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Physics 102 — Midterm Exam 1 — Section D200
2011 October 11
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Physics 102 — Midterm Exam 1 — Section D200
2011 October 11
There are 10 multiple choice questions. Select the correct answer for each one and mark it on the
bubble form on the cover sheet. Each question has only one correct answer. (2 marks each)
1. For questions 1 and 2:
Five small identical metal balls are hung from
insulating silk threads and are handled only by
the threads. They are not allowed to touch one
another during the following experiment. It has
previously been found that none of the balls is affected by a magnet. The balls are brought near
each other, and the following observations are
recorded:
• Metal balls II and V exert no force on each
other.
• Metal balls I and III repel each other.
• All other pairs of metal balls attract one another. For example, ball I attracts balls II,
IV, and V,
(a) kq1 x1 /(x12 + y2a )3/2 —
correct —
(b) kq1 x1 /(x1 + ya )3
(0, ya)
(c) kq1 /x12
(d) kq1 x1 /(x12 + y2a )
(x1,0)
(e) None of the above.
4. A conservative force field is one in which the
work needed to move an object from one point to
another in the field does not depend on the path
taken between the two points. In which case(s) is
it true that the electric field is conservative?
(a) The field of an infinite uniform plane of
charge.
(b) The field of an infinite uniform distribution
of charge on a straight line
(c) The field of any spherically symmetric distribution of charge
(d) None of the above
I
II
III
IV
V
(e) All of the above — correct —
In which pairs of balls do both balls carry charge
of the same sign?
(a) I and II
5. You wish to measure the electric work done on
a heater by a battery, for which purpose you select an ammeter (A) and a voltmeter (V). Which
of the circuits shown below would you use?
(b) I and III — correct —
(c) I and IV
(d) II and IV
(e) II and V
2. In which of the following pairs of balls do the
two balls carry charge of the opposite sign?
(a) III and V
(b) I and V
(c) II and III
(d) I and IV — correct —
(e) IV and V
3. A positive charge q1 is on the x axis at (x1 , 0).
Which formula represents the x component of the
electric field at a point on the y axis at position
(0, ya )? (k is Coulomb’s constant.)
2
6. In the circuit below you don’t know the strength
of the battery or the value of any of the four resistances. Select the expression that will be equal
to the voltage of the battery in the circuit. If 2 or
more are true, or none are true, choose (e).
Physics 102 — Midterm Exam 1 — Section D200
2011 October 11
(a) VA + VB
tion will an electron (q = −1.6 × 10−19 C) experience a force directed toward the top of the
screen?
(b) VA
(c) VA + VB + VC + VD
B
(d) VA +VB +VD — correct
A
—
(e) No single one of the
above.
C
D
V
7. In this diagram a circuit fragment is built from
ten identical 12 Ω resistors. What is the resistance between the points labeled A and B?
Choose A, B, C or
D. If none of them
are right, choose
E. If more than
one are right then
choose E too.
Correct answer is C. (Look at the equipotential
values. They rise to the peak so it’s a positive
charge concentration there, which attracts the
negative charge.)
(a) 15 Ω
(b) 25 Ω — correct —
(c) 48 Ω
(d) 72 Ω
(e) 120 Ω
8. Two small spheres with equal masses m are
suspended by strings of length 1 metre as
shown. Both spheres have a net charge on
them. Each string makes an angle of 30o with
the vertical in equilibrium. In equilibrium,
which one of the following statements is true?
10. Three point charges are arranged as shown in the
figure. The charges are now taken away to infinity, one by one, in the following order: first
A, then B, and finally C. Calculate the work
done by you in removing the charges one by
one. WA is the work done in removing charge
A, WB is the work done in removing charge B,
and WC is the work done in removing charge C.
Grid Spacing is one metre
(a) Q1 and Q2 must have the same magnitudes.
(b) Charges Q1 and Q2 must have opposite
signs. — correct —
(c) Q1 must be a positive charge.
(d) Both (a) and (b)
(e) None of the above
9. The figure shows the equipotential lines of the
electric potential that result from a charge distribution in the plane of the paper. At which loca-
3
(a) WA = 0 J, WB = +1.5 × 10−3 J, WC = 0 J —
correct —
(b) WA = 0 J, WB = −1.5 × 10−3 J, WC = 0 J
(c) WA = −3.6 × 10−3 J, WB
10−3 J, WC = 0 J
= +1.5 ×
(d) WA = −3.6 × 10−3 J, WB
10−3 J, WC = +3.6 × 10−3 J
= +1.5 ×
(e) WA = +3.6 × 10−3 J, WB
10−3 J, WC = −3.6 × 10−3 J
= −1.5 ×
Physics 102 — Midterm Exam 1 — Section D200
2011 October 11
There are two written problems. Show all your work and explain your reasoning to get full credit.
11. Three charges are arranged as shown in the figure. [10 pts]
(a) Calculate the electric field due to charge A at the point X.
EC
EB
EA
ETot
Grid Spacing is one metre
Electric field of A is given by
E A = kc
qA
rA2
where kc = 9 × 109 N-m2 /C. rA = 2 m from the graph. Thus E A = 4500 N/C, down.
(b) Draw vectors on the figure showing the electric fields due to A, B and C separately, approximately
to scale. Show the total electric field vector as well. Label as needed.
(c) Calculate the total electric field at point x due to the three charges A, B and C. Express as x and y
components or in standard vector notation.
E Ax = 0 , E Bx = ECx = −kc qB
(32
3
= −576 N/C
+ 22 )3/2
.
E x tot = 2 × (−576) = −1152 N/C
E Ay = −4500 N/C, , E By = −ECxy
E A tot = −4500 N/C
4
Physics 102 — Midterm Exam 1 — Section D200
2011 October 11
12. Analyse the the circuit shown. [10 pts]
(a) Assume first that E = 0. Find the equivalent resistance of the 3-resistor circuit and calculate the
current I1 . ( When E = 0 the middle battery is like a short.)
I1
100Ω
3V
68Ω
E
I3
150Ω
I2
Note that E is considered replaced by a wire. So the 100-ohm resistor is in series with the parallel
combo of 68-ohm and 150-ohm.
Req = 100 +
68 × 150
= 147Ω
68 + 150
I1 = 3 V/147Ω = 0.0204 A
(b) Now assume E is not zero. Write the Kirchhoff loop equations that will allow solving for the three
currents in terms of E.
Node eqn:
I1 = I2 + I3
For 2 loop equations we have a choice. Since we’re after I1 start with two that contain I1 :
3 − 100I1 − 68I2 + E = 0
3 − 100I1 − 150I3 = 0
(c) Solve the equations for I1 in terms of E.
Get rid of I2 in the first equation: I2 = I1 − I3 :
3 − 168I1 + 68I3 + E = 0
Reduce the second equation by 68/150 so it becomes:
1.36 − 45.3I1 − 68I3 = 0
Now add the equations to get
4.36 − 213.3I1 + E = 0
I1 =
E
4.36 + E
= 0.0204 +
A
213.3
213.3
(d) In your result for the previous part, set E = 0 and compare the result to what you got in part (a).
It is obviously the same: I1 = 0.0204 A.
5