1. 1 Lecture 4 Work, Electric Potential and Potential Energy Ch. 25 Topics Work, electric potential energy and electric potential Calculation of potential from field Potential from a point charge Potential due to a group of point charges, electric dipole Potential due to continuous charged distributions Calculating the electric field from a potential Electric potential energy from a system of point charges Equipotential Surface Potential of a charged isolated conductor Demos Teflon and silk Charge Tester, non-spherical conductor, compare charge density at different radii. Elmo Potential in the center of four charges Potential of a electric dipole Polling

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4. 4 Work, Potential Energy and Electric Potential The electric force is mathematically the same as gravity so it too must be a conservative force. We want to show that the work done is independent of the path and only depends on the endpoints. Then the force is said to be a conservative force. First start with work Work done by the electric force = Then we will find it useful to define a potential energy as is the case for gravity. And the electric potential

5. 5 Lets start with a uniform electric field and find the work done for a positive test charge. b a E P2P2 P3P3 P1P1 c

6. 6 Find work done along path W 12 for a positive test charge b a E P2P2 P3P3 P1P1 c F ds

7. 7 Find Work along path W 23 b a E P2P2 P3P3 P1P1 c F ds

8. 8 W 12 + W 23 = 0 + qEb =qEb Compare this work done along path W 13

9. 9 Work done along path W 13 b a P2P2 P3P3 P1P1 c F   ds

10. 10 Conclusion Work done along path W 12 + W 23 = W 13. Work is independent of the particular path. Although we proved it for a uniform field, it is true for any field that is a only a function of r and is along r. It only depends on the end points i and f. This means we can define a function at every point in space and when we take the difference of that function between any two points, it is equal to the negative of the work done.

11. 11 b a P2P2 P3P3 P1P1 c F  When we go from P 1 to P 3 we evaluate the Work function at P 3 and subtract the value at P 1 and then the a difference equals the negative of the work done in going form P 1 to P 3. This function is called the potential energy function

12. 12 Example of finding the Potential Energy Function U in a Uniform Field What is the electric potential difference for a unit positive charge moving in an uniform electric field from a to b? E E d a b If we set the origin at x b = 0, and measure from b to a, then This is analogy with gravitation where we U =mgh. U b =0 and the potential energy function is U=qEd

13. 13 Now define the Electric Potential Difference Δ V which does not depend on charge. The potential difference is the negative of the work done per unit charge by an electric field on a positive unit charge when it moves from one point to another.

14. 14 For a battery of potential difference of 9 volts you would say that the positive terminal is 9 volts above the negative terminal. Find the potential difference ΔV for a uniform electric field

15. 15 Note relationship between potential and electric field V is a scalar not a vector. Simplifies solving problems. We are free to choose V to be 0 at any location. Normally V is chosen to be 0 at the negative terminal of a battery or 0 at infinity for a point charge.

16. 16 Example for a battery in a circuit In a 9 volt battery, typically used in IC circuits, the positive terminal has a potential 9 v higher than the negative terminal. If one micro-Coulomb of positive charge flows through an external circuit from the positive to negative terminal, how much has its potential energy been changed? q

17. 17 Generalize concept of electric potential energy and potential difference for any electric field

18. 18 (independent of path, ds ) Therefore, electric force is a conservative force. = - Work done by the electric force = y x

19. 19 Find the electric potential when moving from one point to another in a field due to a point charge?

20. 20 eqn 25-26 Replace R with r Potential of a point charge at a distance R

21. 21 Electric potential for a positive point charge V is a scalar V is positive for positive charges, negative for negative charges. r is always positive. For many point charges, the potential at a point in space is the simple algebraic sum (Not a vector sum)

22. 22 Hydrogen atom. What is the electric potential at a distance of 0.529 A from the proton? 1A= 10 -10 m Electric potential due to a positive point charge r = 0.529 A What is the electric potential energy of the electron at that point? U = qV= (-1.6 x 10 -19 C) (27.2 V) = - 43.52 x 10 -19 J or - 27.2 eV where eV stands for electron volt. Note that the total energy E of the electron in the ground state of hydrogen is - 13.6 eV Also U= 2E = -27.2 eV. This agrees with above formula.

23. 23 What is the electric potential due to several point charges? For many point charges, the potential at a point in space is the simple algebraic sum (Not a vector sum) r1r1 x r3r3 y r2r2 q1q1 q2q2 q3q3

24. 24 Four Point charges

25. 25 What is the potential due to a dipole? Two point charges that are opposite and equal

26. 26 Potential for a Continuous Distribution of Charge Point charge For an element of charge Integrate V

27. 27 Chaper 24 Problem 22. With V = 0 at infinity, what is the electric potential at P, the center of curvature of the uniformly charged nonconducting rod?

