PS 250: Lecture 4 Electric Fields, Lines,

and Dipoles

J. B. Snively August 31st, 2015

Today’s Class

Constant Electric Fields Electric Field Lines Demo Electric Dipoles Summary

Electric Fields From rings to disks...

Points Rings DiskZ Z

P

y

xxdEx

r

dr

R

Ring Charge = dQ Disk Charge = Q

Electric Fields From disks to an infinite sheet...

Ex

=⇥

2�o

"1� 1p

(R2/x2) + 1

#

Ex

=⇥

2�o

Electric Field due to Disk:

Let R >> x (i.e., make disk very large!) ...

For an infinite (or very large) sheet of surface charge, the Electric field becomes:

d=1.0cm

+

-100V

What if we have two large, charged, conducting plates (a “Capacitor”)...

E=100V/0.01m =10,000 V/m =10,000 N/C

(Constant E-field Inside)

Today’s Class

Constant Electric Fields Electric Field Lines Demo Electric Dipoles Summary

Today’s Class

Constant Electric Fields Electric Field Lines Demo Electric Dipoles Summary

The Title Page Graphic... Grass seeds in presence of electric field!

From Young and Freedman, Copyright (c) 2008 Pearson Education, Inc.

Electric Dipole

-q

+q

d

Constant E-Field

�p

Electric Dipole ExamplesThe grass seeds in the Title Page Graphic: Exhibit dipole structure as a result of polarization, then align themselves along field lines as a consequence of the resulting torque.

Water molecule:

From Young and Freedman, Copyright (c) 2008 Pearson Education, Inc.

Electric Dipoles

-q

+q

d

�d sin�

Constant E-Field

�p

�F+ = q �E

�F� = �q �E

�p = q�dDipole Moment:⇤� = ⇤p⇥ ⇤E = pE sin⇥Torque on Dipole:

�<90o �<180o90o<0o<

Clockwise / Into-the-board Torques

~⌧~⌧

� >-180o �0o< >-90o-90o>

Counter-Clockwise / Out-of-the-board Torques

~⌧~⌧

Summary / Next Class:

Read Textbook Sections: 22.1–22.2 (feel free to read ahead - I aim to cover more!)

Remember Mastering Physics for Monday; Homework for Wednesday.

Prepare to discuss!

Have Time? ... Electric FluxQuantification of field lines that pass normal (perpendicular) to a surface.

Units: [Newton meter2 / Coulomb]

Basic concept for Gauss’s Law – Which will allow simplified determination of electric fields!

�E = EA

For Constant Electric Field “E”:

Area A

From Young and Freedman, Copyright (c) 2008 Pearson Education, Inc.

From Young and Freedman, Copyright (c) 2008 Pearson Education, Inc.

In the last case, doubling the box dimensions lead to an increase in surface area by factor of 4.

Doubling the dimensions, however, reduces field at the box surface by factor of 4!

Net effect: Unchanged Flux!

�E = EA

� = 0�

E and A Parallel:

�E = 0

� = 90�E and A Normal:

Flux through a Surface:

�

�E = ⇥E · ⇥A = EA cos�

General Case E and A at some angle:

�A�E