0043 Lecture Notes - E&M Multiple Choice Solutions - AP Physics C 1998 Released Exam.docx page 1 of 6

Flipping Physics Lecture Notes: Electricity and Magnetism Multiple Choice Solutions

AP Physics C 1998 Released Exam from the College Board AP® is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product. 36) This is a Kirchhoff’s Rules loop. ΔVt = Vo and, according to their answers, must be positive which defines the direction of the loop.

C = QΔVcapacitor

⇒ΔVc =QC

However, the potential difference

across the capacitor must be negative because the direction of the loop is going from the positive to the negative plate of the capacitor. ΔVResistor = IR and again must be negative because the current direction is in the same direction as the loop and the potential across a resistor goes down as you go in the direction of the current. And remember that, according to Kirchhoff’s Rules, the potential difference around a loop totals zero. Answer is (B).

37) P = I 2R = IΔV = ΔV 2

R⇒ P = ΔV 2

Requivalent⇒ Req =

ΔV 2

P=12( )224

= 6Ω

The correct answer will have a combination of resistors with an equivalent resistance of 6 Ω. As you add resistors in series, the resistances add so (A) and (D) can’t be correct. (R(A) = 8 Ω & R(D) = 12 Ω) As you add resistors in parallel, the resistance decreases so (B) and (C) can’t be correct. (R(B) & R(C) < 4 Ω) The answer must be (E)

38) Charging via Induction!! When the negatively charged rod is brought close to the two conductors it polarizes the two conductors. The conductor #1 on the left has a net positive charge because unlike charges attract and the conductor #2 on the right has a net negative charge because like charges repel. Therefore, after the two conductors are moved apart, conductor #2 retains its net negative charge. Answer is (C).

39) Due to the Law of Charges +q will be repelled from both +Qs; like charges repel. The distances and charges are equal, so both +Qs will repel with the same magnitude force. One force will be down and the other force will be to the left. The net force will then be at a 45° angle down and to the left. Answer is (E).

40) Rather than doing this strictly mathematically, let’s look at the answers and work backwards. Clearly after answering 39, we know that the net forces isn’t zero so (A) can’t be correct. The two forces are at 90° to one another so the net force has to be greater than the original force F, so (B) and (C) can’t be correct. The only way (E) could be correct would be if both forces were in the same direction, so (E) can’t be correct. So (D) must be correct. Okay, now let’s do a mathematical solution. We can use the Pythagorean Theorem:

c2 = a2 + b2 ⇒ Fnet2 = F 2 + F 2 ⇒ Fnet = 2F 2 = F 2 Yep, (D) is the answer.

R E( ) = R + 1R+ 1R

⎛⎝⎜

⎞⎠⎟−1

= 4 + 14+ 14

⎛⎝⎜

⎞⎠⎟−1

= 4 + 2 = 6Ω

page 2 of 6

41) It must be (C) the cube, because there is no Gaussian Surface that one can draw to easily (or conveniently)

integrate E ⋅dA = qinεo!∫ . Why not? Because the angle between E and dA would not be consistent.

42) R1 = R, I1 = I2 = I, R2 = 3R, P1 = P = I 21R1 = I2R, P2 = I2

2R2 = I2 3R( ) = 3 I 2R( ) = 3P Answer is (D)

43) Looks like a job for Dimensions!! An Ampere is a Coulomb per Second and the charge of each proton is 1.6 x 10-19 coulombs.

m109 protons

× 1.6 ×10−3C

s× proton1.6 ×10−19C

= 1.6 ×10−3m

1.6 ×10−10 s= m1×10−7s

=107 ms

Answer is (D)

44) This beam of protons creates the same magnetic field as a current carrying wire and you find it in the same way, using the Right Hand Rule. Point your thumb (of your right hand) in the direction of the current and your fingers will curl in the direction of the magnetic field which will be (A) Concentric circles around the beam. Answer is (A)

45) qI = +3q, qIII = +2q, the electric field caused by a point charge is E = kqr2

& the direction of the electric

field is determined using a small, positive test charge. In order for the E Field to be zero EI and EIII must cancel out. Because like charges repel, to the left of I, both E Fields would be to the left and will not cancel out. To the right of III, both E Fields would be to the right and will not cancel out. Between I & III, EI is to the right and EIII is to the left, so in order to cancel out it must be here. Because qI > qIII then rI > rIII in order for the magnitudes of EI and EIII to be the same. So the point at which Enet = 0 must be closer to III than I, so it must be between II and III. Answer is (C). 46) Unlike Electric Field which is a vector, Electric Potential is a scalar. Therefore, because both charges are positive, both will cause a positive Electric Potential and they will never add up to zero. Answer is (E). 47) The Electric Force is directly related to the Electric Field which is directly related to the change in the

electric potential: Fe = qE⇒ E = − dVdr

Therefore the electric force will be greatest where the slope of the

electric potential as a function of position is the greatest. Answer is (D). 48) The distance is irrelevant, only the charge and the electric potential difference affect the amount of work

necessary. Wexternal agent = ΔUelectric = qΔV ⇒ΔV = Wq

= 50.002

= 2500V Answer is (C).