28. 28 Chapter 24 Problem 26. What is the magnitude of the net electric potential at the center? 1. A thin rod of charge -3.0 µC that forms a full circle of radius 6.0 cm 2. A second thin rod of charge 2.0 µC that forms a circular arc of radius 4.0 cm, subtending an angle of 90° about the center of the full circle 3. An electric dipole with a dipole moment that is perpendicular to a radial line and that has magnitude 1.28 x 10 -21 C·m

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30. 30 Figure 24-44 shows a thin plastic rod of length L and uniform positive charge Q lying on an x axis. With V = 0 at infinity, find the electric potential at point P 1 on the axis, at distance d from one end of the rod.

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32. 32 Potential due to a ring of charge Direct integration. Since V is a scalar, it is easier to evaluate V than E. Find V on the axis of a ring of total charge Q. Use the formula for a point charge, but replace q with elemental charge dq and integrate. Point charge For an element of charge r is a constant as we integrate. This is simpler than finding E because V is not a vector. V

33. 33 Potential due to a line charge We know that for an element of charge dq the potential is For the line charge let the charge density be. Then dq= dx Now, we can find the total potential V produced by the rod at point P by integrating along the length of the rod from x=0 to x=L

34. 34 A new method to find E if the potential is known. If we know V, how do we find E? So the x component of E is the derivative of V with respect to x, etc. –If V = a constant, then E x = 0. The lines or surfaces on which V remains constant are called equipotential lines or surfaces. –See example on next slide

35. 35 Equipotential Surfaces Three examples What is the equipotential surface and equipotential volume for an arbitrary shaped charged conductor? See physlet 9.3.2 Which equipotential surfaces fit the field lines?

36. 36 a) Uniform E field V = constant in y and z directions b)Point charge (concentric shells) c)Electric Dipole (ellipsoidal concentric shells) Blue lines are the electric field lines Orange dotted lines represent the equipotential surfaces x

37. 37 Electric Potential Energy U of a system of charges How much work is required to set up the arrangement of Figure 24-46 if q = 3.20 pC, a = 54.0 cm, and the particles are initially infinitely far apart and at rest? q1q1 q2q2 q3q3 q4q4 W = U

38. 38 E 1 E 2

39. 39 Dielectric Breakdown: Application of Gauss’s Law If the electric field in a gas exceeds a certain value, the gas breaks down and you get a spark or lightning bolt if the gas is air. In dry air at STP, you get a spark when 1 2 V = constant on surface of conductorRadius r 2 r1r1

40. 40 This explains why: Sharp points on conductors have the highest electric fields and cause corona discharge or sparks. Pick up the most charge with charge tester from the pointy regions of the non-spherical conductor. Use non-spherical metal conductor charged with teflon rod. Show variation of charge across surface with charge tester. Van de Graaff + + - - Radius R 1 2 V = constant on surface of conductor Cloud

41. 41 How does a conductor shield the interior from an exterior electric field? Start out with a uniform electric field with no excess charge on conductor. Electrons on surface of conductor adjust so that: 1. E=0 inside conductor 2. Electric field lines are perpendicular to the surface. Suppose they weren’t? 3. Does E =  just outside the conductor 4. Is  uniform over the surface? 5. Is the surface an equipotential? 6. If the surface had an excess charge, how would your answers change?

42. 42 What is the electric potential of a uniformly charged circular disk? We can treat the disk as a set of ring charges. The ring of radius R’ and thickness dR’ has an area of 2  R’dR’ and it’s charge is dq =  dA =  2  R’)dR’ where  =Q/(  R 2 ), the surface charge density. The potential dV at point P due to the charge dq on this ring given by Integrating R’ from R’=0 to R’=R q

43. 43 Chapter 24 Problem 19 The ammonia molecule NH3 has a permanent electric dipole moment equal to 1.31 D, where 1 D = 1 debye unit = 3.34 x10 -30 C·m. Calculate the electric potential due to an ammonia molecule at a point 44.0 nm away along the axis of the dipole. (Set V = 0 at infinity.)

44. 44 Chapter 24 Problem 55 Two metal spheres, each of radius 1.0 cm, have a center-to-center separation of 2.2 m. Sphere 1 has charge +2.0 x 10 -8 C. Sphere 2 has charge of -3.8 x10 -8 C. Assume that the separation is large enough for us to assume that the charge on each sphere is uniformly distributed (the spheres do not affect each other). Take V = 0 at infinity. (a) Calculate the potential at the point halfway between the centers. (b) Calculate the potential on the surface of sphere 1. (c) Calculate the potential on the surface of sphere 2.

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46. 46 Chapter 24 Problem 57 A metal sphere of radius 11 cm has a net charge of 2.0 x10 -8 C. (a) What is the electric field at the sphere's surface? (b) If V = 0 at infinity, what is the electric potential at the sphere's surface? (c) At what distance from the sphere's surface has the electric potential decreased by 500 V?

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