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ELECTROSTATICS - IV- Capacitance and Van de GraaffGenerator
1. Behaviour of Conductors in Electrostatic Field
2. Electrical Capacitance
3. Principle of Capacitance
4. Capacitance of a Parallel Plate Capacitor
5. Series and Parallel Combination of Capacitors
6. Energy Stored in a Capacitor and Energy Density
7. Energy Stored in Series and Parallel Combination of Capacitors
8. Loss of Energy on Sharing Charges Between Two Capacitors
9. Polar and Non-polar Molecules
10.Polarization of a Dielectric
11.Polarizing Vector and Dielectric Strength
12.Parallel Plate Capacitor with a Dielectric Slab
13.Van de Graaff Generator
Behaviour of Conductors in the Electrostatic Field:
E0
Enet =0
EP
1. Net electric field intensity in the interior of a
conductor is zero.
When a conductor is placed in an electrostatic
field, the charges (free electrons) drift towards
the positive plate leaving the + ve core behind.
At an equilibrium, the electric field due to
the polarisation becomes equal to the applied
field. So, the net electrostatic field inside the
conductor is zero.
1. Electric field just outside the charged conductor is
perpendicular to the surface of the conductor.
Suppose the electric field is acting at an angle
other than 90°, then there will be a component E
cos θ acting along the tangent at that point to the
surface which will tend to accelerate the charge on
the surface leading to ‘surface current’.But there is
no surface current in electrostatics.
So, θ = 90° and cos 90° = 0.
n
E cos θ E
θ
•+ q
NOT POSSIBLE
1. Net charge in the interior of a conductor is zero.
The charges are temporarily separated.The total charge of
the system is zero.
S
ΦE = E . dS =
ε0
q = 0
q q
Since E = 0 in the interior of the conductor, therefore q = 0.
1. Charge always resides on the surface of a conductor.
Suppose a conductor is given some excess charge q.
Construct a Gaussian surface just inside the conductor.
Since E = 0 in the interior of the conductor, therefore q = 0
inside the conductor.
1. Electric potential is constant for the entire conductor.
dV = - E . dr
Since E = 0 in the interior of the conductor, therefore dV =
0. i.e. V = constant
q
∮
. Surface charge distribution may be different at different points
σ =
q
S
σmax
σmin
Every conductor is an equipotential volume
(three- dimensional) rather than just an equipotential
surface (two- dimensional).
Electrical Capacitance:-The measure of the ability of a conductor to store charges is known as
capacitance or capacity (old name).
V
,
q α V or q = C V or C = then If V = 1 volt C= q
Capacitance of a conductor is defined as the charge required to raise its potential through one
unit.
SI Unit of capacitance is ‘farad’ (F). Symbol of capacitance:
Capacitance is said to be 1 farad when 1 coulomb of charge raises the potential of conductor
by 1 volt.
Since 1 coulomb is the big amount of charge, the capacitance will be usually in the range of milli
farad, micro farad, nano farad or pico farad.
Depends upon size and shape of the conductor, nature of medium around the conductor, presence
of other conductors in its neighbourhood
q
Capacitance is a property
Capacitor is a device requires very high current 25 ampere -50 ampere to start them.
Capacitor store energy not charge. This energy supply at once.
Comparison with battery
Similarities
Both energy store house.
Can generate PD in a circuit .
Release of energy
Capacitor release energy at once but battery release energy slowly in equal amount. release of energy example
let us consider a dam as a capacitor or battery fill with the water. Dam contains number of gates you can get
water until all the water come out from the dam.
Hence if we add battery in a circuit better release water or energy constantly until all the energy will come out
from the the battery like dam .If we break the dam all the water come out at once from the dam capacitor work
like this process in a small time will produce lots of energy by capacitor .
Differences
Release of energy is different for both.
Way to store energy is different for both.
Amount of energy store is different for both .
Capacitor store less energy in comparison of battery.
Battery provide energy at once after adding in circuit but capacitor provide energy after the charging.
Leyden jar:- insulator placed between two conductors.
Capacitor need charging then we can use it. We can charge it by battery .
Only charge can move from A to B charge will be same as before before the battery connection and after the
battery connection.
Hence capacitor store energy note charge .
Due to pump rubber will stretch and get potential energy.
1L 1L
In capacitor energy will store in the form of electric field.
A
B
+ve Q -ve Q
+ + + +
+ +
+ +
+ +
+ + +
- - - -
- -
- -
- -
- - -
- - - - - -
- - - - - - -
- - - -
- - - - - -
- - - - - - -
- - - -
Principle of capacitor
if we want to transfer more charge from rod to a body.
Charges at infinite distance
V
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
+ + + +
+ +
+ +
+ +
+ + +
- - - -
- -
- -
- -
- - -
+
+
+
+
+
+
+
+
+
+
And bring closer
+ + + +
+ +
+ +
+ +
+ + +
- - - -
- -
- -
- -
- - -
V′
V′ ∠ V
+ + + +
+ +
+ +
+ +
+ + +
- - - -
- -
- -
- -
- - -
+ + + +
+ +
+ +
+ +
+ + +
V
+Q -Q
Capacitor:- In capacitor Battery moves charge only until potential of both the conductor will be same.
V α Q
Q α V
Q = C V
C=Q / V
If Q = C V
If Q = C V
If Q = C V
C is depands upon conductor and properties of medium ,shape and distance b/w plates.
Arrangement of capacitance
1 ( 2 ) more wastage (3) less wastage of electric field
Calculations of capacitance:- Steps
1. Give charge = +Q and –Q to both the plates.
1. Find electric field b/w plates. constant
E
variable
3. Find V b/w plates. constant ΔV=∫E.dr
E
variable ΔV=E.dr
4. C = Q / V
1. Introduction
•A capacitor (formerly known as condenser) is a device that can store electronic energy.
•All capacitors consists of a combination of two conductors separated by an insulator.
•The insulator is called dielectric which could be oil, air or paper and many more such materials
are there which can act as a dielectric medium between conducting plates of a capacitor.
•Now plates of the capacitor are connected to the terminals of a battery, shown below in figure in
order to charge it's conducting plates.
•As soon as capacitor is connected to the battery , charge is transferred from one conductor to another.
•Plate connected to positive terminal of the battery becomes positively charged with charge +Q in it and plate
connected to negative terminal of the battery becomes negatively charged with charge -Q on it i.e. both plates
have equal amount of opposite charge .
•Once the capacitor is fully charged potential difference between the conductors due to their equal and
opposite charges becomes equal to the potential difference between the battery terminals.
•For a given capacitor Q∝V and the ratio Q/V is constant for a capacitor.
Thus,
Q=CV (1)
where the proportionality constant C is called the capacitance of the capacitor.
•Capacitance of any capacitor depends on shape , size and geometrical arrangement of the conductors.
•When Q is in coulombs (C) and V is in volts(V) then the S.I. unit of capacitance is in farads(F) where
1F=1 coulomb/volt
•One farad is the capacitance of very large capacitor and it's sub-multiples such as microfarad(1μF=10-6) or
picofarad(1pF=10-12) are generally used for practical applications.
Standard Units of Capacitance
The basic unit of capacitance is Farad. But, Farad is a large unit for practical tasks. Hence, capacitance is usually measured in
the sub-units of Farads such as micro-farads (µF) or pico-farads (pF).
Most of the electrical and electronic applications are covered by the following standard unit (SI) prefixes for easy calculations:
1 mF (millifarad) = 10−3 F
1 μF (microfarad) =10−6 F
1 nF (nanofarad) = 10−9 F
1 pF (picofarad) = 10−12 F
The distance between the plates
The more distant the plates are, the less the free electrons on the far plate feel the push of the electrons that are being added
to the negative plate. This makes it harder to add more negative charges to the negative plate. If the plates are closer to each
other, the current would flow through a short circuit. This implies that the capacitance of a parallel plate is inversely related to
the plate separation.