49) This problem uses a combination of Right Hand Rules. 1st point your right thumb in the direction of I2 and curl your fingers in the direction of the magnetic field. This means that the B field is into the page above the wire and decreasing with an increased distance from the wire. Next, point your fingers in the direction of I1 (4 different directions, for 4 different wire segments, AB, BC, CD, DA), curl your fingers in the direction of the B field (into the page) and your thumb points in the direction of the magnetic force (again, 4 different directions). The four directions for the magnetic force work out to be AB to the left, BC up the page, CD, to the right and DA down the page. The B field has the same changing magnitude on wire segments AB and CD and the two magnetic forces are 180° from one another and therefore cancel one another out. The magnetic forces on AD and BC are also 180° from one another, however, they do not cancel one another out. AD is closer to I2 and therefore in a stronger B field and therefore has a large magnetic force and therefore the net magnetic force is toward the wire. Answer is (A). 50) Yea! The Right Hand Rule for two problems in a row! Point your right hand fingers in the direction of the velocity (shown in the picture), curl your fingers in the direction of B (+y direction), your thumb points in the +z direction and this is the direction of the magnetic force and therefore the acceleration of the proton. A

0043 Lecture Notes - E&M Multiple Choice Solutions - AP Physics C 1998 Released Exam.docx page 3 of 6

short while later the direction of the velocity of the proton will now also have a component in the +z direction, again using the Right Hand Rule but just looking at the component of the velocity in the +z direction, fingers point with velocity in +z direction, fingers curl with B field in +y direction, thumb points with FB in the –x direction (or inward). You can see that this component of the force will cause the proton to move around the +y axis while it continues to move in the +y direction. This is a helical path with its axis parallel to the y-axis. Answer is (D).

51) The capacitance of a parallel plate capacitor is C = κεAd

and capacitance is defined as C = QΔV

,

therefore we can figure out the electric potential difference between the two plates:

C = κεoAd

= QΔV

⇒ΔV = QdκεoA

We can use the conservation of energy to determine the proton’s Kinetic Energy:

MEi = MEf ⇒Uelectrici = KEf +Uelectric f ⇒ KEf =Uei −Uef = − Uef −Uei( ) = −ΔUe = −qΔV

⇒ KEf = −e QdκεoA⎛⎝⎜

⎞⎠⎟= − eQd

κεoA Answer is (A).

A few points about this answer. Notice that it states that the Kinetic Energy is “proportional to” the answer, therefore the values for the dielectric constant and the permittivity of free space εo are irrelevant. In addition the proton will actually go through a negative electric potential difference because it will lose electric potential energy as it goes from the positive to the negative plate, therefore, in the end the KEf will work out to be positive. 52) Which case has a “nonzero magnetic field that can be conveniently determined by using Amerpe’s Law?”

Ampere’s Law states that B!∫ ⋅dl = µoIin and remember that you have to draw an Amperian Loop when

using Ampere’s Law and, in order to be conveniently determined using Ampere’s Law, that the magnetic field must be constant on the loop and there must be a constant angle between the direction of the magnetic field and the Amperian Loop. (A & B) Magnetic fields are created by moving charges. These charges are not moving, therefore (A) & (B) do not create a nonzero magnetic field. (C) Inside a very long current-carrying solenoid. You should have done this in your studies. This is the answer. (C). I did it in class in my video: http://www.youtube.com/watch?v=fdbIWY0uQ9E. (D) You use Biot-Savart to determine the magnetic field at the center of a current-carrying loop of wire. Again, I did it in class in my video: http://www.youtube.com/watch?v=MelTbwKvFNA. You can’t use Ampere’s Law because you can’t draw an Amperian Loop that would have a constant magnetic field on it. (E) Again, you can’t use Ampere’s Law because you can’t draw an Amperian Loop that would have a constant magnetic field on it. 53) Ahh, the Right Hand Rule, Limber Up! First off realize that because it is an atomic particle that the Force of Gravity is negligible. Let’s start by finding the direction of the magnetic force: Point the fingers of the right hand in the direction of the velocity, +x. Curl your fingers in the direction of the B Field, +y. Stick out your thumb and, because this is a positive particle, your thumb indicates the direction of the magnetic force, which is in the +z direction.

