EXAM 2 April 15, 1988 PS 208
If you need a constant not given here or do not understand
a problem, please ask me about it. No notes, books, etc. may
be used during this exam.
2 -19
g = 9.8 m/sec e = 1.6 x 10 C
9 2 2
1/(4 pi epsilon ) = 9.1 x 10 N m / C
0
-7
mu / (4 pi) = 10 T m/ A
0
-31
The mass of the electron is 9.11 x 10 kg.
1. Give a good argument for remembering the following:
(a). The electric potential caused by a point charge.
(b). The capacitance of a parallel plate capacitor.
(c). The resistance of a wire of given resistivity and size.
2. (From homework,slightly modified) Determine the current
through each of the resistors in the figure below.
_____________________________________
| | |
__|__ | __|__
_________ V R __________ V
| | |
| | |
2R ______|_______ 2R
| _____ 2V |
| | |
|_______ R ________|_________ R ______|
The current in the center branch is 3V/(4R). [Mark the printed
diagram or a redrawn diagram on your paper to show the currents
you are defining. This and the writing of a complete set of
equations (assuming that you do *not* know the middle current) will
get most of the credit for the problem. If you get lost in
algebra, move on to other problems and return if you have time.
Justify your equations.]
3. Prove carefully that the path of a point charge moving in
a constant magnetic field with its initial velocity perpendicular
to the field is a circle, and find an expression for the radius of
the circle in terms of the charge, its mass, its initial speed,
its mass, and the strength of the field.
4. Derive with reasonable care the expression for the
capacitance of a parallel-plate capacitor, whose plates are a
constant distance d from each other, and each of which has the
same shape and a total area of A . Assume d << the length and
width of the plates. The management suggests getting the same
result as that in problem 1 (b). Some extra credit will be given
for accurately including the effects of a dielectric between the
plates.