Download Connection of Capacitors

Electromagnetic 1
Chapter II
CHAPTER II
Capacitors
1
Prof. Dr. T. Fahmy
Electromagnetic 1
Chapter II
Prof. Dr. T. Fahmy
After completing this chapter, the students will know:
 The concept of the capacitor
 The different types of capacitors' connection.
 The energy of charged capacitor.
 The different types of material (insulator, semiconductor and
conductor).
 The dielectric constant.
 The induced charge and induced electric field.
2
Electromagnetic 1
Chapter II
3
Prof. Dr. T. Fahmy
Electromagnetic 1
Chapter II
Prof. Dr. T. Fahmy
CHAPTER II
Capacitors
2.1: Connection of Capacitors
2.1.1: Capacitors in series connection
C1
C2
C3
C4
q
q
q
q
V1
V2
V3
3
V4
3
B
A
V
Note that,
In series connection
a) The potential difference V between the points A and B equals to the
summation of potential difference on all capacitors, i.e.,
V  V1 V2  V3  V4
(2.1)
b) The capacitors have the same charge q, i.e.,
q1  q2  q3  q4  q
(2.2)
q
C
(2.3)
But as known,
V 
Therefore,
q
Ctotal

q
q
q
q
q



     
C1 C 2 C 3 C 4
Cn
Then:
4
(2.4)
Electromagnetic 1
Chapter II
1
Ctotal

Prof. Dr. T. Fahmy
1
1
1
1
1




C1 C 2 C 3 C 4
Cn
(2.5)
The equation (2.5) can be written in general form as follow
1
C total
n
1
i 1 Ci

(2.6)
C1 , q1
2.1.2: Capacitors in parallel Connection
C2 , q2
Note that,
In parallel connection
a) The total charge between points A and B
equals to the summation of charges on all
capacitors, i.e.,
C3 , q3
C4 , q4
q  q1  q2  q3  q4
(2.7)
A
b) The capacitors have the same voltage V, i.e.,
V1 V2 V3 V4  V
(2.8)
Also
q1  C1 V , q 2  C 2 V , q3  C 3 V , qt  C t V
(2.9)
Therefore,
C total V  C1 V  C 2 V  C 3 V  C 4 V      C n V
C total  C1  C 2  C 3  C 4        C n
(2.10)
and in general form, the total capacitance can be written as follow
n
Ctotal   Ci
(2.11)
i 1
5
V
B
Electromagnetic 1
Chapter II
Example (2.1):
In the following figure, calculate
1- The charge on each capacitor
2- Determine the voltage at the
point b
Prof. Dr. T. Fahmy
C2= 4F
a
Va= 1200 V
Solution:
Note that, C2 and C3 are connected in parallel.
c
b
C1= 3F
C3= 2F
Therefore, the resultant
capacitance of these capacitors is:
C2,3  C2  C3  4  2  6 F
Also, the total capacitance of the circuit is
1
1
1 1 1

  
Ct C2,3 C1 3 6
1 1 1 21 3
  

Ct 3 6
6
6
6
Ct   2 F  2 x10 6 F
3
Therefore, the total charge of the circuit is
qt  Ct V  2 x106 x1200
qt  2.4 x103 C
Since, C1 and C2,3 are connected in series, the voltage between the points a and
b is
Vab 
q1 2.4 x10 3

 800 V
C1
3 x10 6
 Va  1200 V
Vb  1200  800  400 V
Also
Vbc  400 V ,
because Vc  0
Also,
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Electromagnetic 1
Chapter II
Prof. Dr. T. Fahmy
q2  C2 x Vbc  4 x10 6 x 400 1.6 x103 C
q3  C3 x Vbc  2 x106 x 400  0.8x103 C
Also
q 2  q3 1.6 x103  0.8x103
 2.4 x103 C
Example (2.2):
Four capacitors are connected as shown
in the figure. Calculate the following:
 The equivalent capacitance between
the points a and b.
 Calculate the charge on each
capacitor if Vab=15 V.
C1=15 F
C2=3 F
C4=20 F
a
b
C3=2.5 F
Solution:
 The equivalent capacitance can be calculated as follow:
Note that:
C1 and C2 are connected in series, so
C1,2=2.5 F
1
1
1
1 1 1 5 6
 
  

C1, 2 C1 C 2 15 3 15 15
C4=20 F
15
C 1, 2   2.5 F
6
a
b
C3=2.5 F
Also, C1,2 and C3 are connected in parallel, so
C1, 2,3  C1, 2  C3  2.5  2.5  5  F
Also, C1,2,3 and C4 are connected in parallel, so
1
1
1 1 1 4 1 5


  

Ceq C1, 2,3 C4 5 20 20 20
Ceq 
C1,2,3=5 F
C4=20 F
a
20
4  F
4
b
Ceq =4 F
 The total charge can be calculated as follow:
Qtotal  Ceq x V  4 x106 x15  6 x105 C
Therefore,
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a
b
Electromagnetic 1
Chapter II
Prof. Dr. T. Fahmy
The charge on Q on 20 F capacitor = 6x10-5 C
and Voltage (V) on that capacitor is
V
Q 6 x105