Because the proton is undeflected, the magnetic force and the electric force must be “balanced.” In other words, the electric force must be opposite the direction of the magnetic force. So the electric force is in the –z direction. Because the proton has a positive charge and an electric field is defined by a small, positive test charge, the electric field is also in the –z direction. The answer is (E).

κ

page 4 of 6

54) Yea! More Right Hand Rule! Luckily we are already limbered up. Point the fingers of your right hand in the direction of the velocity or to the right. Curl your fingers in the direction of the magnetic field … Oh No! We can’t do it!! How could the Right Hand Rule fail us now! Oh yeah, the equation for the magnetic force on a moving point charge is FB = qvBsinθ and the angle between the velocity and the B field is 0° and the sin(0°) = 0. Therefore there is no Magnetic Force. The answer is (E). 55) Hey, isn’t this a Hydrogen Atom? Draw a Free Body Diagram on the electron, the only force acting on it is the Coulomb Force in the in direction. Therefore, we can sum the forces in the in direction:

Fin = mac ⇒kq1q2r2

= m vt2

r∑ ⇒k e( ) e( )R2

= melectronvt2

R⇒ ke2

R= mevt

2 ⇒ KE = 12mevt

2 = ke2

2R

Where k = 14πεo

, so KE = 14πεo

⎛⎝⎜

⎞⎠⎟e2

2R= e2

8πεo, therefore the answer is (B). “But where did the negative

on “-e” go?”, you might ask. Well, when you draw the Free Body Diagram you use the fact that the two charges are unlike and therefore attracted to one another. You have now used the negative on “-e” to determine the direction, so you don’t plug it in again, because that would reverse the direction. Also, the proton does not have any Kinetic Energy because “the proton is stationary”. 56) Square Loop with sides, L, and resistance, R held at rest. B field is uniform and out of the page. The angle between the magnetic field and the area vector is a constant 0°. B = a – bt. We get to use Faraday’s Law to determine the magnitude of the induced current.

ε = IR = − dΦB

dt= − d

dtBAcosθ( ) = − d

dta − bt( ) l2( )cos0( ) = −l2 d

dta − bt( ) = bl2 ⇒ I = bl

2

R

And now Lenz’ Law or the concept of electromagnetic inertia to determine the direction. According to the equation for the magnetic field, B = a – bt, as t increases, B decreases. Therefore the magnetic flux inside the loop is out of the page and decreasing. This means that, according to Lenz’ Law or electromagnetic inertia, the loop tries to maintain the magnetic flux in the loop and will induce a magnetic field out of the page. In order to create this induced magnetic field there is an induced current in the loop that is … well, we need the right hand rule to figure out that direction. Fingers of the right hand point out of the page and the thumb points in the counterclockwise direction on the loop, so the induced current is counterclockwise on the loop. The answer is (E). 57) We need a relationship between frequency of revolution and radius. If you draw a Free Body Diagram on the negatively charged particle, the only force acting on it will be a magnetic force inward.

F∑ in= FB = mac ⇒ qvBsinθ = mrω 2 ⇒ q rω( )Bsin 90( ) = mrω 2 ⇒ qB = mω = m 2π

T⎛⎝⎜

⎞⎠⎟ = m 2π f( )

(The above makes use of vt = rω , ω = ΔθΔt

& ω = ΔθΔt

) and leads to⇒ f = qB2πm

In other words the frequency is independent of the radius. The answer is (A). 58) Because an Electric Field is defined by a small, positive test charge, in a constant Electric Field an electron will experience a constant Electric Force opposite the direction of the Electric Field. According to Netwon’s 2nd Law, if the net force is constant, then so is the acceleration. The answer is (A).

59) V r( ) = kr2 & E = − dVdr

= − ddr

kr2( ) = −2kr ; then the magnitude a distance ro from the origin would be

E ro( ) = 2kro The answer is (C).

0043 Lecture Notes - E&M Multiple Choice Solutions - AP Physics C 1998 Released Exam.docx page 5 of 6

60) Because E = − dVdr

the direction of the Electric Field and the direction of an increasing Electric Potential

are opposite. So, because the Electric Potential, V, is increasing with increasing r, then the Electric Field will be opposite that direction or toward the origin. Because an Electric Field is defined by a small, positive test charge and an electron is negatively charged, an electron will experience a force opposite the direction of the Electric Field or away from the origin. The answer is (B).

