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1.
Fundamental and derivate quantities and units

Fundamental Physical quantities are those that can be directly defined, and for which the
units are chosen arbitrarily, independent of others physical quantities.
No.
Fundamental Quantity
Unit
Symbol
Dimension
1
Length
Meter
m
L
2
Time
Second
s
T
3
Mass
Kilogram
kg
M
4
Current intensity
Ampere
A
I
5
Light intensity
Candela
Cd
J
6
Temperature
Kelvin
K

7
Solid angle
Ste-radian
Sr
8
Plane angle
Radian
rad

The derivative physical quantities are those that are defined indirectly. They have the
measurement units’ functions of fundamental units.
If we have:
2

U = U(L, T, M, I, J, )

[U] = LTMIJ
The homogeneity of physical equations:

3
The physical formulas are invariant to the measurement units’ transformation.
Examples:
3.1 Velocity:
v 
lim
t 0
x dx

t dt
v  x   m  m  s 1  L  T 1
t  s
(1)
v dv

t dt
a   v  m2  m  s 2  L  T 2
t  s
(2)
p  m v  kg m  kg  m  s 1  M  L  T 1
(3)
3.2 Acceleration:
a 
lim
t  0
3.3 Impulse:
p  mv
s
1
3.4 Force:
p dp dmv
dv


m
 ma
lim
t dt
dt
dt
t 0
F 
F  p  kg 2 m  kg  m  s 2  M  L  T 2
t  s
F  1 kg 2 m  1N
s
(4)
3.5 Pressure:
P
P  F  2kg  kg  m 1  s 2  M  L1  T 2
S s  m
dF
dS
P  1 N2
m
 1Pa
(5)
3.6 Mechanical work:
2
L  F s  kg m2
dL  F  ds
s
 kg  m 2  s 2  M  L2  T 2
L  1N  m  1J
(6)
3.7 Kinetic energy:
Ec 
m  v2
;
2
kg  m 2
E c   mv  2  kg  m 2  s 2  M  L2  T 2 ; E c   1kg  m 2  s 2  1J (7)
s
3.8 Gravitational potential Energy:
2
Ep  m  g  h
E   m g h  kg ms
p
2
2
 
 kg  m 2  s 2  M  L2  T 2 ; Ep  1kg  m2  s 2  1J (8)
3.9 Elastically potential Energy:
E p,e 
k  x2
2
E   kx
2
p ,e
3.10

Power:
P
dE
dt
P  E  kg 3m
t 
s
3.11
2
 kg  m 2  s 3  M  L2  T 3
P  1 J  1W
s
(10)
Density:

3.12
 
kg  m 2
 kg  m 2  s 2  M  L2  T 2 ; Ep,e  1kg  m2  s 2  1J (9)
2
s
  m  kg3  kg  m3  M  L3
V m
dm
dV
Angular velocity:
2
(11)

3.13
(12)
Solid angle:
d 
4
    rad  rad  s 1  T 1
t  s
d
dt
2
  A2  m 2  L0  1
R  m
dA
R2
  1L0  1steradian
(13)
Establishes into a constant approximation the formula of a gravitational (mathematical)
pendulum, T using the homogeneity of the dimensional equations.
R: From direct observation the physical quantities that can occur are: the length (l), mass of the
pendulum (m), gravitational acceleration (g).
T = k l m g



1

  2

  0

1
  
2

T1  k 0  L  M  L  T  2

 T1  L    M  T  2  ,
then
by
identifying
the
upper
coefficients:
    0

  0
 2   1

5
T  kl1/ 2 m 0 g 1/ 2  T  k
l
g
The mass of a parachute with jumper (parachutist) is m = 100 kg, and is launched from a tower
completely open with no initial velocity. Find the velocity expression, v(t) and the velocity
limit if we know that the resistance force is proportional with the velocity, R = kv, where k =
500 Ns/m.


R: If the force is defined as: F  ma
Along vertical direction can be rewritten as: F  m
dv
dt
The resultant (net) force is F  G  R , then:
vt 
vt 
dv
dv
m 
k 
m
 mg  kv  
  dt '   ln  g  v 
t
k
dt
k 
m  v 0 
v 0  g 
v 0
m
 vt  
t
 t
mg 
1  e m  .


k 

k
3
v 0  0



k
k
ln 1 
vt    t
m
 mg

If we will replace the physical quantities with values we get:
500

t
100 10 
m
100 

vt  
1

e
 vt   2 1  e 5 t  


500 
s



For the time goes to infinity one obtains:
k

 t 
mg 100 10
 mg 
m 
v lim  lim vt   lim 
1

e

v



lim


k
k
500
t 
t  



 v lim  2
m
s
 7.2 km
h
Finally the velocity expression can be written as:
k
 t

vt   v lim 1  e m 


6.
A skier with the weight G descends a hill that make an angle  with the horizontal plane. The
motion equations are x = Agt2 along the hill land y = 0, where g is the gravitational acceleration
and A a constant coefficient. How much is the friction force between skier and snow on the hill.
(Particular case: the skier mass m = 70 kg, angle  = 300; A = 0.1).
R: The second principle of dynamics can be written as:
  

