Download PES 1120 Spring 2014, Spendier Lecture 5/Page 1 Lecture today

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 << z
2
3
4 0 z 1  d 2 /(4 z 2 ) 
4 0 z 3
for x << 1:
1  x n  1  nx 
n( n  1) 2
x  ....
2!
Dipole moment:

What I did not tell you is that the dipole moment p is actually a vector. The dipole
moment of a dipole is a vector that points in the direction from the negative charge to
the positive charge, and whose magnitude is given by the product of the charge times the
separation distance.

p  qd
unit [C m]
For a water molecule:
What happens when we put a dipole in a E-field?
a) A dipole in an electric field will feel a torque:

When an electric dipole is placed in a region where there is an external electric field, E ,
electrostatic forces act on the charged ends of the dipole. If the electric field is uniform,


those forces act in opposite directions and with the same magnitude F  qE .
PES 1120 Spring 2014, Spendier
Lecture 5/Page 4
Although the net force on the dipole from the field is zero, and the center of mass of the

dipole does not move, the forces on the charged ends do produce a net torque  on the
dipole about its center of mass.
The center of mass lies on the line connecting the charged ends, at some distance x from
one end and a distance d-x from the other end.
The net torque is:
b) It takes work to rotate a dipole in an electric field:
When a dipole rotates from an initial orientation θi to another orientation θf, the work W
done on the dipole by the electric field is
W  U   U f  U i 
So, in an electric field, dipoles acquire potential energy U. In the book you can find the
derivation that the potential energy of the dipole is given as
 
U   pE cos f   p  E
(dot product)
PES 1120 Spring 2014, Spendier
Lecture 5/Page 5
Example 2:
An electric dipole consisting of charges of magnitude 1.50 nC separated by 6.20 μm is in
an electric field of strength 1100 N/C. What are
(a) the magnitude of the electric dipole moment and
(b) the difference between the potential energies for dipole orientations parallel and

antiparallel to E ?