Download ELECTROSTATICS - auroraclasses.org

Coulomb’s Law
The force between two point charges is directly proportional to the product of the charges and
inversely proportional to the square of the distance between them.
This force acts equally on each point charge even if the charges are unequal. i.e. the two
forces are an action-reaction pair. The force acts along the straight line joining the two charges.
Like charges experience forces away from each other while unlike charges experience forces
towards each other
Comparison with gravitational force
Similarities:
1.
2.
Both the forces are directly proportional to the product of the interacting commodities (mass and
charge)
Both the forces are inversely proportional to the square of the distance of separation
Differences:
1.
2.
The Coulombian force can be attractive or repulsive while the gravitational force is attractive only
The magnitude of the Coulombian force depends upon the medium separating the charges while
the gravitational force is independent of the medium.
Permittivity of medium
Coulomb's Law in mathematical form:
The value of the constant K depends upon the nature of the medium. The numerical value of the
constant k depends upon the choice of units.
is an electrical property of the medium surrounding the two charges and it is called the permittivity
of that medium. The units of permittivity are farad/metre.
In case the medium is air/vacuum, it is conventional to use
0
for the permittivity
Relative Permittivity
Thus,
in any medium
and,
in air or vacuum
The ratio of the force in vacuum to the force in any other medium between the same pair of point
charges separated by the same distance is called the relative permittivity of that medium
Relative permittivity has no units.
Electrically conducting media have a higher value of permittivity. Since vacuum has absolutely no
charge carriers, the permittivity is the lowest for vacuum. Every medium therefore has a relative
permittivity higher than unity.
Limitations of Coulomb’s law
1.
The expression
The force is undefined for r = 0
2.
is applicable for point charges only.
Net force due to system of charges
The resultant force on a charge Q due to a system of charges is obtained by adding vectorially all
the individual forces acting on it
Numericals :(1/4
1.
a.
b.
c.
o
= 9  109 SI units)
Two positive point charges of 12 and 8 microcoulomb are separated by a distance of 10cm in
air.
Find the magnitude and direction of the force on the greater charge.
[86.4N]
What is the magnitude and direction of the force on the smaller charge?
[86.4N]
How would the force be affected if :





d.
Each charge is replaced by a negative charge of the same amount
Only the greater charge is replaced by a negative charge of the same amount
The distance separating the charges is doubled
Each charge is doubled and the distance is also doubled
The medium is replaced by another insulating medium
Find the force in water of relative permittivity 80
[1.08N]
2.
Two small masses are 1.5 m apart on a smooth table. They carry identical charges. Each one
experiences a force of 2.0 N. Find the force between them if the distance of separation is
increased to 2.0 m. Find also the force between them if the distance of separation is still 1.5 m
but the medium separating them has a relative permittivity 6.
[1.1N; 0.33N]
3.
Protons have a mass of 1.6 x 10-27 kg and a charge of 1.6 x 10-19 C. Given that G = 6.67 x 1011
SI units and 1/4 o = 9 x 109 SI units, find the ratio of the electrical repulsion to the
gravitational attraction between two protons. Convert this ratio from power of ten notations into
decimal form. Why do you think no one ever mentions gravitational attraction between protons?
[1.4 x 1036 ]
Electric Field
An electric field is a region where electric charges experience Coulombian forces. It exists in the
space around a charged object.
Lines of force
Lines of force are a convenient way of visualizing an electric field.
Lines of force are imaginary lines in an electric field, the tangent to which at any point gives the
direction of the electric field strength at that point.
1.
Lines of force never intersect
2.
Lines of force are in a state of longitudinal tension
3.
Lines of force repel each other laterally
4.
Lines of force commence from a conducting surface normally and they terminate on a conducting
surface normally.
The total number of lines of force (flux) from a charge is directly proportional to the charge.
The total number of lines of force is inversely proportional to the permittivity of the medium in which
the charge is located.
