Download Introductions : - SS Margol College

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CAPACITORS
We cannot handle free point charges and neither can we locate them at specific positions. A
practical method of handling the charges is to put it on some conductor. Thus, a conductor can be
used to store an electric charge. However, every conductor has a last limit to hold the charge. It
cannot be given an infinite charge. If charge is given continuously to a conductor, after some time a
stage is reached where it starts rejecting, the charge. Ability of a conductor to hold the charge
is called as its capacity.
Capacity of a Conductor:
In general, if a charge is given to the conductor then its electric potential increases. The
electric potential of a conductor (v) is directly proportional to the charge (Q) given to it.
i.e.,
V  Q OR Q  V
therefore, Q =C V
Where ‘C’ is the constant of proportionality. Then, C 
Q
---(1)
V
This constant ‘C’ is called as, capacity of a conductor. It is defined as the ratio of charge
given to the conductor to its electric potential. It is a measure of ability of a conductor to
store the charge. SI unit of capacity (C) is farad (F). As this unit is too large smaller units like
mf(millifarad), f( micro farad) etc are used.
1 mf = 10-3 f
1 Coulomb
1f = 10-6 f
1 farad =
1 nf = 10-9 f
1 Volt
1 pf = 10-12 f
Capacity of a conductor is 1 farad if addition of 1 coulomb charge increases its
potential by 1 volt.
Dimensional formula for capacity or capacitance is  [M1, L2 T4 A2]
Capacity of an isolated spherical conductor:
Consider an isolated spherical conductor of radius R. Let charge Q,
is given to it. The electric potential on its surface is given as,
V 
1
40

Q
R
i.e.,
Q
 40 R
V
R
(2)
From the definitions of capacity the ratio (Q/V) in above eqn (2) represents capacity of
spherical conductor i.e. C
C = 40R

……..(3)
Equation (3) states that capacity of a spherical conductor is directly proportionally to its size
and directly proportional to the permittivity of medium. Apart from these, two factors (size and
permittivity), the capacity of a conductor also depends on the presence of charged or uncharged
conductors in its surrounding.
Note: Isolated conductors are not suitable to use as charge storing devices because their charge
storing capacity is very poor. For example to get the capacity of just 1 F, a spherical conductor of
radius 9 km is required.
Capacitor
An isolated conductor has very low capacity. Its capacity can be improved by placing one
more conductor near it at small distance. This system of two conductors separated by small
distance with preferably a dielectric medium between them is called as a Capacitor.
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The first capacitor called Laplesn Jar was constructed in the middle of 18th century. It was
found that, a nail immersed into a glass jar containing mercury ‘stores large charge. In this
capacitor, mercury serves as one conductor and the palm of an experimenter as other conductor.
X
Principle of Capacitor:
Y
X
Y
Consider the conductor plate ‘x.’ let a
charge +Q is given to it. Let V1 be the potential of
x due to the charge on itself.
Therefore, its capacity is given by,
Cx 
Q
V1
Let another conductor plate ‘Y’ is placed
near conductor X. Due to induction, the inner side
of conductor acquires, equal induced negative
charge -Q and its outer side acquires equal positive charge +Q. Induced negative charge on Y, tend
to reduce the potential on X and induced positive charge on Y tend to increase the potential of X.
Now plate X has potential V1 due to its own charge +Q. Similarly due to the induced charges
(i.e. –Q & +Q from inner and outer surface of Y), there is an additional potential at X, say -V2 &
+V3 respectively. Therefore new potential of X is equal to,
V = V1 + (-V2 +V3 )
Then, capacity of X is equal to, C x 
Q
V1  (V2  V3 )
Induced negative charges are closer to X than the induced positive charges. Therefore, the
new potential of X (i.e. V) will be less than its earlier potential V1. Hence its new capacity more than
earlier capacity Cx.
If outer side of conductor Y is provided with earth connection, the induced positive charge
on it vanishes, due to the flow of electrons from earth to it. Therefore, V3 becomes zero. Thus, net
potential at X becomes = (V1 - V2).
Therefore new capacity of X is C 
Q
V1  V2
As, l V2 l  l V1 l but V2  V1. Therefore, (V1 – V2)  V1. Hence new capacity of X i.e. C
 C1. Thus, conductor X can acquire more charge from the source to attain original potential.
Types of Capacitors:
If a capacitor is constructed using two conductors, that are in the form of flat plates, then
the capacitor is called as parallel plate capacitor. If the conductors are in the form of two concentric
spherical shells, then capacitor is called as a spherical capacitor. The coaxial cylindrical conductors
form a cylindrical capacitor.
X
Y
a) Parallel Plate capacitor
b) Spherical capacitor
c) Cylindrical capacitor
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Expression for the capacity of parallel plate capacitor:
Let X & Y are two conductor plates forming a parallel plate capacitor. Conductor X is
charged and conductor Y is earthed.
X
Y
V
Q
E
d
When charge Q is given to conductor X, conductor Y acquires, equal and opposite induced
charge i.e. –Q. These charges on X and Y establish an electric field between the plates. The
intensity of this electric field E between the plate, is given by, E =  / 0 ….(1) where  is, the
surface density of charge on the conductor plates. If ‘A’ is taken as area of the plates, then surface
density of charge  becomes,
 
Q
A
--- (2)
Substituting this in equation (1)
 E = Q /  0A
------------(3)
If V is the potential difference between the plates and d is the separation between the plats,
then electric field between the plates is also given as,
E
V
d
----- (4)
From equations (3) and ( 4) we can write
Q
V

