Download Multiple Choice:

Multiple Choice:
1)
2)
3)
4)
5)
6)
7)
8)
C
D
C
E
E
C
A
D
Part B
1. No, the speed does not change. The force is perpendicular to the velocity and so only
the direction of the velocity changes, not the magnitude. The force is centripetal.
2. While it is true that the acceleration due to gravity is dependent upon distance from
the center of the earth, and so does theoretically change when the height of the object
changes, the earth is so large compared to the additional height that we can throw an
object that this additional distance makes absolutely no detectable difference in the
acceleration due to gravity.
3. The forces are similar in that both are inversely dependent upon the distance between
the two objects squared. That is, they are “r-squared” forces. They both also act along a
line connecting the centers of the objects. Both forces have the same mathematic form.
Both are conservative forces and so have a potential energy associated with them. Both
depend on (quantity 1) time (same quantity for object 2).
The forces are different in that the electrostatic force can be either repulsive or attractive,
while the gravitational force is always attractive. Any two objects interact
gravitationally, while only objects with charge interact electrostatically. The constants
are different for the two forces. Gravitational force depends on the masses of the objects,
electrostatic forces depend on the charges of the objects.
Part C
#1
a) U=qV=(-1.6x10-19)(10)=-1.6x10-18 J
b) U=qV=(-1.6x10-19)(20-10)=-1.6x10-18 J
c)-U=KE so
+1.6x10-18=1/2mv2
v = 1.87 x106 m/s
d) FB=qvxB=qvB=(1.6x10-19)(v)(5x10-4)
=1.50 x 10-16 N
#2
a)



F2net  F1,2  F2,3
q q
qq
 k 1 2  k 3 2
r12 2
r32 2
q
 k 2 (q1  q3 )
16
(8.99x109 )(2 x10 6 )

(3x10 6 )
16
 3.37x103 N
B) Left- no partial credit (+3 points)
C)




E y ,net  E1, y  E2, y  E3, y
q
q
q
 k 1 Sin45o  k 2  k 3 Sin45o
r12
r22
r32
k
Cos45
q
(q1  q3 )  k 2
32
16
(8.99x109 )(.707)(7 x106 ) (8.99x109 )(2 x106 )


32
16
 266 N/C
D)




E x, net  E1, x  E2, x  E3, x
q
q
 k 1 Cos45o  0  k 3 Cos45o
r12
r32
k
Cos45
(q1  q3 )
32
(8.99x109 )(.707)(3x106 )

32
 596 N/C
E) Up and to the right