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Physics 2020 Lectures and Clicker Quizzes Weeks 3-4

2/6/11. Physics 2020 Lectures and Clicker Quizzes Weeks 3-4. M. Goldman Spring, 2011. Upcoming exam. Concepts versus facts = thinking versus remembering

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Physics 2020 Lectures and Clicker Quizzes Weeks 3-4

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  1. 2/6/11 Physics 2020 Lectures and Clicker Quizzes Weeks 3-4 M. Goldman Spring, 2011

  2. Upcoming exam • Concepts versus facts = thinking versus remembering • You do need to remember some facts but we are not so interested in that in this course. (Paper with formulas that you are allowed to bring to exams and formulas given in exam) • In physics you mainly have to think to apply concepts to solve problems you haven't seen before. If you deeply understand concepts you will have a better chance of solving problems. • Seeing analogies is an important ingredient in solving new problems. • Math skills are also important • Knowledge of facts, understanding of concepts and ability to solve unexpected problems are essential for workers in the health/medical industry

  3. Does the force, F on a particle of charge, q, due to an electric field, E point in the same direction as E? - positive charge, q ++ F = qE F + ++ E E F negative charge, q

  4. A positively charged solid metal sphere in equilibrium net charge on surface only E-field  surface E = 0 inside A metal in electrostatic equilibrium

  5. Why does the electric field vector at the surface of a metal point have no component on the surface? The E-field must be perpendicular to the surface (in electrostatic equilibrium), otherwise the component of the E-field along the surface would push electrons along the surface causing movement of charges (and we would not be in equilibrium). On the surface of metal, if Ex was not zero, there would be a force Fx = q Ex = –e Ex on electrons in the metal pushing them along the surface. E Ey Ex metal

  6. How can a uniform electric field be produced? • The electric field between two parallel metal slabs with equal and opposite charges will be almost uniform • This is a capacitor!

  7. Put a conducting slab in an electric field, E. Electronsin slab will distribute themselves so that thereis zero net field inside the slab

  8. Faraday Cage • A metal-screen cage in an external E-field will also distribute its electrons to eliminate any net E-field inside the cage • http://www.youtube.com/watch?v=WqvImbn9GG4

  9. Would you try this?

  10. Van de Graaff demos of "lightning" and of "tassel spreading"

  11. Positively unmanageable hair!

  12. Are you safe in your car during lightning? URBAN LEGEND: The insulating rubber tires on a car protect you from lightning. REAL PHYSICS: The metal body of the car keeps the charges on the outside, away from you! It is essentially a Faraday charge

  13. Explanation Remember that Electric Fields inside conductors (metals) are zero when in electrostatic equilibrium. Inside a hollow metal object (like a car) that has excess charge put on it (from lightning) the charge will be on the outside surface of the object! However, this is not a case of electrostatic equilibrium. The charges are moving, but by similar principle they stay along the outside conductor and then try to leave via the easiest path (in this case to the ground).

  14. Clicker Question 1, Monday, Jan 24 What is the magnitude of the E-field at point P? A) |E| = 2k|Q |/ s2 B) |E| = sqrt(2) k|Q| / s2 C) |E| = k|Q| / ( sqrt(2) s)2 D) zero E) none of the above -|Q| +|Q| s s s P s -|Q| +|Q|

  15. Now begin subject matter of Chapter 17:Potentials and voltage Beware of high voltage

  16. You have already studied gravitational potential energy • When an object has positive gravitational potential energy the force of gravity can do work on it. • Conservation of energy says that as work is done on the body by gravity its potential energy, PE, will decrease and its kinetic energy, KE, will increase but the sum of the two is constant. • Near the surface of Earth the gravitational force is approximately constant and equal to mg, pointing down. • The gravitational potential energy of an object a distance h above theground is the work youdo to lift it there PE = F·h = mgh.

  17. Review of relationship between constant force, work and P.E. m Fhand h m m Choose zeroof P.E. here Fgravity Fgravity Fgravity P.E. = minimum work done by hand (against gravity) to bring m up a vertical distance, h = Fhandh. Minimum |Fhand| = |Fgravity|. P.E. = (mg)hConservative force independent of path. No work to move m horizontally. Only verticalcomponent of Fhand has to balance gravity.