0A d
OR
Q 0 A

V
d
------(5)
From definition of capacity, Q / V = C

C
0 A
d
-----(6)
In case if the space between the conductor plates
is completely filled with a dielectric medium of constant K,
then expression for new capacity is written as,
 if space between the plates filled with a dielectric slab of
thickness ‘t’ ( where ‘t’ less than the distance between
the plates of the capacitor i.e. t d ) then, the capacity of
C
K 0 A
-----(7)
d
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capacitor is given as,
K 0 A
C
[d  t (1 
1
)]
K
From the expression for capacity of a parallel plate capacitor( equation 7), it is clear that,
capacity can be increased by increasing the area of the plates, decreasing the separations between
the plates and increasing the dielectric constant of the dielectric medium between the plates.
Definition of dielectric constant ( K)
Capacity of a parallel plate capacitor without using dielectric medium is given by eqn (6) and
with the dielectric medium is given by equation ( 7)
C
i.e.,
0 A
d
and
Cm 
K 0 A
d
k 0 A
)
C
Cm
d
K  m
Then,
Or

K
 A
C
C
( 0 )
d
Thus, dielectric constant of a medium k is defined as the ratio of capacities of a
capacitor with the dielectric medium and without the dielectric medium.
(
Spherical capacitor :
It consists of two hollow concentric metal spheres X & Y of radii of a & b respectively.
Sphere X is charged with +Q units of charge and sphere Y is earthed. Inner surface of sphere Y
acquires induced charge ‘-Q’.
The Electric Potential on the sphere X due to the positive charges on itself & the negative
charge on Y is given by,
Q
…………Due to its own positive charge
40 a
1 Q
VB 

…………Due to the negative charge
40 b
VA 
and
1

X
Y
Q
on inner surface of B
Then, net Electric potential of the sphere X is given as,
V= (V A  V B ) 

1
Q
1 Q
Q 1 1

(  )
=
40 a 40 b 40 a b
V=
Q
1 1
(  )
40 a b
OR
Q 40 ab

V
(b  a)

C
From definitions of capacity,
C 
Q
V
4 0 ab
`
(b  a )
If a dielectric medium of constant K fills the space between the sphere then new capacity of
spherical capacitor
Is given by
Cm 
40 abK
(b  a )
From the expression for capacity of a spherical capacitor it is clear that, capacity can be
increased by increasing the radii of the spherical conductors, decreasing the separations between
the conductors and increasing the dielectric constant of the dielectric medium between the
conductors.
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Cylindrical Capacitor
It consists of two coaxial metallic cylinders X & Y of equal length L and of radii a and b
respectively. A charge of Q units is given to inner cylinder X and outer surface of cylindrical Y is
earthed.
The strength of electric field E at any point p between the
cylinders, at a distance x meter from the axis, is given by,
Q
E
=
------------- (1)
2  0 x L



Q
Q
 E 




0
A o
(2  L) o 

The potential difference (V) between cylinder X and Y is equal to total amount of work done,
in moving UPC from inner face of Y to the outer face of X.
Which is given as,
x
x a
Y
V=

 dW =
- E . dx
x b
X
OR
L
xa
V=
Q
  2 Lx dx
x b
0
Q
x a
1
dx
20 L x b x
Q
log e xxxba
V=20 L
V = 

V=
as capacity is given by,


log e
20 L 
Q
b
a 
C = Q/ V

2 0 L

b
log e ( )
a
C=
Capacity of cylindrical capacitor with dielectric medium
Of dielectric constant K is given by, Cm =
2K 0 L
b
log e ( )
a
From the expression for capacity of a cylindrical capacitor it is clear that, capacity can be
increased by increasing the length of the cylindrical conductors, decreasing the separations between
the conductors and increasing the dielectric constant of the dielectric medium between the
conductors.
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Energy stored in a charged capacitor ;
In the following fig a capacitor of capacity C is connected across a battery which provides a
constant potential difference of V volts.
As there is no charge on any of capacitor plates initially, the potential difference across its
plates is zero.
When battery is made on, the electrons from plate X of capacitor,
start moving to plate Y through the battery. This makes the plate X
X
Y
positive and plate Y negative. Now, the capacitor is said to be under
charging process. As the time elapses, the positive charge on X and
negative charges on Y goes on increasing. At the same time, Pd across
the plates X & Y increases. This slows down the rate of transfer of
electrons from X to Y. When the potential difference across the plate
becomes equal to the pd across battery terminals, the transfer of
electrons stops completely. At this stage, the capacitor is completely
charged. In transferring a charge from X to Y, the battery does definite
amount of the work. This work done by the battery itself is stored in the
capacitor in the form of electrostatic potential energy (PE).
Let q be the charge on the capacitor at an intermediate stage. Then, V = q/ C
Then the work done in storing an additional charge of ‘dq’ is given by,
dw= V . dq
OR
dw=(q/C).dq .---------------------------(1)
This workdone is stored in the capacitor ,as an electrostatic energy.
Therefore the total energy stored in the capacitor in charging it from 0 to Q charge is
obtained by integrating above equation between the limits 0 and Q.
integrating the equation (1).
Therefore. E =

Q
dw   V  dq -----------(2)
0
As
V = q/ C
Substituting this in equation (2)
Q

Then,
E =  dw  (q / c)  dq
OR
1 q2 
1 Q2
1
E =  q  dq =   = 
c  2 0
c 2
c0
0
Q
Thus,
E=
Q
Q 2 as Q = CV , energy can also
2C
be written as,
E=
1
CV 2
2
E=
1
QV
2
When a charged capacitor is discharged, this energy stored in it is released in the form of
heat or light.
The variation of potential V and the charge Q for a capacitor is as shown in
the graph. From the graph we can calculate the energy stored across the
capacitor by finding half of the area under Q-V graph.
Combinations of capacitor
The capacitors are manufactured with standard values of capacities. In practical applications,
we may need the capacitor of capacity either more than this value or less than this value.
Combinations of capacitors help us to obtain desired value of the capacity from the available ones.
The simple ways in which the capacitors can be connected are,
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1) Series Combination
2) Parallel Combination
If the numbers of capacitors are connected one after another, as in following fig (a), the
connection is called as series combination. If capacitors are connected between two common points,
as in following fig (b), the connection is called as parallel combination.
Effective Capacity of Series Combinations :
Consider three capacitors of capacities C1, C2 and C3. The capacitors are connected in series.
The combination is connected across, a battery of emf ‘V’ volts.
Q
Q
Q
Q
Fig 1
Fig 2
Left plate of C1 receives charges +Q on its left plate, due to which equal –Q charge is
induced on its right plate. At the same time a charge +Q appears on left plate of C2, since the
conductor consisting of right plate of Q, left plate of C2 and the connecting wire all were uncharged
initially, equal negative charge (-Q) is again induced on right plate of C2. The same thing happens
with C3 and the other if any, in the series combinations. Thus, charge acquired by each capacitor
has same magnitude. (It ought to be noted that, total charge received by the arrangement is same
as Q only)
Let V1, V2, & V3 are pd across capacitors C1, C2 & C3 respectively. Using conservations law
for total electric energy are can write
V = V1 + V2 + V3
---(1)
Also, in case of individual capacitors, using the relation, C = Q/V
It can be written as,
V1 
Q
,
C1
V2 
Q
,
C2
& V3 
Q
C3
Substituting these values in equation (1)
V 
Q Q
Q
1
1
1


