Download Series and parallel Connection of capacitors

A dielectric medium is said to be linear when  e is independent of E and the medium is
homogeneous if  e is also independent of space coordinates. A linear homogeneous and
isotropic medium is called a simple medium and for such medium the relative
permittivity is a constant.
Dielectric constant  r may be a function of space coordinates. For anistropic materials,
the dielectric constant is different in different directions of the electric field, D and E
are related by a permittivity tensor which may be written as:
 Dx  11 12 13   Ex 
 D   
 
 y   21  22  23   E y 
 Dz   31  32  33   Ez 
For crystals, the reference coordinates can be chosen along the principal axes, which
make off diagonal elements of the permittivity matrix zero. Therefore, we have
 Dx  1 0 0   Ex 
D    0 
0   E y 
2
 y 
 Dz   0 0  3   Ez 
Media exhibiting such characteristics are called biaxial. Further, if 1   2 , then the
medium is called uniaxial. It may be noted that for isotropic media, 1   2   3 .
Lossy dielectric materials are represented by a complex dielectric constant, the imaginary
part of which provides the power loss in the medium and this is in general dependant on
frequency.
Another phenomenon is of importance is dielectric breakdown. We observed that the
applied electric field causes small displacement of bound charges in a dielectric material
that results into polarization. Strong field can pull electrons completely out of the
molecules. These electrons being accelerated under influence of electric will collide with
molecular lattice structure causing damage or distortion of material. For very strong
fields, avalanche may also occur. The dielectric under such condition will become
conducting.
The maximum electric field intensity a dielectric can withstand without breakdown is
referred to as the dielectric strength of the material.
Boundary Conditions Electrostatic Fields:
Let us consider the relationship among the field components that exist at the interface
between two dielectrics as shown in the figure below. The permittivity of the medium 1
and medium 2 are 1 and  2 respectively and the interface may also have a net charge
density  s Coulomb /m.
We can express the electric field in terms of the tangential and normal components
E1  E1t  E1n
E2  E2t  E2 n
where Et and En are the tangential and normal components of the electric field
respectively.
Let us assume that the closed path is very small so that over the elemental path length the
variation of E can be neglected. Moreover very near to the interface, h  0 .
Therefore,
 E.dl
E1t w  E2t w 
h
h
 E1n  E2 n    E1n  E2 n   0
2
2
Thus, we have,
E1t  E2t or
D1t

1
D2t
2
i.e. the tangential component of an electric field is continuous
across the interface.
For relating the flux density vectors on two sides of the interface we apply Gauss’s law to
a small pillbox volume as shown in the figure. Once again as h  0 , we can write,
 D  ds  ( D .aˆ
1
n2
 D2  aˆn1 )s   s s
i. e. D1n  D2 n  s
or 1E1n   2 E2 n  s
Thus we find that the normal component of the flux density vector D is
discontinuous across an interface by an amount of discontinuity equal to the surface
charge density at the interface.
Example
Two further illustrate these points; let us consider an example, which involves the
refraction of D or E at a charge free dielectric interface as shown in the figure below.
Using the relationships we have just derived, we can write
E1t  E1 sin 1 
D1
1
sin 1  E2t  E2 sin  2 
D1n  D1 cos1  D2 n  D2Cos 2
In terms of flux density vectors,
D1
1
sin 1 
D2
2
sin  2
D2
2
sin  2
D1 cos1  D2Cos 2
Therefore,
tan 1 1  r1
 
tan  2  2  r 2
Capacitance and Capacitors
We have already stated that a conductor in an electrostatic field is an Equipotential body
and any charge given to such conductor will distribute themselves in such a manner that
electric field inside the conductor vanishes. If an additional amount of charge is supplied
to an isolated conductor at a given potential, this additional charge will increase the
surface charge density  s . Since the potential of the conductor is given by
V
ratio
C
1
4 0

