Download Solutions to Practice for Final Exam

PHYS 1215 - Physics II, Solution to Practice Problems for Final exam.
1) EA = kqA / r2 = 9x109 (1 uC) / (1 m)2 = 9000 N/C down
EB = kqB / r2 = 9x109 (4 uC) / (2 m)2 = 9000 N/C left
ET =
( Ex) 2  ( Ey) 2 = 12700 N/C
EA
) = 45 degrees, but should be in quadrant III, so need to add 180 degrees
EB
so θ = 45 + 180 = 225 degrees
θ = tan -1 (
2) U = k q1q2 / r = 32805 J
U is converted to KE = ½ mv2
2 KE
solve for v =
= 5,122 m/s
m
3) U = k q1q2 / r = 1.13x10-16 J.
4) Current, I = q / t
5) V1 = 18 – 5.5
so t = q / I = 128,000 s
I1 = V1 / R1 = 0.1A Down
6) 0.028 N to the right, 0.028 N to the right and zero
7) Clockwise.
8) The index of refraction of blue light is larger than the one of red
light, so blue light is deflected more and white light is dispersed.
9) To correct her vision we need to bring the images of objects that are very far away
( d o   ) to her far point (we want di  0.25m ). Notice that the image distance is
negative. To find the “power” of the lenses we calculate:
" power" 
1
1 1 1
1

  
 -4 D
f d o d i   0.25m
The lenses will be divergent, indicated by the negative power.
do = infinity and di = -25 cm
lens as the object.
Image distance is negative since image is on same side of
10) Energy 
1
2  Energy 2(900 J )
CV 2  C 

 200µF
2
V2
(3000V ) 2
11)
The current should not exceed 0.315 amps and the total resistance is R+1.5ohms so:
0.315 A 
5V
5V
 R  1.5 
 15.87  R  14.37
R  1.5
0.315 A
To be safe it better be more than 15Ω
12)
4.06
 e t /(10M )(18F )
30
 4.06 
 4.06 
ln 
  t /(10M)(18F )  t  (10M)(18F ) ln 
  360 seconds
 30 
 30 
To get the time: VC  Vo e t /  4.06  30e t /(10M )(18F ) 
Or six minutes.
13) Using c  f (fundamental equation of a wave) we get:
0.1578m
3  108 m / s
 0.0394m

 0.1578m and the antenna needs to be
9
4
1.9  10 Hz
Or about 4 cm.
14) According to the problem Φ=50º, so α=40º and we can use Snell’s law at the air-glass
interface to get:


(1) sin   1.54 sin 40    sin 1 1.54 sin 40  77º
15) P = I V
I = P / V = 55 W / 12 V = 4.6 A = 4.6 C / s
#e = Q / e = 4.6 C / 1.6 x 10-19 C/e- = 2.9 x 1019 e-
16) The equation needed is Coulomb’s law: F  k
F  9 x109
q 1q 2
, with the values of the problem:
r2
Nm2 (3.2 x1019 C )( 20.8x1019 C )
 2.4x10-6 N
C2
(0.5x1010 m) 2
17) The definition of electric field is the force per charge: E 
We also know that F  ma or a 
F
, so F  qE .
q
F qE

m m
With the values given:
a
qE (1.6 x1019 C )(250 N / C )

 4.4x1013 m/s2
m
9.1x1031 kg
18) Circular apertures have a distinctive diffraction pattern called “Airy”. As a
consequence the minimum separation between two distinguishable sources of light is:
9
 1.22 
1  1.22(550  10 m) 
  3.8x10-6 degrees
  sin 
diameter
10
m




  sin 1 
19) A sketch of the problem:
The magnitude of each electric field is given by:
E A  EB  K
2
6
qa
9 Nm (30 x10 C )

9
x
10
 2.7  105 N / C
2
2
2
r
C
(1.0m)
Regarding the direction of the vectors:
Notice that q A produces an electric field pointing downwards because the force on a
positive test charge at the origin would be repulsive. Instead q B , which is negative, would
produce an attractive force on a positive test charge at the origin, so its electric field is a
vector that points to the right.

E  (2.7  105 N, - 2.7  105 N)
The magnitude of the vector is:

E  E  (2.7  105 N)2  ( 2.7  105 N)2  3.8x105 N
θ = tan -1 (
EA
) = - 45 degrees
EB
20) The electric field produced by each charge will be the same in magnitude, but one
towards the left and one towards the right, so when you add them together you get zero.
1.8 m
Electric potential is a scalar quantity, so we need to add the potential produced by each
charge. The distance that goes in the formula “r” is the distance from the point midway
between the charges and each charge, so it is 0.9 m (not 1.8 m) and the potential is:
V  9x10 9
2
6
Nm2 18 x106 C
9 Nm 18 x10 C