Area of the plates
It’s a lot easier to add charges to a capacitor if the parallel plates have a huge area. Two wide metal plates would give two
repelling like charges a greater range to spread out across the plate, making it easier to add a lot more negative charge to one
plate. Likewise, a very small plate area would cause the electrons to get cramped together earlier, making it harder to get a
large difference in charge for a given voltage.
Electrostatic Shielding
Electric field inside a cavity in a conductor is always zero. Even if the conductor is charged or charges are
induced on a neutral conductor by an external field, all charges reside only on the outer surface of the
conductor. Hence, the any cavity of any shape and size is always shielded from outer electric influence. This
is called electrostatic shielding.
Principle of Capacitance: A B
A
Step 1: Plate A is positively charged and B is neutral.
Step 2: When a neutral plate B is brought near A, charges are
induced on B such that the side near A is negative and the other
side is positive.
The potential of the system of A and B in step 1 and 2 remains the
same because the potential due to positive and negative charges on
B cancel out.
Step 3: When the farther side of B is earthed the positive charges on
B get neutralised and B is left only with negative charges.
Now, the net potential of the system decreases due to the sum of
positive potential on A and negative potential on B.
To increase the potential to the same value as was in step 2, an
additional amount of charges can be given to plate A.
This means, the capacity of storing charges on A increases.
The system so formed is called a ‘capacitor’.
Potential = V
Potential = V
Potential E
decreases to v
B
Parallel plate capacitor
•A parallel plate capacitor consists of two large plane parallel conducting plates separated by a
small distance d.
•Suppose two plates of the capacitor has equal and opposite charge Q on them. If A is the area of
each plate then surface charge density on each plate is
σ = Q / A
•We have already calculated field between two oppositely charged plates using gauss's law which
is
E = σ /ε0= Q / ε0A
and in this result effects near the edges of the plates have been neglected.
C
A
D
E
+Q
d
A
+
+
+
+
+
+
+
+
+
+
+
-
-
-
-
-
-
-
-
-
-
-Q
Due to plate C
Due to plate D
and the intensities in between the plates due to plate 1 and plate 2.
E1= E2 = Q /2 ε0A
E = E1+E2
E =Q / ε0A
Potential difference between plate A and plate B
V =E.d
=Q d / k A
Capacitance of capacitor
C = Q / V
C= ε0A k / d
# C α A depands upon area.
# C α k depands upon dielectric constant.
# C α1/d depands upon separation between plates.
# C not depands upon potential difference of both the plates.
Calculation of capacitance
•For calculating capacitance of a capacitor first we need to find the potential difference between
it's two conducting plates having charge +Q and -Q.
•For simple arrangements of conductors like two equivalent parallel plates kept at distance d
apart or two concentric conducting spheres etc., potential difference can be found first by
calculating electric field from gauss's law or by Coulomb's law.
•After calculating electric field , potential difference can be found by integrating electric field
using the relation
Va-Vb= ∫ E.dr
where the limits of integration goes from a to b.
•Once we know the potential difference between two conductors of the capacitor , it's
capacitance can be calculated from the relation
C=Q/V (2)
•Calculation of capacitance of some simple arrangements would be illustrated in following few
articles.
•Since electric field between the plates is uniform the potential difference between the plates is
V = Ed = Qd / ε0A
where , d is the separation between the plates.
•Thus, capacitance of parallel plate capacitor in vacuum is
C = Q/V= ε0A/d (3)
•From equation 3 we see that quantities on which capacitance of parallel plate capacitor depends
i.e.,ε0 , A and d are all constants for a capacitor.
•Thus we see that in this case capacitance is independent of charge on the capacitor but depends
on area of it's plates and separation distance between the plates.
Spherical capacitor
•A spherical capacitor consists of a solid or hollow spherical conductor of radius a , surrounded by
another hollow concentric spherical of radius b.
Let draw a gaussian surface of radius r.
•Let +Q be the charge given to the inner sphere and -Q be the charge given to the outer sphere.
•The field at any point between conductors is same as that of point charge Q at the origin and
charge on outer shell does not contribute to the field inside it.
•Thus electric field between conductors is E = Q , E is variable.
2πε0r2
•Potential difference between two conductors is
V = Va − Vb
= − ∫ E.dr
-Q
+Q
r
b
a
where limits of integration goes from a to b.
On integrating we get potential difference between to conductors as
ΔV=∫ 1 Q dr = Q -1 + 1
V= Q (b−a)
4πε0 ba
•Now , capacitance of spherical conductor is
C=QV
or,
C = 4πε0 ba ………………..(1)
(b−a)
Case -1 C for single sphere
a = R
b = ∞
From equation 1 C = 4πε0 b a
b 1−a / b
if a∠∠∠∠b
C = 4πε0 a …………..…(2)
a
b 4πε0r2 4πε0r b a
[ ]
∞
R
( )
again if radius of outer conductor approaches to infinity then from equation 1 we have
C=4πε0a ----(2)
•Equation 2 gives the capacitance of single isolated sphere of radius a.
•Thus capacitance of isolated spherical conductor is proportional to its radius.
Case 2 if a ~ b
a - b = d
From equation 1
C = 4πε0 a2 /d
C = ε0A /d
This is equal to the II plate capacitor.
Note that
(1) As b→∞, the capacitance reduces 4πε0a. This shows that a spherical conductor is a
spherical capacitor with its other plate of infinite radius.
(2) As a and b both become very large, maintaining the difference a−b=d (finite), the
expression for C reduces to C= ε0 A/d. This shows that a spherical capacitor behaves as a
parallel plate capacitor if its spherical surfaces have large radii and are close to each other.
a
b
Capacitance of an Isolated Spherical Conductor:
O •
r
+q
Let a charge q be given to the sphere which is assumed to be
concentrated at the centre. Potential at any point on the surface is
V =
q
C =
4πε0 r
q
V
C = 4πε0 r
1. Capacitance of a spherical conductor is directly proportional to its radius.
2. The above equation is true for conducting spheres, hollow or solid.
3. IF the sphere is in a medium, then C = 4πε0εr r.
4. Capacitance of the earth is 711 μF.
Cylinderical capacitor
•A cylinderical capacitor is made up of a conducting cylinder or wire of radius a surrounded by
another concentric cylinderical shell of radius b (b > a ).
•Let L be the length of both the cylinders and charge on inner cylender is +Q and
charge on outer cylinder is -Q.
•For calculate electric field between the conductors using Gauss's law
consider a Gaussian surface of radius r and length L
•According to Gauss's law flux through this surface is q ε0 where q is
net charge inside this surface.
We know that electric flux is given by
∫E.ds = qin / ε0
E.2ΠrL = qin / ε0
E = Q / 2Π ε0rL
since electric field is constant in magnitude on the Gaussian surface and is perpendicular to this
surface. Thus,
+Q
-Q
r
b a
E is variables
•If potential at inner cylinder is Va and Vb is potential of outer cylinder then potential difference between both
the cylinders isVa and Vb
ΔV = ∫ E.dr
where limits of integration goes from a to b. ΔV= Q ∫1∕r dr = Q [ logr]
2Πε0L 2Πε0L
= Q (logb –loga ) = Q log(b/a)
2Πε0L 2Πε0L
•Potential of inner conductor is greater then that of outer conductor because inner cylinder carries +ve
charge.