Gauss's Law
Gauss's Law states that the flux linked with a closed surface is given by
where Q is the charge enclosed by the surface and e is the permittivity of the medium
Electric field strength or intensity
The electric field strength at a point is the force experienced by a unit positive charge
placed at that point.
Electric field strength is a vector quantity. Its units are newton/coulomb (or volts/metre)
The electric field strength is also given by the number of lines of force per unit area held at right
angles with the lines of force.
This is also called the electric flux density.
Electric field strength at a distance r from isolated point charge
The equation given below is derived from Coulomb's law and is subject to the limitations of
Coulomb's law i.e. it applies ONLY FOR Q BEING AN ISOLATED POINT CHARGE.
Resultant electric field
The resultant electric field due to many point charges would be given by the vector sum of the
individual field strengths E1, E2, E3etc.
A neutral point in an electric field is where the resultant electric field strength is zero
Electric field strength due to charged sphere
The electric field strength inside a charged sphere is zero.
The electric field strength at points outside a charged sphere is to be calculated by considering the
entire charge to be located at the centre of the sphere.
The electric field strength at any point outside the sphere is inversely proportional to the square of
the distance of that point from the centre of the sphere.
Numericals
(1/4
1.
o = 9 x 109 SI units)
Find the magnitude and direction of the electric field strength at a point P distant 10.0 cm from
(i)
an isolated positive point charge of 3.0 C
[2.7 x 106NC-1]
(ii)
an isolated negative point charge of 5.0 C
[4.5 x 106NC-1]
At what distance from the point charge would the electric field strength be zero?
[]
What is the force experienced by a positive charge of 2.0 microcoulomb place at the point P in
each of the cases above?
[5.4N; 9.0N]
2.
3.
4.
An oil drop of radius 0.70 mm containing a charge of -4.8 x 10-19C is placed in an electric field. If
the drop remains stationary at the point find the magnitude and direction of the electric field
intensity there, neglecting the density of air. (Density of oil = 900 kgm-3, g = 10 ms-2).
[2.7 x 1013Vm-1]
If the oil drop now gathers one additional electron of charge -1.6 x 10-19C, describe the
subsequent behaviour of the drop.
A spherical shell of radius 10.0 cm contains a charge of +2 microcoulomb. Find the electric field
intensity
(i)
at a distance of 2 x 10-2 m from the centre,
[0]
-1
4
(ii)
and at a distance of 5 x 10 m from the centre.
[7.2 x 10 N/C]
Find the total flux from a point charge of 10 microcoulomb in
(i)
air
[1.1 x 106 Vm;]
(ii)
water of relative permittivity 80.
[1.4 x 104 Vm]
Hence, for each case, find the intensity of the electric field at a point 12 cm distant from the
charge.
[6.3 x 106 N/C; 7.8 x 104N/C]
5.
A charged particle of -1.6 x 10-19C is placed in a region of uniform electric field of
1200 NC-1.
Find
(i)
the force acting on it,
[1.9 x 10-16N]
(ii)
its acceleration if the mass of the particle is 9.1 x 10-31kg,
[2.1 x 1014 ms-2]
(iii)
the distance covered in 1.5 seconds from rest, and,
[2.4 x 1014 m]
(iv)
the kinetic energy gained in 1.5 seconds.
[4.5 x 10-2 J]
When a charge is displaced in an electric field then work is done
The potential energy of the system changes by the amount of work done when a charge is
displaced in an electric field.
If the direction of the displacement is such that the system's force does the work, then the system
loses potential energy by the same amount. If the work has to be done against the system's force
then there is an external force doing the work and the system's potential energy increases by this
amount.
Thus if the system's force does negative work, the system's potential energy rises by the same
amount.
Since the work done is directly proportional to the force if other things remain constant, and since
force is directly proportional to the charge carried, therefore, the work done is directly proportional
to the charge carried.
Thus if charge Q' is carried from point A to point B,
k
is
therefore
the
work
done
in
carrying
unit
charge
This is called the electric potential difference V between the points A and B.
from
A
to
B
k = VB- VA
It is a scalar quantity.
Its units are joule/coulomb.
