Midterm 2 Solved | Physics for Life Sciences | PHYSICS 6B, Exams of Physics

Material Type: Exam; Class: Physics for Life Sciences Majors: Waves, Electricity, and Magnetism; Subject: Physics; University: University of California - Los Angeles;

Typology: Exams

2011/2012

Uploaded on 06/29/2012

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Physics 6B 2nd Midterm NAME:______________________
Dr. Nassar (1-hour exam -- 3x5 card allowed)
Complete the following problems.
1. Three point charges q1 = 2 µC, q2 = 1 µC, q3 = 2 µC are located at the corners of an
equilateral triangle (a = 1.0 m), as shown below. Another charge q4 = - 1 µC is placed at
the base of the triangle. Calculate the net electric force (magnitude and direction) on q2.
66
92
13 2
2 10 1 10
9 10 1.8 10
1
FF N
−−
××
==×=×
66
92
42
1 10 1 10
9 10 1.2 10
0.87
FN
−−
× −×
=×=×
-------------------------------------------
0
x
F∑=
14
2 22
2 cos30
2(1.8 10 )(0.86) 1.2 10 1.9 10
y
FF F
N
−−
∑= −=
× −×
-------------------------------------------
22
1.9 10 2 10F NN
−−
=××
in the positive y-direction (or 90o)
2. Find the equivalent resistance of the circuit below and the current through
the resistor indicated below.
Equivalent resistance = 1.42 R (or 17/12 R)
Total Current: I = ε/1.42R = 0.71 (ε/R)
0.71 (ε/R) = I+ Iand I= 3I, so I” = 0.17 (ε/R) and I’ = 0.53(ε/R)
I’
I”
I
F1
F3
F4
I
Parallel
Series
pf2

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Physics 6B – 2nd Midterm NAME:______________________

Dr. Nassar (1-hour exam -- 3x5 card allowed)

Complete the following problems.

  1. Three point charges q 1 = 2 μC, q 2 = 1 μC, q3 = 2 μC are located at the corners of an

equilateral triangle (a = 1.0 m), as shown below. Another charge q 4 = - 1 μC is placed at

the base of the triangle. Calculate the net electric force (magnitude and direction) on q 2.

6 6 9 2 (^1 3 )

F F N

− − −

× ×

= = × = ×

6 6 9 2 (^4 )

F N

− − −

× − ×

= × = ×

Fx = 0

1 4 2 2 2

2 cos 30

F y F F

N

− − −

× − × = ×

2 2

F 1.9 10 N 2 10 N

− −

= ×  × in the positive y-direction (or 90

o

2. Find the equivalent resistance of the circuit below and the current through

the resistor indicated below.

Equivalent resistance = 1.42 R (or 17/12 R)

Total Current: I = ε/1.42R = 0.71 (ε/R)

0.71 (ε/R) = I’ + I” and I’ = 3I”, so I” = 0.17 (ε/R) and I’ = 0.53( ε /R)

I’

I”

I

F 3 F 1

F 4

0.87 m

I

Parallel

Series

3. The figure below shows a circuit whose elements have the following

values:

ε 1 = 2.1 V, ε 2 = 6.3 V,

R 1 = 1.7 Ω, R 2 = 3.5 Ω and i 1 = 0.82 A.

Find i 2, i 3 and the potential difference (voltage) between points a and b :

Va – Vb

i 1 (^) + i 2 (^) = i 3

( From a to b and back to a counterclockwise): − i R 1 1 − ε 1 − i R 1 1 + ε 2 + i R 2 2 = 0

( From a to b and back to a clockwise): + i R 3 1 − ε 2 + i R 3 1 + ε 2 + i R 2 2 = 0

1

2

3

0.82 (given in the problem)

i A

i A

i A

Vba = ε 2 + i R 2 2 = 6.3 + −( 0.4)(3.5) =4.9 V