 Q( 
 )
C1 C 2 C3
C1 C 2 C3
---------(2)
Let Cs is the effective capacity of all these capacitors in series. Equivalent capacitor is the
one, which stores the same charge, as stored by a series combination, when connected across same
battery.
From equivalent diagram, C s 
Q
Q
therefore, V 
------------------------(3)
V
Cs
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Comparing equations (2) & (3), it can be written as,
Q
1
1
1
 Q( 
 )
Cs
C1 C 2 C 3
OR
1
1
1
1



C s C1 C 2 C 3
If n number of capacitors with capacity C1, C2 , C3 ……… Cn are connected in series then
the formula for effective capacity is written as,
1
1
1
1


 .......
C s C1 C 2
Cn
Thus, reciprocal of effective capacity of the number of the capacitors connected in
series is equal to the sum of the reciprocals of their individual capacities. Effective
capacity in the series combination is less than the least capacity.
* If only two or only three capacitors are connected in series then the effective
capacity can be found by following formula.
Cs 
C1C 2
(C1  C 2 )
And
Cs 
*
C1C2 C3
(C1C 2  C 2 C3  C3C1 )
If there are ‘n’ number of identical capacitors each of capacity ‘C ‘ , then effective
capacity of series combination is given by, C s 
C
n
Parallel Combinations of capacitors
Consider three capacitors with the capacities C1, C2 & C3 connected to from a parallel
combination. A battery of V volts is connected across the combination as shown in the following in
Q
Q
Fig (a)
Q
Q
Q
Q
Q
As all the capacitors are connected across common points A & B and the points A & B across
the battery terminals, the potential difference across each capacitor is same and is equal to the pd
across battery terminals i.e., V volts. In spite of same Pd across each capacitor, each capacitor
stores different charge due to their different capacities.
Let Q1, Q2 & Q3 are charges on capacitors C1, C2 and C3 respectively,
Then using, C = Q/V or Q = CV we can write,
Q1 = C1V, Q2 = C2V, & Q3 = C3V
------------(1)
 The net charge stored in the parallel combinations is
Q = Q1 + Q 2 + Q 3
= C1V + C2V + C3V
Q = V( C1 + C2 + C3 )
-------------(2)
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If ‘Cp’ is the effective capacity of the parallel combination then, Q capacitor with this capacity
will store the same charge Q as that stored by parallel combination when connected across same
battery. Therefore, it can be written as,
Q = Cp V
--------------(3)
From equations, (2) and (3) it is clear that
CpV = V( C1 + C2 + C3 )
OR
Cp = ( C1 + C2 + C3 )
If there are ‘n’ numbers of capacitors of capacities C1, C2……..Cn in parallel, then effective
capacitor is given as
Cp = C1 + C2 +……….Cn
Thus the effective capacity of the number of capacitors connected in parallel is
equal to the sum of individual capacities.
* The effective capacity in parallel combination is larger than the largest capacity.
* If there are n no. of identical capacitors each of capacity ‘C’ then, effective, capacitor
can be directly written as,
Cp = n C
Differences between series and Parallel Combinations of capacitors
1
2
3
4
5
Series Combinations
If given number of capacitors are
connected one after the another, then this
combinations
is
called
as
series
combinations,
Charge on each capacitor is same
Pd across each capacitor is different
The
effective
capacity
of
series
combinations is always less then individual
capacities.
The formula for effective capacitor of
series combinations is
1
2
3
4
5
Parallel Combinations
If all of the given no. of capacitors are
connected between two fixed points, the
no. this combination is called as Parallel
Combinations.
Charge on each capacitor is different
Pd across each capacitor is same
The effective capacitor of parallel
combinations is always more than
individual capacities.
Formula for effective capacity of parallel
combinations is
1
1
1


 ......
C s C1 C2
Cp = C1 + C2 + ………Cn
*
If ‘n’ number of identical capacitors are first connected in series and then in
parallel then the ratio of capacities in two cases is given as,
Cs = ( 1/n) C
&
Cp = n C
Then,  Cs : Cp = 1 : n2
Uses of Capacitors
1.
2.
3.
4.
5.
Capacitors are used as charge storing or electric energy storing devices.
They can be used to produce strong electric field
They allow the flow of alternating currents of high frequency, hence are used to separate DC
current from AC.
A suitable capacitor clubbed with an inductor form a circuit called tuned circuit. This circuit
can produce as well as detect electromagnetic waves.
They are used to reduce the sparking effect in automobiles.
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7.
8.
9.
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Tiny capacitor can be used in computers as memory cells.
They can be used as potential dividers.
They can be used to find the dielectric constant of unknown medium.
They can be used to control electric currents in an AC circuits.
ADDITIONAL INFORMATION:
Reason for Increase of Capacitance for a dielectric medium:
Suppose a dielectric medium is filled between the plates of a parallel plate capacitor. Every matter is
constituted of molecules or atoms. In an plate capacitor. Every matter is constituted of molecules or atoms. In
an atom positive charge is concentrated at the nucleus and the negatively charged electrons revolve around
the nucleus in orbits. In dielectric medium the electrons are strongly bound to the nucleus and in general the
centre of positive and negative charges in each atom/ molecule coincide. When capacitor is charged, an
electric field is established between the plates of the capacitor. Due to this electric field, the centre of positive
charges are displaced along the direction of electric field or towards plate B, while the centres of negative
charges are displaced opposite to the direction of electric field or towards the plate A. Thus the centres of
positive and negative charges of each atom/ molecule are displaced and the molecule is said to be polarised.

