PHYSICS 222 SPRING 2013

EXAM 1: February 20, 2013; 8:15pm—10:15pm

Name (printed): ______________________________________________ Recitation Instructor: _________________________ Section #_______ INSTRUCTIONS: This exam contains 25 multiple-choice questions plus 2 extra credit questions, each worth 3 points. Choose one answer only for each question. Choose the best answer to each question. Answer all questions. Allowed material: Before turning over this page, put away all materials except for pens, pencils, erasers, rulers and your calculator. There is a formula sheet attached at the end of the exam. Other copies of the formula sheet are not allowed. Calculator: In general, any calculator, including calculators that perform graphing, is permitted. Electronic devices that can store large amounts of text, data or equations (like laptops, palmtops, pocket computers, PDA or e-book readers) are NOT permitted. Calculators with WiFi technology are NOT permitted. If you are unsure whether or not your calculator is allowed for the exam, ask your TA. How to fill in the bubble sheet:

Use a number 2 pencil. Do NOT use ink. If you did not bring a pencil, ask for one. Write and fill in the bubbles corresponding to: - Your last name, middle initial, and first name. - « « Your ID number (the middle 9 digits on your ISU card) « « - Special codes K to L are your recitation section. Always use two digits (e.g. 01, 09,

11, 13). Honors sections: H1 → 02; H2 → 13; H3 → 25. Please turn over your bubble sheet when you are not writing on it. If you need to change any entry, you must completely erase your previous entry. Also, circle your answers on this exam. Before handing in your exam, be sure that your answers on your bubble sheet are what you intend them to be. You may also copy down your answers on a piece of paper to take with you and compare with the posted answers. You may use the table at the end of the exam for this. When you are finished with the exam, place all exam materials, including the bubble sheet, and the exam itself, in your folder and return the folder to your recitation instructor. No cell phone calls allowed. Either turn off your cell phone or leave it at home. Anyone answering a cell phone must hand in their work; their exam is over.

Best of luck,

Dr. Soeren Prell  

1. One of the dangers of tornados and hurricanes is the rapid drop in air pressure that is associated with such storms. Assume that the air pressure inside of a sealed house is 1.02 atm when a hurricane hits. The hurricane rapidly decreases the external air pressure to 0.910 atm. A square window in an outside wall of the house measures 2.02 m on each side. What net force (directed outwards) is exerted on this window?

A) !5.14 × 104 N ! B) ! 5.37 × 105 N ! C) ! 4.78 × 105 N ! D) !4.53 × 104 N E) 6.29 × 104 N

Solution: The force is given by the product of the pressure difference between inside and outside the sealed house and the area of the window. F = !pA = (1.02 " 0.91)atm 1.01#105Pa/atm( )(2.02m)2 = 4.53#104N

2. A 50-cm3 block of wood is floating partially submerged in water, and a 50-cm3 block of iron is totally submerged in water. Which statement is true ? !

A) !The iron block experiences a larger buoyant force. B) !The wood block experiences a larger buoyant force. C) !Both blocks experience the same buoyant force. D) It cannot be determined which block experiences a larger buoyant force without knowing the shapes of the blocks. E) !It cannot be determined which block experiences a larger buoyant force without knowing the densities of the blocks.

Solution: The buoyant force is given by the product of the volume of the displaced water, the density of water and g: Fb =Vsubmerged!g . Since a larger fraction of the volume is submerged for the iron cube, it experiences a larger buoyant force. It still sinks because its weight is larger than the buoyant force. The wood doesn’t sink because its weight is smaller than the buoyant force it experiences.

3. Water is flowing in a horizontal pipe with circular cross section of diameter d. If you want to change the diameter of this pipe so that the speed of the water would be half as great as it was, what should be the new diameter? !

A) !d/4 !B) !2d ! C) !d 2 !D) !d/2 ! E) !d / 2

Solution: Water is an incompressible fluid and thus the continuity equation, which relates pipe cross section to fluid speed applies.

