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Physics 112
Exam I
Summer 2016
1. Two charges are separated by a distance d and exert mutual attractive forces of F on each
other. If the charges are separated by a distance of d/2, what are the new mutual forces?
A)
B)
C)
D)
E)
F/4
F/2
2F
4F
None of these
Solution:
Q1Q2
F k
;
d2
F2  k
Q1Q2
d 2
2
 4k
Q1Q2
 4F
d2
2. Four positive particles of equal charge  12.0 C, are located at the corners of a square of
side 15.0 cm. Calculate the magnitude of the force on the negative charge  13.0 C, placed
at the center of the square.
A)
B)
C)
D)
E)
83.7 N/C
7.6 N/C
3.8 N/C
0.69 N/C
0
Solution:
    
Becouse of symmetry: E  F1  F2  F3  F4  0 .
3. The electric field of two charges is shown below. What statement is correct?
A)
B)
C)
D)
E)
Q1 > Q2 > 0
Q2 > Q1 > 0
Q1 > 0, Q2 < 0
Q1 < 0, Q2 < 0
Q1 < 0, Q2 > 0
Q1
Q2
Solution:
Electric field lines start at positive charge and end on negative charge.
Page 1 of 8
Physics 112
Exam I
Summer 2016
4. Calculate the electric field at the center of a square 30 cm on a side if one corner is occupied
by a charge +25μC and the other three are occupied by charges -15μC.
B)
C)
D)
E)
Q2  15.0  C
Q2
A) 8.0  106 N / C
E1
6.0  106 N / C
1.0  10 6 N / C
8.0  10 4 N / C
1.0  10 4 N / C
d
Q1  25.0  C
E2
Q2
Solution:
Q
Q Q
Q1
 k 2 2  2k 1 2 2 
2
d /2
d /2
d
6
25  15  10 C  8.0  106 N / C
 2  9  109 Nm2 / C 2
2
30  10 2 m
E  E1  E2  k


5. A solid metal sphere of radius 2.0 cm carries a total charge of -0.1 μC. What is the
magnitude of the electric field at a distance from the sphere center of 1.0cm?
A)
B)
C)
D)
E)
0.9 x106 N/C
9.0 x106 N/C
-0.9 x106 N/C
-9.0 x106 N/C
0
Solution:
Inside metal electrostatic field is zero.
6. It takes 50 J of energy to move 10 C of charge from point A to point B. What is the potential
difference between points A and B?
A)
B)
C)
D)
E)
500 V
50 V
5.0 V
0.50 V
0.050 V
Solution:
V  PE Q  50J  10C   5.0V
Page 2 of 8
Physics 112
Exam I
Summer 2016
7. A stationary electron is accelerated through a potential difference of 500 V. What is the
velocity of the electron afterward?
A)
B)
C)
D)
E)
1.3 × 106 m/s
2.6 × 106 m/s
1.3 × 107 m/s
2.6 × 107 m/s
1.3 × 108 m/s
Solution:
mv 2
 eV 
2
v
2eV

m


2 1.6  10 19 C 500V  4 7
 10 m / s  1.310 7 m / s
31
3
9.11  10 kg
8. A parallel-plate capacitor has plates of area 0.20 m2 separated by a distance of 1.0 mm.
What is the strength of the electric field between these plates when this capacitor is
connected to a 6.0-V battery?
A)
B)
C)
D)
E)
1200 N/C
3000 N/C
6000 N/C
8000 N/C
9000 N/C
Solution:
V   Ed ||  E  V / d||  E  6.0V  1.0  10 3 m  6000V / m  6000 N / C
9. A parallel-plate capacitor consists of plates of area 1.5 × 10-4 m2 and separated by 1.0 mm.
The capacitor is connected to a 12-V battery. What is the charge on the plates?
A)
B)
C)
D)
E)
1.6 × 10-11 C
3.2 × 10-11 C
1.6 × 10-14 C
3.2 × 10-14 C
1.6 × 10-19 C
Solution:
C  0
A
A
1.5  10 4 m 2
12V   1.6 1011
; Q  CV   0 V ; Q  CV  8.85  10 12 C 2 / N  m 2
3
d
d
1.0  10 m


