PES 1120 Spring 2014, Spendier

Lecture 5/Page 1

Lecture today: Chapter 22 1) Point particles and electric dipoles moving in an electric field 2) Electric Field Due to a charge distribution 3) HW0 returned, HW1 due this Friday! Extended OFFICE HOURS! (doodle poll) M 2-5pm Tu 2-5pm Wed 10-12am

 Last week we introduced the electric field, E . I defined it as an invisible aura  surrounding every charge. We saw that E is a vector field (we assigned a vector to every point in space) The vector tell us in which direction and with what magnitude a test charge q feels an electrostatic force   F E q The electric field tells us that a positive test charge will always move away from a positive source charge at the origin. A point charge in an Electric Field: What happens to a charged particle when it is in an electric field set up by other stationary or slowly moving charges? What happens is that an electrostatic force acts on the particle, as given by   F  qE

The electrostatic force acting on a charged particle located in an external electric field (electric fieldthat other charges have produced at the location of the particle) has the direction of E if the charge q of the particle is positive and has the opposite direction if q is negative. How was the magnitude of e measured? Millikan's oil drop experiment: The oil drop experiment was an experiment performed by Robert A. Millikan and Harvey Fletcher in 1909 to measure the elementary electric charge (the charge of the electron e). In 1923, Millikan won the Nobel Prize in physics in part because of this experiment.

- slightly negatively charged drop enter a parallel plate capacitor - two oppositely-charge conducting sheets separated by a small distance. [We’ll show in chapter 23 that the field

PES 1120 Spring 2014, Spendier

Lecture 5/Page 2

is uniform inside a parallel plate capacitor. It has same magnitude and direction at every point.] - Use microscope to measure the size of the oil drop. - We know the density of oil and can then calculate the drop's mass. - Gravity acts to accelerate drop down - Interaction between charge -q and the electric field acts to accelerate drop upwards - When forces up and down balance, we can find -q. Example 1: In Millikan’s experiment, an oil drop of radius 1.64 μm and density 0.851 g/cm3 is suspended in the chamber when a downward electric field of 1.92 x 105 N/C is applied. Find the charge on the drop, in terms of e, at equilibrium.

PES 1120 Spring 2014, Spendier

Lecture 5/Page 3

A dipole in an electric field:  Last time we calculated the magnitude and direction of the electric field E at an arbitrary point P along the dipole axis, at distance z from the dipole's midpoint.

Edipole 

1 qd 1 p  3 4 0 z 4 0 z 3

,where p is the dipole moment!

used a binomial expansion to obtain this expression 1 p 1 1 p Edipole   , for d