This unit is given a new name - volt.
1 volt = 1 joule / coulomb
volt x coulomb = joule
In order to calculate the changes of energy at subatomic levels, the joule is found to be too large a
unit. Hence a practical unit of energy is in use and is called the electron volt.
A potential difference of 1 volt between two points means that in carrying a charge of +1 coulomb
from one point to the other 1 joule of work would be done.
If an electronic charge (of 1.6 x 10-19 C) is moved through a potential difference of 1 volt, then the
work done is 1 electron volt.
1 volt x 1.6 x 10-19 coulomb = 1.6 x 10-19 joule
1 volt x 1 electron (unit of charge) = 1 electron volt
Comparison of electrical and gravitational potential differences
The electric potential difference is analogous with the gravitational potential difference between two
points. Both quantities are scalar. Their units are Joule/Coulomb and Joule per kilogram
respectively.
But gravitational potential at a point is independent of properties of the medium whereas electric
potential at a point depends upon the permittivity of the medium.
Electric field strength as potential gradient
Let the point A' be at an infinitesimally small displacement dr from the point A
Let the potentials at A and A' be V and V+dV respectively.
Therefore, work done in carrying unit positive charge from A to A' is given by
dW = (V+dV) – V = dV ..............................(i)
Let the electric field strength at A be E and let it be directed parallel to the displacement dr.
Since the system is gaining in potential energy by the amount dV, the system's force must be
directed against the motion
Hence work done in this displacement is given by:dW = E x dr x cos 180o = - E x dr .................(ii)
From (i) and (ii),
dV = - E dr
or,
The electric field strength at a point is also defined as the potential gradient at that point in the same
direction.
Units of E are therefore, volt/metre
volt/metre = newton/coulomb
Electric potential at a point
The potential of a point A means the potential difference between A and infinity.
i.e. The potential of a point is the work done in carrying unit positive charge from infinity to that
point.
It is a scalar quantity.
The unit is volt.
The potential of a point in an electric field is a characteristic constant of that point. It does not
depend on how much charge is carried. It does not depend upon the distance that is actually
covered but on the displacement of the point from infinity
General expression for work done
The expression for work done is
W = F r cos
where r is the displacement from point A to point B, and is the angle between the force and the
displacement r. But this is applicable only if the force F remains constant throughout the
displacement. Unless the electric field is uniform in the region of the displacement this expression
does not hold.
In a general case whether the field is uniform or non-uniform the expression for work done in a
small displacement dr is given by
dW = F dr cos
The whole work done in displacing the charge through a finite distance from A to B is obtained by
integrating dW.
Potential at a point in a radial field
(a)
Due to isolated point charge +Q
At a variable distance r from an isolated positive point charge Q, a unit positive charge
experiences a force given by
The work done in displacing this unit positive charge from infinity to a distance 'a' from the
point charge Q is given by
This is the electric potential at a distance at a distance 'a' from anisolated positive point charge Q.
(b)
Due to negative point charge
In case of the isolated point charge being negative, the force on the unit positive charge is
attractive. i.e. it is in the same direction as the displacement is. Hence the angle between force F
and displacement dr = 0.
Thus dW = F dr cos 0
dW = F dr.
Hence the final expression for potential at distance a from the point charge -Q becomes
The potential near a positive point charge is positive, while that near a negative point charge is
negative.
Variation of potential due toVariation of potential due to
negative point charge
positive point charge
Variation of potential inside and outside a charged sphere
In case of a charged sphere, for points outisde the sphere, the potential is to be calculated
assuming that the entire charge is concentrated at the centre of the sphere.
V=Q/4 r
(Electric field strengths outside the sphere are also calculated in the same manner)
For points inside the sphere however, the potential at every point is the same as the potential at the
surface
Note that though charges are all at the surface and not in the interior, the potential is the same at
the surface and inside.
When a conductor of irregular shape is charged, even though the charge is concentrated at edges
and corners, the potential is the same at such corners and edges as on the plane surfaces, and as
in the interior where there is acutally no charge.
