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POTENTIAL

POTENTIAL. February 1, 2009. We complete Gauss’s Law We begin the topic of POTENTIAL – Chapter 25. Problem Session Wednesday Morning Examination #1 is on Friday – 2/13/09 Exam includes Chapters 23,24 and as much of 25 as we cover today (Monday) Next Week – More Potential. This Week.

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POTENTIAL

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  1. POTENTIAL February 1, 2009

  2. We complete Gauss’s Law • We begin the topic of POTENTIAL – Chapter 25. • Problem Session Wednesday Morning • Examination #1 is on Friday – 2/13/09 • Exam includes Chapters 23,24 and as much of 25 as we cover today (Monday) • Next Week – More Potential This Week

  3. A particle with charge Q is located at the center of a cube of edge L. In addition, six other identical negative charged particles -q are positioned symmetrically around Q as shown in the figure below. Determine the electric flux through one face of the cube.

  4. A slab of insulating material has a nonuniform positive charge density ρ = Cx2, where x is measured from the center of the slab, as shown in the figure below, and C is a constant. The slab is infinite in the y and z directions. Derive expressions for the electric field for the following regions.

  5. A solid insulating sphere of radius a carries a net positive charge 3Q, uniformly distributed throughout its volume. Concentric with this sphere is a conducting spherical shell with inner radius b and outer radius c, and having a net charge -Q, as shown in the figure below.

  6. A solid insulating sphere of radius a carries a net positive charge 3Q, uniformly distributed throughout its volume. Concentric with this sphere is a conducting spherical shell with inner radius b and outer radius c, and having a net charge -Q, as shown in the figure below. (a) Construct a spherical Gaussian surface of radius r > c and find the net charge enclosed by this surface.(b) What is the direction of the electric field at r > c?(c) Find the electric field at r > c.(d) Find the electric field in the region with radius r where c > r > b.(e) Construct a spherical Gaussian surface of radius r, where c > r > b, and find the net charge enclosed by this surface.(f) Construct a spherical Gaussian surface of radius r, where b > r > a, and find the net charge enclosed by this surface.(g) Find the electric field in the region b > r > a.(h) Construct a spherical Gaussian surface of radius r < a, and find an expression for the net charge enclosed by this surface, as a function of r. Note that the charge inside this surface is less than 3Q.(i) Find the electric field in the region r < a.(j) Determine the charge on the inner surface of the conducting shell.(k) Determine the charge on the outer surface of the conducting shell.(l) Make a plot of the magnitude of the electric field versus r.

  7. ABOUT WebAssign • Check Daily • Note due time • No more extensions because you didn’t know that an assignment was due.

  8. B  q A 

  9. We have a region in space where there is an Electric Field • There is a particle of charge q at some location. • The particle must be moved to another spot within the field. • Work must be done in order to accomplish this.

  10. Electric Potential • We will be dealing with • Work • Energy & Conservation • Work must be done to move a charge in an electric field. • Let’s do a weird demo ….

  11. I need some help. Push vs Pull Mrs. FIELDS vs Mr. External

  12. What we will do …. E • For the moment, assume the charge has MASS. (It may not.) • Assume the charge is initially stationary. • The charge is to be moved to the left. • The charge is to be moved at CONSTANT velocity. + charge Mrs. Fields Mr. External

  13. Start and Stop • ENERGY is required to bring the charge up to speed (if it has mass). • ENERGY is required to bring the particle back to rest (if it has mass). • The sum of these two is ZERO.

  14. Each does the negative amount of work than the other does. WHY ?

  15. So, when we move a charge in an Electric Field .. • Move the charge at constant velocity so it is in mechanical equilibrium all the time. • Ignore the acceleration at the beginning because you have to do the same amount of negative work to stop it when you get there.

  16. Summary-- • When an object is moved from one point to another in an Electric Field, • It takes energy (work) to move it. • This work can be done by an external force (you). • You can also think of this as the FIELD doing the negative of this amount of work on the particle.

  17. And also remember: The net work done by a conservative (field) force on a particle moving around a closed path is ZERO!

