Lectures 5,6 (Ch. 23) Electric Potential

1. Lectures 5,6 (Ch. 23)Electric Potential 1. Work in 2. Electric Potential Energy, U 3. Electric Potential, V 4.V of a point charge 5. A set of the point charges 6. Distributed charges a) superposition principle b) finding V via E 7. Inverse problem: E via V.

2. Work in b q Gravity is a conservative force a Conservative force: does not depend on pass It is reversible Its circulation is zero: Electric force is a conservative force

3. Potential energy b a Electric Potential (Voltage) A work per unit charge

4. SI units of Voltage [V]=[W]/[q]=1J/1C=1V (Volt) [E]=V/m=J/Cm=N/C 1ev (electron-volt) is a unit of energy (not a unit of V!) It is defined as an energy gained by an electron accelerated between two points→ with voltage =1v ΔU=qΔV→1ev=1.6x10-19Cx1V=1.6x10-19J Alessandro Volta (1745-1827) Inventor of the first battery

6. General properties of electric potential 1. V always decreases along a lines: b Vb < Va is always directed from higher to lower potential + + + + + b a b a a b a b

7. 2.Equipotential surfaces (the surfaces where V=const) • No work can be done moving along such surface.Indeed, for arbitrary a and b: Va = Vb→W ab=0 b) Equipotential surfaces are ┴ c) Surface of conductor is equipotential (since E=0, V=Vsurface to this surface) ┴ d) If E=0 in some region arbitrary two points in this region have the samepotential. Indeed, f) V=const inside the volume of the conductor =Vsurface

8. Electric potential and potential energy in the uniform electric field V + + + + + + + + y tan ὰ=V/y=E - - - - - - - - ὰ y 0 V=const→y=const (Equipotential surfaces are planes)

9. Analogy between potential energy in the uniform electric and gravity fields h Energy conservation law: K+U=const

10. Example.An electron was accelerated in a uniform field from v0=0 to vf=106m/s. • Find :a) change of potential energy; • b) work done by the field; • c) change of potential; • d) displacement of the electron, h. • 2. How the answers would change if v0=106m/s in a horizontal direction and vf=√2x106m/s? • Ka+Ua= Kb+Ub; ΔU=Ub-Ua • ΔU=Ka-Kb=-m vf2/2 • ΔU=-9x10-31kgx1012m2/2sec2 • ΔU =-4.5x10-19J vf=106m/s h v0=0 • b) Wab=-ΔU= 4.5x10-19J • c)ΔV= ΔU/q=-4.5x10-19J/(-1.6x10-19C)=2.8V • d)ΔV=Eh→h=ΔV/E= 2.8Vm/900V≈3mm • II. The answers do not change since x component of the velocity does not change and change of the y component is the same as in part I. Indeed, v0y=0, vfy=√(vf2-v02)=106m/s

11. Cathode ray tube Cathode (C) z Deflecting plates (DP) y Anode (A) Fluorescence screen ΔV electrons emitted by cathode are accelerated by voltage between C and A, entering the region between DP with x .Then electrons are deflected by E. In DP region along x: motion with v=v0; along y: motion with acceleration in uniform E E see the previous example and an example in Lect.2

12. V of a point charge b a Equipotential surfaces:V(r)=const → r=const → Spheres ┴ Take ΔV=const between the neighboring equipotenttial surfaces then density of equipotential surfaces shows the steepness of V(r). V 70V 50V 30V r

13. Topographic map analogous to equipotential surfaces V=kq/r h Δh=const : Larger density of the lines →steeper the hill

14. Positive q vs. negative q V decreases as you move outward V=-30V V=-50V V=-70V V r V 0 0 r

15. r+(d/2)cos ὰ r-(d/2)cos ὰ r ὰ d r>>d

16. V of a dipole

17. r r>>d V=2kq/r Figure 23.24c

18. V of a ring of charge

19. Example. A uniformly charged thin ring has a radius10 cm and a • total charge 12 nC. An electron is placed on the ring axis a distance 25 cm • from the center of the ring. The electron is then released from rest. • Find the speed of the electron when it reaches the center of the ring. • Describe the subsequent motion of the electron. 2. Periodic motion back and forth through the center of the ring.

