Benha University
Faculty of Engineering – Shoubra
Engineering Mathematics & Physics
Department – Preparatory year
Final Term Exam
Physics (A)
Date: 31 / 12 / 2016
Duration: 3 hours
 Answer all the following question
 Illustrate your answers with sketches when necessary
 The exam. consists of two pages
Question (1)
 No. of questions: 6 questions
 Total Mark: 115 Marks
 The first page
(17 marks)
(A) A solid conducting ball has a charge of 1.4x10-7 C. What would be its charge if it received 1012 electron?
Answer:
qo = +1.4x10-7 C
q' = 1012 x 1.6x10-19 = 1.6x10-7
q (final) = qo - q'
= 1.4x10-7 - 1.6x10-7 = - 2x10-8 C
(B) In the opposite figure, three charges q1 = q3 = 5 μC and q2= – 3 μC are located at on an arc of radius r = 3
cm, (i) Find the total electric field at the center the arc?
(ii) Use your answer to Calculate the electric force exerted on an electron when placed at point P?
Answer:
q1 = q3 = 5x10-6 C
q2= – 3x10-6 C
E3 sin30
E3
q1
r1 = r2 = r3 = 3 x10-2 m
(i) E1 = K q1 /r12 = 9x109 x 5x10-6 / (3 x10-2)2 = 5x107 N/C
E1= E3 = 5x107 N/C
2
E2 = K q2 /r2 =
9x109
q2
x
3x10-6
/ (3
x10-2)2
=
3x107
E2
30o
r
30o
E3 cos30
30
30 E cos30
1
P
N/C
E1
Ex = E1 cos30 + E3 cos30 – E2 = 5.66x107 N/C
q3
E1 sin30
Ey = E1 sin30 – E3 sin30 = 0
Eres = 5.66x107 N/C
directed along positive X-axis
(ii) qo = 1.6x10-19 C
F = E qo = 5.66x107 x 1.6x10-19 = 9.1x10-12 N
directed in negative X-axis
7. Electric field for two charged parallel plates
that
charge
( σfield
andof–σ
) per unit
(C) Ahave
uniform
electric
magnitude
800area:
N/C exists between two oppositly charged parallel nonconducting plates each of area 10 cm2 and are separated 4 cm apart, (i) Find the surface charge density
+
–
+
–
E1
+
–
– positive plate and at the same instant an electron is released from the
(ii) If a proton is +released from the
+
–
E2
+
–
negative plate. Determine
the distance
from the positive plate at which the two particles pass each other?
+
–
+
–
7. Electric field for two charged parallel
(Ignore the electrical
attraction between
the electron and the proton)
on each plate?
Answer:
E = 800 N/C
1
(i)
A = 102cm2
d = 4 cm
E res  E1  E 2






2 o 2 o
o
σ = εo E = 8.85x10-12 x 800 = 7.08x10-9 C/m2
p
that have charge ( σ and –σ ) per unit area
+
+
+
+
+
+
+
+
E1
E2
–
–
–
–
–
–
–
–
2
1
E res  E1  E 2



7. Electric field for two charged parallel p
that have charge ( σ and –σ ) per unit are
(ii) From Newton’s law,
x = vo t + ½ a t2
x = vo t + ½ (qE/m) t2
+
+
+
+
+
+
+
+
x = vo t + ½ (qE/mp) t2
For the proton,
x = ½ (qE/mp) t2
For the electron,
4 – x = vo t + ½ (qE/me) t2
4 – x = ½ (qE/me)
Dividing the two equations,
x
4-x
2
1
t2
–
–
–
–
–
–
–
–
E res  E1  E 2
x / (4 – x) = me / mp