Aoldvold = Anewvnew ! " d / 2( )2( )vold = " dnew / 2( )2( )vnew ! dnew = dvoldvnew

= d 2

4. Consider a very small hole (1 mm diameter) in the bottom of a bucket

with a 20.0 cm diameter filled with water to a height of 50.0 cm. Find the speed at which the water exits the bucket through the hole.

A) !3.13 m/s B) !9.81 m/s C) 34.9 m/s D) !31.8 m/s E) 0.61 m/s

Solution:

Bernoulli’s equation relates variables at the top (t) and bottom (b) of the bucket.

pt + 12 !vt

2 + !ght = pb + 12 !vb

2 + !ghb The pressure at the top and at the hole is equal to atmospheric pressure p0.

Using the continuity equation we see that vb is neglible compared to vt.

Atvt = Abvb ! vt = vb

AbAt

= vbrbrt

"#$

%&'

2

! vb

p0 + 12 !vt

2 + !ght = p0 + 12 !vb

2 + !ghb" 1

2 ! vt2 + ! ght = 1

2 ! vb2 + ! ghb

" vb2 = vt

2 + 2g ht # hb( ) $ 2g ht # hb( )" vb = 2g ht # hb( ) = 2(9.81m/s2 ) 0.5m( ) = 3.13m/s

5. A rock is suspended from a scale that reads 20.0 N. A beaker of water is

raised up under the rock so that the rock is totally submerged in the water. The scale now reads 12.5 N. What is the density of the rock?

! A) !3.00 × 103 kg/m3 B) !1.60 × 103 kg/m3 C) !2.67 × 103 kg/m3 D) !2.50 × 103 kg/m3 E) !2.33 × 103 kg/m3

Solution: The buoyant force is given by Fb =Vrock!waterg and equal to the difference between the two scale readings. Thus,

Vrock =Fb

!waterg

The density of the rock is given by

!rock =mrock

Vrock= Wrock / g

Fb!waterg

= !waterWrock

Fb= 1000kg/m3( ) 20.0N7.5N

= 2.67 "103kg/m3

6. As shown in the figure, a container has a vertical tube, whose inner radius is 32.00 mm, connected to it at its side. An unknown liquid reaches level A in the container and level B in the tube; level A being 5.0 cm higher than level B. The liquid supports a 20.0-cm high column of oil, between levels B and C, whose density is 460 kg/m3. What is the density of the unknown liquid?

A) !1800 kg/m3 B) !1400 kg/m3 C) !1700 kg/m3 D) !2000 kg/m3 E) !1600 kg/m3

Solution: Using Bernoulli’s equation at level B inside the container and inside the tube we get:

patm + !U g (5.0cm) = patm + !Oil g (20.0cm)

" !U = !Oil(20.0cm)(5.0cm)

= 4 # 460kg/m3 = 1,800kg/m3

7. X and Y are two uncharged metal spheres on insulating stands, and are in

contact with each other. A positively charged rod R is brought close to X as shown in Figure (a). Sphere Y is then moved away from X, as in Figure (b). What are the final charge states of X and Y?

A) !X is positive and Y is neutral. B) !Both X and Y are negative. C) !X is negative and Y is positive. D) !X is neutral and Y is positive. E) !Both X and Y are neutral.

Solution: The positively charged rod will induce a negative charge on X and a positive charge on Y, when the two spheres are in contact with each other. These induced charges stay on the spheres when the spheres are separated. 8. The figure shows three electric charges labeled Q1, Q2, Q3, and electric

field lines in the region surrounding the charges. What are the signs of the three charges? A) Q1 is negative, Q2 is positive, Q3 is negative. B) !Q1 is positive, Q2 is negative, Q3 is positive. C) !Q1 is positive, Q2 is positive, Q3 is negative. D) !All three charges are negative. E) !All three charges are positive.

Solution:

Electric field lines start at positive charges and end at negative charges.

