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Lecture 3: Chapter 19 Cont

Lecture 3: Chapter 19 Cont. Electric Charges, Forces and Electric Fields. Agenda. Compare the electric force to Gravitational force Superposition of forces Electric Field Superposition of Electric Field Shielding and Charging by Induction Electric Flux and Gauss’s Law.

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Lecture 3: Chapter 19 Cont

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  1. Lecture 3: Chapter 19 Cont Electric Charges, Forces and Electric Fields

  2. Agenda • Compare the electric force to Gravitational force • Superposition of forces • Electric Field • Superposition of Electric Field • Shielding and Charging by Induction • Electric Flux and Gauss’s Law

  3. What did we study last lesson? • Electric Charges: positive and neg. Electron (neg) and proton (pos) • Charges are conserved • Polarization • Insulators & Conductors • Coulomb’s Law

  4. Recap • Two rods and a cat fur: • Glass Rod (positive charges) • Rubber Rod (negative)

  5. Charging by Conduction

  6. Recap • Coulomb’s Law • (electrostatic force between point charges)

  7. Recap • Polarization is realignment of charge within individual molecules. • Produces induced charge on the surface of insulators. • how e.g. rubber or glass can be used to supply electrons.

  8. ? + Student Discussion A positively charged object hanging from a string is brought near a non conducting object (ball). The ball is seen to be attracted to the object. • Explain why it is not possible to determine whether the object is negatively charged or neutral. • What additional experiment is needed to reveal the electrical charge state of the object?

  9. - + Cont • Two possibilities: • Attraction between objects of unlike charges. Attraction between a charged object and a neutral object subject to polarization. + - + - + - + + -

  10. ? 0 +- + + + + - + -+- What additional experiment is needed to reveal the electrical charge state of the object? • Two Experiments: • Bring known neutral ball near the object and observe whether there is an attraction. Bring a known negatively charge object near the first one. If there is an attraction, the object is neutral, and the attraction is achieved by polarization. -

  11. Question • Name the first action at a distance force you have encountered in physics so far. Electric Force

  12. Electrical Forces are Field Forces • This is the second example of a field force • Gravity was the first • Remember, with a field force, the force is exerted by one object on another object even though there is no physical contact between them • There are some important similarities and differences between electrical and gravitational forces

  13. Electrical Force Compared to Gravitational Force • Both are inverse square laws • The mathematical form of both laws is the same • Masses replaced by charges • G replaced by ke • Electrical forces can be either attractive or repulsive • Gravitational forces are always attractive

  14. Consider a proton (mp=1.67x10-27 kg; qp=+1.60x10-19 C) and an electron (me=9.11x10-31 kg; qe=-1.60x10-19 C) separated by 5.29x10-11 m. The particles are attracted to each other by both the force of gravity and by Coulomb’s law force. Which of these has the larger magnitude? • Gravitational force • B. Coulomb’s law force Example: Student Participation

  15. Which of these has the larger magnitude? • Gravitational force • B. Coulomb’s law force Pg. 659. By what factor the electric force is greater than the gravitational force? We will mostly ignore gravitational effects when we consider electrostatics.

  16. Problem solving steps • Visualize problem – labeling variables • Determine which basic physical principle(s) apply • Write down the appropriate equations using the variables defined in step 1. • Check whether you have the correct amount of information to solve the problem (same number of knowns and unknowns). • Solve the equations. • Check whether your answer makes sense (units, order of magnitude, etc.).

  17. Superposition Principle • From observations: one finds that whenever multiple charges are present, the net force on a given charge is the vector sum of all forces exerted by other charges. • Electric force obeys a superposition principle.

  18. The Superposition Principle • How to work the problem? • Find the electrical forces between pairs of charges separately • Then add the vectors • Remember to add the forces as vectors

  19. Superposition Principle Example: 3 charges in a line F13 F12 q1 = 6.00 uC q2 = 1.50 uC q3 = -2.00 uC d1 = 3.00 cm d2 = 2.00 cm K = 8.99 x 109 N.m2 /C2

  20. Example: Superposition Principle (Not along a line) Consider three point charges at the corners of a triangle, as shown in the next slide. Find the resultant force on q3 if q1 = 6.00 x 10-9 C q2 = -2.00 x 10-9 C q3 = 5.00 x 10-9 C

  21. Superposition Principle Example • The force exerted by q1 on q3 is • The force exerted by q2 on q3 is • The total force exerted on q3 is the vector sum of and

  22. Consider three point charges at the corners of a triangle, as shown below. Find the resultant force on q3. Solution:

  23. Spherical Charge Distribution • Students need to read and work on pg 663/664

  24. Electric Field - Discovery • Electric forces act through space even in the absence of physical contact. • Suggests the notion of electrical field (first introduced by Michael Faraday (1791-1867).

  25. Electric Field • An electric field is said to exist in a region of space surrounding a charged object. • If another charged object enters a region where an electrical field is present, it will be subject to an electrical force.