This causes an electric field E i between the dielectric medium which is opposite to the direction of electric field
produced due to charges on the plates. Thus, due to presence of dielectric medium, the resultant electric field
between the plates is reduced and hence the potential difference (V=Ed) across the plates is reduced.

Q 
 is increased.
Consequently the capacitance of capacitor  C 
VA B 

Capacitance of a parallel plate capacitor when partly filled with A Dielectric substance:
Consider a parallel plate capacitor, Area of each plate being A, the separation between the plates
being d. Let a dielectric slab of dielectric constant K and thickness t < d be placed between the plates. The
thickness of air between the plates = (d t). If charges on plates be +Q and Q, then surface charge density.
Q
=
A

Q

The electric field between the plates in air, E =
0
0 A
The electric field between the plates in slab, E2 =

Q

K 0 K 0 A
 The potential difference between the plates
VAB = work done is carrying unit positive charge from one plate to another
=  Ex (as field between the plates is not constant).
= E1 (d – t) + E2t
Q
Q
(d  t ) 
t
=
0 A
K 0 A
Q
0 A
t

d  1  K 



VAB =

capacitance of capacitor, C =
or
C
0 A
t
dt 
K

Q
Q

VA B
Q 
1
d  t  
0 A 
K
0 A
1

d  t 1  
K

This is the required expression. As K > 1, it is obvious that due to introduction of slab of thickness t
and dielectric constant K between the plates of a parallel plate capacitor, the effective distance in air is
1

reduced by 1   t; and so the capacitor of capacitor increases.
K

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Special cases:
(i)
If the whole space between the plates be filled with a dielectric substance (i.e. t=d), then capacitance
0 A
K A
of capacitor C =
 0
d
d
dd
K
(ii)
When the whole space between the plates contains air or vacuum, then t=0, so C =
0 A
 A
 0
d
1

d  01  
K

When a metal slab of thickness t be placed between the plates: for metal K = ,
0 A
 A

Capacitance of capacitor, C =
 0
1  dt

d  t 1  


Clearly by the introduction of metal slab, the effective distance between the plates is reduced by the same
amount as the thickness of the slab.
(iii)
A dielectric slab of dielectric constant K introduced between the plates of a charged aircapacitor:
Sl.
No.
1.
2.
3.
4.
Battery disconnected after charging
Battery remains connected
Charge on the capacitor remains unchanged
Potential difference decreases by a factor of K
Capacitance increases by a factor of K
Electrostatic potential energy decreases by a
factor of K
Charge on the capacitor increases by a factor of K.
Potential difference remains unchanged
Capacitance increases by a factor of K
Electrostatic potential energy increases by a factor
of K
Energy density in parallel plate capacitor:
Energy density is the energy stored per unit volume of the capacitor.
Let us now express the total energy of the capacitor in terms of the electric field strength E.
A
Charge,
Q = A = A 0E and C = 0
d
Now,
U=

1 Q2 1 d

(A 0 E) 2  0 E 2 Ad
2 C
2 0 A
2
0 2
E  volume between the plates of the capacitor, i.e., the volume of medium between the two
2
plates of the capacitor.
1

Energy density, u =
0 E2
2
The quantitative relationship between energy density and electric field has been derived for a parallel
plate capacitor. But it holds generally, what ever may e the source of the electric field. So, a general result
may e expressed as under:

If an electric field E exists at any point in space, then that point cane regarded as a seat of stored
1
energy. The amount of energy stored per unit volume is
0E2.
2
Or
U=
Redistribution of charges and concept of common potential:
Consider two insulated conductors A and B capacitances C 1 and C2 carrying charges Q1 and Q2
respectively. Let V1 and V2 be their respective potentials. When the two conductors are joined by a thin
metallic wire, positive charge begins to flow from a conductor of higher potential to a conductor of lower
potential till their potentials are equalised. Thus, the charges are redistributed. But the total quantity of charge
remains (q1 + q2).
12
81898724
SARVAJNYA
If we neglect the capacitance of the connecting wire and assume that the conductors are so much
spaced apart that they do not electrically affect each other, then the combined capacitance of the two
conductors is C1 + C2.
Total ch arg e
Common potential,
V=
Total capaci tan ce
q  q2
C V  C 2 V2
Or
V= 1
or
V 1 1
C1  C 2
C1  C 2
Let Q1’ and Q2’ be the charges on A and B respectively after the redistribution of charges has taken
place.
Then

Q1’ = C1V and Q2’ = C2V
Q1 ' C 1 V
Q1 ' C 1
or


Q2 ' C2
Q2 ' C2 V
Loss of energy in redistribution of charges:
Before connection:
Total potential energy of two conductors.
1
1
Ui =
C1 V12 +
V22
2
2
 C 1 V1  C 2 V2