•Thus capacitance of cylinderical capacitor is C = Q/ ΔV= Q
Q log(b/a)
2Πε0L
C=2Πε0L log(b/a)
•From equation it can easily be concluded that capacitance of a cylinderical capacitor depends on length of
cylinders.
•More is the length of cylinders , more charge could be stored on the capacitor for a given potential difference.
( )
b
a
a
b
Force on II plate capacitor
•A parallel plate capacitor consists of two large plane parallel conducting plates C and D
separated by a small distance d.
•Suppose two plates of the capacitor has equal and opposite charge Q on them. If A is the area of
each plate then surface charge density on each plate is σ = Q / A
•We have already calculated field between two oppositely charged plates using gauss's law which
is
E = σ / ε0= Q/ε0A
= Q/ 2ε0A + Q/ 2ε0A (due to two plates)
C
A
D
E
+Q
d
A
+
+
+
+
+
+
+
+
+
+
+
-
-
-
-
-
-
-
-
-
-
-Q
Due to plate C
Due to plate D
Force on plate 2
F = qE (E due to plate 1 and charge on plate 2)
F = -Q(Q/ 2ε0A)
F = -(Q2/ 2ε0A) (-ve sign due to attraction.)
From this equation force not depands upon separation d.
A A
D
C
D
C
A A
F applied
V is very small
K.E. will be =0
F
w.d. by force
applied
Energy stored in capacitor:- parallel plate capacitor consists of two large plane parallel conducting plates
C and D separated by a small distance d.
# Two plates of the capacitor has equal and
opposite charge Q on them hence net
chargy =0
F =-(Q2/ 2ε0A)
W= F .d =-(Q2/ 2ε0A).d
Work done in sepereting plate will be stored as electric energy in capacitor.
This electric energy will be stored in the form of electric field.
W = - (Q2/ 2ε0A).d
U = Q2 / 2c Q = CV
U = c V2 / 2
U = Q V / 2
This is proved by using mechanical energy.
No type of energy will be entered in this (K.E.=0) only possible if v is very small.
Force applied is equal to force of attraction.
Series Combination of Capacitors:
V1 V2 V3
V
C1 C2 C3
In series combination,
•
•
Charge is same in each capacitor
Potential is distributed in inverse proportion to
capacitances
i.e. V = V1 + V2 + V3
But
q
1
V = 2
V =
q
3
C2 C3
q
, and V =
q
V = ,
(where C is the equivalent capacitance or
effective capacitance or net capacitance
ortotal capacitance)
C
q
=
C1
+
C
q
+
C
2 3
C1
q
q
C
or
The reciprocal of the effective capacitance is the sum of the reciprocals of the individual capacitances.
Note: The effective capacitance in series combination is less than the least of all the individual capacitances.
q q q
1
Ci
1
C
n
= ∑
i=1
=
1
+
C1
1 1
+
C2 C3
1
C
Parallel Combination of Capacitors:
In parallel combination,
•
•
Potential is same across each capacitor
Charge is distributed in direct proportion to capacitances
i.e. q = q1 + q2 + q3
But
(where C is the equivalent
capacitance)
or
The effective capacitance is the sum of the individual capacitances.
Note: The effective capacitance in parallel combination is larger than the largest of all the individual
capacitances.
q1 = C1 V , q2 = C2 V , q3 = C3 V and q = C V
C V = C1V + C2 V + C3 V
n
C = ∑ Ci
i=1
C = C1 + C2 + C3
V q1
C1
C2
C3
V
V
V
q2
q3
+Q -Q
1 2
d
+
+
+
+
+
+
+
+
+
+
-
-
-
-
-
-
-
-
-
-
-
-
Energy Stored in a Capacitor:-
Let both the plates have no charge at all.
Want to charge +Q and –Q on both the plates.
At time t = 0, Q = 0
Let we moved +dq charge from 2nd plate to 1st plate .
+dq charge leave –dq charge behind it.
We are increasing potential difference between the plates.Hance at time t q = Q
1 2
1 2
Energy Stored in a Capacitor:
V
The process of charging a capacitor is equivalent to
transferring charges from one plate to the other of the
capacitor.
The moment charging starts, there is a potential difference
between the plates.Therefore, to transfer charges against the
potential difference some work is to be done.This work is
stored as electrostatic potential energy in the capacitor.
If dq be the charge transferred against the potential
difference V, then work done is
dU = dW = V dq
= dq
The total work done ( energy) to transfer charge q is
U =
0
q
q
C
1 Q2
U =
2 C
U =
1
2
U =
1
2
Q V
dq or or C V2 or
q
C
-Q
+Q
1 2
∫
Energy Density :- Energy Density =Energy / Volume
U =
1
C V2
d
C =
A ε0
V = E d
and
U =
2
1
ε0 Ad E2
1
2 2
ε0 E2
=
U
Ad
ε0 E2
1
U =
2
But
or or
SI unit of energy density is J m-3.
Energy density is generalised as energy per unit volume of the field.
Energy Stored in a Series Combination of Capacitors:
=
1 2 3
1 1
C C C C Cn
1 1 1
+ + + ……….+
1 q2
U =
2 C
U =
1
2
1
q2
[
1
+ +
C1 C2 C3
1
Cn
1
+ ………. + ]
U = U1 + U2 + U3 + ………. + Un
The total energy stored in the system is the sum of energy stored in the individual capacitors.
Energy Stored in a Parallel Combination of Capacitors:
C = C1 + C2 + C3 + ……….. + Cn
U = C V2
U =
1 1
2 2
V2 ( C1 + C2 + C3 + ……….. + Cn)
V =
U = U1 + U2 + U3 + ………. + Un
The total energy stored in the system is the sum of energy stored in the individual capacitors.
Loss of Energy on Sharing of Charges between the Capacitors in Parallel:
Consider two capacitors of capacitances C1, C2, charges q1, q2 and potentials V1,V2.
Total charge after sharing = Total charge before sharing
(C1 + C2) V = C1 V1 + C2 V2
C1 + C2
C1 V1 + C2 V2
The total energy before sharing is
i
U = 1
C1 V 2
1 1
2 2
2 2
C V 2
+
Uf =
The total energy after sharing is
1
2
(C1 + C2) V2
C1 C2 (V1 –V2)2
Ui– Uf =
2 (C1 + C2)
Ui – Uf > 0 or Ui > Uf
Therefore, there is some loss of energy when two charged capacitors are connected
together.
The loss of energy appears as heat and the wire connecting the two capacitors may become hot.
Dielectrics:
Generally, a non-conducting medium or insulator is called a ‘dielectric’.
Precisely, the non-conducting materials in which induced charges are produced on their faces on the
application of electric fields are called dielectrics.
Eg. Air, H2, glass, mica, paraffin wax, transformer oil, etc.
What happened when space between the two plates of the capacitor is filled by a dielectric was first discovered
by faraday.
Faraday discovered that if the space between conductors of the capacitor is occupied by the dielectric, the
capacitance of capacitor is increased.
Dielectric constant of vaccum is unity
If the dielectric completely fills the space between the conductors of the capacitor ,the capacitance is increased by
an factor K which is characterstics of the dielectric and This factor is known as the dielectric constant.
Polarization of Dielectrics:
When a non-polar dielectric slab is subjected to an electric field, dipoles are induced due to separation of
effective positive and negative centres.
E0 is the applied field and Ep is the induced field in
the dielectric.
The net field is EN = E0 – Ep
i.e. the field is reduced when a dielectric slab is
introduced.