  18. Electric forces and fields can be understood in a similar way by defining electric potential energy • The constant electric force, F, experienced by a positive charge, q, at height h inside a capacitor with downward field E is analogous to the "constant gravitational force" Electricpotentialenergy =F·h =q·|E|·h q h E F

  19. Clicker question 2, Jan 24, 2011 • Neglecting gravity, what is the kinetic energy of the particle (with +charge, q) put down at rest when it hits the bottom of the capacitor? A) It cannot be determined without knowing its mass B) It is Eqh C) It is mgh

  20. Clicker question 3, Jan 24, 2011 • What happens if the electric charge put down at rest at the same position is negative, = -|q|? Once again, you can neglect gravity. • The charge moves the same way as the positive charge • The charge will have negative kinetic energy • The potential energy of the charge will decrease as it moves

  21. Explanation - Electricpotentialenergy =F·h =q·|E|·h -|q| E h 21 P.E. is negative. Becomes more negative as chargemoves up from h. More negative means decrease! E.g. -6V < -4V

  22. Electrical P.E.depends on the sign of q. Define a quantity independent of q, called the electric potential, V • Electric potential (voltage), V, is related to electric field, E • The potential (voltage) V is not a vector. It is a ± scalar. • The potential energy (P.E.) of a (test) charge, q, at a point where potential = V is always given by P.E. = qV • For the constant downward electric field, E, inside the capacitor, as drawn, the electric potential (voltage), V, is • V = P.E./q = (Eqh)/q = Eh • V does not depend on the (test) charge, q, which feels force, F, due to E when placed down as shown. • F can be up or down, depending on sign of q

  23. How to visualize electric potential (voltage) • "Connect the dots" to visualize voltage • A curve through all points with the same value of V is called an equipotential curve. • We can draw several of these curves – one for each different value of V (label them by the value of V) • The equipotential lines will always be perpendicular to the electric field lines • For the uniform field lines inside the capacitor the equipotential lines appear as straight lines perpendicular to the field lines

  24. Compare four different ways of understanding the electric force on a test charge, q:

  25. SAT-type analogy question • Electric force is to electric field as electric P.E. is to _______ • Charge • Field line • Voltage • Gravitational P.E. • None of the above

  26. Comments about electrical units • Why are electric field units N/C the same as V/m? • Answer: Because volt has units V = work/charge(Nm)/C (also same as J/C) • What is the energy unit eV (electron volts)? • Answer: K.E. gained by electron in electric field when it loses P.E. associated with 1 Volt.

  27. Use equipotential lines to visualize how a test charge, q, set down at rest will move • A charged particle, q, set down at rest will always move from higher P.E.to lower P.E. as it gains kinetic energy, K.E. (Doesn't matter if q = + or -) • If q is positive it will move from higher V to lower V (V = voltage = electric potential) • If q is negative it will move from lower V to higher V 6 volts 0 volts

  28. You can infer the strength of the electric field from the equipotential lines y-direction 10-3 m V = Eh E = V/h = 6V/10-3m More generally, 6 volts = -6 kV/m 0 volts

  29. There are some conventions for equipotential lines or surfaces The voltage difference between any two drawn equipotentiallines is the same unless they are labeled otherwise y-direction 10-3 m 10-3 m 10-3 m equally spaced The capacitor field, E, is the same betweenany two equipotential lines. It is uniform in this case. 18 volts 12 volts 6 volts 0 volts

  30. Clicker question 1, Wed. Jan 26, 2011 • Some set of charges produce the Voltages (electric potentials) shown below. How much work is required to move the particle • from a to b? • 2 Coulombs • 20 Volts • 40 Joules • 10 Joules • E) None of the Above b Q = 2 C a 10V 20V 30V 40V 50V

  31. What do equipotential curves look like when E has a different magnitude/direction at each point in space (is spatially nonuniform). Consider point (+)charge, Q: Red dashed circles are equipotential curves They are still perpendicular to the electric field lines The voltage difference is the same between any two equipotential circles Thespatial separation, s = rlarger - rsmallerbetween circlesshrinks as V increases 5 V s s The electric field is givenmore and more accurately as s decreases by 10 V 15 V

  32. Equipotential curves are like topo maps used for hiking in the mountains

  33. Relief (3D) map vs topo (2D) map

  34. The equipotential lines are like constant altitude lines on a topo map (but voltage is the altitude) Consider the equipotential lines of a dipole This is a topo map of the following potential terrain (+)charges create ‘mountains’, (-)charges create ‘valleys.’ Potentials from each charge add as ±numbers to get net V. Test (+)charges go ‘downhill’, test ( -)charges go ‘uphill’. Crowded lines indicate higher E-field.

