physics 111 spring 2006 exam # 1
1/8
PHYSICS 1308 SPRING 2008 EXAM # 1 Monday, July 14, 2008 SOLUTIONS
CONCEPTUAL QUESTIONS 1. Imagine a test charge inside a conductor in electrostatic equilibrium. Now imagine that this this charge is moved very slowly. What is the work performed by the electric forces during this process? Explain. The work done is zero. To understand why, first observe that the electric field inside a conductor in electrostatic equilibrium is zero, thus the net electric force on the test charge is also zero. Hence no work is necessary to move the charge inside the conductor.
2. What can you say about the electric field in a region of space that has the same potential throughout? Explain. −V −V −V , E y= , E z= The electric field is given by E x= . If V is constant throughout, we x
can conclude that the electric field is zero.
y
z
3. Current I enters a resistor R shown in the figure. Is the potential higher at point A or B? Explain. I The potential is higher at point A since the current A B moves from points of higher potential to points with lower potential. 4. One terminal of a car battery is said to be connected to “ground.” Since it is not really connected to the ground, what is meant by this expression? This just simply means that the “ground” terminal gives the reference potential. Thus any other potential is measured relative to this value. Remember that absolute potential has no physical meaning only differences are important.
2/8 5. What is the flux through the hemispherical open surface of radius R shown in the figure. The uniform field has magnitude E. Justify your answer. Observe that the same amount of lines that go through the base of the hemisphere go also thorough is surface. Thus 2 the electric flux is E = E⋅A= R E
E
NUMERICAL QUESTIONS 6. Two charges, one of 2.50 nC and the other of 3.50 nC are placed on the xaxis. One at the origin and the other at x= 6.00 m, as shown in the figure below. The net field at x= 32.75 m is equal to ____mN/C. y (m)
A. 42.23 i B. 23.04 i C. 0 D. 42.23 i E. 23.04 i
3.50 nC
+2.50nC
q0
x (m)
32.75 m
9 The net electric field is equal to: 9×10
6.00m
Nm 2 C
2
2.50nC 2
32.75m
− i
3.50nC 32.75m 6m2
i =0
7. The net force exerted by the two charges located on the opposite sides of the xaxis on the positive charge with position x= 10 mm is ______N . (See figure) y (m)
A. 0.48 B. 0.48 C. 1.58 D. 1.58 E. 0.82
j − j i −i sin 17− j
1nC 10 mm
r
10 C 3 mm 3 mm
10 C
x (m)
3/8 From the graphic above we can conclude that the resulting force is along the yaxis, thus :
2 k q Qsin F F− = − j = 2 r
2⋅9×10
9
Nm −9 −6 0 ⋅1×10 C⋅10×10 C⋅sin 17 2 C 2
0.003m 0.01m
where we have used that =arctan
3 mm
10 mm
2
− j=0.48 N −j
=170
8. A capacitor is connected to a 1.5 V battery. The plates are separated by 0.1 mm. An 250 C charge is placed in the middle of the plates of the capacitor. The magnitude of the force exerted on the charge is _____N. A. 2.75
B. 3.00
C. 3.25
The electric field is given by E=
V x
1.5V
=
E. 3.75
D. 3.50
0.1×10 m the force is : F=qE=1.5N /C×250×10 C=3.75N −3
=1.5×104 N /C . Therefore the magnitude of
−6
9. A 0.20 pF capacitor is needed. The area that the plates must have if they are separated by 2.2 mm air gap is _____cm² A. 0.5
B. 1.0
C. 1.5
D. 2.0
The capacitance for a parallel plate capacitor is equal to C= required area we obtain: A=
Cd 0
=
0.20×10−12 F⋅2.2×10−3 m 9×10
−12
2
C / Nm
2
E. 2.5 0 A d
. Therefore, solving for the
=5×10−6 m 2
4/8 10. The electric flux through the closed surfaced shown in the figure is _____kNm²/C A. 888 B. 888 C. 555 D. 111 E. 111
3 mC 7 mC
2 mC
Q
−3 C2 C
encl 3 2 =−111×10 Nm /C By Gauss' Law we can write E = = −12 2 2 0 9×10 C / Nm
11. A large, flat, horizontal sheet of charge has a charge per unit area of 9.00 C/m². The electric field just above the middle of the sheet is ____MN/C A. 0.5 E=
2 0
=
B. 1
C. 1.5
9.00×10−6 C / m 2 2⋅9×10
−12
2
2
C / Nm
D. 2
E. 2.5
=0.5×106 N /C
12. A sphere of radius r = 5 m has a charge of 5 mC distributed uniformly over its surface. The magnitude of the electric field at a distance of 10 m from the surface of the sphere is _____kN/C A. 50
B. 100
D. 200
C. 150
E. 250
We know that by Gauss' Law the field produced by a charged sphere is equal to the one produced by a point charge where the distance to other charges must include the radius of the sphere, hence: ∣E∣=k
∣Q sphere∣ r sphere r sphere− point
=9×10 2
9
N⋅m 2 5×10−3 C C
2
15 m
2
=200000 N /C
5/8 13. Two charged particles Q1 = +5.0 C and Q2 = 3.00 C are separated by 35.0 cm. The electric potential energy of the pair is _____J A. 0.52
B. 0.38
The potential energy is equal to:
C. 0
U =Q1 V = k
Q1 Q2 r
D. 1.23 9
=9×10
E. 2.50
Nm 2 5×10−6 C −3×10−6 C C2
0.35 m
=−0.38 J
14. The potential in a region between x = 0 and x = 6.00 m is V =10.0V −7.00 magnitude and direction of the electric field at x= 3m is ____V/m A. 11 −i
B. 11 i
C. 0
D. 7 −i
E. 7 i
−dV −d V E= = 10.0V −7.00 x =7.00V / m dx dx m
Refer to the following circuit for questions 15 and 16. All capacitances are in mF.