G  R  Ff  ma
The projections along x and y axes are:
G sin    Ff  ma x
R  G cos   ma y
The component of accelerations can be calculated with the following equations:


d 2 x d 2 Agt 2

 2Ag
dt 2
dt 2
d 2 y d 2 0
ay  2 
0
dt
dt 2
ax 
With those:
Ff  G sin   2Amg  Gsin   2A
Ff  mg sin   2A  70  9.81sin 30  2  0.1  686.7  0.3
Ff  206.01 N
7.
A material point with mass m is moving along a trajectory given by the Cartesian components:
xt   A coskt  and yt   B coskt  . Characterize the force F that produces this type of motion
if we know that the force depends only by the material point position. Give some examples.
R: The force components along Ox and Oy axes are:
4
Fx  m  a x  m  x  Fx  m 
d
d
x  mk  A sin kt   mk 2 A coskt   mk 2 x
dt
dt
Fy  m  a y  m  y  Fy  m 
d
d
y  mk  B sin kt   mk 2 B coskt   mk 2 y
dt
dt
The force modulus is:
F  Fx2  Fy2  mk 2 x 2  y2  mk 2 r

where r is the material point position vector.
The ratio:
Fx  mk 2 x  x
  


 cos x , F 
2
F
mk r
r


are the direction cosine of the force F.
2

Fy  mk y  y




 cos y, F 
F
mk 2 r
r


In the same time are the angles between the position vector and OX and OY axes,
respectively. From here we can observe that the force F is anti-parallel with the position vector.
Then the force F is pointing to the centre of the reference frame. Such a force is called
attraction central force.
8.
On a body with a mass m = 2 kg are acting two forces, F1 = 3 N and F2 = 4 N, which are
characterized by the angles 1 = 600, and 2 = 1200 respectively, with the direction of velocity

v 0 . Find the body acceleration, a, velocity, v and the distance covered into a time t = 10 s starting
from the beginning of motion (Particular case v0 = 20 m/s).
R: If we are considering a reference frame related to earth then the acceleration components along
OX and OY axes are:
ax 
F1 cos 1  F2 cos  2 3 cos60  4 cos120 3  0.5  4  0.5
1


   0.25 m / s 2
m
2
2
4
F sin 1  F2 sin  2
ay  1

m
3
3
3
4
2
2  7 3  3.03 m / s 2
2
4
and the acceleration modulus:
a  a 2x  a 2y  0.252  3.032  0.0625  9.1809  9.2434  3.04 m / s 2
We observe that the acceleration is constant in time (a = const), the velocity equation can be
calculated from:
5
dv
a
 dv  a  dt 
dt
vt 
t
v0
0
 vt   v0  at  vt   v0  at
 dv   a  dt'
or by components:
v x t   v0 x  a x t cu v0 x  v0
v y t   v0 y  a y t cu v0  0
and from here:
v x t   v0  a x t  20  0.25  10  17.5 m / s
v y t   a y t  3.03  10  30.3 m / s
and the velocity modulus:
v  v2x  v2y  17.52  30.32  306.25  918.09  1224.34  34.99 m / s


The velocity, v at time t makes an angle  cu v 0 given by the equation:
v 
 30.3 
o
   arctg  y     arctg 
  arctg 1.731  59 59'28" .
v
17
.
5


 x
vy
tg 
vx
In order to calculate the distance we have:
v
ds
dt
 ds  v  dt 
 st   s 0  v 0 t 
2
at
2
s 0 0

st 
t
s0
0
t
 dv   v  dt '  st   s0   v0  at   dt '
st   v 0 t 
0
at
3.04 100
 20 10 
 200  152  352 m
2
2
2
or by components:
x t   v0 t 
yt  
9.
a yt2
2
axt2
0.25  100
 20  10 
 200  12.5  187.5 m
2
2