  18. A nice landscape Work done by external force = mgh How much work here by gravitational field? h  mg

  19. The gravitational case:

  20. Someone else’s path

  21. IMPORTANT • The work necessary for an external agent to move a charge from an initial point to a final point is INDEPENDENT OF THE PATH CHOSEN!

  22. The Electric Field • Is a conservative field. • No frictional losses, etc. • Is created by charges. • When one (external agent) moves a test charge from one point in a field to another, the external agent must do work. • This work is equal to the increase in potential energy of the charge. • It is also the NEGATIVE of the work done BY THE FIELD in moving the charge from the same points.

  23. A few things to remember… • A conservative force is NOT a Republican. • An External Agent is NOT 007.

  24. Electric Potential Energy • When an electrostatic force acts between two or more charged particles, we can assign an ELECTRIC POTENTIAL ENERGY U to the system. • The change in potential energy of a charge is the amount of work that is done by an external force in moving the charge from its initial position to its new position. • It is the negative of the work done by the FIELD in moving the particle from the initial to the final position.

  25. Definition – Potential Energy • PE or U is the work done by an external agent in moving a charge from a REFERENCE POSITION to a different position. • A Reference ZERO is placed at the most convenient position • Like the ground level in many gravitational potential energy problems.

  26. Zero Level Example: E Work by External Agent Wexternal = Fd = qEd= U Work done by the Field is: Wfield= -qEd = -Wexternal d q F

  27. A uniform electric field of magnitude 290 V/m is directed in the positive x direction. A +13.0 µC charge moves from the origin to the point (x, y) = (20.0 cm, 50.0 cm).(a) What is the change in the potential energy of the charge field system?[-0.000754] J

  28. AN IMPORTANT DEFINITION • Just as the ELECTRIC FIELD was defined as the FORCE per UNIT CHARGE: We define ELECTRICAL POTENTIAL as the POTENTIAL ENERGY PER UNIT CHARGE: VECTOR SCALAR

  29. UNITS OF POTENTIAL

  30. Let’s move a charge from one point to another via an external force. • The external force does work on the particle. • The ELECTRIC FIELD also does work on the particle. • We move the particle from point i to point f. • The change in kinetic energy is equal to the work done by the applied forces. Assume this is zero for now.

  31. Furthermore… If we move a particle through a potential difference of DV, the work from an external “person” necessary to do this is qDV

  32. Electric Field = 2 N/C  d= 100 meters 1 mC Example

  33. One Step More

  34. Consider Two Plates OOPS …

  35. Look at the path issue

  36. The difference in potential between the accelerating plates in the electron gun of a TV picture tube is about 25 000 V. If the distance between these plates is 1.50 cm, what is the magnitude of the uniform electric field in this region?

  37. An ion accelerated through a potential difference of 115 V experiences an increase in kinetic energy of 7.37 × 10–17 J. Calculate the charge on the ion.

  38. Important • We defined an absolute level of potential. • To do this, we needed to define a REFERENCE or ZERO level for potential. • For a uniform field, it didn’t matter where we placed the reference. • For POINT CHARGES, we will see shortly that we must place the level at infinity or the math gets very messy!

  39. An Equipotential Surface is defined as a surface on which the potential is constant. It takes NO work to move a charged particle between two points at the same potential. The locus of all possible points that require NO WORK to move the charge to is actually a surface.

  40. Example: A Set of Equipotenital Surfaces

  41. Back To Yesteryear

  42. Field Lines and Equipotentials Electric Field Equipotential Surface

  43. Components Enormal Electric Field Dx Eparallel Work to move a charge a distance Dx along the equipotential surface Is Q x Eparallel X Dx Equipotential Surface

  44. BUT • This an EQUIPOTENTIAL Surface • No work is needed since DV=0 for such a surface. • Consequently Eparallel=0 • E must be perpendicular to the equipotential surface

  45. Therefore E E E V=constant

  46. Field Lines are Perpendicular to the Equipotential Lines

  47. Equipotential

  48. ds V+dV V Consider Two EquipotentialSurfaces – Close together Work to move a charge q from a to b: b a E Note the (-) sign!

  49. Where

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