The dielectric constant is given by
E0
K =
E0 - Ep
Ep
Ep
E0=0
If the dielectric slab occupies the whole space between the plates, i.e. t = d, then
WITH DIELECTRIC SLAB
Physcial Quantity With Battery
disconnected
With Battery
connected
Charge Remains the same Increases (K C0 V0)
Capacitance Increases (K C0) Increases (K C0)
Electric Field Decreases
EN = E0 – Ep
Remains the same
Potential Difference Decreases Remains the same
Energy stored Remains the same Increases (K U0)
K =
C
C0
C = K C0
Dielectric Constant
Polar Molecules:
A molecule in which the centre of positive charges does not coincide
with the centre of negative charges is called a polar molecule.
Polar molecule does not have symmetrical shape.
2 3 2
Eg. H Cl, H O, N H , C O , alcohol, etc.
O
H H
Effect of Electric Field on Polar Molecules:
E = 0 E
p
p = 0
In the absence of external electric field, the
permanent dipoles of the molecules orient in
random directions and hence the net dipole
moment is zero.
When electric field is applied, the dipoles orient
themselves in a regular fashion and hence dipole
moment is induced. Complete allignment is not
possible due to thermal agitation.
105°
p
Non - polar Molecules:
A molecule in which the centre of positive charges coincides with the centre of negative charges is called a
non-polar molecule.
Non-polar molecule has symmetrical shape. Eg. N2 , C H4, O2, C6 H6, etc.
Effect of Electric Field on Non-polar Molecules:
E = 0 E
p
p = 0
In the absence of external
electric field, the effective
positive and negative centres
coincide and hence dipole is
not formed.
When electric field is applied, the positive
charges are pushed in the direction of electric
field and the electrons are pulled in the
direction opposite to the electric field. Due to
separation of effective centres of positive and
negative charges, dipole is formed.
Criteria Polar Non-Polar
Centers of positive and
negative charges
The centers are separated even
in the absence of external
electric field
The centers coincide
Dipole Moment Permanent dipole moment
No permanent dipole
moment
Examples
Hydro Chloric Acid (HCl) and
Water (H2O)
Oxygen(O2) and
Hydrogen(H2)
Dielectrics in External Electric
Field
The individual dipole moments
tend to align with the field.
They develop a net dipole
moment in the direction of the
field.
Positive and negative charges
get displaced. They develop
an induced dipole moment
as a restoring force against
the direction of electric field.
Polarization Vector:
The polarization vector measures the degree of polarization of the dielectric. It is defined as the dipole
moment of the per unit volume in dielectric.
If n is the number of atoms or molecules per unit volume of the dielectric, then polarization vector is
P = n p
SI unit of polarization vector is C m-2.
It is a vector quantity.
Dimension is M0L-2AT.
Dielectric Strength:-Dielectric strength is the maximum value of
the electric field intensity that can be applied to the dielectric without
its electric break down.
Its SI unit is V m-1.
Its practical unit is kV mm-1.
Dielectric constant:- The ratio of external electric field E0 to the resultant
electric field E inside the dielectric substance is known as dielectric constant
of the substance.
k= E0 / E E∠E0 , 1∠k
Breakdown potential difference:- It is that maximum potential difference applied across the dielectric at
which dielectric substance breaks down. Vmax=Emax × d
Dielectric Dielectric
strength
(kV / mm)
Vacuum ∞
Air 0.8 – 1
Porcelain 4 – 8
Pyrex 14
Paper 14 – 16
Rubber 21
Mica 160 – 200
Dielectric in capacitor :- (dielectric slab)
Capacitor partially filled with dielectric slab.
A is the area of plate of capacitor.
d is a separation between two plates .
+ Q charge is given to A plate.
-Q charge is given to be B plate .
k is the dielectric constant of dielectric slab .
t is a thickness of dielectric slab.
x is the distance between plateA and dielectric slab.
(d-x-t) is the distance between dielectric slab and plate B.
E1= Q /ε0A between plate a and dielectric slab .
E2= Q /ε0 Ak of electric slab .
E3= Q /ε0A between plate B and dielectric slab.
ΔV =V12+V23+V34
=E1x + E2t+ E3(d-x-t)
C = Q / ΔV = ε0A / (d-t+t/k ) does not depend upon X hence location of slab not matter.
t
x
d
E2
E1 E3
A
+Q
A
-Q
k
A B
If the entire space between the plates is filled with dielectric
t =d
C = ε0A / (d-d+d/k )
C = ε0A k / d
If the entire space between the plate has a cube
t =0
C = ε0A / (d-0+0/k )
C = ε0A / d
If a slab of metal of thickness t is placed between the the capacitor plates
for metal k= ∞
C = ε0A / (d-t+t/0 )
C = ε0A / (d-t)
More than one slab :-
C = ε0A / (d-t1+t1/k1 -t2 +t2/k2)
Spherical capacitor :-
C1= 4 Π ε0 k1ab /(b-a)
C2= 4 Π ε0 k2bc /(c-b) C= C1 C2
C= 4 Π ε0 k1 k2abc C1 + C2
k2c /(b-a)k1a /(c-b)
A
+Q
t2
t1
k2
k1
-Q
A
d
+ Q
- Q
a
c
b
k1
k2
Capacitance of Parallel Plate Capacitor with Dielectric Slab:
Ep
E0 EN = E0 - E p
t d
V = E0 (d – t) + EN t
E
N
K =
E0
or N
E =
K
E0
0
V = E (d – t) +
E0
t
0
V = E [(d – t) +
K
K
t
]
But 0
E =
σ
=
ε0
q/A
and C =
ε0
q
V
C =
A ε0
[(d – t) +
K
t
]
or C =
A ε 0
d [1 –
K
t
d
t
(1 - ) ]
or C =
C0
[1 –
t
K
t
d
(1 - ) ]
C > C . i.e. Capacitance increases with introduction of
dielectric slab.
0
does not depend
upon X hence location
of slab not matter.
Van de Graaff Generator:
T
D
C1
P1
P2
C2
M
S
I S
HVR
S – Large Copper sphere
C1, C2 – Combs with sharp points P1, P2 – Pulleys
to run belt
HVR – High Voltage Rectifier M – Motor
IS – Insulating Stand
D – Gas Discharge Tube T - Target
Principle:
Consider two charged conducting spherical shells such that one is smaller and the other is larger.
When the smaller one is kept inside the larger one and connected together, charge from the
smaller one is transferred to larger shell irrespective of the higher potential of the larger shell.
i.e. The charge resides on the outer surface of the outer shell and the potential of the outer
shell increases considerably.
Sharp pointed surfaces of a conductor have large surface charge densities and hence the
electric field created by them is very high compared to the dielectric strength of the
dielectric (air).
Therefore air surrounding these conductors get ionized and the like charges are repelled by the
charged pointed conductors causing discharging action known as Corona Discharge or Action of
Points.The sprayed charges moving with high speed cause electric wind.
Opposite charges are induced on the teeth of collecting comb (conductor) and again opposite
charges are induced on the outer surface of the collecting sphere (Dome).
Construction:
Van de Graaff Generator consists of a large (about a few metres in radius) copper spherical
shell (S) supported on an insulating stand (IS) which is of several metres high above the
ground.
A belt made of insulating fabric (silk, rubber, etc.) is made to run over the pulleys (P1, P2 )
operated by an electric motor (M) such that it ascends on the side of the combs.
Comb (C1) near the lower pulley is connected to High Voltage Rectifier (HVR) whose other end
is earthed. Comb (C2) near the upper pulley is connected to the sphere S through a conducting
rod.
A tube (T) with the charged particles to be accelerated at its top and the target at the bottom
is placed as shown in the figure. The bottom end of the tube is earthed for maintaining lower
potential.
To avoid the leakage of charges from the sphere, the generator is enclosed in the steel tank
filled with air or nitrogen at very high pressure (15 atmospheres).