 C1  C 2
Subtracting (2) from (1), we get
=
1
(C1 + C2=)
2
Ui – Uf =
=



2
2
1
1  C 1 V1  C 2 V2
(C1 V12 + C2V22) 
2
2  C 1  C 2




1
[(C1 + C2) (C1V12 + C2V22)  (C1V1 + C2V2)2]
2(C 1  C 2 )
On simplification
Clearly, (Ui  Uf) > 0
Ui  Uf =
C1 C 2
( V1  V2 ) 2
2(C1  C 2 )
13
81898724
One-mark questions:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
SARVAJNYA
QUESTION BANK
What is meant by capacity of a conductor?
Define capacity of a conductor.
Define Farad.
Define dielectric constant in terms of capacity of a capacitor.
What is a capacitor?
If earth were a solid conducting sphere of radius 6400Km, What would be its capacity?
A piece of paper is introduced between the plates of parallel plate capacitor. What will
happen to its capacity?
If the potential of the conductor increases then what happens to its capacity?
What is fringe effect in case of parallel plate capacitor?
What is a guard ring arrangement?
What is the significance of combining the capacitors?
Two-mark questions:
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
Write any four uses of capacitor.
Write the formula for the capacity of parallel plate capacitor & define the terms.
Write the formula for the capacity of spherical capacitor & define the terms.
Write the formula for the capacity of cylindrical capacitor & define the terms.
Write the formula for effective capacity of series combination & define the terms.
Write the formula for effective capacity of parallel combination & define the terms.
What are the different factors governing the capacity of a conductor?
Why a single conductor is not suitable to use as a charge-storing device?
Mention the types of capacitors.
How do you increase the capacity of parallel plate capacitor?
How do you increase the capacity of spherical capacitor?
How do you increase the capacity of cylindrical capacitor?
Write any four differences between series and parallel combination of capacitor.
Five-mark questions:
25.
26.
27.
28.
29.
30.
31.
32.
What is a capacitor? Explain the principle of capacitor.
Draw a parallel plate capacitor & deduce the formula for its capacity.
Draw a spherical capacitor. Write the formula for its capacity & define the terms used in it.
Draw a Cylindrical capacitor. Write the formula for its capacity & define the terms used in
it.
Explain the charging of capacitor. What is meant by charged capacitor?
Obtain an expression for the energy stored across a charged capacitor.
What is a series combination of capacitors? Derive an expression fro the effective capacity
of number of capacitors connected in series.
What is a parallel combination of capacitors? Derive an expression fro the effective capacity
of number of capacitors connected in parallel.
14
81898724
SARVAJNYA
NUMERICALS:
1.
A capacitor has parallel plates of dimensions 0.05 m x 0.02 m. The plates are separated by 1.2 mm
and the space between the plates is filled with paper of relative permittivity 4.2. What is the capacity
of the capacitor?
[30.99 pF]
2.
A spherical capacitor consists of two concentric spherical shells of radii 0.03 m and 0.05 m
respectively. The space between the shells is filled with an insulator of dielectric constant 8. Calculate
its capacitance.
[66.7 pF]
3.
The inner diameter of outer cylinder and the outer diameter of inner cylinder of a cylindrical capacitor
are 0.02 m and 0.03 m respectively. the space between the cylinder is filled with an insulator of
dielectric constant 12. The length of the capacitor is 0.5 m. Calculate its capacitance.
[246.8 pF]
4.
An electron is shot horizontally along +x axis with a velocity v. A uniform electric field E directed
vertically downwards is applied to it. (i) Show that the vertical displacement of electron after t second
is given by y =
eEt2
, (ii) Show that the trajectory of electron is a parabola.
2m
[y =
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
2
 eE
1 eE  x 

    
2 m  v 
 2mv 2
 2
 x  Kx 2 ]