3.03 100
 151.5 m
2
A tractor is traveling with a velocity of v0 = 36 km/h. If the radius of the wheel is R = 0.5 m find
out:
a)
The parametric motion equations of a point from the external wheel circumference.
b)
The tangential velocity components and the value of velocity.
c)
The path distance by a point between two contacts with the road.
R:
a)
v  R
 
v0
rad
 20
R
s
6
x  v 0 t  x W

y  0  t  y w
x w  x CW  x R

y w  yCW  y R
x R  R sin t 

y R  R cost 
x CW  0

yCW  R  y R
x R  v 0 t  R sin t   10t  0.5 sin 20t 

y R  R  R cost   0.5  0.5 cos20t 
this is the equations of a cycloid.
b) v x 
vy 
dx
 v 0  R cost   v 0 1  cost   101  cos20t 
dt
dy
 R sin t   v 0 sin t   10 sin 20t 
dt
v  v 2x  v 2y  v 02 1  cost   v 02 sin 2 t   v 0 21  cost   2v 0 sin
2
t
2
v  20 sin 10t
c)
v
ds  vdt
ds
dt
 ds  2R sin
t
dt
2
t
ddt

ds  2R sin

d
2
2
2



0 ds'  0 2R sin 2 d  s  4R   cos 2  0  8R  4 m
s
10. A body with a mass m = 5 kg can slide with friction on a horizontal surface if is pushed by a
spiral spring, with an elastic constant k = 200 N/m, and which was compressed at half of his
length l = 20 cm. Find out:
a) the mechanical work of friction force during expansion.
b) how much must is the friction coefficient if in the final position the spring in not
stressed.
R:
a) Fe  kx
W
dW  F  dx 
W  200
0
0
x2
0 dW'  l 2F  dx  W  l 2 kx  dx  k 2
0
k
l 2
l
8
20 2 10 4
 1J
8
b) E t  Lf
E f  E i  Lf
 0k
l2
kl
 Nl   
  Nl
2
2mg
7

11.
200  0.1
 0.204
2  5  9.81
A concrete cube with the side of a = 0.8 m and a density of 2500kg/m3 must be tipped around
an edge. Calculate:
a)
Point of application direction and value of the minimal force needed to turn turn-over;
b)
The expression of a horizontal force that can turn-over the concrete block if this is applied
on superior edge as function of rotation angle.
c)
R:
The mechanical work spent for turning over the concrete block.

 
  
a) M F  r  F  r  F  cos rF 
 
 
cos rF   1
  

r  a 2


M F max  r  Fmax  cos rF   a 2Fmax
 

MF max  MG
Fmax
 a 2Fmax  G
a
a
Vg ga 3
 Fmax  mg


2
2a 2 2 2 2 2
2500  9.81  0.83

 4439.5 N
2 2
b)
M F  aF   a  F  cos
2


MG  G  a
 cos   
2
4

F  a 3g
F 
 aF cos  Ga
2


cos   
2
4


2  
a 3g





1  tg 
cos
cos


sin
sin


F






2 cos   4 
2
4

0.83  2500  9.81
1  tg   6278.41  tg  N
2
W  MF  
c)
dW  MF  d 
a
0
W
4
0
0
 dW '   M
4
F
 d  W 

0
a3g
1  tg d
2
4
4
W
MF  MG
4

2

2

2
2



g cos     d  a 4
g  sin      a 4
g  1 

2
4
2
40
2
2




8
W
a 4g

2


2 1
W
0.84  2500  9.81

2


2  1  2080.48 J
On an inclined plane with inclination  = arcsin(3/5) and length l = 2.1 m is rolling with no
12.
sliding friction an homogeneous sphere of mass m = 40 g and radius R = 10 mm. How much is
the transversal velocity, angular velocity and frequency at the base of the plane.
R: E t ,A  E t ,B
E t ,A  E p,A  mgh  mgl sin  
E t ,B  E kin,tr,B  E kin,rot ,B
I
2
mR 2
5
E t ,B 
v  R
 

13.
E kin ,rot ,B
I2

2
v
R
mv 2 I2 mv 2 2
v2
mv 2 2 mv 2 7 mv 2


 mR 2



2
2
2
5
2R 2
2
5 2
5 2
7 mv2
mgl sin   
5 2

E kin,tr,B
mv 2

2
 v
10
10
3
gl sin    v 
9.81  2.1   4.20
7
7
5
m
s
v 4.2
rad

 420.2
R 0.01
s
 420.2

 66.88 Hz
2
2
Demonstrate that after a perfect elastic collision of two hockey pucks of the same mass,
initially the first one being at rest, the angle between the directions of pucks is 90 0. How much
are the velocities and scatter angle 2 if the initial velocity of projectile puck is v = 40 km/h
and 1 = 600?
R:
Ec,i  Ec,f
 