x / (4 – x) = 9.1x10-31 / 1.67x10-27





2 o 2 o
o
x / (4 – x) = 5.45x10-4
x / (4 – x) = 1/1835
1835 x = 4 – x
x = 0.0022 cm
Question (2)
(18 marks)
(A) Define the following:
Answer:
(i) Conservative force.
A conservative force is the force by which a body would move from one point to another point and the
work done does not depend on the path between the two points.
Or,
When an electrical force acts on a test charge (qo) that moves from point (a) to point (b) , the work
done does not depend on the path between the two points
(ii) Electron Volt.
The change in potential energy when an electron would move between two points that differ in potential
by one volt
1 eV = 1.6 x 10-19 J
(B) Explain Gauss’s law? Apply Gauss’s law to Calculate the electric field near an infinite line of charge of
circular cross-section?
Answer: Gauss’s law states that: the surface integral of the normal component of the electric field
intensity over a closed surface is equal to (q / εo)
   E cos θ dS 
q
o
Where, q is the charge enclosed by the surface.
Consider point p at distance R from a charged rod
E
R
of charge λ per unit length.
- apply Gauss’s law:
E
E
 E dS cosθ 
q
o
+ + + + +
+ + + + +
L
The above equation can be calculated by dividing the
integration into three parts as:
E
dS cosθ 
surface
E
cap 1

q in
 E dS cos(90
)
o
)
cap 1
cap 1
 E 2πR L 
o
cap 2
 E dS (1)   E dS (0)   E dS (0) 
surface
o
cap 2
 E dS cos(0)   E dS cos(90
surface
dS cosθ 
E
dS cosθ 
cap 2
L
o
L
o
E 

L
o
1 λ
2π o R
q1
15 cm
B
(C) In the figure shown, q1 = – 5 µC and q2 = 2 µC are held at two corners of
a rectangle whose lengths are 15 cm and 5 cm.
(i) Calculate the potentials at corner A and corner B?
5 cm
A
q2
(ii) Find the work required in moving a proton from B to A along the diagonal of the rectangle?
Answer:
q1 = – 5x10-6 C
q2= 2x10-6 C
r1 = 5 x10-2 m
r2 = 15 x10-2 m
(i) VA = K q1 /r1 + K q2 /r2
= 9x109 x – 5x10-6 / 5 x10-2 +
= – 9x105 + 1.2x105 = – 7.8x105
9x109 x 2x10-6 / 15 x10-2
VB = K q1 /r2 + K q2 /r1
= 9x109 x – 5x10-6 / 15 x10-2 +
= – 3x105 + 3.6x105 = 0.6x105
9x109 x 2x10-6 / 5 x10-2
(i) W = q (VB – VA)
= 1.6x10-19 [0.6x105 – (– 7.8x105)] = 1.344x10-13 J
Question (3)
(23 marks)
(A) Define the capacitance? Derive an expression for the capacitance of parallel plate capacitor?
Answer:
The capacitance (C) of a capacitor is defined as: the ratio of the magnitude of the charge on either
conductor to the potential difference between them.
Coulmb
Q
(Farad)
C 
Volt
V
The electric field for conducting
plates is σ/εo
b
Va  Vb 
E
dx
a
Where, a is a point at the first
plate and b is a point at the
second plate. So,
b
Vab  E
 dx
a
+
+
+
+
+
+
+
+
E
a
b
d
–
–
–
–
–
–
–
–
X
 Vab 
σ
εo
x
b
a
 Vab 
Q A
b - a 
εo
 Vab 
Q
d
A εo
Then,
+Q
A
-Q
d
Q
A
 εo
Vab
d
 C  εo
A
d
Or , another accepted proof
E 
And,
Vab
d
σ
Q A

