Air University
Mid Term Examination, Spring-2014
PH-106- Physics 106 Max. Marks: 20 Date: 27-03-2014 Time
Allowed: 2 hrs
Instructors: A. Sadiq, M. Anwar, M. Atif, R. Nasir, A. Zulqarnain, K. Ali
Note: Attempt all 5 questions. All questions carry equal marks. Please encircle your final answers. You may use basic scientific calculator. DO NOT write in pencil or red ink on answer sheet .
1. Fig 1 shows two point charges distance of
Q

4 nC and q
 
2 nC that are fixed in their places at a a

20 mm from each other. Take the location of charge Q as the origin. a) At what point does the electric field intensity vanish? b) At what point does the electric potential vanish? c) Are these the only points where these quantities vanish?
1.5
1.5
1.0
Solution : a) The location of the zero electric field intensity for two similar and unequal charges lies outside on the line joining and them and closer to the smaller of the two charges. Let x be its distance from charge Q then its distance from charge q is x
 a . This gives,
Q q
Fig 1 x
2
Q
2
 x

 q
 a

2 a x

2

 x
2

0 , or
, giving, x
2 x


4 ax
4 a


2 a
2 a
2
2
4


2
0

, or,

2

2
 a

68 .
3 mm b) We first look for a point on the line joining the two charges where the electric potential might vanish. Such a point is expected to lie between the two charges. Again let x be its distance from charge Q . Then its distance from charge q will be a
 x , giving,
Q
2 x
 a

 q
 a
 x
 x

 x
3 x

2 a , or,

0 , or,
, giving, x

40

13 .
3 mm
3 c) The point where the electric field vanishes is the only such point. However there are infinitely many points where the electric potential vanishes. All such points lie on a surface known as the equipotential surface.
2. Fig 2 shows a thin uniformly charged ring of radius R

10 cm of total charge q

0 .
1

C .
a) What is its charge density? 0.5 b ) Derive the expression for magnitude and direction of electric field intensity E at the point P on its axis at a distance c) Find the values of E for i) z

0 z from its center
, ii) z
 
10 cm
O . and iii)
 z

100 cm
2.0
0.5 d) Sketch E as a function of z . 1.0
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Solution : a) Charge density
  q
2

R

10

7
2

3 .
14

0 .
1

0 .
16

Cm

1 b) A point on the axis of the ring at a distance x from its center is at the same distance r from all points on the ring that is given by, r
 z
2 
R
2
The magnitude of the electric field intensity due to two infinitesimally small parts of the ring lying at the opposite ends of its diameter is given by, dE

2 dq
4

0 r
2 r z

4

0
2 dqz
 z
2 
R
2
3

/ 2
Its direction is along the axis of the ring. The total electric field intensity at this point due to the charged ring is obtained by integrating the above expression over the entire ring, giving,
E

4

0
2 qz
 z
2 
R
2
3

/ 2
 k c) From the above expression we have,
E
 z

0


4

0
 z
2
2 dqz

R
2
3

/ 2

0
E
 z

10 cm


9

10
9
2

2

10

7
2

10

3

0 .
1

6 .
4

10
5
NC

1
E
 z

100 cm


9

10
9 
2

10

7 
1
1

640 NC

1
2 2

1
Fig 3
E
  z d) Fig 3 shows a sketch of the electric field as a function of the distance of the point from the center of the ring.
3. Consider a large flat uniform sheet of charge of charge density
 
10

Cm

2 . a) What is the direction of the electric field at a point near the sheet and how does its magnitude vary as this point is moved, i) parallel, and ii) perpendicular to the sheet? 1.0 b) Find the magnitude of the electric field at a point near the sheet. c) Calculate the work done in placing a charge Q

10 nC
1.0 on a conducting sphere of radius R

2 cm . 2.0
Solution : a) The direction of the electric field intensity at a point near a large sheet of charge is perpendicular to the sheet. Its magnitude doesn’t change as this point is moved either parallel or perpendicular to the sheet so long as this point is far from the edges of the sheet and its distance is much smaller than the linear dimensions of the sheet. b)
Let’s consider a Gaussian surface in the shape of a cylindrical pill box with each of its two flat surfaces of area A that are parallel to and lie on the opposites sides of the sheet of charge. Electric flux through its cylindrical surface is zero and that through its two flat surfaces is given by,
 
2 EA
Also the charge enclosed inside this cylindrical surface is given by, q encl
 
A
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P z
From Gauss’s law, therefore, we have,
2
E
EA

 q

 encl
0
2

0
 n
, or,

2

4

0
 n

9

10 9 
6 .
28

10

5 
5 .
6

10 5 NC

1 Fig 2
O
R c) Electric potential at the surface of a conducting sphere of radius R and charge q is given by,
V
 q
4

0
R
Work done to put an additional charge dq on its surface is given by, dW

Vdq
 qdq
4

0
R
,
Therefore total work done to put charge Q on the sphere,
W
 
Q
0 qdq
4

0
R

Q
2
8

0
R

9

10
2


9 
10

16
2

10

2

0 .
25

10

5
J
4. A parallel plate capacitor of disconnected from a V 10 V
A

0 .
01 m
2 with air gap between the plates
 power source after it is fully charged. d

0 .
1 mm is a) Find its capacitance and the charge on it. 1.0 b) Find its capacitance, the charge on and the potential difference across its plates after a dielectric material of
 
5 is inserted between its plates. 1.5 c) What will be your answer to part (b) if the dielectric sheet is inserted while the capacitor remains connected to the power source?
Solution : a) q

CV

8 .
9
C

10



10
0
A
 d

10

8 .
9

10
10
8 .
9 nC

12

4

10

2

8 .
9

10

10
F
1.5 b) Capacitance with the dielectric material between its plates
C '
 k
 d
0
A

5

8 .
9

10

10
F

4 .
45 nF
Since the capacitance is disconnected from the power source the charge on it remains the same.
V '
 q
C '

5
8 .
9

10

9

8 .
9

10

10

2 V c) If the dielectric sheet is inserted while the capacitor remains connected to the power source then the potential difference across its plates remains the same. He charge stored in it changes to, q '

C ' V

4 .
45

10

9 
10

44 .
5 nC
5 . a) What is the magnitude and direction of the electric field E

3 i

4 j ? 1.0 b) Sketch at least three lines of force corresponding to this field. 1.0 c) Find the potential difference between two points with coordinates ( 0 , 0 ) and ( 1 , 0 ) . 1.0 d) Sketch at least three equipotential surfaces corresponding to this field. 1.0
Solution : a) For E

3 i

4 j
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E

3
2 
4
2 
5

E

3 i

4 j
5

0 .
6 i i

0 .
8 j b) Fig. 5 gives a sketch of the lines of force
Fig 4 c) The potential difference between the two points
A ( 0 , 0 ) and B ( 1 , 0 ) is given by,

V
 
B
A
E .
d l
 
1
0
E x dx

3 V d) Since the given electric field is uniform its equi-potential surfaces are planes perpendicular to its lines of force. Fig 6 shows a sketch of the projection of these surface in the x-y plane.
Fig 6
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