pi  p f

mv 2  mv 12  mv 22
  
p  p1  p 2
  
p 2  p12  p 22

  
p

p

p
1
2


m2 v 2  m2 v12  m2 v 22
  
p  p1  p 2
p1  p 2 2  p12  p 22  2p1p 2  p12  p 22
cos1  2   0  1  2  900


 p1  p 2
2  900  1  900  600  300
9
 2p1p 2 cos1  2   0
cos1  
mv 1 v1

mv
v

v1  v cos1   40 
cos2  
mv 2 v 2

mv
v
 v 2  v cos2   40 
1
m
 20
2
s
3
m
 34.64
2
s
14. On the deflection plates of a cathodic oscilloscope
with sensibility of 2 cm/V on both axes Ox and Oy it
is applied simultaneously the electric voltage
U x  2 cos100t  and U y  4 sin 2 50t . Calculate:
i. The spot equations of motion on Ox and Oy axes.
ii. Equation of the trajectory.
a) Sx 
R:
Sy 
y
Uy
x
Ux
 x  Sx U x  2
 y  Sy U y  2
cm
2V cos100t   x  4 cos100t cm 
V
cm
4V sin 2 50t   y  8 sin 2 50t cm
V
x
x
 4  cos100t 
x  4 cos100t 
  cos100t 
b) 
 
 4
2
x  8 sin 50t 
 y  8 1  cos100t 
 y  41  cos100t 

2
 x
y  41    4  x  x  y  4
 4


15. A material point execute a motion described by the equation: xt   2 sin 2  3t   . Show that
2

this is a harmonic oscillatory motion. Find the amplitude and period of motion, the velocity and
acceleration of the material point.
R:


2
x t   2 sin  3t  2 

  x t   2 1  cos6t     1  cos6t   1  sin  6t   



2
2

sin 2    1  cos2 

2
x t   x ' A sin t  0 
x '  1 A  1
2
1


 
 6  T   0.33 s
      6

x t   1  sin  6t   
T
3
   2
2 

 0
v
dx d


 x ' A sin t   0   A cost   0   6 cos 6t  
dt dt
2

10
a
16.
A
dv


 36 2 sin  6t  
dt
2

material
point
with
the
mass
m
=
10
g
is
oscillating
under
the
law:


xt   10 cos 2  t    5 . Calculate:
4
 12
a. The time t1 necessary to reach the maximum velocity, and the time t2 necessary to reach
the maximum acceleration;
b. The maximum value of the elastic force that act on the material point;
c. Expression of kinetic, potential and total energy.
R:




2 
1  cos t  
x t   10 cos  12 t  2   5
2
 
6


 x t   10
 5  5 sin  t 
a) 
2
6 
cos 2    1  cos2 

2


v max  5

6
dx


 
v
 5 cos t   

dt
6
6 
cos  t 1   1
  6 

2
a

5
 max
dv
2
36

 
a
 5 sin  t   


dt
36  6 


sin  t   1
  6 2 

t 1  n  t 1  6n s
6


t 2  2n  1
6
2
 t 2  32n  1 s
b) Fe  kx  kA sin t  0   Fmax sin t  0 
Fmax  kA  m2 A  10 10 3 
2
 5 10  2  1.37 10  4 N
36
4
mv 2 mA 2 2 cos 2 t  10  25 10  36
 
 
c) E c 


cos 2  t   3.43 10 6 cos 2  t  J
2
2
2
6 
6 
2
kx 2 mA 2 2 sin 2 t  10  36  25 10
 
 
Ep 


sin 2  t   3.43 10 6 sin 2  t  J
2
2
2
6 
6 
2
Et 
4
mv 2 kx 2

 3.43 10 6 J
2
2
11
17.
A harmonic oscillator which oscillates with the amplitude, A of 8 mm is after   0.01 s from
the beginning of motion (from equilibrium position) at the distance 4 mm measured from
equilibrium position. Calculate:
a. oscillatory pulsation;
b. oscillatory period;
c. oscillatory frequency;
d. velocity of oscillation in the given position;
e. acceleration of oscillation in the given position.
R:
a) yt   A sin t  0 
y0  0  A sin   0  0   sin 0   0  0  0
A
1



50
y  4   A sin   sin  
  
 

 

2
2
6
6 6  0.01
3
2
2
2
 T

 T  0.12 s
b)  
50
T


3
1
1
   8.33 Hz
c)   
T 0.12
dy
50
 50


 A cost   v 
  8 10 3 cos   0.01  0.419 cos  
d)
dt
3
 3

6
v  0.36 m s
v
e) a 
dv
2500 2
m
 50

 2 A sin t   a  
  8 10 3 sin    0.01   a   10.97 2
dt
9
s
 3

18. The sonic boom created by an airplane it
is heard in the moment in which the
direction observer–airplane makes an
angle  = 600 with the vertical. How much
is the airplane speed? (vs = 340 m/s)
R:   90  
sin  
A0 v s  t

