5) O r cross section D dr dA E act = E tot - E self = 1 / 2 E tot E tot E tot = total field, just outside conducting sphere: E tot =  /    E self E self = 1 / 2  /    E act Electric field acting on charge in pill box : (apparently due to all “other” charges)

14. Energy = f ( D,E ) (6) O r cross section D E dr dA Gauss pill box:  f .dA = D.dA Acting E act = 1 / 2 E tot ( = 1 / 2 E ) dW = 1 / 2 D.E . dA . dr dW = 1 / 2 D.E .dv Work to replace dA from r to r+dr : dW =  f .dA. E.dr = E.  f . dv

15. Energy = f ( D,E ) (6) O r 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)

16. Conclusions Following expressions are derived: (  f = density of free charges ; V = potential function) E E volume v surface A  ( x,y,z )

17. Example: Parallel-plate capacitor Gauss: D  f E = V/d the end V ++++++++++ -------------- E D +Q - Q d A