An uncharged sphere isolated from all other charges has zero potential. The earth is always
considered an uncharged isolated sphere even when charges are added to it or removed from it.
Equipotential surface and lines of force
The locus of all points having the same electric potential is called an equipotential surface.
Equipotential surfaces due to a point charge or a charged spherical conductor would be spherical.
Equipotential surfaces due to a very long charged rod would be cylindrical.
The direction of the electric field is always perpendicular to the equipotential surface at that point.
Numericals
(1/4
o = 9 x 109 SI units)
1.
i.
ii.
Find the electric potential at a distance of 5.0 cm from a positive point charge of 4.0 x10-8 C
(i)
in air, and,
[7200V]
(ii)
in a medium of relative permittivity 3.0.
[2400V]
2.
Charges of +1 C, -2 C, and -3 C are located at the vertices ABC of an equilateral triangle of
side 20 cm. Find the electric potential at X, the mid-point of BC.
[-4.0 x 105V]
3.
A heavily charged cloud is at a potential of -5000 V. A bolt of lightning transfers 50 C of charge
between the cloud and the ground. Find the work done in transfering the charge. [25 x 104J]
4.
The high voltage terminal of a Van de Graaf generator consists of a spherical conducting shell
of radius 0.50 m. Estimate the maximum potential to which it can be raised in air for which
electrical breakdown occurs when the field intensity exceeds 3x 106 Vm-1
[1.5 x 106 V]
5.
A spherical shell of radius 10 cm is given a charge of 4 micro Coulomb. Find the electric
potential at a distance
(i)
20 cm from the centre,
[18 x 104V]
(ii)
3 cm from the centre.
[36 x 104V]
i.
6.
Two positive point charges of 8.0 C and 12 C are separated by a distance of 12 cm in a
medium of relative permittivity 5. Find the work done in reducing the distance of separation by 4
cm.
[7.2 x 10-1J]
Dipole Axis
An electric dipole is a system of two equal and opposite point charges separated by a small distance.
The extended straight line joining the two point charges in a dipole is called the dipole axis.
The perpendicular bisector of the dipole axis is called the neutral axis.
Dipole moment
An electric field is a region where an electric charge experiences a Coulombian Force.
A uniform electric field is one in which the magnitude and direction of the field strength is the same
everywhere. Hence lines of force are parallel in such a field. If an electric dipole is placed in a
uniform electric field:
 Since net force acting on the dipole is zero, there is no linear acceleration of the dipole.
 Since, as shown in the figure, there is an anticlockwise torque acting on the dipole, there is an
anticlockwise angular acceleration.
The two poles experience equal forces EQ in opposite directions The external unbalanced force is
zero and therefore there is no linear acceleration.
Thus when the dipole axis makes an angle with respect to the direction of the uniform field, there
is an external unbalanced torque acting on the dipole whose magnitude is given by
= EQ x 2l sin
= 2lQ x E x sin
= P x E x sin
..............(i)
where P is called the dipole moment. It is the product of the length 2l of the dipole and the charge
Q at any one end
P = 2l x Q
units : coulomb metre.
Dipole moment is a vector quantity.
Its direction is from the negative charge to the positive charge.
From equation (i) above,
If E = 1
(i.e. if the uniform electric field is of unit strength)
and
if sin = 1,
(i.e. the dipole is held perpendicular to the field)
Then,
=P
The dipole moment is the torque experienced if the dipole is held perpendicular to a uniform electric
field of unit strength.
Oscillation of Dipole in Uniform Electric Field
Since the torque acting on the dipole is variable, (being proportional to sin ,) the angular
acceleration produced is also not constant and is proportional to sin . For = 0, the torque
vanishes and the dipole placed in this position therefore remains in equilibrium. At any other
position torque is not zero and hence there is an angular acceleration.