AA ' v a  t
 va 
vs
v
340
m
 s 
 2  340  680  2 Mach
1
sin  cos 
s
2
19. How much is the fundamental frequency of vibration of a copper wire ( = 8.9 kg/dm3) with
section of 2 mm2 and the length of 1 m when is stress with a force of 17.8 daN.
12
R:   vT

v 



 

 
v

 
T

m V Sl


 S
l
l
l
v

 v
T
;   2l
S
1 T
1
17.8  10
100


 50 Hz
6
2l S 2  1 8900  2  10
2
20. Calculate the wavelength value of a sound of 440 Hz in air (va = 340 m/s) and into a tram line
(vt = 5000 m/s). Calculate the steel Young modulus if the density is (t = 8000 kg/m3).
R:  a 
t 
vt 
v a 340

 0.773 m  77.3 cm
 440
v t 5000

 11.36 m

440
E


 E  v 2t  8  103  5  103

2
 2  1011
N
m2
21. In the liquid Helium (below 4.2 K) the sound velocity is 220 m/s. If we know the He density 
= 0.15 g/cm3 find the compressibility modulus. Compare this result with the water
compressibility modulus if we know that the sound velocity is 1460 m/s.
R: v He 
 He

  He  v 2He  150  220 2  7.26 106
 W  v 2W  1000 1460 2  2.1316  109
N
m2
N
m2
 W 2.1316 109

 293.6
 He
7.26 106
22. By a copper wire with the length l = 160 cm and diameter  = 1 mm is suspended a body with
mass 10 kg. If the copper wire elongation is 2.56 mm find out:
a. The stress,  
F
l
, strain,  
and elastic constant, k;
S
l
b. The Young modulus, E;
o
c. How much is the inter-atomic distance if in the unstressed material is R 0  2.56 A ;
13
F mg 4mg 4 10  9.81
N
 2 

 1.25 108 2
2
6
S r
d
 10
m
3
l 2.56 10


 1.6 10 3  0.16%
l
1.6
mg
10  9.81
N
F  Fe  kl  mg  k 

 3.83 10  4
3
l 2.56 10
m
8
F
l
 1.25 10
N
b)  E
   E  E  
 7.8 1010 2
3
S
l
 1.6 10
m
R l
l

 R  R 0  R 0  1.6 10 3  2.56 10 10  4.096 10 13 m .
c)
R0
l
l
R: a)  
23. In front of a microphone a sound has the pressure of 0.441 Pa (N/m2). How much we have to
amplify the sound in order to reach the pain threshold, Ipain = 102 W/m2? We know the air
acoustic impedance Za = c = 441 kg/m2s. How much is the pressure and the sound level?
R:
Nm  NA  NL
 NA  NL  Nm
 Im
N m  10 lg
I
 0
NA  NL  Nm




I 
N L  10 lg  L 
 I0 
I 
I 
I 
 N A  10 lg  L   10 lg  m   10 lg  L 
 Im 
 I0 
 I0 
I L  I pain


p 2 p 2 0.4412
4 W
I

 m c  Z  441  4.41 10 m 2
a

 IL 
 102

 1 


  60 lg 
N A  10 lg    10 lg 

4 
 4.41 
 4.41 10 
 Im 
N A  53.5 dB
24.
Find the relation between the intensity of wave into an absorbed media as a function of depth
of penetration and absorption coefficient.
R:
dI  Idx 
25.
dI
 dx
I

Ix 

I0
x
dI
  dx
I 0
 Ix  
  x
 ln 
 I0 
 Ix   I 0 e x
Hydrogen atom is constituted from a proton and an electron with equal electrical charges but
opposite sign, and absolute value e = 1.610-19 C. In the fundamental state, the radius of electron
first orbit around proton is r0 = 0.5310-10 m. Calculate the attraction force between proton and
electron.
14
R:
FC  


19
1 e2
9 1.6  10


9

10
40 r02
5.3 1011


2
2
 9 
1.62
1093822  0.82 107
2
5.3
FC  8.2 10 8 N
26.
What is the value of equal electrical charges, whit which two identical balls, having the same
mass 1 kg, must be charged, if these are situated in air at 1 mm distance, and the Coulombian
force that acts on each ball is equal in value with the ball weight into a place in which the
gravitational acceleration is g = 9.8 m/s2.
R:
e2
FC   k e 2
rpp
FG  k g
FC
e2
  ke 2
FG
rpp
m 2p
 kg
rpp2
m 2p
rpp2




k e2
9  109
1.6 10 19
 e 2 
k g m p 6.67  10 11 1.67 10 27
2
2

9  1 .6 2
 109381154
6.67  1.67 2
FC
 1.238 1036
FG
27.
Two small identical spheres placed in void, having the same
mass m = 0.1 kg are suspended from a single point with the help
of two isolated wires, not extensible, of neglected mass and with
the same length l = 20 cm. What is the value of the equal charges
with which the two spheres must be charged is such way that the
angle between the fires should be 2 = 900? The gravitational
accelerations is considered g = 9.8 m/s2. (sin 45 = 0.707; tg 45 =
1).
R:
FC  k e
q2
d2
G  mg d  2l sin 
q2
2
 F
tg  C  d
2 G
mg
ke
 
q  2  0.2  sin 45
0
 qd
m  g  tg
 
0.1  9.8  tg 450
9 109
ke


m  g  tg

2  2l  sin
2
2
ke
q  0.4  0.707
q  2.95 106 C
15
0.98 1 4
10  2.95 106
90
28.
Calculate the dependence of electrical field intensity and
electrical potential on the center of a sphere having the
electrical charge uniform distributed in the entire volume. The
total charge is denoted by Q and the sphere radius by R.
R: a) r < R;
  Qi
E
  d s  0