S'
 S ds '
r
, the potential of the conductor will also increase maintaining the
Q
same. Thus we can write
V
Q
, where the constant of proportionality C is called the capacitance of the isolated
V
conductor. SI unit of capacitance is Coulomb/ Volt also called Farad denoted by F. It can
be seen that if V  1 , C  Q . Thus capacity of an isolated conductor can also be defined
as the amount of charge in Coulomb required to raise the potential of the conductor by 1
Volt.
Of considerable interest in practice is a capacitor that consists of two (or more)
conductors carrying equal and opposite charges and separated by some dielectric media
or free space. The conductors may have arbitrary shapes. A two-conductor capacitor is
shown in the figure below.
Two-conductor capacitor
When a d-c voltage source is connected between the conductors, a charge transfer occurs
which results into a positive charge on one conductor and negative charge on the other
conductor. The conductors are Equipotential and the field lines are perpendicular to the
conductor surface. If V is the mean potential difference between the conductors, the
capacitance is given by C 
Q
. Capacitance of a capacitor depends on the geometry of
V
the conductor and the permittivity of the medium between them and does not depend on
the charge or potential difference between conductors. The capacitance can be computed
by assuming Q (at the same time Q on the other conductor), first determining E using
Gauss’s theorem and then determining V    E.dl . We illustrate this procedure by
taking the example of a parallel plate capacitor.
Example: Parallel plate capacitor
For the parallel plate capacitor shown in the figure above, let each plate has area A and a
distance h separates the plates. A dielectric of permittivity  fills the region between the
plates. The electric field lines are confined between the plates. We ignore the flux
fringing at the edges of the plates and charges are assumed to be uniformly distributed
over the conducting plates with densities  s and  s ,  s 
By Gauss’s theorem we can write E 
Q
.
A
s Q


A
As we have assumed  s to be uniform and fringing of field is neglected, we see that E is
constant in the region between the plates and therefore, we can write V  Eh 
Thus, for a parallel plate capacitor we have, C 
hQ
.
A
Q
A
 .
V
h
Series and parallel Connection of capacitors
Capacitors are connected in various manners in electrical circuits; series and parallel
connections are the two basic ways of connecting capacitors. We compute the equivalent
capacitance for such connections.
Series Case: Series connection of two capacitors is shown in the figure. For this case we
can write,
V  V1  V2 
Q Q

C1 C2
V
1
1
1

 
Q Ceqs C1 C2
Series Connection of Capacitors
The same approach may be extended to more than two capacitors connected in series.
Parallel Case: For the parallel case, the voltages across the capacitors are the same.
Capacitors in parallel
The total charge Q  Q1  Q2  C1V  C2V
Therefore, Ceqp 
Q
 C1  C2
V
Electrostatic Energy and Energy Density
We have stated that the electric potential at a point in an electric field is the amount of
work required to bring a unit positive charge from infinity (reference of zero potential) to
that point. To determine the energy that is present in an assembly of charges, let us first
determine the amount of work required to assemble them. Let consider a number of
discrete charges Q1 , Q2 , ...............QN are brought from infinity to their present position
one by one. Since initially there is no field present, the amount of work done in bring
Q1 is zero. Q2 is brought in the presence of the field of Q1 , the work done W1  QV
1 21
where V21 is the potential at the location of Q2 due to Q1 . Proceeding in this manner, we
can write, the total work done
W  V21Q2  (V31Q3  V32Q3 )  ...............  (VN 1QN  ........  VN ( N 1)QN )
Had the charges been brought in the reverse order,
W  (V1N Q1  ........  V12Q1 )  ............................  (V( N 2)( N 1)QN 2  V( N 2) N QN 2 )  V( N 1) N QN 1
There fore,
2W  (V1N  V1( N 1)  .......  V12 )Q1  (V2 N  V2( N 1)  .......  V23  V21 )Q1......
..........................  (VN 1  .......VN 2  VN ( N 1) )QN
Here VIJ represent voltage at the I th charge location due to J th charge. Therefore,
N
2W  V1Q1  .................  VN QN  VI QI
I 1
Or, W 
1 N
VI QI .
2 I 1
If instead of discrete charges, we now have a distribution of charges over a volume v then
we can write,
W
1
V v dv
2 v
Where  v is the volume charge density and V represents the potential function.
Since v  .D , we can write,
W


1
.D Vdv
2 v
Using the vector identity,
.(VD)  D.V  V .D we can write,
W

In
1
2


1
.(VD)  D.V dv
2 v
1
2
 VD .ds  2   D.V  dv
1
s
v
 VD .ds , for point charges, since V
s
varies as
varies as
1
1
and D varies as 2 , the term VD
r
r
1
1
while the area varies as r 2 . Hence the integral term varies at least as and
3
r
r
as the surface becomes very large ( r   ), the integral term tends to zero. Thus the
equation for W reduces to
W 




1
1
1
D
.

V
dv

D
.
E
dv

 E 2 dv   we dv




2v
2v
2v
v
1
we   E 2 is called the energy density in the electrostatic field.
2