1. 1/22/07184 Lecture 81 PHY 184 Spring 2007 Lecture 8 Title: Calculations on Electrostatics

2. 1/22/07184 Lecture 82AnnouncementsAnnouncements  Homework Set 2 is due Tuesday morning, January 23, at 8:00 am.  Honors Option students will provide help in the SLC starting this week.  Today we will finish the electric field and begin electric potential.  We will start clicker questions today. More details later during the lecture.

3. 1/22/07184 Lecture 83 Review – Gauss’s Law q = net charge enclosed by S

4. 1/22/07184 Lecture 84 Review - Electric Fields from Charge Distributions  The electric field E at distance r from a wire with charge density is  The electric field E produced by an infinite non-conducting plate with charge density  is

5. 1/22/07184 Lecture 85 Review - Electric Fields from Charge Distributions (2)  The electric field E produced by an infinite conducting plane with charge density  is  The electric field inside a spherical shell of charge q is zero  The electric field outside a spherical shell of charge q is the same as the field from a point charge q.

6. 1/22/07184 Lecture 86 Review - Spherical Charge Distributions r R + + + + + + + + R Q Q Conducting sphere E + + + + + + + + + + + + Non-conducting sphere E=0 r R E

7. 1/22/07184 Lecture 87 Review - Electric Fields from a Ring of Charge  The electric field E resulting from a ring of charge (radius R, charge density =q/(2  R)) on the axis  Strategy: Imagine the ring is divided into differential elements of charge dq= ds. Use the electric field of a point charge for every one of them. ds  kq/z 2 for large z

8. 1/22/07184 Lecture 88 Example - Charge in a Cube  Q=3.76 nC is at the center of a cube. What is the electric flux through one of the sides?  Gauss’ Law:  Since a cube has 6 identical sides and the point charge is at the center Q

9. 1/22/07184 Lecture 89 Example - E Field and Force  The figure shows the defecting plates of an ink-jet printer. A negatively charged ink drop (q=1.5 x 10 -13 C) enters the region between the plates with a velocity of v=18 m/s along x. The length L of each plate is 1.6 cm. The plates are charged to produce an electric field at all points between them (E=1.4 x 10 6 N/C). The vertical deflection of the drop at x=L is 0.64 mm. What is the mass of the ink drop?  Idea: A constant electrostatic force of magnitude qE acts upward on the drop. … constant acceleration

10. 1/22/07184 Lecture 810 Example - E Field and Forces (2)  What is the mass of the ink drop?  Idea: Let t be the time required to pass through the plates. Then…

11. 1/22/07184 Lecture 811 Clicker Quizzes Starting Today  You need a registered HITT clicker.  Get up to 5% (but not more) extra credit according to Clicker’s Law (you can miss 20% of the quizzes and still get the full extra credit)  You can expect clicker questions each lecture.  If you missed the clicker registration, fill in the clicker sheet.

12. 1/22/07184 Lecture 812 Induction, Conduction and Polarization Which diagram best represents the charge distribution on the spheres when a positively charged rod is brought near the leftmost sphere (without touching it)? Consider three neutral metal spheres in contact and on insulating stands. - + - + - + ++++++++ C + - + - + - ++++++++ A - - - + ++ ++++++++ D ++ + - - - ++++++++ B

13. 1/22/07184 Lecture 813  Shown is an arrangement of five charged pieces of plastic (q 1 =q 4 =3nC, q 2 =q 5 =-5.9nC and q 3 =-3.1nC). A Gaussian surface S is indicated. What is the net electric flux through the surface? Clicker Question - Enclosed Charge A:  =-6 x 10 -9 C/  0 = -678 Nm 2 /C B:  =  x  10 -9 C/  0 = -1356 Nm 2 /C C:  =0 D:  =  x 10 -9 C/  0 = 328 Nm 2 /C

14. 1/22/07184 Lecture 814  Shown is an arrangement of five charged pieces of plastic (q 1 =q 4 =3nC, q 2 =q 5 =-5.9nC and q 3 =-3.1nC). A Gaussian surface S is indicated. What is the net electric flux through the surface? Clicker Question - Enclosed Charge enclosed charge A:  =-6x10 -9 C/  0 = -678 Nm 2 /C

15. 1/22/07184 Lecture 815  Shown is a Gaussian surface in the form of a cylinder of radius R and length L immersed in a uniform electric field E. What is the flux  of the electric field through the closed surface? Clicker Question - Flux A:  =2  R 2 E B:  =  R 2 E C:  =0 D:  =(2  RL+2  R 2 )E

16. 1/22/07184 Lecture 816  Shown is a Gaussian surface in the form of a cylinder of radius R and length L immersed in a uniform electric field E. What is the flux  of the electric field through the closed surface? Clicker Checkpoint - Flux C:  =0 Fluxes: …left end =  R 2 …right end = +  R 2 …around cylinder = 0 …full flux = 0

17. 1/22/07184 Lecture 817 The Electric Potential

18. 1/22/07184 Lecture 818 Electric Potential  We have been studying the electric field.  Next topic: the electric potential  Note the similarity between the gravitational force and the electric force.  Gravitation can be described in terms of a gravitational potential and we will show that the electric potential is analogous.  We will see how the electric potential is related to energy and work.  We will see how we can calculate the electric potential from the electric field and vice versa.

19. 1/22/07184 Lecture 819 Electric Potential Energy  The electric force, like the gravitational force, is a conservative force. (‡)  When an electrostatic force acts between two or more charges within a system, we can define an electric potential energy, U, in terms of the work done by the electric field, W e, when the system changes its configuration from some initial configuration to some final configuration. (‡) Conservative force: The work is path-independent.

20. 1/22/07184 Lecture 820 Electric Potential Energy (2)  Like gravitational or mechanical potential energy, we must define a reference point from which to define the electric potential energy.  We define the electric potential energy to be zero when all charges are infinitely far apart.  We can then write a simpler definition of the electric potential taking the initial potential energy to be zero,  The negative sign on the work: If E does positive work then U < 0 If E does negative work then U > 0

21. 1/22/07184 Lecture 821 Constant Electric Field  Let’s look at the electric potential energy when we move a charge q by a distance d in a constant electric field.  The definition of work is  For a constant electric field the force is F = qE …  … so the work done by the electric field on the charge is Note:  = angle between E and d.

22. 1/22/07184 Lecture 822 Constant Electric Field - Special Cases  Displacement is in the same direction as the electric field A positive charge loses potential energy when it moves in the direction of the electric field.  Displacement is in the direction opposite to the electric field A positive charge gains potential energy when it moves in the direction opposite to the electric field.

23. 1/22/07184 Lecture 823 Definition of the Electric Potential  The electric potential energy of a charged particle in an electric field depends not only on the electric field but on the charge of the particle.  We want to define a quantity to probe the electric field that is independent of the charge of the probe.  We define the electric potential as  Unlike the electric field, which is a vector, the electric potential is a scalar. The electric potential has a value everywhere in space but has no direction. Units: [V] = J / C, by definition, volt “potential energy per unit charge of a test particle”