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Equipotentials and Energy

Equipotentials and Energy. Electric Potential. Today…. Text Reference: Chapter 24.3,5 and 25.1 . examples: 24.7,11,13,15 and 25.1. Equipotentials and conductors E from V Calculate electric field of dipole from potential Electric Potential Energy of charge in external electric field

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Equipotentials and Energy

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  1. Equipotentials and Energy Electric Potential

  2. Today… Text Reference: Chapter 24.3,5 and 25.1 examples: 24.7,11,13,15 and 25.1 • Equipotentials and conductors • E from V • Calculate electric field of dipole from potential • Electric Potential Energy • of charge in external electric field • stored in the electric field itself (next time) • Appendix: • Example calculation of a spherical charge configuration • Calculate electric field of dipole from potential

  3. Dielectric Breakdown Insulator Conductor Arc discharge equalizes the potential Sparks • High electric fields can ionize nonconducting materials (“dielectrics”) • Breakdown can occur when the field is greater than the “dielectric strength” of the material. • E.g., in air, What is ΔV? Ex. Note: High humidity can also bleed the charge off  reduce ΔV.

  4. Question 1 r2 r1 Ball 1 Ball 2 (a)Ball 1 (b)Ball 2 (c)Same Time Two charged balls are each at the same potential V. Ball 2 is twice as large as ball 1. • As V is increased, which ball will induce breakdown first?

  5. Question 1 r2 r1 Ball 1 Ball 2 (a) Ball 1 (b) Ball 2 (c) Same Time \ Smaller r higherE closer to breakdown Ex. High Voltage Terminals must be big! Two charged balls are each at the same potential V. Ball 2 is twice as large as ball 1. • As V is increased, which ball will induce breakdown first?

  6. Lightning! _ _ _ + + + Collisions produce charged particles. The heavier particles (-) sit near the bottom of the cloud; the lighter particles (+) near the top. Stepped Leader Negatively charged electrons begin zigzagging downward. Attraction As the stepped leader nears the ground, it draws a streamer of positive charge upward. Flowing Charge As the leader and the streamer come together, powerful electric current begins flowing Contact! Intense wave of positive charge, a “return stroke,” travels upward at 108 m/s Factoids:

  7. A B Question 2 Two spherical conductors are separated by a large distance. They each carry the same positive charge Q. Conductor A has a larger radius than conductor B. Compare the potential at the surface of conductor A with the potential at the surface of conductor B. a) VA > VBb) VA = VBc) VA < VB

  8. Er Equipotential (where ) Potential from a charged sphere • The electric field of the charged sphere has spherical symmetry. • The potential depends only on the distance from the center of the sphere, as is expected from spherical symmetry. • Therefore, the potential is constant along a sphere which is concentric with the point charge. These surfaces are called equipotentials. • Notice that theelectric field is perpendicular to the equipotential surface at all points. Last time…

  9. Equipotentials • Defined as: The locus of points with the same potential. • Example: for a point charge, the equipotentials are spheres centered on the charge. • The electric field is always perpendicular to an equipotential surface! Why?? From the definition of potential Along the surface, there is NO change inV(it’s an equipotential!) Therefore, We can conclude then, that is zero. If the dot product of the field vector and the displacement vector is zero, then these two vectors are perpendicular, or the electric field is always perpendicular to the equipotential surface.

  10. + + + + + + + + + + + + + + Conductors • Claim The surface of a conductor is always an equipotential surface (in fact, the entire conductor is an equipotential). • Why?? If surface were not equipotential, there would be an electric field component parallel to the surface and the charges would move!!

  11. B A Question 3 The same two conductors that were in Question 2 are now connected by a wire, before they each carried the same positive charge Q. How do the potentials at the conductor surfaces compare now ? a) VA > VB b) VA = VB c) VA< VB

  12. B A Question 4 The same two conductors that were in Question 2 are now connected by a wire, before they each carried the same positive charge Q.What happens to the charge on conductor A after it is connected to conductor B ? a) QAincreases b) QAdecreases c) QAdoesn’t change

  13. + + + + + - - - - - + - + - - +q + - - + - + - + - - - + + + + + Charge on Conductors? • How is charge distributed on the surface of a conductor? • KEY: Must produce E=0 inside the conductor and E normal to the surface . Spherical example (with little off-center charge): E=0 inside conducting shell. charge density induced on inner surface non-uniform. charge density induced on outer surface uniform Eoutside has spherical symmetry centered on spherical conducting shell.