15 V
¼ ¾
½
15. The equivalent capacitance of the circuit above is _____mF A. 13/6
B. 11/12
C. 19/12
D. 27/4
E. 6/13
D. 55/4
E. 68/4
16. The charge in the ¾ mF is _____mC A. 10/4
B. 25/4
C. 45/4
V x . The m
6/8 Notice that the voltage across the ¾ mF is the same voltage of the battery, therefore 3 45 Q=V C= mF 15 V = mC 4 4
17. The power delivered in 0.2 ms by the total discharge of a 50 mF capacitor connected to 25 V battery is ______W A. 78
P=
E t
B. 67
=
1 C V2 2 t
=
C. 52
1 50×10−6 F 25V 2 2
0.2×10 s −3
D. 49
E. 32
=78W
Refer to the circuit below for questions 18, 19 and 20 10.0 V
I1
+
I2
b
2.0
5.0
I3
a
+ 10.0
15.0 V
18. The current through the 2.0 Ω resistor is _____A A.
5 16
B.
15 16
C.
25 16
D.
35 16
E.
45 16
D.
31 16
E.
41 16
19. The current through the 10.0 Ω resistor is _____A A.
1 16
B.
11 16
C.
21 16
7/8 20. V b −V a is ______V 30
A.
8
B.
35 8
C. 0
D.
−30 8
E.
−35 8
One set of equations for the circuit is: I 1=I 2I 3 5.0 I 2 2 I 1−10=0 10.0 I 3−5.0 I 2 −15=0
solving for the currents we obtain: I 1=
45
7 31 A , I 2 = A ,I 3= A 16 8 16
Notice that the current I 2 is positive, therefore moves from b to a and thus the point b is at 7 8
higher voltage that a. So V b −V a = A×5 =
35 V 8
21. How many 100W lightbulbs, connected in parallel to 120 V source, can be used with a maximum current of 2.5 A? A. 7
B. 6
C. 5
D. 4
E. 3 P
100W
V
120 V
The current for one light bulb working between operating parameters is I = =
=
10 A 12
.
Since they are connected in parallel the total current is the sum of each individual current. Thus 10 the number of light bulbs is n=2.5/ =3 12
22. A 4.5 V battery is connected to a bulb whose resistance is 1.6 . The charge that leaves the battery per minute is ______ C A. 75
B. 92
C. 125
D. 154 E. 169
8/8 V 4.5 V I= = =2.81 A thus, Q=I t=2.81 A∗60 s=169 C R 1.6
23. Complete the following memorable quote from Finding Nemo. Dude? Dude? Focus dude... Dude? [Marlin wakes up] Oh, he lives. Hey, dude! Marlin: Oh... What happened? Saw the whole thing, dude. First you were all like "whoa", and we were like "whoa", and you were like "whoa..." Marlin: What are you talking about? You, MiniMan, takin' on the jellies. You've got serious thrill issues, dude. Awesome. Marlin: Oh, my stomach. Ohh. Oh, man. Hey, no hurling on the shell, dude, ok? Just waxed it. Marlin: So, Mr. Turtle? Whoa, Dude. Mister Turtle is my father. The name's_____.