Page 3 of 8
Physics 112
Exam I
Summer 2016
10. Two capacitors 10 μF and 20 μF are connected in parallel, and this pair is then connected in
series with a 30-μF capacitor. What is the equivalent capacitance of this arrangement?
A)
B)
C)
D)
E)
10 μF
15 μF
25 μF
60 μF
66 μF
Solution:
C1|  10 F
C 2|  20 F
C 3  30 F
C12|  C1  C 2|  10F  20F  30F
C123  30F / 2  15F
11. The length of a wire is doubled and the radius is doubled. By what factor does the resistance
change?
A)
B)
C)
D)
E)
Increase 4 times
Increase 2 times
Decrease 2 times
Decrease 4 times
Remains the same
Solution:
L
L
2L
R    R   2 ; R2  
 12 R
A
r
 2r 2
12. Three resistors of 12 Ω, 12 Ω, and 6.0 Ω are connected in parallel. A 12-V battery is
connected to the combination. What is the current through the 6.0-Ω resistor?
A)
B)
C)
D)
E)
1.0 A
2.0 A
3.0 A
4.0 A
6.0 A
Solution:
V 12V
I 
 2.0 A
R 6.0
Page 4 of 8
Physics 112
Exam I
Summer 2016
13. Three resistors of 12, 12, and 6.0 Ω are connected in series. A 12-V battery is connected to
the combination. What is the current through the battery? Neglect internal resistance of the
battery.
A)
B)
C)
D)
E)
0.10 A
0.20 A
0.30 A
0.40 A
0.60 A
Solution:
V
12V
I 
 0.4 A
R 12  12  6.0
14. If the current flowing through a circuit of constant resistance is doubled, the power dissipated
by that circuit will
A)
B)
C)
D)
E)
quadruple
double
decrease to one half
decrease to one fourth
remain the same
Solution:
2
P  I 2 R  P2  2 I1  R  4 I12 R  4 P1
15. A 500-W device is connected to a 120-V ac power source. What is the peak voltage across
this device?
A)
B)
C)
D)
E)
4.2 V
5.9 V
120 V
170 V
240 V
Solution:
I 0  2I rms  2 120V  170V
Page 5 of 8
Physics 112
Exam I
Summer 2016
16. You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the
normal way, you devise a circuit that places them in series across normal household voltage.
Which statement is correct?
A)
B)
C)
D)
E)
Both bulbs glow at the same reduced brightness
Both bulbs glow at the same increased brightness
Both bulbs glow at the same brightness as connected in parallel
The 100-W bulb glows brighter than the 50-W bulb
The 50-W bulb glows brighter than the 100-W bulb
Solution:
Use the following equations: P  V 2 R  I 2 R
17. An electric bulb with resistance of 48 Ω is connected to the battery with emf of 12 V and
internal resistance 2 Ω. Find current across the bulb.
A)
B)
C)
D)
E)
6.0 A
4.0 A
0.50 A
0.25 A
0.24 A
Solution:

12V
I

 0.24 A
R  r 48  2
18. What statement is wrong?
A) At any junction point in an electric circuit, the sum of all currents entering the junction must
equal to the sum of all currents leaving the junction.
B) The sum of the changes in potential around any closed path of a circuit must be zero.
C) The ammeter should be inserted in series with the resistance, to measure the current in this
resistance.
D) The voltmeter should be connected in series with the resistance, to measure potential across
the resistor.
E) Resistance of an ideal ammeter is zero.
Solution:
The voltmeter should be connected in parallel with the resistance, to measure potential across the
resistor.
Page 6 of 8
Physics 112
Exam I
Summer 2016
19. Which of the equations here is valid for the circuit shown? (Current I is in amperes)
A)
B)
C)
D)
E)
2 - I1 - 2I2 = 0
2 - 2I2 - 4I3 = 0
4 - I1 + 4I3 = 0
2 - I1 - 2I3 = 0
6 - I1 - 2I2 = 0
Solution:
Use second Kirchhoff’s rule for the left loop.
  I1R1  I 2 R2
20. A 2.0-μF capacitor is charged to 12 V and then discharged through a 4.0 × 106 Ω resistor.
What will be the voltage across the capacitor after 11 seconds?
A)
B)
C)
D)
E)
1.0 V
3.0 V
5.0 V
7.0 V
9.0 V
Solution:
VC  V0 e t / RC VC  12V  exp  11s  4.0  106  2.0  10 6 F   3.0


Page 7 of 8
Physics 112
Exam I
Summer 2016
Answer Sheet
1
D) 4F
11
C) Decrease 2 times
2
E) zero
12
B) 2.0 A
3
E) Q1 < 0, Q2 > 0
13
D) 0.40 A
4
A)
14
A) quadruple
8.0  10 N / C
6
5
E) 0
15
D) 170 V
6
C) 5.0 V
16
E) The 50-W bulb glows
more brightly than the
100-W bulb
17
E) 0.24 A
7
C) 1.3 × 107 m/s
8
C) 6000 N/C
9
A) 1.6 × 10-11 C
18
D) The voltmeter should
be connected in series
with the resistance, to
measure potential across
the resistor.
(wrong statement)
19
B) 2 - 2I2 - 4I3 = 0
10
B) 15 μF
20
B) 3 V
Page 8 of 8