A parallel plate capacitor of 0.2 F is connected to a dc supply of 500 V. What is the energy stored by
it? If the capacitor is completely immersed in an oil of relative 3.6. What is the energy stored? Assume
that insulation is perfect throughout.
[0.09 J]
A 2 F capacitor is charged to 4 V and a 5F capacitor is charged to 6 V. They are then connected in
parallel. Find the common p.d. of the combination.
[5.42 V]
Four capacitors 2 F, 4F, 6 F and 8 F are connected first in series and then in parallel. Find the
ratio of effective capacities.
[6:125]
The effective capacity of two capacitors is 12/7 F when in series and 7 F when in parallel. Calculate
the individual capacities.
[C1 = 3F]
Three capacitors of 2F, 3F and 5F are connected in series. The free end of 2F capacitor is
connected to 20 V and the free and end of 5 F is earthed. Find the p.d. across each capacitor. [3.870 V]
Three capacitors of 10 F, 20 F and 30 F are connected in parallel to a 120 V dc source. Find the
charge on each capacitor.
[7.2x103C]
Calculate the equivalent capacity between A and B in the figure shown.
[3.317 F]
Two capacitors 2 F and 1.5 F are connected in series and two other capacitors of 3 F and 6 F are
connected in series. The series combinations are then connected in parallel with each other. Find
effective capacitance of the combination.
[2.857 F]
Find the effective capacity between A and B in the figure shown.
[0.806 pF]
A parallel plate capacitor of 8.5 nF has to be constructed using sheets of tinfoil of dimensions of 0.3 m
x 0.2 m separated by a paraffin waxes paper 0.015 cm thick. Find how many sheets of metal are
required. Dielectric constant of paper is 4.2.
[n=6+1=7]
Two capacitors are rated as 3F  110 V and 6 F – 110 V respectively. They are connected in series
to a 220 V source. Which one is likely to be damaged?
[73.3 V]
A parallel plate capacitor consists of two plates, each of dimensions 0.1 m x 0.2 m and separated by
1:2 mm. Calculate the capacity. If a dielectric material of relative permittivity 6 in introduced between
the plates, what is the new capacity?
[0.148 nF, 0.885 nF]
A parallel plate capacitor consists of circular plates of radius 1 cm. The plates are separated BY 0.8
mm. If the space between the plates si filled with dielectric medium of relative permittivity 8, calculate
the capacity.
[111.2 pF]
Two capacitor plates, each of area 6x104 m2 are separated by 0.8 mm. Find its capacity. If the space
between the plates is filled with a dielectric of relative permittivity 12, what will be the new capacity?
[6.64 pF, 79.68 pF]
The separation between the plates of a parallel plate capacitor is trebled and area of each plate is
halved. Find the ratio of new capacity to the initial capacity.
[1:6]
n identical drops, each of capacity C, merge to form a single drop. What is the capacity of the bigger
drop?
[n(1/3)/ C]
The charge of plates of a capacitor increases by 50 C when the p.d. between them increases from 16
V to 32 V. What is the capacity?
[3.125 F]
15
81898724
22.
23.
24.
25.
26.
27.
28.
29.
30.
SARVAJNYA
A parallel plate capacitor of 17.71 pF has to be constructed with paper sheets of 0.05 mm thick as
dielectric. Find the number of sheets of paper needed for this purpose. The area of each plate is
6x104 m2 and dielectric constant of paper is 5.
[30]
The inner and outer radii of conductors of a spherical capacitor are 2 mm and 3 mm respectively. The
space between the conductors is filled with an insulator of dielectric constant 5. What is the capacity
of the capacitor?
[0.667 pF]
The inner and outer diameter of a cylindrical capacitor are 0.01 m and 0.014 m respectively. The
space between the cylinders is filled with a dielectric of relative permittivity 3.6. If the length of the
cylinder is 0.3 m, calculate the capacity.
[178.5 pF]
The area of each plate of a parallel plate capacitor is 40x104 m2 and the plates are separated by 1
mm. What is its capacity? It is connected to a 20 V battery. What is the charge on each plate? What is
the energy stored in it?
[70.8x1011 C, 7.08x109 J]
A capacitor of 0.47 F is charged to a potential of 20 V. If the potential is charged to 40 V, find the
change in energy stored.
[2.82x104 J]
Three capacitors 1 F, 3 F and 5 F are connected first in series and in parallel to a voltage source of
10 V. Find the energy stored in each case.
[32.6 J, 450 J]
Three capacitors 2 F, 4 F and 6 F are connected in parallel to a 50 V source. (i) What is the p.d.
across each capacitor? (ii) What is the charge on each capacitor?
[50 V, 100 C, 200 C, 300 C]
Four capacitors 1 F, 2 F, 3 F and 4 F are connected first in series and then in parallel. Find the
ratio of effective capacities.
[6:125]
Four identical capacitors are connected in the form of a square. What is the effective capacity between
(i) any two adjacent and (ii) any two diagonally opposite corners?
31.
33.
35.
36.
37.
38.
[
15
F, 39.13V, 13.04 V, 7.83 V]
23
Three capacitors 20 F, 40 F and 60 F are connected in series. The free end of 20 F is at +200 V
and the free end of 60 F is earthed. Find the charge on each capacitor and the p.d. across each
capacitor.
[2.182x103 C, 109.1 V, 54.55 V, 36.37 V]
How many different combinations are possible from 3 capacitors of 2 F each. Calculate the effective
capacity in each case.
34.
4C 0
, C0]
3
Three 1 F, 3 F and 5 F are connected in series. A p.d. of 60 V is applied to the combination. (i)
Find the total capacitance, (ii) What is the p.d. across each capacitor? (iii)What is the charge on each
capacitor?
32.
[
[4,
2
4
F, 6 F, 3 F,
F]
3
3
A capacitor is filled with two dielectrics of relative permittivity K 1 and K2 respectively as shown. Area of
each plate is 2x104 m2 and separation between the plates is 5 mm. If K 1=2 and K2=5, what is the
effective capacity/
[1.107 pF]
A capacitor is filled with dielectrics of dielectric constants 4 and 10 respectively. The area of each plate
is 2x104 m2 and the plates are separated by 2 mm. Find the effective capacity of the composite
capacitor.
[6.198 pF]
Three capacitors 2 F, 4F and 6F are connected so that the first two are in series and the third is in
parallel. Find the effective capacity. If a p.d. of 5V is applied to the combination, find the energy
stored by it.
[91.67 J]
Two capacitors of 6 F are charged to the same potential of 100 V separately. The charged capacitors
are connected between two points A and B so that oppositely charged plates are connected to the
same point. Find the p.d. between A and B. What is the total energy stored in the two capacitors
before and after connection?
[50 V, 10 mJ, 10 mJ]
A capacitor of 3 F is charged to a potential of 5 volt. Another capacitor of 6 F is charged to a
potential of 8 volt. If they are connected in parallel, find the common p.d. across the combination.[7 V]
16
81898724
SARVAJNYA
CET QUESTIONS FROM PREVIOUS EXAMS
1.
In the circuit below, capacitors A and B have
identical geometry but a material of dielectric
constant is present between the plates of B.
The potential differences across A and B are
respectively.
(CET Karnataka 1996)
7.
8.
2.
3.
4.
5.
6.
a) 2.5 V, 7.5 V
b) 2 V, 8 V
c) 7.5 V, 2.5 V
d) 8 V, 2 V
The electric field intensity due to a hollow
spherical conductor is maximum. (CET
Karnataka 1996)
a) outside the sphere
b) on the surface of the sphere
c) at any point inside the sphere
d) only at the centre of the sphere
1000 drops of water of radius 1 cm each
carrying a charge of 10 esu combine to form
a single drop. The capacitance of combined
drop increases. (CET Karnataka 1996)
a) 1 time
b) 10 times
c) 100 times
d) 1000 times
A spherical conductor of radius R, placed in
air, is given a charge Q. Then the potential at
a point inside the conductor and at a
distance R/2 from its centre is. (CET
Karnataka 1997)
1 Q
1 Q
a) V =
b) V=
4  0 R
4  0 2R
c) VC = 4R
d) V = 40R
A parallel plate air capacitor is charged by
connecting its plates to a battery. Without
disconnecting the battery, a dielectric is
introduced between its plates. As a result
(CET Karnataka 1997)
a) p.d. between the plates increase
b) charge on the plates decreases
c) capacitance of the capacitor decreases
d) none of the above
The effective capacitance between the points
A and B in the circuit shown in (F) is
(capacitance of each capacitor is 1F) (CET
Karnataka 1997)
9.
10.
11.
12.
13.
14.
a) 2
b) 4
c) 3
d) 0.4
125 water drops of equal radius and equal
capacitance C, coalesce to form a single drop
of capacitance C’. The relation between C
and C’ is (CET Karnataka 1997)
a) C’ = 125
b) C’ = C
c) C’ = C/125 d) C’ = 5C
The two spherical conductors of radii 4 m
and 5 m are charged to the same potential.
If 1 and 2 be the respective values of the
surface density of charge on the two
conductors, then the ratio is.
(CET
Karnataka 1998)
a) 16/25 b) 25/16
c) 4/5
d) 5/4
In the electric circuit givenelow, capacitance
of each capacitor is 1F. The effective
capacitance between the points A and B is (in
F) (CET Karnataka 1998)
a) 3/2 b) 2/3 c) 6
d) 1/6
A point charge A of charge +4C and
another point charge B of charge 1 C are
placed in air at a distance 1 metre apart.
Then the distance of the point on the line
joining the charges and from the charge B,
where the resultant electirc field is zero, is (in
metre) (CET Karnataka 1998)
a) 0.5 b) 1.5 c) 2
d) 1
Electric charges +10C, +5C, 3C and
+8C are placed at the corners of a square
of side 2 m. The potential at the centre of