9. One very small uniformly charged plastic ball is located directly above another such charged ball in a test tube as shown in the figure. The balls are in equilibrium a distance d apart. If the charge on each ball is doubled, the distance between the balls in the test tube would become

A) !8d ! B) !2d !C) !4d D) !d !E) d/2

Solution:

In both situations the ball is in equilibrium, i.e. no net force is acting on it. The weight of the ball pointing down is balanced by the repulsive Coulomb force pointing up. The weight is the same in both situations and thus the Coulomb force is the same, too. The Coulomb force is given by F = keQ1Q2 / d

2 . Doubling each charge increases the numerator by a factor of 4. To keep F the same, we have to increase the denominator by a factor 4, too, by doubling the distance d.

10. A very long, thin rod parallel to the y-axis is centered at x = −1.0 cm and z

= 0.0 cm and carries a uniform linear charge density of +1.0 nC/m. A second very long, thin rod parallel to the z-axis is centered at x = +1.0 cm and y = 0.0 cm and carries a uniform linear charge density of −1.0 nC/m. What is the x component of the net electric field due to these rods at the origin?

A) !−1.8 × 103 N/C B) !−3.6 × 103 N/C C) !+1.8 × 103 N/C D) !+3.6 × 103 N/C E) !zero Solution: From symmetry considerations, at the origin each wire can only have a component of the electric field in the direction of the positive x-axis. The electric fields from the two wires have the same magnitude and direction at the origin.

!E = 2 ! 2ke" / d = 4(9 !10

9Nm2 /C2 )(1.0nC/m) / (0.01m) = 3600N/C

11. The figure shows two unequal point charges, q and Q, of opposite sign.

Charge Q has greater magnitude than charge q. In which of the regions X, Y, Z can there be a point at which the net electric field due to these two charges is zero?

A) !only regions X and Z B) !only region Y C) !only region X D) !only region Z E) !all three regions

Solution:

The electric fields of both charges point in the same direction in region Y, thus the net electric field cannot be zero there. In region Z, the electric fields point in opposite direction and the magnitude of the net field is

given by E = keQrQ2 ! ke

qrq2 . Since q < Q, E can only be zero for a point closer

to q than to Q. There may or may not be a point in region X where E is zero, but there is no such point in regions Y or Z.

12. An electric dipole of dipole moment ! = (5.00 × 10-10 C · m) ! is placed in

an electric field ! = (3.00 × 106 N/C) ! + (2.00 × 106 N/C) ! . What is the magnitude of the torque that the electric field exerts on the dipole? !

A) !1.80 × 10-3 Nm B) !0.00 Nm C) !1.50 × 10-3 Nm D) !2.50 × 10-3 Nm E) !1.00 × 10-3 Nm

Solution:

!! = !p "

!E

! z = pxEy # pyEx = pxEy = 5 "10#10Cm( ) 2 "106N/C( ) = 10#3Nm

The x and y components of τ are zero.

13. In the figure, a thin ring 0.71 m in radius carries a charge of + 580 nC uniformly distributed over it. A point charge Q is placed at the center of the ring. The electric field is equal to zero at field point P, which is on the axis of the ring, and 0.73 m from its center. The point charge Q is closest to

A) !−300 nC !B) !+300 !nC C) !−210 !nC D) !+210 nC ! E) !−420 nC

Solution: The electric field from the ring has to cancel the electric field from charge Q at point P. All points on the ring are a distance d = r2 + a2 away from P, where r is the radius of the ring and a is the distance between Q and P. The net electric field from the ring has only a component in the direction of the symmetry axis Q-P (x axis) and is given by Ex = keqring cos! / d

2 , where α is the angle between the x axis and d.