Also, as the diagram shows, a dipole displaced clockwise will experience an anticlockwise torque,
which will accelerate it and tend to restore it to its equilibrium position. Though at this position the
torque is zero, the dipole has been accelerated by the torque during its motion into this position. It
therefore has inertia of motion and will swing past the rest position in an anticlockwise direction
even as the torque becomes clockwise and starts to decelerate it to a momentary rest. The whole
process then begins again in the opposite direction. The resulting motion of the dipole is thus an
oscillatory one. This oscillation however is not necessarily simple harmonic.
Angular Harmonic Motion
It is only if is a small angle that sin may be replaced by (in radian measure) without incurring
significant error. Thus the torque is directly proportional to the angular displacement , this makes
the angular acceleration created by this torque also directly proportional to .
Thus the angular acceleration
,
is proportional to the angular displacement q and is also oppositely directed. The dipole
undergoes angular harmonic motion in this case.
Its time period is derived below.
In the event of the angle
being not small, there is still oscillatory motion but it is not simple harmonic.
Behaviour of dipole in non-uniform field
In case of the external field being non-uniform, the two poles experience unequal forces EQ and
E'Q. Thus in addition to the torque of the previous case, there is also anexternal unbalanced force
acting on the dipole. Hence there would be translatory as well as rotatory motion. The dipole would
move from the weaker to the stronger parts of the non-uniform field, as it oscillates.
Work is done in turning a dipole in an uniform electric field
Due to dissipation of energy, the oscillating dipole in a uniform field would ultimately settle in its
position of minimum potential energy. This is the position parallel to the direction of the electric field.
Work has to be done by an external agent in order to rotate the body in any direction from this
position of rest. This work is the product of the system's torque and the angle through which the
dipole is turned. But since the torque varies at different positions of the dipole, being proportional to
sin , the work done has to be found for infinitesimal displacements d and then summed as
shown below:
This is the work done by the external agent since the direction of the displacement
torque are both assumed to be the same.
and the
The above expression also gives the gain in potential energy which is equal to the work done
against the system's torque. However to find the total potential energy of the dipole in this position
one has to add to this the work done in bringing the dipole from infinity to this point in the electric
field.
In order to bring the dipole from an infinite distance (i.e. beyond the electric field), to a position
perpendicular to the field lines, no work has to be done since there is no displacement of either pole
in the direction of the force. Hence the potential energy of the dipole in this position is zero . But in
this position the dipole has more energy than in the position of alignment with the field.
To turn the dipole into this position from its rest position, the external agent would have to do work
given by:
W1 = PE (1-cos 90o)
= PE
In this position, the dipole has more potential energy by the amount PE than it has in its rest
position aligned with the electric field. Since work done by an external agent to bring the dipole from
infinity to this position is zero, the potential energy in the rest position is less than zero by the
amount PE. Thus, its potential energy in the rest position is given by
W0 = - PE
Hence the potential energy of the dipole placed at an angle
amount PE(1-cos ) than W0
Hence potential energy at an angle of inclination
with the field lines is higher by the
to the field lines is given by
W = -PE + PE (1-cos )
= -PE cos
Electric field strength
(a)
At end-on position (or tan A position)
A test charge placed on the dipole axis at a distance d from the midpoint of the dipole would
experience unequal forces in opposite directions along the dipole axis. This would give rise
to a net force. This net force per unit positive charge placed at that point is calculated below:
If d >>>> l, l2 is negligible in comparison with d2.
b)
At broadside-on position or tan B position
In case of the test charge being located at a distance d from the midpoint of the dipole but on the
neutral axis of the dipole, it would experience forces of equal magnitude but not acting parallel to
each other. Hence the resultant electric field must be calculated by summing vectorially the forces
experienced by unit positive charge placed at that point.
E1 = E 2
The components E1 sin and E2 sin directed along the neutral axis nullify each other but the
components E1 cos and E2 cos both act in the same direction parallel to the dipole axis. Hence
the net electric field strength at this position is given by :-
E = 2 E1 cos
If l <<<< d then
It should be noted that the electric field strength in the broadside-on position is exactly half of that in
the end-on position, distance being the same.
Derivation of electric potential
(a)
at end-on position
Potential due to the dipole is given by the algebraic sum of the electric potentials due to
each of the charges of the dipole.