εo
εo
E 
Vab
Q A

d
εo
Then,

Q
A
 εo
Vab
d
C  ε
o
A
d
(B) A 100 pF capacitor is charged to a potential difference of 50 V and the charging battery is disconnected.
The capacitor is then connected in parallel with a second (initially uncharged) capacitor. If the measured
potential difference drops to 35 V. What is the capacitance of the second capacitor?
Answer:
C1 = 100x10-12 F
V = 50 V
With C1 and C2 in Parallel , V’ = 35 V
C1 = Q / V
100x10-12 = Q / 50
Q = 5x10-9 C
In case of C1 and C2 in Parallel, then
And,
C’ = C1 + C2
C’ = Q / V’
C1 + C2 = Q / V’
100x10-12 + C2 = 5x10-9 / 35
C2 = 4.29x10-11 = 42.9 pF
(C) An aluminum wire 6.75 m long and 4 mm in diameter carries a current of 2 A. The wire dissipates energy
at the rate of 60 mW. (i) Find the conductivity of aluminum?
(ii) What is the strength of electric field present in the wire?
Answer:
L = 6.75 m
and,
D = 4x10-3 m
so,
r = 2x10-3 m
A = πr2 = π (2x10-3)2 = 1.26x10-5 m2
P = 60x10-3 W
i=2A
(i) P = i2R
R = P / i2
= 60x10-3 / (2)2 = 1.5x10-2 Ω
σ = 1/ρ = L / R A
= 6.75 / [1.5x10-2 x 1.26x10-5] = 3.6x107 Ω-1m-1
(ii) E = V / L
= iR / L = 2 x 1.5x10-2 / 6.75 = 4.44x10-3 V/m
(D) (i) In the circuit shown, use Kirchhoff’s rules to Calculate the current in each branch?
1818V,V,1 1ΩΩ
(ii) If the potential of point (e) is taken to be zero, Find the potential
of point (f)?
Answer: (i) from Kirchhoff’s rules,
i1 = i2 + i3
14 i1 + 18 i2 = 30
10 Ω
…..(2)
14 i3 – 18 i2 = 24
…..(3)
Solving the three equations,
14 (i2 + i3) + 18 i2 = 30
14 i3 + 32 i2 = 30
……(4)
From (3) and (4),
14 i3 + 32 i2 = 30
……(4)
14 i3 – 18 i2 = 24
…..(3)
50 i2 = 6
so,
14 i1 + 18 (0.12) = 30
i1 = 1.989 A
1.99 = 0.12 + i3
so,
i3 = 1.869 A
(ii) Using Kirchhoff’s rules, for Ve = 0
Ve –10 i2 + 12 – 1 i2 – 7 i2 = Vf
Vf = 9.84 V
f
I2
12 V, 1 Ω
d
13 Ω
–10 i2 + 12 – 1 i2 – 7 i2 – 36 + 1 i3 + 13 i3 = 0
so,
7Ω
a
Take loop, efdae
From (1) in (2),
c
e
+ 8 i1 – 18 + 1 i1 + 5 i1 + 7 i2 – 12 + 1 i2 + 10 i2 = 0
From (1),
5Ω
8Ω
….(1)
Take loop, ebcfe
From (2),
I1
b
i2 = 0.12 A
I3
36 V, 1 Ω
Part (2): Properties of Matter
Question 5 (15 Marks)
(A) (4 Marks)
i- Poisson's ratio:
ii- Ultimate strength (The Breaking Stress):
Is the stress required to cause actual breaking of a material
(B) (6 Marks)
Suppose that a wire has original length Li and is stretched by a length ΔL when a
force F is applied at one end. If the elastic limit is not exceeded (the extension is
directly proportional to the applied force)
This work is stored in the wire as elastic potential energy (E.P.E)
(C) (5 Marks)
Given [Amplitude A=2cm, Vmax=12 cm/sec, T=?]
The speed is maximum when the body path through the center of the motion
12 cm / sec  ω x 2cm
 ω  6 sec-1
2

ω

Question 6 (20 Marks)
T
T 
2 2 x3.14
Center

 1.047 sec
ω
6
(A) (5 Marks)
By using the parallel axis theorem
I axis  I center  Mh 2
1
L
ML2  M( ) 2
12
6
1
 ML2
9
I axis 
I axis
L/6
(B) (5 Marks)
 y ( x,0)  A sin(
2

x)
(C) (5 Marks)
Given [y = (2mm) sin [(20 m-1) x - (600 s-1) t , T=15 N.]
 y  A sin( kx  t )
 A  2 mm, k  20 m 1 ,   600 sec 1
 

k
v 
 

600
 30 m / sec
20
(i )
T

T
15

 0.017 Kg / m  17 g / m
2
v
30 2
(ii )
(D) (5 Marks)
Given [A=9.8 mm, φ=1000]
Because they are identical, the waves have the same amplitude A=9.8 mm. Thus, the amplitude
of the resultant wave is given by equation

100 0
)  13mm
Amplitude of the resultant wave= 2 A cos( )  (2)(9.8mm) cos(
2
2