Write your answers here: 6. C
7. B
8. E
9. A
10. D
11. A
12. D
13. B
14. E
15. B
16. C
17. A
18. E
19. D
20. B
21. E
22. E
23. Crush
PHYSICS 1308 SPRING 2008 EXAM # 1 Monday, July 14, 2008 SOLUTIONS
CONCEPTUAL QUESTIONS 1. Imagine a test charge inside a conductor in electrostatic equilibrium. Now imagine that this this charge is moved very slowly. What is the work performed by the electric forces during this process? Explain. The work done is zero. To understand why, first observe that the electric field inside a conductor in electrostatic equilibrium is zero, thus the net electric force on the test charge is also zero. Hence no work is necessary to move the charge inside the conductor.
2. What can you say about the electric field in a region of space that has the same potential throughout? Explain. −V −V −V , E y= , E z= The electric field is given by E x= . If V is constant throughout, we x
can conclude that the electric field is zero.
y
z
3. Current I enters a resistor R shown in the figure. Is the potential higher at point A or B? Explain. I The potential is higher at point A since the current A B moves from points of higher potential to points with lower potential. 4. One terminal of a car battery is said to be connected to “ground.” Since it is not really connected to the ground, what is meant by this expression? This just simply means that the “ground” terminal gives the reference potential. Thus any other potential is measured relative to this value. Remember that absolute potential has no physical meaning only differences are important.
2/8 5. What is the flux through the hemispherical open surface of radius R shown in the figure. The uniform field has magnitude E. Justify your answer. Observe that the same amount of lines that go through the base of the hemisphere go also thorough is surface. Thus 2 the electric flux is E = E⋅A= R E
E
NUMERICAL QUESTIONS 6. Two charges, one of 2.50 nC and the other of 3.50 nC are placed on the xaxis. One at the origin and the other at x= 6.00 m, as shown in the figure below. The net field at x= 32.75 m is equal to ____mN/C. y (m)
A. 42.23 i B. 23.04 i C. 0 D. 42.23 i E. 23.04 i
3.50 nC
+2.50nC
q0
x (m)
32.75 m
9 The net electric field is equal to: 9×10
6.00m
Nm 2 C
2
2.50nC 2
32.75m
− i
3.50nC 32.75m 6m2
i =0
7. The net force exerted by the two charges located on the opposite sides of the xaxis on the positive charge with position x= 10 mm is ______N . (See figure) y (m)
A. 0.48 B. 0.48 C. 1.58 D. 1.58 E. 0.82
j − j i −i sin 17− j
1nC 10 mm
r
10 C 3 mm 3 mm
10 C
x (m)
3/8 From the graphic above we can conclude that the resulting force is along the yaxis, thus :
2 k q Qsin F F− = − j = 2 r
2⋅9×10
9
Nm −9 −6 0 ⋅1×10 C⋅10×10 C⋅sin 17 2 C 2
0.003m 0.01m
where we have used that =arctan
3 mm
10 mm
2
− j=0.48 N −j
=170
8. A capacitor is connected to a 1.5 V battery. The plates are separated by 0.1 mm. An 250 C charge is placed in the middle of the plates of the capacitor. The magnitude of the force exerted on the charge is _____N. A. 2.75
B. 3.00
C. 3.25
The electric field is given by E=
V x
1.5V
=
E. 3.75
D. 3.50
0.1×10 m the force is : F=qE=1.5N /C×250×10 C=3.75N −3
=1.5×104 N /C . Therefore the magnitude of
−6
9. A 0.20 pF capacitor is needed. The area that the plates must have if they are separated by 2.2 mm air gap is _____cm² A. 0.5
B. 1.0
C. 1.5
D. 2.0
The capacitance for a parallel plate capacitor is equal to C= required area we obtain: A=
Cd 0
=
0.20×10−12 F⋅2.2×10−3 m 9×10
−12
2
C / Nm
2
E. 2.5 0 A d
. Therefore, solving for the
=5×10−6 m 2
4/8 10. The electric flux through the closed surfaced shown in the figure is _____kNm²/C A. 888 B. 888 C. 555 D. 111 E. 111
3 mC 7 mC
2 mC
Q
−3 C2 C
encl 3 2 =−111×10 Nm /C By Gauss' Law we can write E = = −12 2 2 0 9×10 C / Nm
11. A large, flat, horizontal sheet of charge has a charge per unit area of 9.00 C/m². The electric field just above the middle of the sheet is ____MN/C A. 0.5 E=
2 0
=
B. 1
C. 1.5
9.00×10−6 C / m 2 2⋅9×10
−12
2
2
C / Nm
D. 2
E. 2.5
=0.5×106 N /C
12. A sphere of radius r = 5 m has a charge of 5 mC distributed uniformly over its surface. The magnitude of the electric field at a distance of 10 m from the surface of the sphere is _____kN/C A. 50
B. 100
D. 200
C. 150
E. 250
We know that by Gauss' Law the field produced by a charged sphere is equal to the one produced by a point charge where the distance to other charges must include the radius of the sphere, hence: ∣E∣=k
∣Q sphere∣ r sphere r sphere− point
=9×10 2
9
N⋅m 2 5×10−3 C C
2
15 m
2
=200000 N /C
5/8 13. Two charged particles Q1 = +5.0 C and Q2 = 3.00 C are separated by 35.0 cm. The electric potential energy of the pair is _____J A. 0.52
B. 0.38
The potential energy is equal to:
C. 0
U =Q1 V = k
Q1 Q2 r
D. 1.23 9
=9×10
E. 2.50
Nm 2 5×10−6 C −3×10−6 C C2
0.35 m
=−0.38 J
14. The potential in a region between x = 0 and x = 6.00 m is V =10.0V −7.00 magnitude and direction of the electric field at x= 3m is ____V/m A. 11 −i
B. 11 i
C. 0
D. 7 −i
E. 7 i
−dV −d V E= = 10.0V −7.00 x =7.00V / m dx dx m
Refer to the following circuit for questions 15 and 16. All capacitances are in mF.