  35. Clicker question 1, Fri., Jan 28 The equipotential surfaces around a line of charge (viewed end-on) are shown. Each equipotential line is 2 meters from the nearest-neighbor equipotential. What is the approximate magnitude of the electric field at point A? • 0.1 Volts/m • B) 0.2 Volts/m • C) 0.4 Volts/m • D) 0.7 Volts/m • E) None of the above A -1.0V -2.2V -1.4V -1.8V

  36. Clicker Question 2, Fri, Jan 28, 2011 The equipotential lines are labeled in volts. Spatial distances can be determined from the x and y axes. What is the approximate magnitude and direction of the electric field, E, at point P? • 2.7 V/m, in the x-direction • 3.2 V/m, in the y-direction • 1 V/m, in the negative y-direction • .5 V/m, in the y-direction • None of the above P Correct answer is: .25 V/m up (negative y direction

  37. Further questions to nail down your understanding of the meaning of potential • Additional questions: • How can you visualize the electric field lines and vectors? • What are the signs of the two point charges which create the electric field. • Which charge has the greater magnitude? • What is the P.E. of a charge q = 3 C put down at P? • What is the force on it? • What is the P.E. of a charge of q = -3 C put down at P? • What is the force on it? P

  38. Clicker Question 3, Friday, Jan 28, 2011 • If you know the voltage at a single point only, can you determine the electric field there? • No • Yes • Magnitude, but not direction • Direction, but not magnitude Voltage is like altitude, and E-field is like slope

  39. What is the formula for voltage, V (electric potential) surrounding a point (+)charge, Q? First we find the potential energy, P.E. of another positivecharge, q, put down at the tip of the position vector, r. P.E. = Work to bring charge q from infinity, where P.E. is defined = 0 to position r. Path doesn't matter for conservative force. Work NOT EQUAL to force times distance because this force depends on position. y q Q r x

  40. But circle of radius, r, isequipotential (voltage is same everywhere on circle) Voltage caused by point charge, Q, anywhere on circle of radius r is V = kQ/r Q r

  41. Visualizationsof voltage V = kQ/rsurroun-ding (+) point charge, Q > 0 at different x and y r = √(x2+y2) y V y x V y x

  42. If charge Q = -|Q| is negative the formula for voltage is still correct but values of V are negative. V = -k|Q|/r r = √(x2+y2) V y x x y V 42 x

  43. Visualizationof voltage V = kQ/rsurroun-ding (+) point charge, Q > 0 at different x, y, and z r = √(x2+y2+z2) Equipotential surfaces areconcentric spheres ratherthan circles z y x

  44. Can the electric field from a source charge or charges be "blocked?" • The field goes right through any charges put in its way • But you must add the field of the charges in its way to get the net field on the other side of the charges in its way. E of (+)charge here (-) charge Net E (+) charge E of (-)charge

  45. What can you learn about the electric field if you know the equipotentials (equal voltage curves) • Step 1: visualize the mountains and valleys of voltage; imagine you are standing on skis at some point on voltage terrain • Step 2: electric field, E, is like downhill slope of mountain or valley at that point. Steeper slope means stronger E. • Step 4:Your ski trail along the steepest slope (no turns) becomes an electric field line when projected onto the contour map Contour map Relief map

  46. Clicker Question 1, Monday, Jan 31, 2011 A set of equipotential lines are shown below. Each equipotential line is 2 meters from the nearest-neighbor equipotential. What is the approximate electric field at point A? • 1.0 V/m in –x direction • B) 1.0 V/m in +y direction • C) 2.0 V/m in +x direction • D) 2.0 V/m in –y direction • E) None of the above 10V A y 8V 6V x 4V 1 V/m in +x direction

  47. Clicker Question 2, Monday, Jan 31, 2011 An electron of mass, m, is fired away (to the left) from a large positive charged point starting at a distance r. Its initial velocity (to the left) is v +Q -e r The electron will escape (go out to -infinity) if: A) B) C) D) It can never escape.

  48. Roughguide togrades A = 84-96 A- = 80 B+ = 76 B = 72 B- = 68 C+ = 64 C = 56-60 C- = 52 D = 40-48 F = 24-36

  49. Questions 20-25

  50. Clicker question 3 • An electron is set down with an initial velocity, v, near a proton What determines the trajectory (path the electron follows at later times) • The initial velocity, v together with the distance from the proton • The initial velocity alone • The force due to the proton alone • The distance from the proton alone • It cannot be determined

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