PHY 184

1. PHY 184 Spring 2007 Lecture 8 Title: Calculations on Electrostatics 184 Lecture 8

2. Announcements • 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. 184 Lecture 8

3. Review – Gauss’s Law q = net charge enclosed by S 184 Lecture 8

4. 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 184 Lecture 8

5. Review - Electric Fields from Charge Distributions (2) • The electric field E produced by an infinite conducting plane withcharge density  is • The electric field inside aspherical 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. 184 Lecture 8

6. Review - Spherical Charge Distributions + Q + + + + R + + + + + + + + + + + + + + + E r R r R Non-conducting sphere Conducting sphere Q E E=0 184 Lecture 8

7. Review - Electric Fields from a Ring of Charge ds • 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.  kq/z2 for large z 184 Lecture 8

8. Example - Charge in a Cube Q • 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 184 Lecture 8

9. 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 106 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 184 Lecture 8

10. 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… 184 Lecture 8

11. 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. 184 Lecture 8

12. Induction, Conduction and Polarization C D B ++++ ++++ ++++ ++++ ++++ - - - - + -+ - + -+ - + -+ - + -+ - A - - - - - + - + ++++ - + - + - + - + - + - + Consider three neutral metal spheres in contact and on insulating stands. Which diagram best represents the charge distribution on the spheres when a positively charged rod is brought near the leftmost sphere (without touching it)? 184 Lecture 8

13. Clicker Question - Enclosed Charge • Shown is an arrangement of five charged pieces of plastic (q1=q4=3nC, q2=q5=-5.9nC and q3=-3.1nC). A Gaussian surface S is indicated. What is the net electric flux through the surface? A: =-6 x 10-9C/0= -678 Nm2/C B: = x10-9C/0= -1356 Nm2/C C: =0 D: =  x 10-9C/0= 328 Nm2/C 184 Lecture 8

14. Clicker Question - Enclosed Charge enclosed charge • Shown is an arrangement of five charged pieces of plastic (q1=q4=3nC, q2=q5=-5.9nC and q3=-3.1nC). A Gaussian surface S is indicated. What is the net electric flux through the surface? A: =-6x10-9C/0= -678 Nm2/C 184 Lecture 8

15. Clicker Question - Flux • 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? A: =2R2E B: =R2E C: =0 D: =(2RL+2R2)E 184 Lecture 8

16. Clicker Checkpoint - Flux • 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? C: =0 Fluxes: …left end = -pR2 …right end = +pR2 …around cylinder = 0 …full flux = 0 184 Lecture 8

17. The Electric Potential 184 Lecture 8

18. 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. 184 Lecture 8

19. 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, We, when the system changes its configuration from some initial configuration to some final configuration. (‡)Conservative force: The work is path-independent. 184 Lecture 8

20. 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 > 0 184 Lecture 8

21. 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 theforce is F = qE … • … so the work done by the electric field on the charge is Note: q = angle between E and d. 184 Lecture 8

22. Constant Electric Field - Special Cases • Displacement is in the samedirection 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. 184 Lecture 8

23. 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. “potential energy per unit charge of a test particle” Units: [V] = J / C, by definition, volt 184 Lecture 8