E(P) = keQ / a2 + keqring cos! / d

2 = 0" Qa2

= #qring cos!

d 2" Q

a2= #

qringad 3

"Q = #qringa

3

d 3= #

qringa3

a2 + r2( )3/2= # (580nC)(0.73m)2

0.73m( )2 + 0.73m( )2( )3/2= #210nC

14. Four dipoles, each consisting of a +10-µC charge and a −10-µC charge, are

located in the xy-plane with their centers 1.0 mm from the origin, as shown. A sphere passes through the dipoles, as shown in the figure. What is the electric flux through the sphere due to these dipoles?

!A) 11 × 105 N m2/C !B) !9.0 × 106 N m2/C C) !0.00 N m2/C D) !4.5 × 106 N m2/C E) !−9.0 × 106 N m2/C

Solution:

The electric flux through a closed surface is equal to the enclosed charge divided by ε0.

!E =qenclosed"0

= 4 #10µC8.85 #10$12C2 /(Nm2 )

= 4.5 #106Nm2 /C

15. An uncharged conductor has a hollow cavity inside of it. Within this

cavity there is a charge of +10 µC that does not touch the conductor. There are no other charges in the vicinity. Which statement about this conductor is true?

A) !The inner and outer surfaces of the conductor each contain charges of −5 µC. B) !The net electric field within the material of the conductor points away from the +10 µC charge. C) !The outer surface of the conductor contains +10 µC of charge and the inner surface contains −10 µC. D) !The inner surface of the conductor carries a charge of −10 µC and its outer surface carries no excess charge. E) !Both surfaces of the conductor carry no excess charge because the conductor is uncharged.

Solution: There is no electric field inside the conductor. Thus, inside a Gaussian surface that lies between the inner and the outer surface of the conductor, there can be no net charge. Consequently, the charge on the inside surface of the conductor has to be −10 µC to cancel the charge inside the cavity. Since the conductor is uncharged there has to be a charge of +10 µC on its outside surface.

16. The cross section of a long coaxial cable is shown in the

figure, with radii as given. The linear charge density on the inner conductor is −30 nC/m and the linear charge density on the outer conductor is −70 nC/m. The inner and outer cylindrical surfaces are respectively denoted by A, B, C, and D, as shown. The radial component of the electric field at a point that lies 34 mm from the axis is closest to

A) !+16,000 N/C B) !+37,000 N/C C) −16,000 N/C D) !−37,000 N/C E) !zero

Solution: Exploiting the cylindrical symmetry and using a cylindrical Gaussian surface with radius r = 34mm, we get

!E !da = qencl / "0 #cylinder"$ E 2%r( ) L = & L / "0

# E = ' &inner2%"0r

= ' 2ke&innerr

= '2(9.0 (109Nm2 /C2 ) '30nC/m( )

0.034m( ) = '16,000N/C

17. !Suppose a region of space has a uniform

electric field, directed towards the right, as shown in the figure. Which statement about the electric potential is true?

A) !The potential at point A is the highest, the potential at point B is the second highest, and the potential at point C is the lowest. B) !The potential at points A and B are equal, and the potential at point C is higher than the potential at point A. C) !The potential at all three locations (A, B, C) is the same because the field is uniform. D) !The potential at points A and B are equal, and the potential at point C is lower than the potential at point A. E) The potential at point A is the highest, the potential at point C is the second highest, and the potential at point B is the lowest.

 Solution: Since

!E =!!V a uniform electric field in positive x direction originates from a

potential of the form V = !ax +C , where a is positive and C is a constant. Thus, points at smaller x are at a higher potential. The potential does not depend on y.    

18. An isolated, charged capacitor stores energy U. Without connecting this capacitor to anything, a dielectric having dielectric constant K is now inserted between the plates of the capacitor, completely filling the space between them. How much energy does the capacitor store now?

A) !U/2K B) !KU ! C) !U/K !D) !2KU ! E) !U

Solution: The energy stored in the capacitor is given by U = Q2/(2C) where Q is the stored charge of the capacitor and C is its capacitance. Putting a dielectric between the capacitor plates changes C to KC and thus U to U/K.