The process continues for a longer time to store more and more charges on the sphere and the
potential of the sphere increases considerably. When the charge on the sphere is very high, the
leakage of charges due to ionization of surrounding air also increases.
Maximum potential occurs when the rate of charge carried in by the belt is equal to the rate at
which charge leaks from the shell due to ionization of air.
Now, if the positively charged particles which are to be accelerated are kept at the top of the tube
T, they get accelerated due to difference in potential (the lower end of the tube is connected to the
earth and hence at the lower potential) and are made to hit the target for causing nuclear
reactions, etc.
Van de Graaff Generator is used to produce very high potential difference (of the order of several
million volts) for accelerating charged particles.
The beam of accelerated charged particles are used to trigger nuclear reactions.
The beam is used to break atoms for various experiments in Physics.
In medicine, such beams are used to treat cancer.It is used for research purposes.
Uses:

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Capacitor

  • 1. ELECTROSTATICS - IV- Capacitance and Van de GraaffGenerator 1. Behaviour of Conductors in Electrostatic Field 2. Electrical Capacitance 3. Principle of Capacitance 4. Capacitance of a Parallel Plate Capacitor 5. Series and Parallel Combination of Capacitors 6. Energy Stored in a Capacitor and Energy Density 7. Energy Stored in Series and Parallel Combination of Capacitors 8. Loss of Energy on Sharing Charges Between Two Capacitors 9. Polar and Non-polar Molecules 10.Polarization of a Dielectric 11.Polarizing Vector and Dielectric Strength 12.Parallel Plate Capacitor with a Dielectric Slab 13.Van de Graaff Generator
  • 2. Behaviour of Conductors in the Electrostatic Field: E0 Enet =0 EP 1. Net electric field intensity in the interior of a conductor is zero. When a conductor is placed in an electrostatic field, the charges (free electrons) drift towards the positive plate leaving the + ve core behind. At an equilibrium, the electric field due to the polarisation becomes equal to the applied field. So, the net electrostatic field inside the conductor is zero. 1. Electric field just outside the charged conductor is perpendicular to the surface of the conductor. Suppose the electric field is acting at an angle other than 90°, then there will be a component E cos θ acting along the tangent at that point to the surface which will tend to accelerate the charge on the surface leading to ‘surface current’.But there is no surface current in electrostatics. So, θ = 90° and cos 90° = 0. n E cos θ E θ •+ q NOT POSSIBLE
  • 3. 1. Net charge in the interior of a conductor is zero. The charges are temporarily separated.The total charge of the system is zero. S ΦE = E . dS = ε0 q = 0 q q Since E = 0 in the interior of the conductor, therefore q = 0. 1. Charge always resides on the surface of a conductor. Suppose a conductor is given some excess charge q. Construct a Gaussian surface just inside the conductor. Since E = 0 in the interior of the conductor, therefore q = 0 inside the conductor. 1. Electric potential is constant for the entire conductor. dV = - E . dr Since E = 0 in the interior of the conductor, therefore dV = 0. i.e. V = constant q ∮
  • 4. . Surface charge distribution may be different at different points σ = q S σmax σmin Every conductor is an equipotential volume (three- dimensional) rather than just an equipotential surface (two- dimensional). Electrical Capacitance:-The measure of the ability of a conductor to store charges is known as capacitance or capacity (old name). V , q α V or q = C V or C = then If V = 1 volt C= q Capacitance of a conductor is defined as the charge required to raise its potential through one unit. SI Unit of capacitance is ‘farad’ (F). Symbol of capacitance: Capacitance is said to be 1 farad when 1 coulomb of charge raises the potential of conductor by 1 volt. Since 1 coulomb is the big amount of charge, the capacitance will be usually in the range of milli farad, micro farad, nano farad or pico farad. Depends upon size and shape of the conductor, nature of medium around the conductor, presence of other conductors in its neighbourhood q
  • 5. Capacitance is a property Capacitor is a device requires very high current 25 ampere -50 ampere to start them. Capacitor store energy not charge. This energy supply at once. Comparison with battery Similarities Both energy store house. Can generate PD in a circuit . Release of energy Capacitor release energy at once but battery release energy slowly in equal amount. release of energy example let us consider a dam as a capacitor or battery fill with the water. Dam contains number of gates you can get water until all the water come out from the dam. Hence if we add battery in a circuit better release water or energy constantly until all the energy will come out from the the battery like dam .If we break the dam all the water come out at once from the dam capacitor work like this process in a small time will produce lots of energy by capacitor . Differences Release of energy is different for both. Way to store energy is different for both. Amount of energy store is different for both . Capacitor store less energy in comparison of battery. Battery provide energy at once after adding in circuit but capacitor provide energy after the charging.
  • 6. Leyden jar:- insulator placed between two conductors. Capacitor need charging then we can use it. We can charge it by battery . Only charge can move from A to B charge will be same as before before the battery connection and after the battery connection. Hence capacitor store energy note charge . Due to pump rubber will stretch and get potential energy. 1L 1L In capacitor energy will store in the form of electric field. A B +ve Q -ve Q + + + + + + + + + + + + + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  • 7. Principle of capacitor if we want to transfer more charge from rod to a body. Charges at infinite distance V - - - - - - - - - - - - - - - + + + + + + + + + + + + + - - - - - - - - - - - - - + + + + + + + + + +
  • 8. And bring closer + + + + + + + + + + + + + - - - - - - - - - - - - - V′ V′ ∠ V + + + + + + + + + + + + + - - - - - - - - - - - - - + + + + + + + + + + + + + V +Q -Q Capacitor:- In capacitor Battery moves charge only until potential of both the conductor will be same. V α Q Q α V Q = C V C=Q / V If Q = C V If Q = C V If Q = C V C is depands upon conductor and properties of medium ,shape and distance b/w plates.
  • 9. Arrangement of capacitance 1 ( 2 ) more wastage (3) less wastage of electric field
  • 10. Calculations of capacitance:- Steps 1. Give charge = +Q and –Q to both the plates. 1. Find electric field b/w plates. constant E variable 3. Find V b/w plates. constant ΔV=∫E.dr E variable ΔV=E.dr 4. C = Q / V
  • 11. 1. Introduction •A capacitor (formerly known as condenser) is a device that can store electronic energy. •All capacitors consists of a combination of two conductors separated by an insulator. •The insulator is called dielectric which could be oil, air or paper and many more such materials are there which can act as a dielectric medium between conducting plates of a capacitor. •Now plates of the capacitor are connected to the terminals of a battery, shown below in figure in order to charge it's conducting plates.
  • 12. •As soon as capacitor is connected to the battery , charge is transferred from one conductor to another. •Plate connected to positive terminal of the battery becomes positively charged with charge +Q in it and plate connected to negative terminal of the battery becomes negatively charged with charge -Q on it i.e. both plates have equal amount of opposite charge . •Once the capacitor is fully charged potential difference between the conductors due to their equal and opposite charges becomes equal to the potential difference between the battery terminals. •For a given capacitor Q∝V and the ratio Q/V is constant for a capacitor. Thus, Q=CV (1) where the proportionality constant C is called the capacitance of the capacitor. •Capacitance of any capacitor depends on shape , size and geometrical arrangement of the conductors. •When Q is in coulombs (C) and V is in volts(V) then the S.I. unit of capacitance is in farads(F) where 1F=1 coulomb/volt •One farad is the capacitance of very large capacitor and it's sub-multiples such as microfarad(1μF=10-6) or picofarad(1pF=10-12) are generally used for practical applications.
