1. Energy in Electric Field1 Energy in the Electric Field © Frits F.M. de Mul

2. Energy in Electric Field2 Energy in the Electric Field Available: Electric Field E produced by a distributed charge density  [C/m 3 ] E E volume V surface A  (x,y,z) Question: How much energy did it cost to build up the electric field E by positioning the charge density  

3. Energy in Electric Field3 Expressions for the Energy E E Following expressions will be derived: (  f = density of free charges ; V = potential function) Following expressions will be derived: (  f = density of free charges ; V = potential function) volume v surface A  (x,y,z)

4. Energy in Electric Field4 Energy = f (  f,V ) (1) Q P E Assume : charge Q in O produces E-field in P ; potential in P : V P (Q) How much energy W is needed to bring charge q from infinity to P ? q q q W = q.V P (Q) Suppose N charges Q i (i=1..N) in O : Each charge produces its own V in P

5. Energy in Electric Field5 Energy = f (  f,V ) (2) Q1Q1 P Q3Q3 Q2Q2 QiQi 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 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 :

6. Energy in Electric Field6 Energy = f (  f,V ) (3) Q1Q1 PNPN QNQN Q2Q2 QiQi 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 in Electric Field7 Suppose all charges are distributed as charge density  [C/m 3 ] : Energy = f (  f,V ) (4) Q1Q1 PNPN QNQN Q2Q2 QiQi 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)

8. Energy in Electric Field8 Energy = f (  f,V ) (5) E E volume v surface A  (x,y,z) If dielectric material present: V originates from all charges (free and bound) ;  originates from free charges only (being “transportable”) :

9. Energy in Electric Field9 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 in Electric Field10 Energy = f (D,E) (2) 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 Questions: 1. how much energy is involved in increasing radius from r to r+dr ?? Questions: 1. how much energy is involved in increasing radius from r to r+dr ?? 2. integrate answer from r=0 to infinity.

11. Energy in Electric Field11 Energy = f (D,E) (3) O r Question 1: How much energy is involved in increasing radius from r to r+dr ?? 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: dQ f =  f.dA Work to shift dA from r to r+dr: dF dW = dF.dr E = dQ f.E.dr dW =  f.dA.E.dr = E.  f. dv dr

12. Energy in Electric Field12 Energy = f (D,E) (4) O r cross section E dr dA Work to shift dA from r to r+dr: dW =  f.dA.E.dr = E.  f. dv Question: what are E and  f ?? D D Gauss pill box:  f.dA = D.dA Gauss pill box:  f.dA = D.dA What E acts on dA? see next slide

13. Energy in Electric Field13 Energy = f (D,E) (5) O r cross section D dr dA 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) E act = E tot - E self = 1 / 2 E tot

14. Energy in Electric Field14 Energy = f (D,E) (6) O r cross section D E dr dA Work to replace dA from r to r+dr: dW =  f.dA.E.dr = E.  f. dv 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

15. Energy in Electric Field15 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. Energy in Electric Field16 Conclusions E E volume v surface A  (x,y,z) Following expressions are derived: (  f = density of free charges ; V = potential function) Following expressions are derived: (  f = density of free charges ; V = potential function)

17. Energy in Electric Field17 Example: Parallel-plate capacitor V ++++++++++ -------------- E D +Q - Q d A Gauss: D  f E = V/d the end