19. A conducting sphere of radius R carries an excess positive charge and is very far from any other charges. Which one of the following graphs best illustrates the potential (relative to infinity) produced by this sphere as a function of the distance r from the center of the sphere?

Solution: Outside the sphere (r > R), the potential is that of a positive point charge (V = keQ/r) according to the spherical shell theorem. Since the sphere is conducting all charge sits on its outer surface. Thus there is no electric field inside the sphere and the potential inside the sphere is the same as on the surface.

20. A parallel-plate capacitor with plate separation of 1.0 cm has square plates,

each with an area of 6.0 × 10-2 m2. What is the capacitance of this capacitor if a dielectric material with a dielectric constant of 2.4 is placed between the plates, completely filling them?

A) !1.3 × 10-12 F B) !15 × 10-12 F C) !1.3 × 10-10 F D) !15 × 10-14 F E) !64 × 10-14 F

Solution:

C = KC0 = K!0A / d

= 2.4 8.85 "10#12C2N#1m#2( ) 6 "10#2m2( ) / (0.01m) = 1.3"10#10F

21. !Consider the group of three +2.4 nC point charges shown in the figure. What is the electric potential energy of this system of charges relative to infinity?

A) !4.4 × 10-6 J ! B) !4.1 × 10-6 J ! C) !4.6 × 10-6 J ! D) !4.8 × 10-6 J E) 3.0 × 10-6 J

Solution: We look at the work it takes to bring the three charges from infinity to their final positions. It takes no work to bring in the first charge. The work it takes to bring in the second charge is equal to the potential energy between charges 1 and 2:U2 = keQ

2 / d12 . The work it takes to bring in the third charge is equal to the potential energies between charges 1 and 3 and between charges 2 and 3: U3 = keQ

2 / d13 + keQ2 / d23 . The total potential energy is the sum of the two:

Utotal =U2 +U3 = keQ2 1d12

+ 1d23

+ 1d13

!"#

$%&

= 9.0 '109Nm2 /C2( )(2.4nC)2 10.03m

+ 10.04m

+ 10.05m

!"#

$%&

= 4.1'10(6 J

22. When a potential difference of 10 V is placed across a certain solid cylindrical

resistor, the current through it is 2 A. If we replace this resistor with another cylindrical resistor of the same material with triple the diameter and twice the length of the original one, the current will be !

A) ! 9 A! B) ! 18 A ! C) ! 2/3 A ! D) ! 4/9 A ! E) ! 3 A Solution: The original current is with . The new resistor has three times larger diameter d and twice the length L and thus 2/9 of the original resistance. Therefore, the current through the new resistor is 9/2 times larger than the current through the original resistor.

I =V / R0 R0 = !L / "d 2 / 4( )

23. The potential as a function of

position x is shown in the graph in the figure. Which statement about the electric field is true?

A) !The electric field is zero at x = 5 cm, its magnitude is at a maximum at x = 0, and the field is directed to the right there. B) !The electric field is zero at x = 10 cm, its magnitude is at a maximum at x = 5 cm, and the field is directed to the left there. C) !The electric field is zero at x = 0, its magnitude is at a maximum at x = 5 cm, and the field is directed to the right there. D) !The electric field is zero at x = 0, its magnitude is at a maximum at x = 10 cm, and the field is directed to the left there. E) The electric field is non-zero and directed to the right everywhere.

Solution: For a one-dimensional potential E = − dV/dx. Thus, the electric field is zero where the slope of the potential is zero (x = 0 and 10 cm and at very large x). The electric field is directed to the right (positive x direction), where the slope is negative i.e. between x = 0 and x =10 cm.

24. A silver wire has a cross sectional area 2.0 mm2. A total of 9.4 × 1018 electrons

pass through the wire in 3.0 s. The conduction electron density in silver is 5.8 × 1028 electrons/m3. What is the drift velocity of these electrons?