 3 volt
C 20 x10 6
Therefore, the voltage on the rest of the circuit is
15-3= 12 volt
Also,
Charge on Q on 6 F capacitor = 6x10-6 x12= 7.2x10-5 C
Also,
Charge on Q on 15 F capacitor = 6x10-5 - 7.2x10-5 = -1.2x10-5 C
Then
Charge on Q on 3 F capacitor = -1.26x10-5 C
2.2: Energy of a Charged Capacitor
The work done to transport a charge of dq through a capacitor is
expressed as follow
dwV dq
(2.12)
Where V is the potential difference
but, as known
V
q
C
Then
dw
q
dq
C
(2.13)
Therefore, the total work done (W) or the energy (U) of the charged capacitor is
w
w   dw 
0
w
1 Qq
dq
C 0 C
1 Q2
U
2 C
(2.14)
Hence,
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Electromagnetic 1
Chapter II
Prof. Dr. T. Fahmy
1 Q2
2 C
1
U  QV
2
1
U  CV 2
2
U
(2.15)
If the capacitor is considered as two parallel plates with cross-sectional
area of A, the distance between the plates is d and the total surface charge
density = Q/A.
This leads to the capacitance of the capacitor can be
expressed as follow:
d
A
d
C  0
and
A
Q  A
1 Q2
2 C
1  2 A2 1  2 A2 d
U 

2 C
2 0 A
U 
U 
1  2 Ad
2 0
(2.16)
But as known,
E

0

  E 0
(2.17)
Then, from equation (2.17) in (2.18), we can obtain
U
1  02 E 2
Ad
2 0
1
U  0 E2 Ad
2
( 2.18)
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Electromagnetic 1
Chapter II
Prof. Dr. T. Fahmy
Therefore, the energy density is u
U
, where, V is the volume  A d
V
1  E2
u  0
Ad
2 Ad
u
1
u  0 E2
2
(2.19)
C3= 3 F
Example (2.3):
For the system of capacitors shown in the figure,
Calculate the following:
 The equivalent capacitance
 The potential across each capacitor
 The charge on each capacitor
 The total energy stored by the group
Solution:
 The equivalent capacitance can be calculated as follow:
Note that:
C1 and C2 are connected in parallel, so
1
1 1 1 1 2 1 3
 
  

C1, 2 C1 C2 2 4
4
4
C1, 2
C1= 2
F
C
C11=
=2
2
F
F
C2= 4 F
90 V
C3,4= 2 F
C1= 2 F
C1,2=C1.3
FF
1= 2
4
 1.3 F
3
Also,
C3 and C4 are connected in parallel, so
1
1 1 1 1 2 1 3
 
  

C3, 4 C 3 C4 3 6 6
6
C3, 4
C4= 6 F
90 V
Ceq= 3.3 F
6
  2 F
3
Therefore, the equivalent capacitance (Ceq) is:
Ceq  C1, 2  C3, 4 1.3  2  3.3 F
90 V
 The total charge can be estimated as follow:
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Electromagnetic 1
Chapter II
Prof. Dr. T. Fahmy
Qtotal V Ceq
 90 x 3.3x10 6  2.97 x10 4 C
 The total energy stored by the group
1
U total  C V 2
2
1
 x 3.3 x10  6 x (90) 2
2
1.3 x10  2 J
Example (2.4):
C1=20 F
V1=1000 v
C2=10 F
V2=100 v
Determine the energy of these two capacitors
before and after the connection?
Solution:
1- Before connection
q1  C1 V1  20 x106 x103  2 x102 C
q2  C2 V2 10 x10 6 x102 10 3 C
Also, the total energy (U) can be calculated as follow
1
1
U  C1 V12  C2 V22
2
2
 