61) The equation for the Electric Field caused by a point charges is E = kqr2

. Therefore as you get closer to

a point charge the Electric Field gets closer and closer to infinity. Therefore the Electric field at 2 & 4 meters should approach infinity. Also, the Electric Fields caused by the two point charges will be equal in magnitude and opposite in direction at 3 meters and therefore will cancel one another out and cause an zero Electric Field at 3 meters. The only answer that fits these criteria is (A). 62) This is the definition of an equipotential surface or line. An equipotential surface (or line) is a surface on which no work needs to be done to move a charge from one point on the surface to another point on the same surface. The answer is (D). 63) The work done on the point charge is W = ΔUele = qΔV = q Vf −Vi( ) and the equation for the electric

potential due to a point charge is V = kqr

and outside a uniformly charged sphere, it will act like a point

charge. Therefore W = q kQrf

− kQri

⎛

⎝⎜⎞

⎠⎟= kqQ 1

R− 1r

⎛⎝⎜

⎞⎠⎟

. Note that this is from an initial location r distance

from O to a final location R distance from O. We also need to include the work going from R to O. However, because all of the charges are uniformly distributed throughout the hollow sphere, this will act just like a solid conductor in electrostatic equilibrium and we know the Electric Field inside the sphere will be zero and it takes no work to move a charged particle in a location of zero Electric Field. Therefore the total work is

W = kqQ 1R− 1r

⎛⎝⎜

⎞⎠⎟ . The answer is (E).

64) The two capacitors in parallel sum: C12 = C1 + C2 = 3 + 3 = 6 µF

Adding the third capacitor in series is: Ceq =1C12

+ 1C3

⎛⎝⎜

⎞⎠⎟

−1

= 16+ 13

⎛⎝⎜

⎞⎠⎟−1

= 16+ 26

⎛⎝⎜

⎞⎠⎟−1

= 2µF

The answer is (B). 65) The potential difference from X to Z must be the same as the battery. So we can figure out the total charge delivered by the battery to the capacitors:

Ceq =Qtotal

ΔVt⇒Qt = CeqΔVt = 2 ×10−6( ) 12( ) = 24 ×10−6C = 24µC

In series the charge stored on the capacitors in the same so Qt = 24µC =Q12 =Q3 The potential difference between Y and Z is the potential difference on that capacitor:

C3 =Q3

ΔV3⇒ΔV3 =

Q3

C3= 24µC3µF

= 8V The answer is (D).

page 6 of 6

66) The switch S has “zero” resistance so all of the charges will flow through the switch and not flow through bulb 2, therefore bulb 2 will go out. The addition of the switch will also reduce the resistance of the circuit, however the electric potential difference across the battery will remain the same, therefore because ΔV = IR the current through bulb 1 will increase. Therefore because the power output of the bulb is determined by P = I 2R , bulb 1 will convert more electrical energy per second, in other words it will get brighter. The answer is (B). 67) First, the magnetic field goes from North Pole to South Pole and is in loops. 1 Therefore to the left of the magnet, inside the current loop, the magnetic field will be almost exactly to the left. Now we need to use the Right Hand Rule on the current carrying loop. Point your fingers in the direction of the current on the loop, which is Into the Page in the part of the loop that is at its highest point (or most near the top of the page). Your fingers curl in the direction of the magnetic field, which is almost exactly to the left, however, not quite. The magnetic field is actually directed slightly toward the top of the page at this point. This means your thumb, which is pointing in the direction of the magnetic force, is almost pointing directly toward the top of the page, however, it has a small component to your right. If you repeat this process on the whole loop, you will see that all of the radially outward components of the magnetic field cancel out, however, each one has a small magnetic field component to your right or toward the magnet. Therefore the net magnetic field is (A) toward the magnet.

68) ε ind = bAt12( ) = − dφB

dt= − d

dtBAcosθ( ) = −Acos 0°( ) d

dtB( ) = −A dB

dt⇒ bAt

12( ) = −A dB

dt

⇒ −bt12( ) = dB

dt⇒ dB = −bt

12( )dt⇒ dB∫ = − bt

12( ) dt∫ ⇒ B = − bt

32( )

32

+C = − 23bt

32 +C

Because the “magnetic field could be given by” means that the constant C “could” be zero and we could

have used 180° for our angle which would give us B = + 23bt

32 . The answer is (E).

69) Parallel plate capacitor. ΔV = Vo. The electric field in a parallel plate capacitor is constant so:

ΔV = −Ed⇒ E = − ΔVd

⇒ E = Vod

The question just asks for “magnitude”. The answer is (B).

70) Adding the dielectric increases the capacitance of the capacitor because the dielectric constant, k, of the insulating material is, by definition, greater than 1 and the capacitance of a parallel plate capacitor is given by

C = κεoAd

. Therefore, the capacitance will increase. The answer is (A).

1 “Magnetic field of an ideal cylindrical magnet with its axis of symmetry inside the image plane.” http://commons.wikimedia.org/wiki/File:VFPt_cylindrical_magnet_thumb.svg This file is licensed under the Creative Commons Attribution-Share Alike 3.0 Unported license.