Instructor: Tomoyuki Nakayama
Monday, June 7, 2010
PHY 2005 Applied Physics 2  Summer C 2010
Exam 1 Solution
________________________________________________________________________________
1. Since the two metal spheres are identical, they have equal amounts of charge after touching. Each
will have (2 – 10 )/2 = 4
μ
C. According to Coulomb’s law, the electric force between them is
proportional to the product of the two charges and inversely proportional to the square of the
separation distance. The initial and final separation distances are the same. Therefore, the magnitude
of the electric force after touching is
F/F
0
= (4
2
)/(2×10) = 4/5
⇒
F = (4/5)F
0
=
4.0 × 10
4
N. The
force is repulsive because they have the same sign of charge.
2. Electric field is directed away from a positive charge and directed toward a negative charge. If you
consider the region between the two particles, electric fields due to individual particles are both
directed to the left. Thus the net field
cannot
be zero. In the region to the left of the origin, the electric
field due to the 8
μ
C charge is always larger than that due to 2
μ
C charge; it is because any point in
this region is closer to the 8
μ
C charge that has a larger amount of charge. Therefore, the net field
cannot be zero in this region either. If there is a position where the net field is zero, then it must be in
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 Summer '08
 Lee
 Applied Physics, Charge, Electric charge, loop rule, Tomoyuki Nakayama

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