the square (V) is (CET Karnataka 1999)
a) 18x105
b) 1.8x106
c) 1.8
d) 1.8x105
When 1019 electrons are removed from a
neutral metal plate, the electric charge on its
is (coulomb) (CET Karnataka 1999)
a) 1019
b) +1.6
c) 1.6 d) 1010
Two metal spheres of radii R1 and R2 are
charged to the same potential. The ratio of
the charge on the two spheres is. (CET
Karnataka 1999)
a) R1/R2
b) 1
c) ½
d) R1 – R2
When a p.d. of 103 V is applied between A
and B, a charge of 0.75 mC is stored in the
system of capacitors. The value of C is (F)
(CET Karnataka 1999)
17
81898724
SARVAJNYA
22.
15.
16.
17.
18.
19.
20.
21.
a) 3
b) 2.5 c) 2
d) 1/2
The equivalent capacitance of three
capacitors of capacitance C1, C2 and C3
connected in parallel is 12 units and the
product C1 C2 C3 = 48. When the capacitors
C1 and C2 are connected in parallel the
equivalent capacitance is 6 units. Then the
capacitance’s are (CET Karnataka 1999)
a) 1, 5, 6
b) 1.5, 2.5, 8
c) 2, 3, 7
d) 2, 4, 6
A parallel plate air capacitor is charged to a
p.d. of V. After disconnecting the battery the
distance between the plates of the capacitor
is increases using an insulating handle. As a
result the p.d. between the plates
(CET
Karnataka 1999)
a) decreases
b) does not change
c) becomes zero
d) increases
n identical mercury droplets charged to the
same potential V coalesce to form a single
bigger drop. The potential of the new drop
will be (CET Karnataka 2000)
a) nV2 b) n2/3V
c) V/n
d) nV
Two capacitors with capacitance’s C1 and C2
are charged to potentials V1 and V2
respectively. when they are connected in
parallel the ratio of their respective charges
is (CET Karnataka 2000)
C
V
C2
V2
a) 1 b) 1 c) 1
d) 1
2
2
V
C2
C2
V2
In a charged capacitor the energy is stored in
(CET Karnataka 2000)
a) the edges of the capacitor plates
b) the electric field between the plates
c) both in positive and negative charges
d) positive charges
A charge q is placed at the centre of the line
joining two equal charges Q. The sytem of
three charges will be in equilibrium if q is
equal to (CET Karnataka 2001)
Q
Q
Q
Q
a) 
b) 
c) 
d) 
4
2
2
4
Two equal negative charges q are fixed at
the points (0, a) and (0, a) on the y-axis. A
positive charge Q is released from rest at the
point (2a, 0) on the x-axis. The charge Q will
(CET Karnataka 2001)
a) execute simple harmonic motion about
the origin
23.
24.
25.
26.
27.
28.
b) execute oscillatory but not simple
harmonic motion
c) move to the origin and remain at rest
d) move to infinity
A ball of mass 1 g and charge 108 C moves
from a point A whose potential is 600 V to
the point B whose potential is zero. Velocity
of the ball at the point B is 20 cm s1. The
velocity of the ball at the point A is (CET
Karnataka 2001)
a) 2.8 cm s1
b) 2.8 ms1
1
c) 16.7 cm s
d) 16.7 ms1
IN a parallel plate capacitor of capacitance C,
a metal sheet is inserted between the plates,
parallel to them. The thickness of the sheet is
half of the separation between the plates.
The capacitance now becomes (CET
Karnataka 2001)
C
C
a)
b)
c) 4C d) 2C
2
4
While a capacitor remains connected to a
battery, a dielectric slab is slipped between
the plats. (CET Karnataka 2001)
a) the electric field between the plates
increases
b) the energy stored in the capacitor
decreases
c) the potential difference between the
plates is changed
d) charges flow from the battery to the
capacitor
Two nucleons are at a separation of 1x1015
m. The net force between them is F1 if both
are neutrons, F2 if both are protons and F3 if
one is a proton and the other is a neutron.
Then (CET Karnataka 2001)
a) F1 = F3 > F3
b) F1 = F2 > F3
c) F1 = F2 = F3
d) F1 > F2 > F3
64 small drops of mercury, each of radius r
and charge q coalesce to form a big drop.
The ratio of the surface density of charge of
each small drop with that of the big drop is
(CET Karnataka 2002)
a) 1:64 b) 64:1
c) 4:1
d) 1:4
Two capacitors of capacitance 3F and 6F
are charged to a potential of 12V each. They
are now connected to each other, with the
positive plate of each joined to the negative
plate of the other. The potential difference
across each will be. (CET Karnataka 2002)
a) 6V b) 4V c) 3V d) zero
Three point charges are placed at the corners
of an equilateral triangle. Assuming only
electrostatic forces are acting (CET Karnataka
2002)
a) the system can never be in equilibrium
18
81898724
29.
30.
31.
32.
b) the system will be in equilibrium if the
charges rotate about the centre of the
triangle
c) the system will be in equilibrium if the
charges have different magnitudes and
different signs
d) the system will be in equilibrium if the
charges have the same magnitude but
different signs
What fraction of the energy drawn from the
charging battery is stored in a capacitor?
(CET Karnataka 2002)
a) 100%
b) 75%
c) 50%
d) 25%
A charge q is placed at the centre of line
joining two equal point charges each equal to
+Q. The system of 3 charges will be in
equilibrium if q is equal to (CET Karnataka
2003)
Q
Q
Q
Q
a)
b)
c)
d)
4
2
2
4
The inward and outward electric flux from a
closed surface are respectively 8x103 and
4x103 unit. Then the net charge inside the
closed surface is (CET Karnataka 2003)
 4  10 3
a)
coulomb
0
b) 4 x 103 0 coulomb
c) 4 x 103 coulomb
d) 4 x 103 coulomb
In the circuit shown, the effective
capacitance between A and B is. (CET
Karnataka 2003)
SARVAJNYA
35.
34.
a) 8 F b) 4 F
c) 2 F d) 3 F
Capacitance of a parallel plate capacitor
4
becomes
times its original value if a
3
d
dielectric slab of thickness t=
is inserted
2
between the plates (d is separation between
the plates). The dielectric constant of the
slab is. (CET Karnataka 2001)
a) 2
b) 6
c) 4
d) 8
An uncharged sphere of metal is placed
inside a charged parallel plate capacitor. The
lines of force will look like (CET Karnataka
2004)
3
14
F b)
F c) 21 F d) 23 F
3
14
The electric flux for Gaussian surface A that
encloses the charged particle in free space is
(given q1 = 14 nC, q2 = 78.85 nC, q3 = 56
nC) (CET Karnataka 2005)
a) 103 CN1 cm2
b) 103 Nm2C1
c) 6.32 x 103 Nm2 C1 d) 6.32x103 CN1 m2
A comb, when run through one’s dry hair,
attracts small bits of paper. This is due to
(CET Karnataka 2002)
a) comb is a good conductor
b) paper is a good conductor
c) the atoms in the paper get polarised by
charged comb
d) the comb possesses magnetic properties
In a charged capacitor the energy resides
a) on the positive plate
b) on both the positive and negative plates
c) in the field between the plates
d) around the edge of capacitor plates
A parallel plate capacitor is charged and the
charging battery is then disconnected. If the
plates of the capacitor are moved farther
apart by means of insulating handles.
a) the charge on the capacitor increases
b) the voltage across the plate increases
c) the voltage across the plate increases
d) the electrostatic energy stored in the
capacitor decreases
A parallel plate air capacitor is connected to a
battery. The quantities charge, voltage,
electric field and energy associated with the
capacitor are given by Q0, V0, E0 and U0
respectively. A dielectric slab is now
introduced to fill the space between the
a)
36.
37.
38.
33.
Effective capacitance between A and B in the
figure (all capacitances are in F) is (CET
Karnataka 2004)
39.
40.
19
81898724
41.
42.
43.
44.
45.
46.
plates with a battery still in connection. The
corresponding quantities now given by Q.V, E
and U are related to previous ones as.
a) Q = Q0
b) V > V0
c) E > E0
d) U > U0
A condenser is charged through a p.d. of 200
volts and possesses charge of 0.1 coulomb.
When discharge it would release an energy
of
a) 400 J
b) 0.02 J
c) 0.04 J d) 0.08 J
Two identical capacitors are joined in parallel,
charged to a potential V, separated and then
connected in series i.e., the positive plate of
one is connected to the negative plate of
other.
a) the charges on the free plates connected
together are destroyed
b) charges on the free plates are enhanced
c) the energy stored in the system
increases
d) the potential difference between the free
plates is 2V
A parallel plate air condenser is immersed in
an oil of dielectric constant 2. The field
between the plates is
a) increased proportional to 2
b) decreased proportional to 1/2
c) increased proportional to 2
d) decreased proportional to 1/ 2
A number of condensers, each of capacitance
1F and each one of which gets punctured if
a p.d. just exceeding 500 V is applied, are
provided. Then an arrangement suitable for
giving a capacitor of capacitance 1F across
which 3000 volts may be applied requires at
least.
a) 6 component capacitors
b) 12 component capacitors
c) 72 component capacitors
d) 36 component capacitors
An air filled parallel plate capacitor has a
capacitance 1 pf. The separation of the plats
is doubled and wax inserted between them,
which makes the capacitance 2pf. This
implies that dielectric constant of wax is
a) 2.0
b) 4/3
c) 4.0
d) 8.0
A parallel plate condenser is filled with two
dielectrics as shown. Area of each plate is A
metre2 and the separation of d metre. The
dielectric constants are K1 and K2
respectively. Its capacitance in farad will be
SARVAJNYA
0 A
 A K  K2
(K1+K2)
b) 0 . 1
d
d
2
0 A
 0 A K 1 .K 2
c)
2 (K1K2)
d)
.
d
d
2
A parallel plate condenser with plate area a
and separation d is filled with dielectrics as
shown. The dielectric constants are K1 and K2
respectively. The capacitance will be.
a)
47.
a)
0 A
(K 1  K 2 )
d
b)
0 A
d
 K1  K 2