! A) !9.1 × 10-5 m/s B) !1.1 × 10-5 m/s C) !7.4 × 10-5 m/s D) !2.7 × 10-5 m/s E) !5.2 × 10-5 m/s

Solution:

vd =Jqn

= IqnA

= !Q / !tqnA

=9.4 "1018( )e / (3.0s)

e 5.8 "1028m#3( ) 2 "10#6m2( ) =9.4 "1018( ) / (3.0s)

5.8 "1028m#3( ) 2 "10#6m2( ) = 2.7 "10#5m/s

25. Two large conducting parallel plates A and B are separated by 2.4 m. A uniform field of 1500 V/m, in the positive x-direction, is produced by charges on the plates. The center plane at x = 0.00 m is an equipotential surface on which V = 0. An electron is projected from x = 0.00 m, with an initial velocity of 1.0 × 107 m/s perpendicular to the plates in the positive x-direction, as shown in the figure. What is the kinetic energy of the electron as it reaches plate A?

A) !4.4 × 10-16 J B) !3.3 × 10-16 J C) !2.9 × 10-16 J D) !5.9 × 10-16 J E) !The electron does not reach plate A. It will hit plate B and will be absorbed.

Solution: The electron will initially move to the right and be slowed down by the electric force. It will come to a rest when its kinetic energy is zero and turn around. Then it will accelerate until it hits plate A. Since the electric force is conservative total energy is conserved. Ui + Ki =U f + K f

! K f = Ui "U f( ) + Ki = q Vi "Vf( ) + Ki = "e("EAd / 2)+ Ki

= (1.6 #10"19C)(1500V/m)(1.2m)+ 12 9.1#10

"31kg( ) 107m/s( )2= 3.3#10"16 J

26. The capacitors C1, C2, C3 and C4 in the network

shown in the figure all have a capacitance of 5.0 µF. What is the equivalent capacitance, Cab, of this capacitor network? A) !10 µF ! B) !20 µF ! C) !1.0 µF ! D) !3.0 µF !E) !5.0 µF Solution: 1/C34 = 1/C3 +1/C4 = C / 2!C234 = C2 +C34 = 3C / 2!1/C1234 = 1/C1 +1/C234 = 5 / (3C)!C1234 = 3C / 5 = 3.0µF

27. The four identical capacitors in the circuit shown in

the figure are initially uncharged. Let the charges on the capacitors be Q1, Q2, Q3, and Q4 and the potential differences across them be V1, V2, V3, and V4. The switch is thrown first to position A and kept there for a long time. It is then thrown to position B. Which of the following conditions is true with the switch in position B? A) !V1 = V0 B) !V1 + V2 + V3 + V4 = V0 C) ! V1 = V2 = V3 = V4 D) !Q1 = Q2 E) ! Q1 = 3 Q2 Solution: While the switch is in position A capacitor C1 is charged with charge Q0 = C1 V0. The other capacitors remain uncharged. When the switch is moved to position B, the potential difference across capacitor C1 is the same as over the equivalent capacitance C234 = C1/3 for capacitors C2, C3 and C4. When the switch is flipped, the charge on the top plate of C1 is distributed between C1 and C2 and the charge on the bottom plate of C1 is distributed between C1 and C4. The charge on capacitances C2, C3 and C4 connected in series is the same and equal to the charge stored on the equivalent capacitance:

V1 =Q1C1

=Qeq

C234

= Q2

C1 / 3!Q1 = 3Q2

Since Vi =Qi /Ci !V1 = 3V2 . Also, since Q0 =Q1 +Q2 !V0 = Q1 +Q2( ) /C1 >V1 . Finally, V1 +V2 +V3 +V4 = 3V2 + 3V2 = 6V2 > 4V2 =V0 .  

Physics 222 midterm exam 1 - KEY  1      D     11      C     21      B  

2      A     12      E     22      A  

3      C     13      C     23      C  

4      A     14      D     24      D  

5      C     15      C     25      B  

6      A     16      C     26      D  

7      C     17      D     27      E  

8      B     18      C      

9      B     19      B      

10  D     20      C