 
Q
E
  n  ds  0i
 Q
 E   ds  i

0

Qi
0
 E  4  r 2 
Q
Q
Qi


3
4r 3
V 4R
3
3
 Qi 
r3
Q
R3


1 r3
Q
r
E  4  r 2 
Q

E

3
0 R
4   0 R 3

dV
E  V  
dV  E  dr 
dr
Vr   
Q 1
4   0 R 3

r
0
V r 
r
V 0
0
   dV   E  dr'
Vr   
r' dr'
 Vr   V0   
r
0
Q 1 r2
4   0 R 3 2
Vr   
Q r'
 dr'
4   0 R 3
Q r2
8   0 R 3
b) r > R;


Qi
 E  ds  


0

dV
E  V  
dr
Vr   VR    
 

V r  
29.
0
 dV  E  dr 
Q
 dr'
R 4 r' 2
0
Vr   VR  
Qi
 E  n  ds  
r
Q
40 r 2
r
1
 d r' 
R

 Q
 E   ds 

0
V r 

V R

Q
 E  4  r 2 
0
r
dV   E  dr'
R
 Vr   VR   
 Vr   VR  
Q
40 r
Calculate the electric field intensity at a
distance a from a rod, uniformly charged with a
total charge Q.
16
Q
40 r 2

1
 dr'
R r' 2

r
Q
Q

40 r 40 R

E
Q
40 r 2
dE 
R:

1 dQ
40 r 2
Q dQ

l
dl
Et
1 dx
40 r 2
l2
0

d x a2  x2

 12
  dx  a
2
 Et  2
 x2

a2  x2  x2

a
2
 x2

3
2
 12
a
40
dx
a
0 r 3  2 40
l2

 1
 x   a 2  x 2
 2
dx 
a
a2
2
 x2

3
2

 32
2 x  dx 
Et 
a 2  dx
l2

l2
 a
0
a
2

two conductors having the radius a and b, with a <
b. Use the Gauss theorem to determine the
dependence of electrical field intensity and electric
potential from the coaxial cable axes. Calculate the
coaxial cable capacity. The cable length is labeled
by l is charged with an electrical charge q.
R: a) r < a;


Qi


0

dV
E  V  
dr


0
 E  ds  


 E0
0
 dV  E  dr  dV  0  V  const
17
 x2

1
2

32

x2
1


a2  x2



1 
x
2



40 a  a 2  x 2

1
Q
40 2a 4a 2  l 2
 E  ds  
 x2
1
Let’s consider a coaxial cable constituted from
30.
dx
2
dx
l2
1  
x
0 a 2  x 2 3 2  2 40 a 0 d a 2  x 2
1  l2
1 Q
l
2

2
40 a
40 a  l
l
l2
a2 
a2 
4
4
1 
Et  2
40 a
a
r
r2  a2  x2
dQ  dl  dx
 dE t   2 sin 
0
sin  
dE t  2dE y  2dE  sin 
l2



0

 dx


b) r > b;
  Qi
E
  d s  0


dV
E  V  
dr
  0
E
  ds  0

 E0
 dV  E  dr  dV  0  V  const
c) a < r < b;
  Qi
E
  d s  0

 
Q
E
  n  ds  0i
 Q
 E   ds  i

0

Q Qi

L
l
 E  2r  l 
 E  2r  l 
l
0

dV
E  V  
dV  E  dr 
dr
Va   Vb  
Vr  
31.
Q
2l 0
Q
ln r 
2l 0

b
a
1
 dr'
r'
C
 
 
 
Q
E

n

ds

E

n

ds

E
1
2
3  n  ds  0i

Qi
0
 E
V b 

V a

2r 0
dV   
b
a
 E
Q 1
 dr'
2l 0 r'
Va   Vb   U 
Q
b
ln 
2l0  a 
Q 2l 0

U
b
ln  
a
Use the Gauss theorem to calculate the expression of a
plane capacitor capacity.
R:
 
  Qi
Q
E

d
s


E
 n  ds  i



0
0
 
 
 