  14. Lecture 6, ACT 1 -q An uncharged spherical conductor has a weirdly shaped cavity carved out of it. Inside the cavity is a charge -q. How much charge is on the cavity wall? 1A (a) Less than< q(b) Exactly q(c) More than q 1B How is the charge distributed on the cavity wall? (a) Uniformly (b) More charge closer to –q (c) Less charge closer to -q 1C How is the charge distributed on the outside of the sphere? (a) Uniformly (b) More charge near the cavity (c) Less charge near the cavity

  15. Lecture 6, ACT 1 -q An uncharged spherical conductor has a weirdly shaped cavity carved out of it. Inside the cavity is a charge -q. How much charge is on the cavity wall? 1A (a) Less than< q(b) Exactly q(c) More than q By Gauss’ Law, since E=0 inside the conductor, the total charge on the inner wall must be q (and therefore -q must be on the outside surface of the conductor, since it has no net charge).

  16. Lecture 6, ACT 1 -q The induced charge will distribute itself nonuniformly to exactly cancel everywhere in the conductor. The surface charge density will be higher near the -q charge. 1B How is the charge distributed on the cavity wall? (a) Uniformly (b) More charge closer to -q (c) Less charge closer to -q

  17. Lecture 6, ACT 1 -q 1C How is the charge distributed on the outside of the sphere? (a) Uniformly (b) More charge near the cavity (c) Less charge near the cavity As in the previous example, the charge will be uniformly distributed (because the outer surface is symmetric). Outside the conductor the E field always points directly to the center of the sphere, regardless of the cavity or charge. Note: this is why your radio, cell phone, etc. won’t work inside a metal building!

  18. r L But: r S Smaller sphere has the larger surface charge density ! Þ Charge on Conductor Demo • How is the charge distributed on a non-spherical conductor?? Claim largest charge density at smallest radius of curvature. • 2 spheres, connected by a wire, “far” apart • Both at same potential

  19. Equipotential Example • Field lines more closely spaced near end with most curvature – higher E-field • Field lines ^ to surface near the surface (since surface is equipotential). • Near the surface, equipotentials have similar shape as surface. • Equipotentials will look more circular (spherical) at large r.

  20. Electric Dipole Equipotentials • First, let’s take a look at the equipotentials:

  21. Some fish have the ability to produce & detect electric fields • Navigation, object detection, communication with other electric fish • “Strongly electric fish” (eels) can stun their prey Electric Fish Black ghost knife fish Dipole-like equipotentials More info: Prof. Mark Nelson, Beckman Institute, UIUC -Electric current flows down the voltage gradient -An object brought close to the fish alters the pattern of current flow

  22. E from V? V+dV V • Expressed as a vector, E is the negative gradient of V • Cartesian coordinates: • Spherical coordinates: • We can obtain the electric field E from the potential V by inverting our previous relation between E andV:

  23. Preflight 6: This graph shows the electric potential at various points along thex-axis. 8) At which point(s) is the electric field zero? A B C D

  24. E from V: an Example • Consider the following electric potential: • Something for you to try: Can you use the dipole potential to obtain the dipole field? Try it in spherical coordinates ... you should get (see Appendix): • What electric field does this describe? ... expressing this as a vector:

  25. The Bottom Line If we know the electric field E everywhere, Þ If we know the potential function V everywhere, allows us to calculate the electric field E everywhere • Units for Potential! 1 Joule/Coul = 1 VOLT allows us to calculate the potential function V everywhere (keep in mind, we often define VA = 0 at some convenient place)

  26. Lecture 6, ACT 2 a Q b 1B The electric potential in a region of space is given by The x-component of the electric field Ex at x= 2 is (a)Ex = 0(b)Ex>0(c)Ex< 0 1A • A point charge Q is fixed at the center of an uncharged conducting spherical shell of inner radius a and outer radius b. • What is the value of the potential Vaat the inner surface of the spherical shell? (b) (a) (c)

  27. Eout Lecture 6, ACT 2 • How to start?? The only thing we know about the potential is its definition: • To calculate Va, we need to know the electric field E • Outside the spherical shell: • Apply Gauss’ Law to sphere: • Inside the spherical shell: E = 0 1A • A point charge Q is fixed at the center of an uncharged conducting spherical shell of inner radius a and outer radius b. • What is the value of the potential Va at the inner surface of the spherical shell? a Q b (b) (a) (c)

  28. Lecture 6, ACT 2 We know V(x) “everywhere” To obtain Ex “everywhere”, use 1B The electric potential in a region of space is given by The x-component of the electric field Ex at x= 2 is (a)Ex= 0(b)Ex> 0(c)Ex< 0

  29. Therefore, a particle moving under the influence of the Coulomb force is said to have an electric potential energy defined by: this “q” is the ‘test charge” in other examples... Electric Potential Energy • The Coulomb force is a CONSERVATIVE force (i.e. the work done by it on a particle which moves around a closed path returning to its initial position is ZERO.) • The total energy (kinetic + electric potential) is then conserved for a charged particle moving under the influence of the Coulomb force.