  • 13. Standard Units of Capacitance The basic unit of capacitance is Farad. But, Farad is a large unit for practical tasks. Hence, capacitance is usually measured in the sub-units of Farads such as micro-farads (µF) or pico-farads (pF). Most of the electrical and electronic applications are covered by the following standard unit (SI) prefixes for easy calculations: 1 mF (millifarad) = 10−3 F 1 μF (microfarad) =10−6 F 1 nF (nanofarad) = 10−9 F 1 pF (picofarad) = 10−12 F The distance between the plates The more distant the plates are, the less the free electrons on the far plate feel the push of the electrons that are being added to the negative plate. This makes it harder to add more negative charges to the negative plate. If the plates are closer to each other, the current would flow through a short circuit. This implies that the capacitance of a parallel plate is inversely related to the plate separation. Area of the plates It’s a lot easier to add charges to a capacitor if the parallel plates have a huge area. Two wide metal plates would give two repelling like charges a greater range to spread out across the plate, making it easier to add a lot more negative charge to one plate. Likewise, a very small plate area would cause the electrons to get cramped together earlier, making it harder to get a large difference in charge for a given voltage. Electrostatic Shielding Electric field inside a cavity in a conductor is always zero. Even if the conductor is charged or charges are induced on a neutral conductor by an external field, all charges reside only on the outer surface of the conductor. Hence, the any cavity of any shape and size is always shielded from outer electric influence. This is called electrostatic shielding.
  • 14. Principle of Capacitance: A B A Step 1: Plate A is positively charged and B is neutral. Step 2: When a neutral plate B is brought near A, charges are induced on B such that the side near A is negative and the other side is positive. The potential of the system of A and B in step 1 and 2 remains the same because the potential due to positive and negative charges on B cancel out. Step 3: When the farther side of B is earthed the positive charges on B get neutralised and B is left only with negative charges. Now, the net potential of the system decreases due to the sum of positive potential on A and negative potential on B. To increase the potential to the same value as was in step 2, an additional amount of charges can be given to plate A. This means, the capacity of storing charges on A increases. The system so formed is called a ‘capacitor’. Potential = V Potential = V Potential E decreases to v B
  • 15. Parallel plate capacitor •A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance d. •Suppose two plates of the capacitor has equal and opposite charge Q on them. If A is the area of each plate then surface charge density on each plate is σ = Q / A •We have already calculated field between two oppositely charged plates using gauss's law which is E = σ /ε0= Q / ε0A and in this result effects near the edges of the plates have been neglected. C A D E +Q d A + + + + + + + + + + + - - - - - - - - - - -Q Due to plate C Due to plate D
  • 16. and the intensities in between the plates due to plate 1 and plate 2. E1= E2 = Q /2 ε0A E = E1+E2 E =Q / ε0A Potential difference between plate A and plate B V =E.d =Q d / k A Capacitance of capacitor C = Q / V C= ε0A k / d # C α A depands upon area. # C α k depands upon dielectric constant. # C α1/d depands upon separation between plates. # C not depands upon potential difference of both the plates.
  • 17. Calculation of capacitance •For calculating capacitance of a capacitor first we need to find the potential difference between it's two conducting plates having charge +Q and -Q. •For simple arrangements of conductors like two equivalent parallel plates kept at distance d apart or two concentric conducting spheres etc., potential difference can be found first by calculating electric field from gauss's law or by Coulomb's law. •After calculating electric field , potential difference can be found by integrating electric field using the relation Va-Vb= ∫ E.dr where the limits of integration goes from a to b. •Once we know the potential difference between two conductors of the capacitor , it's capacitance can be calculated from the relation C=Q/V (2) •Calculation of capacitance of some simple arrangements would be illustrated in following few articles.
  • 18. •Since electric field between the plates is uniform the potential difference between the plates is V = Ed = Qd / ε0A where , d is the separation between the plates. •Thus, capacitance of parallel plate capacitor in vacuum is C = Q/V= ε0A/d (3) •From equation 3 we see that quantities on which capacitance of parallel plate capacitor depends i.e.,ε0 , A and d are all constants for a capacitor. •Thus we see that in this case capacitance is independent of charge on the capacitor but depends on area of it's plates and separation distance between the plates.
  • 19. Spherical capacitor •A spherical capacitor consists of a solid or hollow spherical conductor of radius a , surrounded by another hollow concentric spherical of radius b. Let draw a gaussian surface of radius r. •Let +Q be the charge given to the inner sphere and -Q be the charge given to the outer sphere. •The field at any point between conductors is same as that of point charge Q at the origin and charge on outer shell does not contribute to the field inside it. •Thus electric field between conductors is E = Q , E is variable. 2πε0r2 •Potential difference between two conductors is V = Va − Vb = − ∫ E.dr -Q +Q r b a
  • 20. where limits of integration goes from a to b. On integrating we get potential difference between to conductors as ΔV=∫ 1 Q dr = Q -1 + 1 V= Q (b−a) 4πε0 ba •Now , capacitance of spherical conductor is C=QV or, C = 4πε0 ba ………………..(1) (b−a) Case -1 C for single sphere a = R b = ∞ From equation 1 C = 4πε0 b a b 1−a / b if a∠∠∠∠b C = 4πε0 a …………..…(2) a b 4πε0r2 4πε0r b a [ ] ∞ R ( )
  • 21. again if radius of outer conductor approaches to infinity then from equation 1 we have C=4πε0a ----(2) •Equation 2 gives the capacitance of single isolated sphere of radius a. •Thus capacitance of isolated spherical conductor is proportional to its radius. Case 2 if a ~ b a - b = d From equation 1 C = 4πε0 a2 /d C = ε0A /d This is equal to the II plate capacitor. Note that (1) As b→∞, the capacitance reduces 4πε0a. This shows that a spherical conductor is a spherical capacitor with its other plate of infinite radius. (2) As a and b both become very large, maintaining the difference a−b=d (finite), the expression for C reduces to C= ε0 A/d. This shows that a spherical capacitor behaves as a parallel plate capacitor if its spherical surfaces have large radii and are close to each other. a b
  • 22. Capacitance of an Isolated Spherical Conductor: O • r +q Let a charge q be given to the sphere which is assumed to be concentrated at the centre. Potential at any point on the surface is V = q C = 4πε0 r q V C = 4πε0 r 1. Capacitance of a spherical conductor is directly proportional to its radius. 2. The above equation is true for conducting spheres, hollow or solid. 3. IF the sphere is in a medium, then C = 4πε0εr r. 4. Capacitance of the earth is 711 μF.