At a distance d in the end on position the potential is given by :-
(b)
at broadside-on position
At a distance d in the broadside position the potential is given by:-
Thus at any point along the neutral axis the electric potential is zero.
Electric Field Strength and Potential Gradient on Neutral Axis
Note that the potential gradient along the neutral axis is zero and the electric field strength has no
component in this direction.
The electric field strength is directed parallel to the dipole axis. Along this direction the potential
gradient is not zero.
Numericals
1.
An electron of charge 1.6 x 10-19 C and a proton are 5.3 x 10-11 m apart.
(i)
What is their dipole moment if they are at rest?
[8.5 x 10-30 Cm]
(ii)
What is the average dipole moment if the electron revolves about the proton in a circular
orbit?
[0]
A capacitor is a device that stores charge.
When a body is charged, its potential changes.
Capacitance
The amount of charge required to change the potential of the body by one unit is called the
capacitance of the body.
Thus, if charge Q changes the potential of the body by an amount V then capacitance C is given by
The unit of capacitance is farad (F).
farad = coulomb/volt
Thus the capacitance of a capacitor is 1 farad if it requires 1 coulomb of charge to change its
potential by one volt.
Since the electrostatic potential is proportional to the charge contained by the body, capacitance is
independent of the charge given or the potential of the body. It depends only on the shape and size
of the body and on the permittivity of the surrounding medium.
Capacitance of an isolated sphere
If a sphere of radius r is given a charge Q then the potential anywhere on it is given by :-
Thus, the capacitance of the sphere is
It should be noted that the capacitance of the sphere depends only on its radius and on the
permittivity of the surrounding medium.
Methods of increasing capacitance
The presence of a second uncharged or oppositely charged body near the charged conductor
increases the capacitance of the conductor. There is opposite charge induced on the second body
at its near end and similar charge at the far end. Both these types of induced charges create
electrostatic potentials in the neighborhood in which the first conductor lies, as shown in the
diagram below.
The first body already had an electrostatic potential V due to the charge stored on itself. It now has
additionally, an increase in this existing potential due to the similar charge at the far end of the
second body, and a reduction in potential due to the opposite charge at its near end. These two
amounts of induced charges are the same in magnitude. However their distances from the first
conductor are unequal. The opposite charge being nearer, its effect on the potential is more. Thus
the net effect of the proximity of the second conductor is to reduce the electrostatic potential of the
system without changing the charge stored, thereby enhancing the charge storing capacity.
If the space between the two conductors is filled with an insulating medium whose molecules
undergo polarisation then the capacitance of the system is further increased. Such a medium is
called a dielectric.Each polarized molecule has its oppositely charged pole nearer to the conductor,
thereby reducing the effective potential in a manner similar to the action of the second conductor in
the vicinity.
In order to increase the charge storing capacity capacitors are therefore always designed as two
pieces of metal separated by an insulating medium.
Capacitance of concentric shells.
Let a and b be the radii of the two concnetric shells such that a<b. The inner shell is given a charge
+Q while the outer shell is earthed.
Hence, charge -Q is induced on the inside of the outer sphere and +Q on its outside. This free
charge on the outside of the outer sphere is neutralised due to the earth connection.
Since the potential (VA) of the inner sphere is partly due to its own positive charge and partly due to
the negative charge on the inside of the outer sphere, this potential is given by,
The potential difference V between the two spheres is therefore given by:
Thus the capacitance of this system is greater than that of a single sphere.
Note once again that the capacitance of the capacitor depends only on the radii of the two spheres
and on the permittivity of the medium, and not on the amount of charge stored or the potential
difference between its two parts.
Capacitance of cylindrical capacitor
Let charge per unit length of a long cylinder be . Threrfore the charge contained on a length l of the
cylinder is given by x l.
The total flux from this charge is
Flux =
Thus, at a distance x from the axis of the cylinder, the electric field strength E is given by:
since the area through which this normal flux is linked is 2 rl.