20. V of the conducting sphere 1. r>R (as for the point charge) 2. r<R

21. Lighteningelectric discharge in atmosphere Benjamin Franklin (1706-1790) Lightening rod controls conducting pass: E is stronger near the surface with small radius: - - - - - V +++ + + E=Ecr dielectric strength (dielectric material becomes conducting due to ionization). In air Ecr=3x106V/m. When E>3MV/m accelerated electrons collide with molecules in air producing avalanche ionization (conducting pass). E is also stronger closer to the cloud

22. Van der Graaf Generator It produces high charge or voltage, limited by Ecr=3x106V/m.

23. Scanning tunneling microscope (STM),Nobel Prize 1986 Very sharp tip -e Dielectric breakdown of vacuum In very strong fields e-p pairs are produced and vacuum becomes a conductor Recombination of e and p (annihilation) results in emission of electromagnetic radiation (0.5 MeV annihilation line from the center og our Galaxy): e+p=2mc2 → E=mc2/e=2x9x10-31kgx9x1016m/s/1.6x10-19 C=0.5Me

24. V of the dielectric sphere 1. r<R q R 2. r>R , the same as for conducting sphere:

25. Very long conducting cylinder, L>>R, r L • r<R: E(r)=0→V(r)=V(R) • r>R: lnx x 1 For infinitely long charged objects never choose V(r→∞)=0 !!! Choose V=0 at some finite distance, for example, V(R)=0. V(r) r

26. Long conducting cylindrical shell a Choose V(a)=0. Then 1. r<a: V(r)=0. 2. a<r<b: b V a b r 3. r>b: A Geiger counter counts the particles which ionize the air . The emitted electrons accelerated by high E lead to avalanche ionization resulting in detected current.

27. Conducting slabs - - - - - - - - - - + + + + + + + + + + E=const Choose V(a)=0 1.x<a →V(x)=0 2. x>a → V(x)=E(x-a) E=const Choose V(a)=0 1.x<a →V(x)=0 2. x>a → V(x)=E(a-x) - - - - - + + + + + v v x -a a x -a a V=Ed V=Ex x d

28. Conclusions 1. For a finite size objects: Choose V(∞)=0 Calculate the potential first outside the object then inside the object 2. For infinite size objects: Choose V on some surface=0 Calculate potential inside and outside

29. 3. Potential varies continuously on the surface of both conductor and insulator 4. Inside conductor V=const=V on the surface Insulator V ~ r2 5. Outside both conductor and insulator sphere ~ 1/r Cylinder ~ ln r Plane ~ r

30. Electric potential energy of two charges → Choose Vb=0 (and hence Ub=0) when rb→∞ then Thus U is = the work done by electric force produced by charge q on charge q0→to move q0→from r to infinity.

31. The same sign of q and q0 : U>0 The work by electric force to bring q0 to ∞ is positive: (force is in the direction of motion) Opposite sign of q and q0 : U<0 The work by electric force to bring q0 to ∞ is negative: (force opposes motion)

32. Ionizationenergy(disassembling H atom) F r=0.5x10-10m Ionizationenergy=E=13.6eV= Wexternal force to move e from r=0.5x10-10m to ∞

33. Nuclear energy He4 r=10-10m

34. Nuclear radius Au, Z=79 ὰ particle (4He2+), Z=2 K0=6MeV r Ernest Rutherford (1871– 1937), Nobel Prize 1908 Head on collision: Some ὰ particles were scattered back which contradicted the “plum pudding” model of atom and led to the planetary model with tiny nucleus in the center.

35. Several point charges q0 r01 = work which was done by external force to bring an electric charge q0from infinity to a given position in the presence of all other charges q1 qN q2 = work which was done by external force to bring all electric chargesfrom infinityto given positions

36. Example. Three point charges q0 r01 r02 q1 q2 r12

37. Inverse problem: Finding E via V Example 1. Example 2. Example 3. q

38. General case:Finding E via V

39. Example. Knowing potential of the ring on x axis find E at P. NB: the above expression for V is valid only on the x-axis. It does not allow to derive Ey andEz. The obtained expression for Ex is valid also only on the x-axis.

40. Example.