  26. Electric Field • A charged particle, with charge Q, produces an electric field in the region of space around it • A small test charge, qo, placed in the field, will experience a force. • “The electric field is the force per charge at a given location”

  27. Electric Field • Mathematically, • SI units are N / C Given: One finds:

  28. Electric Field • Use this for the magnitude of the field • The electric field is a vector quantity • The direction of the field is defined to be the direction of the electric force that would be exerted on a small positive test charge placed at that point

  29. Direction of Electric Field • The electric field produced by a negative charge is directed toward the charge • A positive test charge would be attracted to the negative source charge

  30. Direction of Electric Field, cont • The electric field produced by a positive charge is directed away from the charge • A positive test charge would be repelled from the positive source charge

  31. More About a Test Charge and The Electric Field • The test charge is required to be a small charge • It can cause no rearrangement of the charges on the source charge • The electric field exists whether or not there is a test charge present • The Superposition Principle can be applied to the electric field if a group of charges is present

  32. Electric Fields and Superposition Principle • The superposition principle holds when calculating the electric field due to a group of charges • Find the fields due to the individual charges • Add them as vectors • Use symmetry whenever possible to simplify the problem

  33. Electric Field, final

  34. Problem Solving Strategy • Draw a diagram of the charges in the problem • Identify the charge of interest • You may want to circle it • Units – Convert all units to SI • Need to be consistent with ke

  35. Problem Solving Strategy, cont • Apply Coulomb’s Law • For each charge, find the force on the charge of interest • Determine the direction of the force • Sum all the x- and y- components • This gives the x- and y-components of the resultant force • Find the resultant force by usingthe Pythagorean theorem and trig

  36. Problem Solving Strategy, Electric Fields • Calculate Electric Fields of point charges • Use the equation to find the electric field due to the individual charges • The direction is given by the direction of the force on a positive test charge • The Superposition Principle can be applied if more than one charge is present

  37. Examples of Electric Field • E will be constant (a charged plane) • E will decrease with distance as (a point charge) • E in a line charge, decrease with a distance 1/r

  38. Example: • An electron moving horizontally passes between two horizontal planes, the upper plane charged negatively, and the lower positively. - - - - - - - - - - - - - - - - - - - - - - vo - + + + + + + + + + + + + + + + + + + + + + +

  39. Example: • A uniform, upward-directed electric field exists in this region. This field exerts a force on the electron. Describe the motion of the electron in this region. - - - - - - - - - - - - - - - - - - - - - - vo - + + + + + + + + + + + + + + + + + + + + + +

  40. - - - - - - - - - - - - - - - - - - - - - - vo - + + + + + + + + + + + + + + + + + + + + + + Observations: • Horizontally: • No electric field • No force • No acceleration • Constant horizontal velocity

  41. - - - - - - - - - - - - - - - - - - - - - - vo - + + + + + + + + + + + + + + + + + + + + + + Observations: • Vertically: • Constant electric field • Constant force • Constant acceleration • Vertical velocity increase linearly with time.

  42. - - - - - - - - - - - - - - - - - - - - - - - + + + + + + + + + + + + + + + + + + + + + + Conclusions: The charge will follow a parabolic path downward. Motion similar to motion under gravitational field only except the downward acceleration is now larger.

  43. y E1 E P E2 0.400 m x q1 0.300 m q2 Example: Electric Field Due to Two Point Charges (Superposition of Fields) Question: Charge q1=7.00 mC is at the origin, and charge q2=-10.00 mC is on the x axis, 0.300 m from the origin. Find the electric field at point P, which has coordinates (0,0.400) m.

  44. Question: Charge q1=7.00 mC is at the origin, and charge q2=-10.00 mC is on the x axis, 0.300 m from the origin. Find the electric field at point P, which has coordinates (0,0.400) m. Observations: • First find the Electric field at point P due to charge q1 and q2. • Field E1 at P due to q1 is vertically upward. • Field E2 at due to q2 is directed towards q2. • The net field at point P is the vector sum of E1 and E2. • The magnitude is obtained with

  45. Question: Charge q1=7.00 mC is at the origin, and charge q2=-10.00 mC is on the x axis, 0.300 m from the origin. Find the electric field at point P, which has coordinates (0,0.400) m. Set up the problem: q1=7.00 mC q2=-10.00 mC K = 8.99 x 109 N. m2 /C2 r1 = 0.400m r2= ? What do we do?

  46. Question: Charge q1=7.00 mC is at the origin, and charge q2=-10.00 mC is on the x axis, 0.300 m from the origin. Find the electric field at point P, which has coordinates (0,0.400) m. • Calculate r2 • Calculate the electric field at P • I will work set it up, but you will need to finish (it looks like example in pg. 669)

  47. Question: Charge q1=7.00 mC is at the origin, and charge q2=-10.00 mC is on the x axis, 0.300 m from the origin. Find the electric field at point P, which has coordinates (0,0.400) m. Solution:

  48. Electric Field Lines • A convenient aid for visualizing electric field patterns is to draw lines pointing in the direction of the field vector at any point • These are called electric field lines and were introduced by Michael Faraday

  49. Electric Field Lines, cont. • The field lines are related to the field in the following manners: • The electric field vector, , is tangent to the electric field lines at each point • The number of lines per unit area through a surface perpendicular to the lines is proportional to the strength of the electric field in a given region

  50. Electric Field Line Patterns • Point charge • The lines radiate equally in all directions • For a positive source charge, the lines will radiate outward

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