Capacitance of a parallel plate capacitor
Let A be the area of one surface of a metal plate and Q be the charge contained by that one
surface. Thus charge per unit area of the plate is
p = Q/A
The flux from this unit area is given by
The flux density or the electric field strength is given by:
Units of permittivity
Therefore, units of
are farad/metre (Fm-1)
Dielectric constant of a medium
This is defined as the ratio of the capacitance of a capacitor with the medium between the plates to
the capacitance of the same capacitor with vacuum between the plates
In practice every device used as a capacitor comprises two metal conductors separated by a
dielectric medium, i.e. a medium that permits its molecules to be polarized (induced into electric
dipoles).
In an electric circuit the symbol of a capacitor is therefore as shown below:
Charging a capacitor.
When an uncharged conductor is connected to a charged conductor the charges redistribute
bringing both conductors to the same electroststic potential which lies between the initial potentials
of the two conductors. The process is instantaneous.
In the circuit diagram shown above, each conducting plate of the capacitor is similarly
connected to the corresponding terminal of the cell, and must therefore instantly attain the same
potential as that terminal of the cell. Since chemical action within the cell keeps each terminal at a
fixed electrostatic potential despite flow of charge into or from the terminal, the potential difference
between the cell terminals remains at the original value V. The capacitor plates are thus charged to
the same potential difference V. It should be clearly understood that a cell cannot manufacture
electric charge. The electrons that are removed from the positively charged plated of the capacitor
are despatched to the negative terminal of the capacitor. The positive and negative charge stored
on the two plates of the capacitor must be of the same magnitude.
When a capacitor is connected to a cell as shown in the diagram alongside, flow of charge in the
circuit occurs for a very short while until each plate of the capacitor is charged to the same potential
as the plate of the cell to which it is connected. Thereafter there is no further flow of charge in the
circuit as there are equal voltages oppositely directed. Thus a filament bulb connected to a circuit
comprising cell and capacitor would not light up at all.
The graphs above shows the variation of charge on any one plate of the capacitor with respect to
time variation of current in the circuit with respect to time.
Work done in charging.
During the process of transference of charge the potential difference between the plates of the cell
and of the capacitor undergo change. Thus the work done in carrying a total charge Q against a
potential difference varying from V to 0 is given by
This gives the work done by the battery in charging the capacitor. It is also the energy stored on the
capacitor. If the capacitor is discharged through an external resistor then this is the amount of
energy it will release in the circuit.
In case there is a resistance in series with the capacitor in the charging circuit, then the final
charge stored (Q=CV) and the final energy stored (W=QV/2) are still unchanged but the time taken
for charging increases as some fraction of the energy supplied by the battery in every unit time is
used to produce heat in the circuit.
Energy released in Discharging of a capacitor:
The two plates of a charged capacitor are at two different potentials. The process of
discharging involves joining with a conducting material (a wire) the two oppositely charged metal
plates of the capacitor. This results in flow of charge in the wire until the two plates of the capacitor
come to the same potential. The electric potential energy (W=CV2/2 or W= QV/2 or W=Q2/2C) that
was stored on the plates of the charged capacitor does the work in driving the charges around the
circuit during the process of discharging. The energy stored is therefore converted to heat in
overcoming the resistance of the wire as the charges flow through it.
Connecting capacitors in parallel
Across the plates of each of the capacitors is the potential difference V. Thus charge stored
on each is given by the product of the potential difference V and its capacitance.
Hence total charge Q stored is given by
Q = Q1 + Q2 + Q3
= C1V + C2V + C3V
= V(C1 +C2 +C3)
=VxC
Where
C = C1 + C2 +C3
C is the effective capacitance of the combination.
Capacitors are connected in parallel when maximum charge is to be stored.
Connecting capacitors in series
When capacitors are connected in series the potential difference V is applied across the
entire combination. The p.d. across each is a part of V.
Thus,
V = V1 + V2 + V3
The charge Q on the entire combination is actually stored on the outer plates of C 1 and C3.
On the plates not directly connected to the cell, equal and opposite charges are induced.
