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Exam 1, Phys208Honors
October 13, 2003
Name: ___SOLUTION___________
Please give me a word or number to post your grade____________________
1. (30 pts) Please choose T (True) or F (False) or proper answers to the following
questions.
1) A neutral object carries no charges.
T
F
2) Gauss’ law is only valid in problems with symmetry.
T
F
3) The potential on a conducting object is the same everywhere, i.e. the conducting
object is an equal potential object.
T
F
4) The capacitance of a capacitor depends on the geometry of the capacitor and is
independent of the applied voltage.
T
F
5) If electric field is zero everywhere on an object, the potential must be zero.
T
F
6) A charge neutral object can produce an electric field when it is polarized. T F
7) Superposition principle is valid in calculating electric static forces, electric fields, and
potentials.
T
F
8) For a positive charge in an electric field, it always moves from high potential point to
low potential points. T F
9) What is the electric field halfway to the center of a conducting sphere charged to a
potential of 15 V? What is the potential halfway to the center?
There is no field inside a conductor.
15V. Conductors are equipotential at all points.
10) You are sitting inside an uncharged hollow spherical shell. Suddenly someone dumps
a billion Coulombs of charges on the shell, distributed uniformly. What happen to the
electric field at your location?
Nothing, the uniform charge does not create any field inside the shell.
2. (35pts) The spherical capacitor is often considered because it is finite-sized but
without the complication of a fringing field, often existed in the parallel-plate capacitor .
Find an expression for its capacitance.
a. (10pts) Use Gauss's law to determine the electric field at a
point between the two conducting shells when oppositely
charged. Be sure to show the Gaussian surface used and
indicate the direction of the resulting field.
We select a sphere such that b < r < c.
Note that we use a sphere that is concentric
with the capacitor and the enclosed charge is
only on the inner conducting sphere.
 E dA 

qenc

E 4 r 2 
0
E
qenc
0
qenc
4 0 r 2
b. (15 pts) Evaluate the capacitance.
q=CV
b
V    E  dr
a
b
V  
a
q
4 0 r 2
 ds
ds   dr
b
q 1
V 
 
4 0  r  a
q 1 1
q ba 
V
V
  


4 0  a b 
4 0  ab 
4 0 ab
q
q
C
C

q ba 
V
ba


4 0  ab 
c. (10 Pts) A simple check of your true understanding of conductors and Gauss's law:
If identical charges are deposited on both conductors, say +1 microcoulombs, how are the
charges distributed among the four surfaces?
q
b
dr
V

4 0 a r 2
#1, inner surface
#1, outer surface
#2, inner surface
#2, outer surface
- none (No field inside a conductor!)
+1 microcoulombs
-1 microcoulombs (No field inside conductor #2!)
+2 microcoulombs
3. (35 pts)
An electron (circle) is accelerated from still by a potential difference of
100V, which does not affect the electron after it escapes from the hole on the second
plate. The electron enters a uniform field of 10V/m generated by another two parallel
plates. (Assuming E becomes zero outside and no edge effect). The separation of two
plates is 2.0 cm. Calculate (a) (10pts) the velocity of the electron before it enters E field;
(b) (10 pts) the acceleration of the electron inside E, (c) (10 Pts) the position of the
electron (with respect to its original path) after it exits E field, and (d) (5 pts)the final
velocity of the electron. (e=1.6 x 10-19 C, me=9.1 x 10-31 kg).
0
100V
E
L=5 cm
a)
1 2
mv  qVacc
2
U  qVacc


2 1.6 10 19 C 100V 
2qVacc
v

 5.9 106 m / s
m
9.1 1031 kg
The electron is moving to the right.
b)
E  10 V / m
F  qE  ma


1.6 10 19 C 10 V / m 
qE
a

 1.8 1012 m / s 2
m
9.1 10 31 kg
The acceleration is upward as q is negative.
c)
2qVacc
m
vx 
L  vx t
t
 ~ 8.5 10 s 
L
vx
9
2
1
1 L
qE
L2 m
EL2
Vertical " lift "  at 2  a   


2
2  vx 
2m 2qVacc 4Vacc
10V / m .05 m 
Vertical " lift " 
4 100 V 
2
 6.3 105 m  63  m
The electron moves up as it has negative charge and moves in the
opposite directions of E.
d)
v y  at 
qE
m
q
L
 EL
m
2qVacc
2mVacc
v y  10V / m .05 m 
1.6 10

2 9.1 10
31
19
C


kg 100 V 
 1.4 104 m / s
v  5.9 106 i  1.4 104 j
v 
 5.9 10   1.4 10 
6
2
4
2
 5.9 106