Resolution for Exam 1 - Algebra-based Physics II | PHYS 2020, Exams of Physics

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Name Feb. 21, 2008

Phys 2020, NSCC Exam #1 — Spring 2008

1. (8)

MC (10)

Total (100)

Multiple Choice

Choose the best answer from among the four! (2) each.

-q +q

A B C^ D

1. A negative charge −q is in the vicinity of two charges^ -q ±q, as shown. Which vector gives the direction of the force on the charge −q?

a) A b) B c) C d) D

2. At the point which is precisely between two equal point charges^ Q^ Q

a) The electric field is zero, but the electric potential is not zero.

b) The electric field is not zero, but the electric potential is zero. c) Neither the field nor the potential is zero. d) Both the field and potential are zero.

3. A Farad is equal to

a) C V b) V C c) N V d) V N

4. If we put a resistor R and a resistor 2R in parallel, the equivalent resistance of the combi- nation is

a) 13 R b) 23 R c) 32 R d) 3 R

v

B

5. A negative charge moves in the plane of the page, in the “up” direction in a uniform magnetic field which points to the right. The force on the charge is

a) Into the page. b) Out of the page. c) Up. d) Down.

3. Two conducting parallel plates, each of area 20.0 cm^2 are separated by 5.00 mm. (There is air between the plates.) A potential difference of 100.0 V is applied across the plates. Find:

a) The charge that collects on each plate. (8)

The capacitance of the plates is C =  0

A

d Since the area is A = (20. 0 × 10 −^4 m)^2 = 2. 00 × 10 −^3 m^2

we get

C = (8. 85 × 10 −^12 )

(2. 00 × 10 −^3 m^2 ) (5. 00 × 10 −^3 m)

= 3. 54 × 10 −^12 F = 3.54 pF

Then we can use Q = CV to get the charge:

Q = (3. 54 × 10 −^12 F)(100 V) = 3. 54 × 10 −^10 C

b) The magnitude of the electric field between the plates. (2)

Use E = ∆ ∆Vs to get the magnitude of E:

E =

(100 V)

(5. 00 × 10 −^3 m)

= 2. 00 × 10 4 N C

c) The magnitude of the force on a particle of charge +e if it is in the space between the plates. (4)

The magnitude of the force (on charge q = e) is

F = qE = (1. 60 × 10 −^19 C)(2. 00 × 10 4 N C ) = 3. 2 × 10 −^15 N

V = 0 V

p

V = 30.0 kV

p

v

4. A proton is accelerated by having it move through a potential difference of 30.0 kV, starting from rest. In doing so, it loses potential energy and gains kinetic energy.

a) Find the loss in electric potential energy the proton undergoes. (4)

The change in potential is − 30 .0 kV, so the change in po- tential energy is

∆EPE = q∆V = (1. 60 × 10 −^19 C)(− 30. 0 × 103 V) = − 4. 8 × 1015 J

that is, it loses 4. 8 × 1015 J of potential energy.

b) Find the final speed of the proton. (The mass of the proton is 1. 67 × 10 −^27 kg.) (6)

From (a), the proton must gain 4. 8 × 1015 J of kinetic energy. Then:

KE = 12 mv^2 =⇒ v^2 = 2KE m

= 2(4.^8 ×^10

− 15 J)

(1. 67 × 10 −^27 kg)

= 5. 75 × 10 12 m s 22

Then: v = 2. 40 × 10 6 m s

2.0 x 10-15m

5. Find the work required to bring a proton from far away up to 2. 0 × 10 −^15 m away from another proton. (Hint: Find the change in electrical potential energy of the second proton.) (8)

With the first proton fixed at the origin, we find the electrical potential of the second proton when it is at infinity and when it is near the other proton. In both cases, use V = k q r = +k e r. At infinity, the potential is zero. At the given distance it is

V 2 = k

e r

= (8. 99 × 10 9 N·m

2 C^2 )

(1. 60 × 10 −^19 C)

(2. 0 × 10 −^15 m)

= 7. 19 × 105 V

The work required is the change in potential energy, so

W = q∆V = (1. 60 × 10 −^19 C)(7. 19 × 105 V − 0) = 1. 15 × 10 −^13 J

6. We need to make a resistor of resistance 5.00 Ω from a length of copper wire which has a circular cross-section of radius 0.200 mm. What length should we take? (The resistivity of copper is 1. 72 × 10 −^8 Ω · m.) (8)

The cross-sectional area of the wire is A = πr^2 = π(0. 200 × 10 −^3 m)^2 = 1. 26 × 10 −^7 m^2

We are given R and the resistivity of copper, so we use

R = ρ

L

A

=⇒ L =

RA

ρ

(5.00 Ω)(1. 26 × 10 −^7 m^2 ) (1. 72 × 10 −^8 Ω · m)

= 36.6 m

40.0 V

300 kW

500 kW

400 kW

30.0 V

• 20.0 V

• (^) - +

i 1^ i^2

i 3

8. Consider the messy circuit shown at the left; it has three batteries (note the direction of their polar- ities!) and several resistors. We would like to solve for the currents in all three branches. For your con- venience I have assigned and labelled these currents in the figure.

Write down (clearly, please) three equations that would allow us to solve for i 1 , i 2 and i 3. You don’t need to solve these equations for these currents. (8)

Using the currents as defined in the diagram, the junction rule (at either the top or bottom junction point) gives: i 1 + i 3 = i 2

Choosing a clockwise path round the left loop, the loop rule gives us

+40.0 V + 20.0 V − i 1 (300 kΩ) = 0

Choosing a counterclockwise path around the right loop, we get

+20.0 V − i 3 (400 kΩ) + 30.0 V − i 3 (500 kΩ) = 0

Actually these equations are quite easy to solve the for three currents. We get:

i 1 = 0.200 mA i 3 = 0.055 mA i 2 = 0.255 mA

v = 6.5 x 10^6 m/s

7.3 x 10-13^ N

9. A proton is moving to the right in a uniform magnetic field with speed 6. 5 × 10 6 m s. The force on the proton has magnitude
10. 3 × 10 −^13 N and is directed out of the page. Find the magnitude and direction of the magnetic field. (6)

Since the proton has a positive charge, we find from the right-hand rule for magnetic forces that the field must be pointing up (in the plane of the page) in order for there to be a force out of the page. To get the magnitude of the field (since v is perpendicular to B), use

F = qvB =⇒ B =

F

qv

(7. 3 × 10 −^13 N)

(1. 6 × 10 −^19 C)(6. 5 × 10 6 m s )

= 0.702 T

You must show all your work and include the right units with your answers!

F = ma Fc =

mv^2 r

KE = 12 mv^2

F = k

|q 1 q 2 | r^2

k =

4 π 0

F = qE E = k

|q| r^2

Eplates =

σ  0

Q

 0 A

∆EPE = ∆Uelec = q∆V Es = −

∆V

∆s

V = k

q r

1 eV = 1. 602 × 10 −^19 J

q = CV C =  0

A

d

Energy = 12 CV 2 Cdiel = κCvac

V = IR R = ρ

L

A

P = IV = I^2 R =

V 2

R

Rser = R 1 +R 2...

Rpar

R 1

R 2

∑ Iin =

∑ Iout

∑ loop

V = 0 F = qvB sin θ

r =

mv |q|B

m =

( qr^2 2 V

) B^2 F = ILB sin θ τ = NIAB sin φ

k = 8. 99 × 10 9 N C·m 2 2  0 = 8. 854 × 10 −^12 NC·m^22 e = 1. 602 × 10 −^19 C A = πr^2