9x10
 360,000 V
C2
0.9m
C2
0.9m
21) Formula for the capacitance of a pair of parallel plates:
A
C  εo
d
This is valid in vacuum and very approximately, also in air. In water it will have to be
corrected by multiplying this value by the dielectric constant, so:
A
1.55m2
12
C  Kε o  81x(8.85 x10 F / m)
 4.4 μF
d
0.00025m
Notice that we converted 0.25 mm to meters in the above calculation.
.
22) E = -13.6 Z2 / n2 so E2 = -13.6 12 / 22 = -3.4 eV and E4 = -13.6 12 / 42 = -0.85 eV
ΔE = E4 - E2 = 2.55 eV
λ = hc / ΔE = 1240 nm eV / 2.55 eV = 486 nm
23)
Each charge will contribute a potential and we need to add those two values. Recall that
potential is a scalar and so there is no need to find direction of forces or other vectors.
Q1
Q
Nm2 (2  10 6 C )
Nm2 (1  10 6 C )
 k 2  9  109 2
 9  109 2
 -22,500 volts
r1
r2
C
1m
C
2m
Q
Note: When you find the potential due to a point charge using the formula V  k
you
r
implicitly assume it is the difference in potential between the point at a distance r and a
point at infinity.
V k
24)
C  Kε o
A
0.24m 2
 3.4  (8.85  10 12 F / m)
 0.020 μF
d
0.00036m
25) The charge accumulated will be: Q  (1.6  10-19 C )(6.022  1023 )  96,500C
And to get that amount of charge with a current of 5.5 you need a time of:
Q
Q 96,500C
I t  
 17,500 s
t
I
5.5 A
26) (i) Electric potential:
(ii) Electric field
(iii) Magnetic field
scalar measured in volts or J/C
Vector measured in V/m or N/m
Vector measured in tesla (T) or gauss (G)
27) If two adjacent parallel wires carry electric current in the same direction, they are
attracted to each due to the magnetic force.
28) The particle shown in the figure is positive because it follows the right hand rule
without reversing the direction of the force. (F = qv x B)
29) Each wire produces a magnetic field at point P as follows:
The magnitudes of the vectors are:
7 Tm
 o I1 4  10 A (15 A)
B1 

 10 6 T
2r1
2 (3m)
7 Tm
o I 2 4  10 A (20 A)
B2 

 10 6 T
2r2
2 (4m)
Since the vectors are at 90º to add them we calculate: B  B12  B22  1.41x10-6 T
θ = tan -1 (
B2
) = 45 degrees in quadrant I.
B1
30) The angle of incidence of light in water should be smaller than in air because of the
index of refraction of water, so the approximate position of the image will be as indicated
in the figure:
Notice that the real trajectory of light is indicated with a solid line. The person looking at
the fish will have the illusion that the fish is at the position of the image.
31) The angle with respect to the vertical in air is 33º, the angle in water can be calculated
using Snell’s law:
 sin 33 
  24.2º
nwater sin  water  nair sin  air  1.33 sin  water  (1) sin 33   water  sin 1 
 1.33 
32) Polarization.
Half of the intensity of un-polarized light will go through the first polarizer, of this
intensity a fraction equal to cos2(37º)=0.64 will go through the second polarizer, and
finally a fraction equal to cos2(37º)=0.64 will go through the third polarizer, so the final
intensity will be only
1
(0.64)(0.64)  0.203 of the original intensity
2
33) Wave nature of light, interference. The density of lines implies a distance between
slits of:
1inch
2.54 x10 2 m
d

 1.6933x10-6 m
15000lines 15000lines
To get the angular separation we will find the angle for each wavelength and calculate the
difference:

d sin(  )      sin 1  
d 
9
 589.0 x10 m 
  20.355º
1  sin 1 
6 
 1.6933x10 
 589.6 x10 9 m 
  20.376º
6 
 1.6933x10 
 2  sin 1 
So the difference is 0.022º
34) Interference pattern. The angle between the central maximum and the first bright
fringe in a double slit interference pattern is given by:

sin(  ) 
d
So, if the distance d is reduced, the angle will be larger, spreading the pattern.
35) According to the equation that describes the diffraction grating:
9
1   
1  530  10 m 

  8.7º for green light
  sin    sin 
6
3
.
5

10
m
d 


9

1  650  10 m 

  10.7º for red light
  sin    sin 
6
3
.
5

10
m
d 


1
36) We use the equation that we learned in Physics I, but we use the dilated time in the
equation:
d
d
v2 d
v 
 1 2
t To
c To
From the point of view of physics the problem is solved. If you wrote this equation you
will get most of the credit for the problem.
What comes after is just algebra:
1.6
v2
v2
8
1  2  8  10 1  2
With the numbers of the problem: v 
2  109
c
c
8
To simplify the problem notice that 8  10  2.67c , so:
v2
v
v2
v  2.67c 1  2   2.67 1  2
c
c
c
It is easier to calculate β, where β is v/c, so writing the equation as:   2.67 1   2
Solving for  , we get:  2  7.11 (1   2 )   
So v = 0.936c or 2.81x108 m/s
7.11
v
 0.936 
8.11
c
37) We find gamma:  
1

1
v2
1  0.62
1 2
c
L
8m
Length contraction: L  o 
 6.4 m
 1.25
 1.25
38) Δ p = m Δv
Δv = 0.015 x 105 m/s = 1500 m/s
ΔxΔp >= ћ
Δx m Δv >= ћ
Δx >= ћ / (mp Δv) = 1.055 x 10-34 J / {(1.673 x 10-27 kg) (1500 m/s)} = 4.2 x 10 -11 m