Q
  E  n  ds   E  n  ds   E  n  ds  i
S1
S2
Sl
0
 E   ds  E   ds  0 
S1
S2
 2E  S1 
Qi
;
0
 2E  S1 
  S1
0
Qi
0

 E
 2E   ds 
S1
Qi
0
Q Qi

;
S S1

2 0
Qi    S1
E C  2E 

dV
E  V  
dx
18
Q 1
2l 0 r

0

dV  E  dx 
V b 
b
V a
a
   dV  E   dr'
Va   Vb  U  E  d
32.
U
Va   Vb  Eb  a 


Q
d 
d
0
0  S
C
Q 0  S

U
d
Apply the Ampere’s law to Calculate the magnetic induction a) at a distance r from an
infinitely long conductor b) and inside of an infinitely long solenoid (coil).
R:
a)
 
 I
B
C  d l  0  I  B  C dl  0  I  B  2r  0  I  B  20r
b)
b 
 c  d  a 
 
B

d
l



I

B

d
0
t

 l   B  d l   B  d l   B  d l  0  N  I 
C
a
b
c
d
b
 
B

d
l

0

0

0



N

I

B

0

 dl   0  N  I 
b
a
a
B  l  0  N  I  B 
33.
0  N  I
 0  n  I
l
Two electric currents I1 = 5 A and I2 = 10 A of the same sense are flowing through two parallel
conductors, situated at a distance d1 = 20 cm one from each other. Calculate:
a) The force per unit length with which the two conductors are mutually attracting;
b) Calculate the mechanical work per unitary length spend to move the conductors at distance
d2 =30 cm one from each other. (0 = 410-7 H/m).
R:
F12  B1  I 2  L

a)
 0  I1  
B1 
2  d1 
F12 
 0  I1
 I2  L 
2   d1
F12 4 107  5 10 4  5


105  5 105
2
L
2  20 10
22
b) dL  Fx   dx 
L
d2
 dL'   Fx   dx
0
 I I
F12
 0 1 2
L
2   d1
N
m

d1
19
L
 0  I1  I 2
L  dx 
2  x
d1
d2


2
 0  I1  I 2
 0  I1  I 2
dx  0  I1  I 2
d
L
L  dx 
L 

L  ln x  d2
1
2  x
2
x
2
d1
d1
d2
d

L
L 0  I1  I 2  d 2  4 107  5 10  30 

 ln  
 ln   0.405465 105
L
2
d
2

 20 
 1
34.
d 
 0  I1  I 2
L  ln 2 
2
 d1 
L
 4.05465 10 6
L
An electric current with I = 10 A is flowing through a straight line
infinite long conductor which is placed in the plane of a rectangular frame
parallel with two of his sides, like in the figure. We know the a = c = 20
cm and b = 30 cm. Calculate: a) the magnetic flux through the rectangular
frame; b) the force that acts on each side of the frame; c) the total force
that acts on the frame when through this a I’ = 5 A electric current is
passing through. What is the sense of this electric current to observe an
attraction force?
R:
a)
 
d  B  dS 

b)

ca
ca
ca
 
0 I
0 I
dr
0 d'  S B  dS    c B  b  dr  b  c 2r  dr  b  2  c r
 0 I  b  c  a  4  107  10  0.3  40 
 ln
 ln   4.15888  10-7 Wb

2
2
 c 
 20 
dF  I' B  dl 
a c
F12  F34  I'  
c
b
F23  I' 
0
F 

0

 dF   I' B  dl

 F  I'  B  dl

0 I
 I' I  a  c  4  107  10  5  40 
 dr  0  ln
 ln   6.93147  106 N

2r
2
c
2



 20 
0 I
 I' I b
4  107  5  10 30
 dr  0 


 7.5  106 N
2c  a 
2 c  a
2
40
0 I
 0 I' I b 4 107  5 10 30
F41  I' 
 dr 
 

 15 106 N
2c
2 c
2
20
0
b
 



F  F12  F23  F34  F41
c)
b
 I 'I
a
 I 'I
a
 b

  0
b 
 0
b 2
2
cc  a 
2
2a
c  a c 
7
 I 'I b
4  10  5  10 30
F  0
 

 7.5  10 6 N
4 a
4
20
F  F23  F41 
 0 I 'I
2
20
J
m
35.
The electric current intensity that flow through a conductor is time dependent according to the
following law: I(t) = 2+3t+t2. Calculate:
a)
The electric charge that pass through circuit during the time from t1 = 2 s to t2 = 5 s.
b)
The electric current intensity of a constant current, which leads to the same amount of
electrical charge in the given time.
R:
dQ
dt
I
a)
t2
t2
t1
t1


