Physics 213
Homework #3
Spring 2008
Read:
Chapter 22, section 22.5;
Chapter 25, intro., sections 25.1, 2, 3, and 6
Chapter 23, sections 23.1 (thru mid of p. 782), 23.2 (thru p. 791, not Ex. 23.3), 23.5 (all)
Handouts:
Conductors in Electrostatic Equilibrium
;
For study:
Chap. 22:
Q's #Q22.10, 11, 12, 14, 15, 16, 17
E's & P's #22.19, 31, 37, 43, 44, 49
Chap. 25:
Q's #Q25.1, 2, 3, 9, 10, 11, 12, 13
E's & P's #25.3, 8, 17, 27
Chap. 23:
Q's #Q23.5, 7, 10, 12, 13, 17, 20
To be prepared for Wednesday-Friday, Feb. 13,
±
, 15, at your 2nd weekly recitation session:
#22.31
[Charge in Metal Hollow]
#22.38
[Line Charge & Hollow Conducting Tube]
#22.42
[Sphere in a Sphere]
#22.51
[Charged Conducting Plate]
#25.61
[E-fields in a Wire]
#1.
[Thunderstorm Electric Fields]
A simple model for the
electric charges in a thunderstorm is shown at the right:
two
spherically symmetric regions of equal and opposite charge +Q =
40 C and -Q = –40 C above one another at heights H = 5.0 km and
2H = 10.0 km above the ground.
To determine the electric field
here, we must remember that the ground is a conductor, so there
will be induced charges on the ground surface.
To account for
these we can use the
Method of Image Charges
discussed in lecture.
The image charge configuration is shown below.
The electric field
due to this charge configuration in the region above the ground
plane is identical to the field above the ground due to the real
charges + the real induced charges on the ground.
(a)
Write an algebraic expression for the electric field E(x) at any
point P just outside the ground as a function of the distance x from
the point on the ground directly under the thunderstorm charges.
What is the direction of this field?
What does this expression for
E(x) simplify to at x = 0?
Does this simplified result make sense?
[HINT: