Physics 106 Homework #2
Submitting Answers
Remember to be careful about units.
Implicit multiplication is not understood. Type 3*x, not 3x
Defined Constants
The following constants can be used by name throughout this assignment:
ke = 8.99×10
9
Nm
2
/C
2
qe = 1.6×10
–19
C
mp = 1.67×10
–27
kg
me = 9.11×10
–31
kg
eps0 = 8.85×10
–12
C
2
/(Nm
2
)
g = 9.80 m/s
2
.
Problem 2.1
Purpose: To calculate the electric field of a
series of
point
charges.
Problem
with data for try #1
:
Consider the picture to the right. The charge of
the middle charge is 2.49 μC. Determine the
electric field strength at a point 1.00 cm to the
left of the middle charge.
Additional input values q2 (C): 2.23, 2.62
Defined symbols:
q2
charge of the middle charge in μC
q1
charge of the left chargein μC
q3
charge of the right charge in μC
d12
distance between
q1
and
q2
in m
d23
distance between
q2
and
q3
in m
The answer is
E
electric field strength at a
point 1.00 cm to the left of the middle charge in
N/C
Hints:
You will have to calculate the electric field of each of the three charges at the point 1.00 cm left
of the middle charge and add the results together. You add the electric field just like you added the
forces in problem 4.
It can be helpful to invent a small positively charged ball and place it at the point where you
wish to find the field. The only thing that's different between finding the force on this ball and the
electric field at this point is that the electric field doesn't include the charge of the ball.
Range of answers: –1.15×10
8
N/C to +2.00×10
7
N/C
.
Problem 2.2
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Purpose: This is a continuation of Problem 1.
The input values are the same as for Problem 1.
Problem
with data for try #1
:
Consider the picture below. The charge of the
middle charge is 2.49 μC. If a charge of +2.00
μC is placed at a point 1.00 cm to the left of the
middle charge, what are the magnitude and
direction of the force on it? (Let a force to the
right be positive and a force to the left be
negative.)
Additional input values q2 (C): 2.23, 2.62
Defined symbols:
q
charge of the test charge in μC
q2
charge of the middle charge in μC
q1
charge of the left chargein μC
q3
charge of the right charge in μC
d12
distance between
q1
and
q2
in m
d23
distance between
q2
and
q3
in m
The answer is
F1
magnitude of the force on
the +2.00 μC charge in N
Hints:
Since we know the electric field from the previous problem (Be sure to use the E from the
correct input!), the force is just
F=qE
. Because you know the electric field strength at that point,
you can recalculate the force on any charge placed there without having to add all the vectors
again.
Range of answers –230 N to +40.0 N
.
Problem 2.3
Purpose: This is a continuation of Problem 1.
The input values are the same as for Problem 1.
Problem
with data for try #1
:
Consider the picture below. The charge of the
middle charge is 2.49 C. If a charge of –2.00 μC
is placed at a point 1.00 cm to the left of the
middle charge, what are the magnitude and
direction of the force on it? (Let a force to the
right be positive and a force to the left be
negative.)
Additional input values q2 (C): 2.23, 2.62
Defined symbols:
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 Fall '09
 Electric Potential, Energy, Potential Energy, Work, Electric charge, N/C, Additional input values

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