 
2
1
1
 x 20 x10 6 x 103  x10 x10 6 x 102
2
2
10  0.05 10.05 J
2- After connection
The total energy (U) can be calculated as follow
11
2
Electromagnetic 1
Chapter II
Prof. Dr. T. Fahmy
1 Q2
,
2 C
Q (total ch arg e)  q1  q2  0.02  0.021  0.021C
U
C (total capaci tan ce)  C1  C2  20  10  30 F  30 x10 6 F
1 0.021
 7.35 J
2 30 x10 6
2
U
Example (2.5):
Squared parallel plate capacitor with capacitance of 4.65x10-10 F and having
0.4 mm gap between the plates is filled with a material of dielectric constant k=
2.1 and E= 6x107 V/m. Calculate the following:
 The voltage (V)
 The energy (U)
Solution:
 The voltage (V)= E x d= 6x107 x0.4x10-3 = 2400 V
 The energy (U) = ½ CV2
= ½ x 4.65x10-10 x (2400)2
= 0.13 J
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Electromagnetic 1
Chapter II
Prof. Dr. T. Fahmy
Dielectric Materials
The materials can be classified according to its electrical conductivity
into different types:
1- Insulators or Dielectrics
These materials are characterized by very low electrical conductivity
and very high resistance. These materials do not have free electrons.
2- Semiconductors
Semiconductor materials are classified into two types as follows:
a) Intrinsic semiconductors
These materials are characterized by electrons and holes as charge
carriers.
b) Extrinsic semiconductors
These materials are classified into two types
 n- type
The majority of charge carriers are electrons in this type, whereas, the minority
of charge carriers are holes.
 p- type
The majority of charge carriers are holes in this type, whereas, the minority of
charge carriers are electrons.
3- Conductor Materials
These materials are characterized by high electrical conductivity due to free
electrons.
4- Superconductor Materials
These materials are characterized by very high electrical conductivity and
resistance equals to zero.
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Electromagnetic 1
Chapter II
Prof. Dr. T. Fahmy
2.4: Dielectric Constant
Dielectric constant is defined as the ratio between the capacitance of a
capacitor ( C) to the capacitance of the same capacitor when it contains a space
between its plates (C0)
K
C
C0
(2.20)
C Q
Also, as known
V E
C
Q
V
C0 
Q
V0
(2.21)
d
Substituting from equation (2.21) in equation (2.20)
C0 Q
Q V0
x
V Q
V
K 0
V
K
V0 E0
d
(2.22)
Also,
V
d
and
E
E0 
 V  E d
V0
 V0  E0 d
d
(2.23)
Then from equation (2.23) in equation (2.22), we can get
14
Electromagnetic 1
K
Chapter II
Prof. Dr. T. Fahmy
E0 d
Ed

K
E0
E
(2.24)
Also, as we know
E
q
4r

2
q 1


2
4r 


E
q
40 r
2

q 1


2
4r  0
0
(2.25)
Then from equation (2.25) in equation (2.24), we can get
K


x
0 

K

0
(2.26)
Finally, the dielectric constant can be expressed as follow:
K
C
C0
or
V0
V
E0
E
or
or

0
2.5: The relation between the induced charge and dielectric
constant
Suppose that, the electric field in case of space
between the plates of the capacitor is E0, whereas, will
be E in case of existence of a dielectric material between
qi
the plates.
15
q
Electromagnetic 1
Chapter II
Prof. Dr. T. Fahmy
Therefore:
E  E0  Ei
(2.27)
where, Ei is the induced electric field due to the induced charge qi inside
the capacitor.
By substituting from equations (2.25) and (2.26) in equation (2.27), we
can write the following equation:

K 0



i

0
0
   i
K
i   
i

K
(2.28)

1 
 
1  K 



But,
i 
qi
A

and

q
A
Where A is the area
Then
qi
q

A
A
qi
q

1 

1  K 



(2.29)

1 

1  K 



Example (2.6):
16
Electromagnetic 1
Chapter II
Prof. Dr. T. Fahmy
A capacitor with two parallel of charge equals to 30 C, the area of each
plate is 1 m2. If a dielectric with permittivity () of 15x10-12 C2/N. m2 filled the
region between the plates, calculate the following:
1- The resultant electric field of the dielectric ‘E’.
2- The density of the induced surface charge (i).
3- The resultant of the electric field due to the induced charge ‘Ei’.
Knowing that,  = 8.85x10-12 C2/N. m2
Solution:
q= 30 C
A= 1m2
= 15x10-12 C2/N. m2
 = 8.85x10-12 C2/N. m2
1- As known,
q
30 x10 6

 30 x10  6
C /m2
A
1

30 x10  6


 2 x10 6 V / m

15 x10 12
 
E
2
1 



 i   1 
K 


 
8.85 x10 12 

 i   1  0   30 x10  6 1 

15 x10 12 


6
 i  12.3 x10 C / m 2

317

Electromagnetic 1
Ei 
Chapter II
Prof. Dr. T. Fahmy
i
12.3x10 6

 0 8.85 x10 12
 1.39 x10 6
V /m
Example (2.7):
An air filled capacitor consists of two parallel plates each with an area of 7.6
cm2, separated by a distance of 1.8 mm. If a 20 V potential difference is
applied to these plates, calculate the following:
 The electric field between the plates (E).
 The surface charge density ()
 The capacitance
 The charge on each plate
Solution:
 The electric field between the plates (E)
V
20

d 1.8 x10 3
 1.11x10 6 V / m
E 
 The surface charge density ()
E 

    E x0
0
 8.85 x10 12 x1.11x10 4  9.83 x10 8 C / m 2
 The capacitance
C
 0 A 8.85 x1012 x7.6 x104
d

1.8 x10
3
 3.47 x1012 F
 The charge on each plate
18
Electromagnetic 1
V
Chapter II
Q
 Q  C xV
C
 3.47 x1012 x 20  7.48 x1011 C
19
Prof. Dr. T. Fahmy