Thus each capacitor stores the same charge Q.
But Q = CV
where C is the effective capacitance.
Hence,
When capacitors are connected in series the total charge stored is less than the amount
that could have been stored on any one of them connected singly to the same cell.
Capacitors are connected in series so as to protect the dielectric from damage due
to excess field.
Work done in charging.
During the process of transference of charge the potential difference between the plates of
the cell and of the capacitor undergo change. Thus the work done in carrying a total charge
Q against a potential difference varying from V to 0 is given by
This gives the work done by the battery in charging the capacitor. It is also the energy
stored on the capacitor. If the capacitor is discharged through an external resistor then this
is the amount of energy it will release in the circuit.
In case there is a resistance in series with the capacitor in the charging circuit, then the
final charge stored (Q=CV) and the final energy stored (W=QV/2) are still unchanged but
the time taken for charging increases as some fraction of the energy supplied by the battery
in every unit time is used to produce heat in the circuit.
Energy released in Discharging of a capacitor:
The two plates of a charged capacitor are at two different potentials. The process of
discharging involves joining with a conducting material (a wire) the two oppositely charged
metal plates of the capacitor. This results in flow of charge in the wire until the two plates of
the capacitor come to the same potential. The electric potential energy (W=CV2/2 or
W= QV/2 or W=Q2/2C) that was stored on the plates of the charged capacitor does the
work in driving the charges around the circuit during the process of discharging. The energy
stored is therefore converted to heat in overcoming the resistance of the wire as the
charges flow through it.
Connecting capacitors in parallel
Across the plates of each of the capacitors is the potential difference V. Thus charge stored
on each is given by the product of the potential difference V and its capacitance.
Hence total charge Q stored is given by
Q = Q1+ Q2 + Q3
= C1V + C2V + C3V
= V(C1 + C2 + C3)
=VxC
where C = C1 + C2 + C3
C is the effective capacitance of the combination.
Capacitors are connected in parallel when maximum charge is to be stored.
Connecting capacitors in series
When capacitors are connected in series the potential difference V is applied across the entire
combination. The p.d. across each is a part of V.
Thus,
V = V1 + V2 + V3
The charge Q on the entire combination is actually stored on the outer plates of C 1 and C3.
On the plates not directly connected to the cell, equal and opposite charges are induced.
Thus each capacitor stores the same charge Q.
But Q = CV
where C is the effective capacitance.
Hence,
When capacitors are connected in series the total charge stored is less than the amount
that could have been stored on any one of them connected singly to the same cell.
Capacitors are connected in series so as to protect the dielectric from damage due to
excess field.
Numericals
(
1.
)
A spherical conductor of radius 0.50m is placed in air and charged to a p.d. of 1000V.
Find its capacitance, the charge stored on it and its electric potential energy.
[ 5.5 x 10-11F; 5.5 x 10-8 C; 2.8 x 10-5 J]
2.
Two concentric shells have radii 4.0 cm and 3.0 cm. Find the capacitance of the
system if they are placed in air. What is the electric potential energy stored if the
potential difference between the two spheres is 200 V?
[1.3 x 10-11F; 2.7 x 10-7J]
3.
The parallel plates of a capacitor have an area of 0.1 m 2 and are 2 x 10-2 m apart. The
plates are charged to a p.d. of 1000 V. Find the capacitance and the energy stored on
the capacitor (a)if the medium between the plates is air, and (b) if the medium between
the plates is of dielectric constant 6. [4.5x10-11F; 2.2x10-5J;2.6x10-10F;1.3x10-4J]
4.
For each of the following figures, find the effective capacitance of the capacitors
C1 and C2. Find also the charge stored on each capacitor, the p.d. across each
capacitor and the energy stored on each capacitor.
(a) 1.2 F; 2.4 x 10-4C; 120V; 80V; 1.44 x 10-2 J;0.96 x 10-2 J]
(b) 5.0 F; 4 x 10-4C;4 x 10-4C; 200V; 200V; 4 x 10-2 J; 6 x 10-2 J]