 K 1K 2



 0 A  2K 1K 2 
2 0 A  K 1  K 2 




d)
d  K 1  K 2 
d  K 1K 2 
Three capacitors of capacitance 3F, 9F and
18 F are connected first in series and then
in parallel. The ratio of equivalent
capacitance in two cases (C s/ Cp) will be.
a) 1:15
b) 15:1
c) 1:1
d) 1:3
When air in a capacitor is replaced by a
medium of dielectric constant K, the
capacitance.
a) decreases K times
b) increases K times
c) increases K2 times
d) remains constant
In a parallel plate capacitor, the distance
between the plates is ‘d’ and potential
difference across plates is V. Energy stored
per unit volume between the plates of
capacitor is:
1
V2
Q2

a)
b)
0
2
d2
2V 2
c)
48.
49.
50.
c)
51.
1 V2
2  0 d2
d)
1 2 V2
0
2
d2
Each of two identical capacitors has
capacitance C. One of them is charged to a
potential V1 and the other to a potential V2.
The negative terminals of capacitors are
connected together. When their positive
terminals are also connected together, then
energy loss of whole system is.
1
1
a)
C (V12 – V22)
b)
C (V12 + V22)
4
4
1
1
c)
C (V1 – V22)
d)
C (V1 + V2)2
4
4
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