15 V
¼ ¾
½
15. The equivalent capacitance of the circuit above is _____mF A. 13/6
B. 11/12
C. 19/12
D. 27/4
E. 6/13
D. 55/4
E. 68/4
16. The charge in the ¾ mF is _____mC A. 10/4
B. 25/4
C. 45/4
V x . The m
6/8 Notice that the voltage across the ¾ mF is the same voltage of the battery, therefore 3 45 Q=V C= mF 15 V = mC 4 4
17. The power delivered in 0.2 ms by the total discharge of a 50 mF capacitor connected to 25 V battery is ______W A. 78
P=
E t
B. 67
=
1 C V2 2 t
=
C. 52
1 50×10−6 F 25V 2 2
0.2×10 s −3
D. 49
E. 32
=78W
Refer to the circuit below for questions 18, 19 and 20 10.0 V
I1
+
I2
b
2.0
5.0
I3
a
+ 10.0
15.0 V
18. The current through the 2.0 Ω resistor is _____A A.
5 16
B.
15 16
C.
25 16
D.
35 16
E.
45 16
D.
31 16
E.
41 16
19. The current through the 10.0 Ω resistor is _____A A.
1 16
B.
11 16
C.
21 16
7/8 20. V b −V a is ______V 30
A.
8
B.
35 8
C. 0
D.
−30 8
E.
−35 8
One set of equations for the circuit is: I 1=I 2I 3 5.0 I 2 2 I 1−10=0 10.0 I 3−5.0 I 2 −15=0
solving for the currents we obtain: I 1=
45
7 31 A , I 2 = A ,I 3= A 16 8 16
Notice that the current I 2 is positive, therefore moves from b to a and thus the point b is at 7 8
higher voltage that a. So V b −V a = A×5 =
35 V 8
21. How many 100W lightbulbs, connected in parallel to 120 V source, can be used with a maximum current of 2.5 A? A. 7
B. 6
C. 5
D. 4
E. 3 P
100W
V
120 V
The current for one light bulb working between operating parameters is I = =
=
10 A 12
.
Since they are connected in parallel the total current is the sum of each individual current. Thus 10 the number of light bulbs is n=2.5/ =3 12
22. A 4.5 V battery is connected to a bulb whose resistance is 1.6 . The charge that leaves the battery per minute is ______ C A. 75
B. 92
C. 125
D. 154 E. 169
8/8 V 4.5 V I= = =2.81 A thus, Q=I t=2.81 A∗60 s=169 C R 1.6
23. Complete the following memorable quote from Finding Nemo. Dude? Dude? Focus dude... Dude? [Marlin wakes up] Oh, he lives. Hey, dude! Marlin: Oh... What happened? Saw the whole thing, dude. First you were all like "whoa", and we were like "whoa", and you were like "whoa..." Marlin: What are you talking about? You, MiniMan, takin' on the jellies. You've got serious thrill issues, dude. Awesome. Marlin: Oh, my stomach. Ohh. Oh, man. Hey, no hurling on the shell, dude, ok? Just waxed it. Marlin: So, Mr. Turtle? Whoa, Dude. Mister Turtle is my father. The name's_____.
Write your answers here: 6. C
7. B
8. E
9. A
10. D
11. A
12. D
13. B
14. E
15. B
16. C
17. A
18. E
19. D
20. B
21. E
22. E
23. Crush