  • 23. Cylinderical capacitor •A cylinderical capacitor is made up of a conducting cylinder or wire of radius a surrounded by another concentric cylinderical shell of radius b (b > a ). •Let L be the length of both the cylinders and charge on inner cylender is +Q and charge on outer cylinder is -Q. •For calculate electric field between the conductors using Gauss's law consider a Gaussian surface of radius r and length L •According to Gauss's law flux through this surface is q ε0 where q is net charge inside this surface. We know that electric flux is given by ∫E.ds = qin / ε0 E.2ΠrL = qin / ε0 E = Q / 2Π ε0rL since electric field is constant in magnitude on the Gaussian surface and is perpendicular to this surface. Thus, +Q -Q r b a
  • 24. E is variables •If potential at inner cylinder is Va and Vb is potential of outer cylinder then potential difference between both the cylinders isVa and Vb ΔV = ∫ E.dr where limits of integration goes from a to b. ΔV= Q ∫1∕r dr = Q [ logr] 2Πε0L 2Πε0L = Q (logb –loga ) = Q log(b/a) 2Πε0L 2Πε0L •Potential of inner conductor is greater then that of outer conductor because inner cylinder carries +ve charge. •Thus capacitance of cylinderical capacitor is C = Q/ ΔV= Q Q log(b/a) 2Πε0L C=2Πε0L log(b/a) •From equation it can easily be concluded that capacitance of a cylinderical capacitor depends on length of cylinders. •More is the length of cylinders , more charge could be stored on the capacitor for a given potential difference. ( ) b a a b
  • 25. Force on II plate capacitor •A parallel plate capacitor consists of two large plane parallel conducting plates C and D separated by a small distance d. •Suppose two plates of the capacitor has equal and opposite charge Q on them. If A is the area of each plate then surface charge density on each plate is σ = Q / A •We have already calculated field between two oppositely charged plates using gauss's law which is E = σ / ε0= Q/ε0A = Q/ 2ε0A + Q/ 2ε0A (due to two plates) C A D E +Q d A + + + + + + + + + + + - - - - - - - - - - -Q Due to plate C Due to plate D
  • 26. Force on plate 2 F = qE (E due to plate 1 and charge on plate 2) F = -Q(Q/ 2ε0A) F = -(Q2/ 2ε0A) (-ve sign due to attraction.) From this equation force not depands upon separation d. A A D C D C A A F applied V is very small K.E. will be =0 F w.d. by force applied Energy stored in capacitor:- parallel plate capacitor consists of two large plane parallel conducting plates C and D separated by a small distance d. # Two plates of the capacitor has equal and opposite charge Q on them hence net chargy =0 F =-(Q2/ 2ε0A) W= F .d =-(Q2/ 2ε0A).d
  • 27. Work done in sepereting plate will be stored as electric energy in capacitor. This electric energy will be stored in the form of electric field. W = - (Q2/ 2ε0A).d U = Q2 / 2c Q = CV U = c V2 / 2 U = Q V / 2 This is proved by using mechanical energy. No type of energy will be entered in this (K.E.=0) only possible if v is very small. Force applied is equal to force of attraction.
  • 28.
  • 29. Series Combination of Capacitors: V1 V2 V3 V C1 C2 C3 In series combination, • • Charge is same in each capacitor Potential is distributed in inverse proportion to capacitances i.e. V = V1 + V2 + V3 But q 1 V = 2 V = q 3 C2 C3 q , and V = q V = , (where C is the equivalent capacitance or effective capacitance or net capacitance ortotal capacitance) C q = C1 + C q + C 2 3 C1 q q C or The reciprocal of the effective capacitance is the sum of the reciprocals of the individual capacitances. Note: The effective capacitance in series combination is less than the least of all the individual capacitances. q q q 1 Ci 1 C n = ∑ i=1 = 1 + C1 1 1 + C2 C3 1 C
  • 30.
  • 31. Parallel Combination of Capacitors: In parallel combination, • • Potential is same across each capacitor Charge is distributed in direct proportion to capacitances i.e. q = q1 + q2 + q3 But (where C is the equivalent capacitance) or The effective capacitance is the sum of the individual capacitances. Note: The effective capacitance in parallel combination is larger than the largest of all the individual capacitances. q1 = C1 V , q2 = C2 V , q3 = C3 V and q = C V C V = C1V + C2 V + C3 V n C = ∑ Ci i=1 C = C1 + C2 + C3 V q1 C1 C2 C3 V V V q2 q3
  • 32. +Q -Q 1 2 d + + + + + + + + + + - - - - - - - - - - - - Energy Stored in a Capacitor:- Let both the plates have no charge at all. Want to charge +Q and –Q on both the plates. At time t = 0, Q = 0 Let we moved +dq charge from 2nd plate to 1st plate . +dq charge leave –dq charge behind it. We are increasing potential difference between the plates.Hance at time t q = Q 1 2 1 2
  • 33. Energy Stored in a Capacitor: V The process of charging a capacitor is equivalent to transferring charges from one plate to the other of the capacitor. The moment charging starts, there is a potential difference between the plates.Therefore, to transfer charges against the potential difference some work is to be done.This work is stored as electrostatic potential energy in the capacitor. If dq be the charge transferred against the potential difference V, then work done is dU = dW = V dq = dq The total work done ( energy) to transfer charge q is U = 0 q q C 1 Q2 U = 2 C U = 1 2 U = 1 2 Q V dq or or C V2 or q C -Q +Q 1 2 ∫
  • 34. Energy Density :- Energy Density =Energy / Volume U = 1 C V2 d C = A ε0 V = E d and U = 2 1 ε0 Ad E2 1 2 2 ε0 E2 = U Ad ε0 E2 1 U = 2 But or or SI unit of energy density is J m-3. Energy density is generalised as energy per unit volume of the field. Energy Stored in a Series Combination of Capacitors: = 1 2 3 1 1 C C C C Cn 1 1 1 + + + ……….+ 1 q2 U = 2 C U = 1 2 1 q2 [ 1 + + C1 C2 C3 1 Cn 1 + ………. + ] U = U1 + U2 + U3 + ………. + Un The total energy stored in the system is the sum of energy stored in the individual capacitors.
  • 35. Energy Stored in a Parallel Combination of Capacitors: C = C1 + C2 + C3 + ……….. + Cn U = C V2 U = 1 1 2 2 V2 ( C1 + C2 + C3 + ……….. + Cn) V = U = U1 + U2 + U3 + ………. + Un The total energy stored in the system is the sum of energy stored in the individual capacitors. Loss of Energy on Sharing of Charges between the Capacitors in Parallel: Consider two capacitors of capacitances C1, C2, charges q1, q2 and potentials V1,V2. Total charge after sharing = Total charge before sharing (C1 + C2) V = C1 V1 + C2 V2 C1 + C2 C1 V1 + C2 V2
  • 36. The total energy before sharing is i U = 1 C1 V 2 1 1 2 2 2 2 C V 2 + Uf = The total energy after sharing is 1 2 (C1 + C2) V2 C1 C2 (V1 –V2)2 Ui– Uf = 2 (C1 + C2) Ui – Uf > 0 or Ui > Uf Therefore, there is some loss of energy when two charged capacitors are connected together. The loss of energy appears as heat and the wire connecting the two capacitors may become hot.
  • 37. Dielectrics: Generally, a non-conducting medium or insulator is called a ‘dielectric’. Precisely, the non-conducting materials in which induced charges are produced on their faces on the application of electric fields are called dielectrics. Eg. Air, H2, glass, mica, paraffin wax, transformer oil, etc. What happened when space between the two plates of the capacitor is filled by a dielectric was first discovered by faraday. Faraday discovered that if the space between conductors of the capacitor is occupied by the dielectric, the capacitance of capacitor is increased.
  • 38. Dielectric constant of vaccum is unity If the dielectric completely fills the space between the conductors of the capacitor ,the capacitance is increased by an factor K which is characterstics of the dielectric and This factor is known as the dielectric constant. Polarization of Dielectrics: When a non-polar dielectric slab is subjected to an electric field, dipoles are induced due to separation of effective positive and negative centres. E0 is the applied field and Ep is the induced field in the dielectric. The net field is EN = E0 – Ep i.e. the field is reduced when a dielectric slab is introduced. The dielectric constant is given by E0 K = E0 - Ep Ep Ep E0=0
  • 39. If the dielectric slab occupies the whole space between the plates, i.e. t = d, then WITH DIELECTRIC SLAB Physcial Quantity With Battery disconnected With Battery connected Charge Remains the same Increases (K C0 V0) Capacitance Increases (K C0) Increases (K C0) Electric Field Decreases EN = E0 – Ep Remains the same Potential Difference Decreases Remains the same Energy stored Remains the same Increases (K U0) K = C C0 C = K C0 Dielectric Constant
  • 40. Polar Molecules: A molecule in which the centre of positive charges does not coincide with the centre of negative charges is called a polar molecule. Polar molecule does not have symmetrical shape. 2 3 2 Eg. H Cl, H O, N H , C O , alcohol, etc. O H H Effect of Electric Field on Polar Molecules: E = 0 E p p = 0 In the absence of external electric field, the permanent dipoles of the molecules orient in random directions and hence the net dipole moment is zero. When electric field is applied, the dipoles orient themselves in a regular fashion and hence dipole moment is induced. Complete allignment is not possible due to thermal agitation. 105° p
  • 41. Non - polar Molecules: A molecule in which the centre of positive charges coincides with the centre of negative charges is called a non-polar molecule. Non-polar molecule has symmetrical shape. Eg. N2 , C H4, O2, C6 H6, etc. Effect of Electric Field on Non-polar Molecules: E = 0 E p p = 0 In the absence of external electric field, the effective positive and negative centres coincide and hence dipole is not formed. When electric field is applied, the positive charges are pushed in the direction of electric field and the electrons are pulled in the direction opposite to the electric field. Due to separation of effective centres of positive and negative charges, dipole is formed.