  30. Preflight 6: A E C B 6) If a negative charge is moved from point A to point B, its electric potential energy a) increases b) decreases c) doesn’t change

  31. r q A Q • charge +q is brought to pt A, a distance r from Q. • charge +2q is brought to pt B, a distance 2r from Q. Q B 2r (b)UA= UB (c)UA> UB (a)UA< UB • Compare the potential energy of q (UA) to that of 2q(UB): 2q 3B • Suppose charge 2q has mass m and is released from rest from the above position (a distance 2r from Q). What is its velocity vf as it approaches r =¥? (c) (a) (b) Lecture 6, ACT 3 Two test charges are brought separately to the vicinity of a positive charge Q. 3A

  32. r Q q A Q 2q 2r B (b)UA= UB (c)UA> UB (a)UA< UB 3A Lecture 6, ACT 3 • Two test charges are brought separately to the vicinity of positive charge Q. • charge +q is brought to pt A, a distance rfrom Q. • charge +2q is brought to pt B, a distance 2r from Q. • Compare the potential energy of q (UA) to that of 2q (UB): • Look familiar? • This is ALMOST the same as ACT 2 from the last lecture. • In that ACT, we discovered that the potential at A was TWICE the potential at B. The point was that the magnitudes of the charges at A and B were IRRELEVANT to the question of comparing the potentials. • The charges at A and B are NOT however irrelevant in this ACT!! • The potential energy of q is proportional to Qq/r. • The potential energy of 2q is proportional to Q(2q)/(2r). • Therefore, the potential energies UA and UB are EQUAL!!!

  33. 3B • Suppose charge 2q has mass m and is released from rest from the above position (a distance 2r from Q). What is its velocity vf as it approaches r =¥? (c) (a) (b) • What we have here is a little combination of 111 and 112. • The principle at work here is CONSERVATION OF ENERGY. • Initially: • The charge has no kinetic energy since it is at rest. • The charge does have potential energy (electric) = UB. • Finally: • The charge has no potential energy (Uµ 1/R) • The charge does have kinetic energy = KE Lecture 6, ACT 3

  34. Accelerators • Electrostatic: Van de Graaff • electrons ®100 keV ( 105 eV) • Electromagnetic: Fermilab • protons ®1TeV ( 1012 eV) Energy Units MKS: U = QV1 coul-volt = 1 joule for particles (e, p, ...) 1 eV = 1.6x10-19 joules

  35. Reading assignment: 25.2, 4 Examples: 25.2,3,5,6 and 7 Summary • Physically, DV is what counts • The place where V=0 is “arbitrary” (at infinity) • Conductors are equipotentials • Find E from V: • Potential Energy • Next time, capacitors:

  36. z r 1 +q r r 2 a q r2-r1 a -q • Rewrite this for special case r>>a: Þ Now we can use this potential to calculate the E field of a dipole(after a picture) (remember how messy the direct calculation was?) Appendix A: Electric Dipole The potential is much easier to calculate than the field since it is an algebraic sum of 2 scalar terms.

  37. z r 1 +q r r 2 a q a -q • Calculate E in spherical coordinates: the dipole moment Þ Appendix A: Electric Dipole

  38. Appendix A: Dipole Field z +q r q a a -q

  39. (b) qmax= 45° (c) qmax= 90° (a) qmax= 0 • The expression for the electric field of a dipole (r >> a) is: • The polar component of E is maximum when sinq is maximum. • Therefore, Eq has its maximum value when q= 90°. z Sample Problem r • Consider the dipole shown at the right. • Fixr = r0>> a • Define qmax such that the polar component of the electric field has its maximum value (for r = r0). 1 +q r r 2 a q a -q What is qmax?

  40. - + - + - + - + + - - - - - - + + + + + + + + + + Appendix B: Induced charge distribution on conductorvia“method of images” • Consider a source charge brought close to a conductor: + • Charge distribution “induced” on conductor by source charge: • Induced charge distribution is “real” and sources E-field that is zero inside conductor! • resulting E-field is sum of field from source charge and induced charge distribution • E-field is locally perpendicular to surface • just like the homework problem. • With enough symmetry, can solve fors on conductor • how? Gauss’ Law

  41. - - - - - Appendix B: Induced charge distribution on conductorvia“method of images” • Consider a source charge brought close to a planar conductor: • Charge distribution “induced” on conductor by source charge • conductor is equipotential • E-field is normal to surface • this is just like a dipole + - • Method of Images for a charge (distribution) near a flat conducting plane: • reflect the point charge through the surface and put a charge of opposite sign there • do this for all source charges • E-field at plane of symmetry - the conductor surface determines s.

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