 Q   dQ   I  dt   2  3t  t 2  dt
t2


3t 2 t 3 
3t 22 t 32  
3t12 t13 
Q   2t 
    2t 2 
    2t1 
 
2
3
2
3
2
3

 t1 
 

3  52 53  
3  2 2 23 
  2  5 
    2  2 
   89.166  12.666  76.5
2
3
2
3

 
Q  76.5 C
I0 
b)
36.
Q
Q
76.5


 25.5 A
t t 2  t1 5  2
The wave function associated with a particle which describe the quantum behavior inside of an
 
box with infinite walls is: n  x   A sin  n
 a

x  , where a is the box width. Using the

probabilistic interpretation of the wave function and the atemporal Schrödinger equation
calculate: a) the normalization constant; b) the particles energy for a = 2 Å, m = 9.110-31 kg
and n = 1, 2, 3, …
R: a) The normalized probability is mathematically given by:
 dP  1 and using the relation dP   x 

2
dx (Copenhagen interpretation of wave function)


  

2
2 
0 dP  1   x  dx  1   A sin  n a x dx  A sin  n a x dx
1
2
2
2
 2 n 
1  cos
x
a
a
2
A2 
a
a 
 2n   A 
 2n  

2
A 
dx 
x dx  
sin 
x 
a 
a  cos
2
2  0
2 n
 a   2 
 a  0 
0
a

A2
a 1 
2
A
2
a
21
2
2

a
2 10 10
A

A  105 m 1/ 2
d 2  2m
b)
E  U   0

dx 2  2
d
n
 n 
A
cos
x
dx
a
 a 
Schrödinger’s equation
d 2
 n 
 n 
  A
x
 sin 
2
dx
 a 
 a 
2
and U = 0
Schrödinger’s equation became:
 n 
 n
 A
 sin 
 a 
 a
2
 2m
 
x   2 EA sin  n
 
 a
 n  2m
 A
  2 EA  0 

 a 
2m
n 2 2
E

a
2
2


2
h2
6.6 10 34

8ma 8  9.110 31  2 10 10

 2 2 2
E1 
1  9.35 eV 
2ma
37.

x  0


2


En 
 2 2 2
n
2ma
6.6 2 10 68
 1.495 10 18 J   9.35eV 
 51
8  9.1 2  10
 2 2 2
E2 
2  37.4 eV 
2ma
 2 2 2
E3 
1  84.1 eV 
2ma
An atom from a Si crystalline network can be considered as a quantum particles inside a
potential field of elastic forces U x   kx2 / 2 which oscillate with the frequency  = 2.51013
Hz. The wave function in fundamental state is x  Aex . Use the Schrödinger equation to
2
calculate: a) the constant  and b) the particles energy if the atom mass is m = 38.2  10-28 kg.
x  Aex
R: a)
2 x 2
4 A x e
2
 2 Ae
2
x 2
2
2
2
d
d 2
 2 A    x  e x and
 4 A 2 x 2 e x  2 Ae x
2
dx
dx
2m 
kx2  x 2

 Ae
 2 E 
0
2 
 
 2 2m k  2  2m

 4  2  x   2 E  2   0 for any x values.
 2



 2 2m k 
 4   2 2   0



 2m E  2   0
  2


mk
 
2 and
from where 
2
 E  

m
22
k
m m2 2 2 m



h
m
2
h
2
2
2
 27
13
2 38.2  10 2.5  10
2 2 38.2  2.5 10 27  1013


 2.856  10 22 m 2
6 .6
6.6  10 34
10 34

mk
m

2
2
E
mk  2  k  h



2 m 2 m
2
2
The expression of energy of quantum oscillator is:
1

E n   n   h
2

E
results the fundamental state with n = 0.
h 6.6 10 34  2.5 1013

 8.25 10  21 J   5.15 10  2 eV 
2
2
A particle of mass m = 10-28 kg is described by the wave function: x, t   Ae i t  kx  , with the
38.
normalization constant A = 105 (m1/2),  = 1015 (rad/s) and k = 1010 (m-1). Applying the
quantum mechanical operators calculate the particles: energy, momentum (impulse),
kinetically and potential energy.

R: a) Eˆ  i
t
i
and Ê  E

 i 2 Ae i t  kx   Ae i t  kx     E
t
E   
E  3.310 19 J   2.06eV 
b) pˆ 
 
i x
pˆ  
  
 Aike i t  kx   kAe i t  kx   k  p
i x i
and p̂  p
 1010  6.6 10 34
m
p  k 
 3.3 10  24 kg
2
s
mv 2
pˆ 2
2 2
ˆ
Ec 
 T

2
2m
2m x 2

p2
3.3 10 24
T

2m
2 10  28

2

3.32
10  20  5.445 10  20 J   0.34eV 
2
E  T  U  U  E  T  2.06  0.34  1.72eV 
23
6.6 10 34  1015
2