PHY2054 Spring 2007
1
Prof. Eugene Dunnam
Prof. Paul Avery
Feb. 6, 2007
Exam 1 Solutions
1. A charge
Q
1
and a charge
Q
2
= 1000
Q
1
are located 25 cm apart. The ratio of the electrostatic
force on
Q
1
to that on
Q
2
is:
(1) none of these
(2) 1000
(3) 0.0032
(4) 0.001
(5) 316
The ratio of forces is 1.0, as can be seen from Coulomb’s law or Newton’s third law.
2. The charge on each of two tiny spheres is doubled while their separation distance is tripled.
The ratio of the new electrostatic force to the old force is:
(1) 4/9
(2) 2/3
(3) 9/4
(4) 4/3
(5) 1
Doubling the charge makes the numerator 4 times larger, while tripling the distance makes the
denominator 9 times larger. Together they yield a force that is 4/9 of the original force.
3. An electron circles a carbon nucleus (6 protons, 6 neutrons) at a distance of 0.0353 nanome-
ters. What is its velocity in km/s?
(1) 6560
(2) 9280
(3) 2680
(4) 153
(5) 1950
Here we balance centripetal force with the coulomb force, or
(
)
2
2
/
6
/
mv
r
k
e e r
=
, where 6e is
the charge of the carbon nucleus, r is the orbital radius and m is the mass of the electron. Solv-
ing for v gives
2
6
6
/
6.56
10
/
6560
v
ke
mr
m s
=
=
×
=
km/s.

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