  • 42. Criteria Polar Non-Polar Centers of positive and negative charges The centers are separated even in the absence of external electric field The centers coincide Dipole Moment Permanent dipole moment No permanent dipole moment Examples Hydro Chloric Acid (HCl) and Water (H2O) Oxygen(O2) and Hydrogen(H2) Dielectrics in External Electric Field The individual dipole moments tend to align with the field. They develop a net dipole moment in the direction of the field. Positive and negative charges get displaced. They develop an induced dipole moment as a restoring force against the direction of electric field.
  • 43. Polarization Vector: The polarization vector measures the degree of polarization of the dielectric. It is defined as the dipole moment of the per unit volume in dielectric. If n is the number of atoms or molecules per unit volume of the dielectric, then polarization vector is P = n p SI unit of polarization vector is C m-2. It is a vector quantity. Dimension is M0L-2AT. Dielectric Strength:-Dielectric strength is the maximum value of the electric field intensity that can be applied to the dielectric without its electric break down. Its SI unit is V m-1. Its practical unit is kV mm-1. Dielectric constant:- The ratio of external electric field E0 to the resultant electric field E inside the dielectric substance is known as dielectric constant of the substance. k= E0 / E E∠E0 , 1∠k Breakdown potential difference:- It is that maximum potential difference applied across the dielectric at which dielectric substance breaks down. Vmax=Emax × d Dielectric Dielectric strength (kV / mm) Vacuum ∞ Air 0.8 – 1 Porcelain 4 – 8 Pyrex 14 Paper 14 – 16 Rubber 21 Mica 160 – 200
  • 44. Dielectric in capacitor :- (dielectric slab) Capacitor partially filled with dielectric slab. A is the area of plate of capacitor. d is a separation between two plates . + Q charge is given to A plate. -Q charge is given to be B plate . k is the dielectric constant of dielectric slab . t is a thickness of dielectric slab. x is the distance between plateA and dielectric slab. (d-x-t) is the distance between dielectric slab and plate B. E1= Q /ε0A between plate a and dielectric slab . E2= Q /ε0 Ak of electric slab . E3= Q /ε0A between plate B and dielectric slab. ΔV =V12+V23+V34 =E1x + E2t+ E3(d-x-t) C = Q / ΔV = ε0A / (d-t+t/k ) does not depend upon X hence location of slab not matter. t x d E2 E1 E3 A +Q A -Q k A B
  • 45. If the entire space between the plates is filled with dielectric t =d C = ε0A / (d-d+d/k ) C = ε0A k / d If the entire space between the plate has a cube t =0 C = ε0A / (d-0+0/k ) C = ε0A / d If a slab of metal of thickness t is placed between the the capacitor plates for metal k= ∞ C = ε0A / (d-t+t/0 ) C = ε0A / (d-t)
  • 46. More than one slab :- C = ε0A / (d-t1+t1/k1 -t2 +t2/k2) Spherical capacitor :- C1= 4 Π ε0 k1ab /(b-a) C2= 4 Π ε0 k2bc /(c-b) C= C1 C2 C= 4 Π ε0 k1 k2abc C1 + C2 k2c /(b-a)k1a /(c-b) A +Q t2 t1 k2 k1 -Q A d + Q - Q a c b k1 k2
  • 47. Capacitance of Parallel Plate Capacitor with Dielectric Slab: Ep E0 EN = E0 - E p t d V = E0 (d – t) + EN t E N K = E0 or N E = K E0 0 V = E (d – t) + E0 t 0 V = E [(d – t) + K K t ] But 0 E = σ = ε0 q/A and C = ε0 q V C = A ε0 [(d – t) + K t ] or C = A ε 0 d [1 – K t d t (1 - ) ] or C = C0 [1 – t K t d (1 - ) ] C > C . i.e. Capacitance increases with introduction of dielectric slab. 0 does not depend upon X hence location of slab not matter.
  • 48. Van de Graaff Generator: T D C1 P1 P2 C2 M S I S HVR S – Large Copper sphere C1, C2 – Combs with sharp points P1, P2 – Pulleys to run belt HVR – High Voltage Rectifier M – Motor IS – Insulating Stand D – Gas Discharge Tube T - Target
  • 49.
  • 50. Principle: Consider two charged conducting spherical shells such that one is smaller and the other is larger. When the smaller one is kept inside the larger one and connected together, charge from the smaller one is transferred to larger shell irrespective of the higher potential of the larger shell. i.e. The charge resides on the outer surface of the outer shell and the potential of the outer shell increases considerably. Sharp pointed surfaces of a conductor have large surface charge densities and hence the electric field created by them is very high compared to the dielectric strength of the dielectric (air). Therefore air surrounding these conductors get ionized and the like charges are repelled by the charged pointed conductors causing discharging action known as Corona Discharge or Action of Points.The sprayed charges moving with high speed cause electric wind. Opposite charges are induced on the teeth of collecting comb (conductor) and again opposite charges are induced on the outer surface of the collecting sphere (Dome).
  • 51. Construction: Van de Graaff Generator consists of a large (about a few metres in radius) copper spherical shell (S) supported on an insulating stand (IS) which is of several metres high above the ground. A belt made of insulating fabric (silk, rubber, etc.) is made to run over the pulleys (P1, P2 ) operated by an electric motor (M) such that it ascends on the side of the combs. Comb (C1) near the lower pulley is connected to High Voltage Rectifier (HVR) whose other end is earthed. Comb (C2) near the upper pulley is connected to the sphere S through a conducting rod. A tube (T) with the charged particles to be accelerated at its top and the target at the bottom is placed as shown in the figure. The bottom end of the tube is earthed for maintaining lower potential. To avoid the leakage of charges from the sphere, the generator is enclosed in the steel tank filled with air or nitrogen at very high pressure (15 atmospheres).
  • 52. The process continues for a longer time to store more and more charges on the sphere and the potential of the sphere increases considerably. When the charge on the sphere is very high, the leakage of charges due to ionization of surrounding air also increases. Maximum potential occurs when the rate of charge carried in by the belt is equal to the rate at which charge leaks from the shell due to ionization of air. Now, if the positively charged particles which are to be accelerated are kept at the top of the tube T, they get accelerated due to difference in potential (the lower end of the tube is connected to the earth and hence at the lower potential) and are made to hit the target for causing nuclear reactions, etc. Van de Graaff Generator is used to produce very high potential difference (of the order of several million volts) for accelerating charged particles. The beam of accelerated charged particles are used to trigger nuclear reactions. The beam is used to break atoms for various experiments in Physics. In medicine, such beams are used to treat cancer.It is used for research purposes. Uses: