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Transcript
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ELECTRIC FIELD
A charged particle does not interact directly with another charged particle kept at a distance. They
interact through an invisible communicating medium, called electric field. It is present every where in the
space surrounding the charge. The concept of electric field present around the charge, is analogous to the
gravitational field produced by a mass. Electric charge produces an electric field and electric field exerts a
force on electric charges kept in it.
Strength of Electric Field (Electric Intensity)
The strength of an electric field at a point is described in terms of force experienced by a charge
kept at that point.
Electric intensity or electric field strength at any point is defined as the force acting on
an unit positive charge(UPC) kept at that point. It is donated by E. It is a vector quantity. The
direction of electric field at a place is same as the direction of the force experienced by a test charge (an
unit positive charge) kept at that point.
If ‘F’ is the force experienced by a charge ‘q’ kept in the electric field then the strength of electric
field ‘E’ at that point is written as,
E = F/q or F = q E
An electric field in the space is said to be uniform, if a charge kept in it experiences same force at all
points & the field is said to be non-uniform, if the charge kept in it experiences different force at different
points. In S.I., the electric field strength is measured with unit NC-1
Expression for the electric field at a point due to an isolated charge :
+Q
X
q
Field due to + ve charge
Y
d
(a)
Q
a
X
Y
Field due to a – ve charge
Let a positive charge of magnitude Q is kept at X. Let Y is the point at a distance d from the charge,
where intensity of electric field is to be determined. An arbitrary charge say q is placed at Y. The force
acting on this q can be calculated using Coulomb’s law as follows:
1
Qq
F=
 2
4 0 d
1
Qq
 2
F
4 0 d
But from the definitions of electric field strength As E =  E 
q
q
i.e.,
1
Q
E=

4 0 d 2
2
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The direction of E is taken from X to Y for a positive charge[ i.e. away form the charge] & for the
negative charge field is taken along Y to X [ i.e. towards the charge].
The electric field due to an isolated charge varies with distance ‘d’ as follows:
d2
d2
An electric field at a point due to large number
of charges can be calculated by taking vector sum of
individual electric fields.



En

Ei

E2

If E1 , E2 , E3 ,...... are fields at a point due to charges q1,
 

q2, q3 Then net field at that point is E  E1  E2 + ……. This is

E3
also called as super position principle of electric fields.
ELECTRIC DIPOLE:
Electric dipole is an arrangement of two equal and opposite point charges separated by
a small distance. Many molecules exist as an electric dipoles. For example the hydrogen chloride
molecule, in which hydrogen part shows positive charge and chloride the negative charge, this constitute an
electric dipole. The line passing through the charges forming electric dipole is called as a axial line and the
line passing through the mid point of dipole axis and perpendicular to the axial line is called as equatorial
line of dipole.
2d
O
The strength of electric dipole is measured in the terms of a quantity called Electric Dipole moment,
which is denoted by ‘P’. The magnitude of electric dipole moment of an electric dipole is expressed as
product of magnitude of the charge possessed by the particles forming a dipole and the separation between
them.
If ‘q’ is the magnitude of charge and ‘2d’ is the separations between the, electric dipole moment of
dipole ‘P’ is written as,
P = q . 2d
Dipole moment is a vector quantity. Its direction is taken from negative charge to positive charge. In
S.I. it is measured in Coulomb meter ( C-m).
The molecules like H-Cl, show permanent
dipolement. These molecules are called as Polar
molecules. Their dipole moment arises due to
asymmetric distribution of positive and negative
charges in the molecules. In these molecules, the
center of all positive charges do not coincide with
the center of all negative charges. However, there
are other molecules like H2, N2 etc in which the
centers of positive and negative charges coincide. This is due to the symmetric distribution of charges +ve
& –ve in the molecules. Such molecules do not exhibit dipolement. They are called as Non-polar
molecules.
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The molecules that are non polar basically can be forced to become a polar by subjecting them to
an electric field. The force exerted by applied electric field disturbs the coincidence of centers of positive
and negative charges of the molecule.
“The molecule which exhibits dipole property under the action of external field is called as
induced dipole”.
Electric Filed due to an electric Dipole:
Field on axial point of a dipole:
Consider an electric dipole consisting of charges +q & -q which are d distance apart. A line passing
through the charges forming a dipole is called as its axial line. And a line that acts as a perpendicular
bisector of axial line, which passes through center of a dipole is called as its equatorial line.
 The intensity of electric field at a point on the axial line of an electric dipole is given by,
E
For a short dipole, x >> d
=
axial
 Eaxial =
Where ‘P’ represents dipolement and
where the field is expected.
1
40

2 Px
(x  d 2 )2
(1)
2
1
2P
 3
4  0
x
‘x’ represent the distance from the center of dipole to the point
Electric field on the equatorial point of a dipole:
Electric field at a point on equatorial line of a dipole, which is
E
For short dipole x >> d.
equit
=
Eequit =
1
40
‘x’ m apart from center of dipole is given by,
P

(x  d )
2
2
(2)
3
2
1
P
.
4  0 x 3
From equations (1) & ( 2) it can be shown that
E axial = 2 E
equit
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Also in both cases, E 
1/x3
Electric field due to an electric dipole at any point:
Consider an electric dipole made of charges +q and q separated
by a distance 2d. Let P be a point at a distance x from the centre of the
dipole. Let  be the angle made by the axis of the dipole and the line
joining the point and the midpoint of the dipole.
The magnitude of electric field at P is given by
E=
1 P
4  0 x 3
3 cos 2   1
Behavior of an electric dipole kept in External uniform Electric
Field:
Consider an electric dipole consisting of charges +q & -q which are distance 2d apart. Let the dipole
is kept in uniform electric field of strength E as shown in fig. Axis of dipole makes an angle  with the field.
(dipole moment vector making angle  with the field)
Each of charge forming the dipole, experience the force of ‘qE’ amount. These forces are oppositely
directed.
Although the net force on dipole is zero, the dipole
experiences the torque due to non zero perpendicular
distance between these forces. This torque tends the
dipole to rotate and align itself parallel to the applied field.
The torque actions on dipole can be calculated as follows:
From definition : Torque = Force x perpendicular
distance between the force
 = q E . 2d sin
 = q .2d . E sin
 = PE sin …………(1) (  P = q.2d)
i.e.

= P.E
Definitions of Dipole moment:
From above expression for the torque,  = PE sin
Dipole is kept at right angles to the uniform field of unit strength then ,
for  = 90 & E = 1 NC-1
 = p x1 x sin 900 = p
OR  = P
Thus dipole moment of an electric dipole is defined as numerical value of, torque
experienced by it when it is kept at right angles to an uniform electric field of unit strength
Polarization of Dielectric Medium :
A dielectric material is one which has negligible number of free
charges. They are non-conducting electrically. Dielectric substances are
composed of either polar atoms or non-polar atoms. In case of a polar
dielectric material, although the individual atoms or molecules exhibit
finite non zero dipole moment, because of random orientation of all
dipoles, the net dipole moment of entire material turns to be zero.
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It is called as an un polarized dielectric material, see fig
Un polarized dielectric
polarized dielectric
When such un polarized dielectric is kept in external electric field, the dipoles of dielectric material
experience torque. This torque compels them to align parallel to the
directions of field as in fig off course alignment is not 100% perfect,
because the thermal agitations in the material tends to disalign some
dipoles. This is called as polarization of a dielectric. The polarized
dielectric exhibits finite dipole moment.
As a result of polarization, there is net negative charge on one
side of the dielectric and equal net positive charge on its other side. This
establishes small induced electric field ‘Ep’ inside the dielectric (due to
polarization) whose direction is opposite to the external field ‘E0’. Thus
resultant field in the space, where dielectric is kept, is less than ‘E0’,
and it is written as, Ed= E0 – Ep.
The quantity dielectric constant (K) for a medium
describes, how many fold the electric field in the space is reduced due to introduction of
dielectric.
IAs, E0  field in the space in the absence of dielectric
Ep  induced electric field in a dipole due to polarisation.
Then the dielectric constant k is written as,
K = E0/Ed= E0 /
(E0-Ep)
Note: If a conductor is kept is electric field, the free electrons in the conductor experience the force and
move opposite to the external field. The movement of electrons from one side of the conductor to the other
side, establish reverse electric field in conductor. This reverse field grows till it turns equal and opposite to
external field. Thus, net field in the space where conductor is kept is zero. Therefore, Ep=E0 OR Ed=E0E0=0.
Thus, for a good conductor of electricity, its dielectric constant is, K = E0/0 = 
In case of a perfect insulator, Ep=0, OR Ed=E0-0= E0.
therefore, its dielectric constant is, K = E0/ E0 = 1
Polarizability:

For a single molecule induced electric dipole moment ( p ) is proportional to the applied electric field



(E) i.e., p  0 E


p = 0  E
Or
where  is a constant and is called atomic or molecule polarizability and is given by
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=
P
0 E
SI Unit of  =
Unit of p
Cm
 2 1 2
 m3
Unit of 0  unit of E C N m  NC 1
Thus unit of  is m3 (i.e., the unit of volume).
Dielectric Strength of a medium :
‘The electric dipole moment acquired per unit volume of a dielectric is called as
polarization’. It is denoted by ‘p’ The polarization ‘p’ of a dielectric can be increased, with the strength
of external field. This continues till the external field takes a critical value. Beyond this value, the electrons
from the molecules or atoms of dielectric material, get themselves detached, because of strong force
exerted by field. This stage of dielectric is called as breakdown. As result of breakdown, dielectric loses its
insulation property and starts conducting.
The maximum strength of external field that can be applied to a dielectric without
causing a breakdown is called as dielectric strength of the medium. Esafe =
Vsafe
d
Where Vsafe is the maximum potential that can be applied to a dielectric of thickness d without its
electric breakdown.
*
Dielectric strength of air medium is 3x106vm-1
Electric Lines of Force :
Electric lines of Force are imaginary, hypothetical invisible lines, which are used to visualize the
electric field due to a charge or group of charges. The concept of lines of force was introduced by Faraday.
They exist in 3 dimensional space around the charge.
Electric line of force in an electric field is an imaginary curved line such that the tangent drawn to it
at a given point gives, the direction of electric field at that point.
Properties of Lines of Force:
1. They originate from the surface of an isolated positive point charge and are directed radially
outward. The lines of force, terminate on an isolated negative point charge and are directed radially
inwards.
2. The tangent drawn to the electric line of force at any point gives the
directions of electric field at that point. Line of force represents a path
on which a test positive charge moves or tends to move.
3. The no. of lines of forces per unit area perpendicular to the
direction Electric field, is equal to the intensity of electric field at
that point.
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4. Electric lines of force, do not intersect with each other. That is, they do not meet at a common
point. If it happens so, then it leads to the following contradiction. If two lines of forces intersect at
a point, then at the point of intersection, one can draw two tangents, one tangent to each line of
force. It means that, a test charge kept at this point will move simultaneously along two directions
which is impossible!
5. We know that, electric field inside good conductor is zero. Therefore the lines of force do not exist
inside it. However, in case of an insulator or a dielectric, the electric field does exist in it. Hence the
lines of force also can exist in it.
6. Electric lines of force are perpendicular to the surface of a charged body.
7. Electric lines of force exert, lateral pressure on one another. This explains repulsion between like
charges.
8. Lines of force shows longitudinal contractions. This explains attraction between unlike charges.
Lateral Pressure
Longitudinal Contractions.
Electric Flux:
Electric Flux is the measure of the no. of electric lines of force, penetrating some surface. The no. of
lines of force penetrating a surface, in turn is directly proportional to the charge within that surface.
Electric Flux of electric field through a surface is defined as the no. of lines of force passing through
it. If the lines of force are passing normally then it called as Normal flux. It is denoted by .
Consider small plane area ‘ds’ in the electric field, which is normal to the direction of electric field of
strength E. The Normal flux through ds is given as
d= E. ds
……….(1)
If  is angle between normal to the surface and
Field E, then flux through the surface ds is,
d= E . ds Cos
………….(2)
Electric Flux is a scalar quantity.
If the directions of lines of force, are directed out of the given surface then the flux is taken positive and if
lines of force are directed into the given surface flux is taken negative.
Area vector:
Area is a scalar quantity. But in certain situations are of a surface has to be represented by a vector
so as to specify it completely. The area vector is represented by a vector along the normal to the area. The
length of the vector is proportional to the magnitude of the area. Its direction indicates the orientation of
the area.
Gauss Theorem :
Gauss Theorem
relates the Electric Flux of electric field and the charge. It states that total
normal Electric Flux through any closed surface is equal to
enclosed by it.
(1 / ) times the net charge
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=
1

xQ
Where Q = q i = total charged enclosed.
Explanations of Gauss Theorem :
Consider a closed surface of any shape S as shown in the
adjacent fig, which contains the charges q1, q2…..qn. Let
surface S is divided into large number of elementary areas
ds1, ds2, ds3 ……..etc. Let E1, E2….. etc are the strengths of
electric fields on these surfaces respectively. The individual
normal flux through these elementary areas can be given as,
d1 = E1ds1 cos1
d2 = E2ds2 cos2
..
..
..
and so on.
Then the total normal Electric Flux through the entire surface is given by
total = d1 + d2 +…. = dn1
According to Gauss theorem this total normal flux ie. total is equal to ( 1/
enclosed.
i.e.,  total =
0)
times net charge
1
. Q , Where Q = q1 +q2 +…..

 E ds Cos =
OR
1
.Q

Applications of Gauss theorems
1.
Electric intensity near a charged conductor:
Consider a charged conductor of any shape. It is
because of its irregular shape, the density of charges
at various places, on its surface, is different Let p is
the point close to the surface at which intensity of
electric field E is to be calculated. Let a cylindrical
Gaussian surface is constructed, which consists of
point p on its circular top surface, of area ds. part of
cylinder lies inside the charged conductor and the
other part outside. As ‘p’ is close to the charged
conductor, the electric field at point p is perpendicular
to the surface of conductor just below it.
As there is no electric field inside the conductor the flux of electric field through the part of cylinder
inside the charged conductor is zero.
The electric flux through the cylindrical region of Gaussian surface is also zero, as there is no any
component of electric field perpendicular to it.
Thus, net normal flux through entire cylindrical Gaussian surface is ,
= E . ds
… (1)
Using Gauss theorem
 = [ 1/] . ( charge enclosed say Q )
…. (2)
if ‘’ is surface density of the charge on the conductor at a place just below point P then,
charge enclosed by Gaussian surface will be,
Q =  . ds
 Substituting in equation (1)
=
1
. (  . ds)

E=


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comparing equation (1) and (3)
E . ds =
1
0
(  . ds) , Or
Thus electric intensity at a point near charged conductor is directly proportional to the
charge density at that point.
1. Electric Intensity near a charged spherical conductor :
Consider a spherical conductor of radius R which is given with a charge Q. the charge gets
uniformly distributed over the surface of the sphere because of its symmetrical shape.
Let P is the point at distance d from the sphere where electric intensity is to be determined.
Case1:
Point is outside the conducting sphere.
Let us construct a spherical Gaussian surface of radius d, that encloses the charged sphere and
contains the point P. Because of symmetrical shape of charged sphere, the electric field at all points on
Gaussian surface is same in magnitude and is radially outwards from the centre O.
The electric flux through small area ds around point p is given as
d= Edscos
as the field every where is perpendicular, = 00,
d = Eds … (1)
The total normal electric flux through the spherical
Gaussian surface is equal to,
 = E.∑ds
or  = E . ∑ds
but for spherical Gaussian surface of radius d
∑ ds = 4d2
= E . (4d2) …. (2)
As we know, total charge enclosed by Gaussian surface is equal to the charge given to the sphere
i.e. Q. then the above total normal flux can be also obtained by Gauss theorem as,
 = (1/ 0 ). Q …. (3)
Comparing (2) and (3)
 E . 4d2 =
1
.Q
0
or
E =
1
4 0

Q
d2
This expression is same as expression for the field at a point due to a point charge Q kept
at O. Thus it can be concluded that, for a point outside the charged sphere, it appears that as if the
total charge on sphere is concentrated at its centre.
Case 2 :
Point is outside & close to surface of the conducting sphere
If P is close to surface of charged spherical
conductor , then , the distance of point P can be taken equal
to the radius of the sphere( this is due to the reason that
the radius of the Gaussian surface drawn in this case, will be
same as the radius of the sphere). Thus substituting d = R
, expression for the electric field at a point close and outside
the charged spherical conductor is
written as,
E =
Case 3:
1
40

Q
R2
Point is inside the conducting sphere
Let point P be inside the sphere which is at a distance d
its center, as shown in the fig. A Gaussian surface is constructed
consists of point P. if E is the electric field intensity at P, then field
from
which
at
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SARVAJNYA
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every point on Gaussian surface is,
 = E . 4d2
From the Gauss theorem,  =
1
0
(ch arg e  enclosed )
As in this case Gaussian surface does not enclose the charge,
therefore Q = 0

=0
 E . 4d2 = 0 OR
E=0
“Thus electric field at every point inside the charges
conductor is zero “
The adjacent figure shows variation of electric field at a point
near a charged spherical conductor:
3.
Electric field near a charged plate
Consider thin plate with surface area A.
If charge Q is given to it, then the surface density of
Q
charge on the plate is given as,
 =
. As the total charge given to it is equally distributed on its
A

both surfaces the charge density on each surface is equal to,
.
2
Electric intensity at a point close to such plate is given as,
E 

2 0
Electric potential:
In the topic of gravitation, it is have learnt that, a mass kept in gravitational field experiences force.
If the test mass is moved in the gravitational field, the work is done. This work done is stored in it, in the
form of gravitational potential energy. ( even in the absence of test mass, every point in gravitational field,
has capacity to exert force on a mass kept in it, thus every point in gravitational field is said to have
gravitation potential )
In analogy to gravitation, it is observed that, every point in electric field has a capacity to exert
force on a test charge kept in it. Thus a charge kept in electric field acquires an energy called electric
potential energy. If a test charge is moved from one place to another place in the electric field, against the
direction of field, then there is a work done. This work done itself is stored in the charge, in the form of its
potential energy. The potential energy stored in a charge at infinite distance from the source of electric field
is taken zero, since it does not experience force. This place of zero potential acts as a reference level for
measurement of potential energy of a charge.
The electric potential at a point in the electric field, is defined as amount of work done
in bringing an unit positive charge, from infinity to the concerned point in the electric field. If
the work is done against the field, then the Potential at that point is taken positive and if work
is done by the field itself then potential is taken negative.
Electric potential at a point in the electric field is also defined as, the potential energy
stored in an unit charge kept at that point
If w is the work done in bringing a charge Q from infinity to the point in the electric field the from
definitions he electric potential at that point is, v = w/Q
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In S.I. it is measured with the unit volt. Electric potential at a point is said to be one volt, if
one joule of work is done in bringing an Unit Positive Test Charge from infinity to that point
against the field.
Electric Potential at a point due to an isolated point charge.
Let Q be a positive point charge placed at O. Its electric field is outwardly directed and extends
towards infinity as shown in the fig 1. P is the point where electric potential is desired. X and Y are two
points in the electric field which are separated by a small distance dx such that an unit positive charge
experiences almost same force at X and Y. Let OX=x
Let an unit positive charge is kept at X. The force F that acts on this UPC is,
F=
1
40

Q
X2
…(1)
In order to displace it against this force, an external force of same magnitude but in opposite
direction is to be applied on the UPC ( i.e. – F ). This applied force is such that there is no acceleration of
UPC. The work done in displacing UPC from X to Y is given as,
dw = - F . dx
..(2)
Negative sign in expression also represent that, the UPC is displayed against the electric field.
Using (1)
dw= -
1
40

Q
. dx
X2
…(3)
Electric potential at Point P is the total work done in bringing UPC from  to point P. To find this
total work done, equations (3) can be integrated between the limits x =  and x =d
p
W=  dW =


dW =

dW =

dW =

xd

x 
1
40
Q
40
xd

x 

Q
dX
X2
1
dX
X2
1  1 


4 0  X 
Q 1 1 

40  d  
Q
1
x
dW =
4 0 d
p
But from the definition of electric potential ,
 dW = V, therefore,
V=
1
4 0

Q
d

Relation between electric filed and electric potential :
Let A and B are the two points in electric field, produced by point charge +Q. the distance between
A& B points (dx) is so small that UPC kept at A and B experiences same force (F) i.e. Electric field E from A
to B is same.
From definition, the potential difference between A and B points ( dv) is equal to work done in
moving UPC from A to B.
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Or

dv = dw = F . dx
dv = - E . dx
= - E . dx
E = - ( dv/dx)
The quantity (dv/dx) is called as potential gradient. It is defined as rate of fall of potential with the
distance.
Thus electric intensity at a point in an electric field is equal to the negative potential
Differences between electric field and electric potential :
Electric field
1. Electric field at a point is defined as force
experienced by UPC kept at that point.
2. Electric field is a vector quantity.
3. It is measured in NC-1
4. Electric field at point due to a point
charge is E= 1/40 x Q/d2
5. Electric field inside charged spherical
conductor is zero.
Electric Potential
1. Electric potential at point is defined as
amount of work done in bringing UPC from
infinity to that point.
2. Electric potential is a scalar quantity
3. It is measured in volt or JC-1
4. Electric potential at a point due to point
charge is V = ¼0 x Q/d
5. Electric potential inside charged conductor is
constant.
Electric potential at a point due to large no. of charges.
Electric potential is a scalar quantity. Therefore the total electric potential at a point due to group of
charges can be obtained by adding the electric potentials due to individual charges.
Let V1 is the potential at a point P due to charge Q1 at a distance r1, V2 due to charge Q2 at distance
R2 and so on at P is, V = V1+V2+…..Vn
i.e.
V

q1
1 q2
1 qn
.

 ....... 
40 r1 40 r2
40 rn
1
V
1
40
i n
qi
i 1
i
r
Electric potential at a point due to an electric dipole:
Electric potential at a point, at a distance of ‘x’ from the center and on the axis of electric dipole is
given by the following formula,
V=
1
40

P
(x  d 2 )
2
Electric potential at any point on its equatorial line is zero. It is due to the reason that every point
on equatorial line of electric dipole is equidistant from the opposite point charges forming the dipole. Hence
equatorial line of dipole is also called as zero potential line.
Equipotential surface:
Any surface which has the same electrostatic potential at every point is known as equipotential
surface. An equipotential surface may e the surface of a body or simply a surface in space near a charged
body. Equipotential surfaces cane drawn properties of an equipotential surface:
1.
No work is done in moving a test charge from one point to another in an equipotential surface.
2.
Electric field is always perpendicular to the equipotential surface.
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840942174
3.
4.
The equipotential surfaces help in determination of strong electric fields from weak electric fields.
Equipotential surfaces will never intersect.
Potential energy of system of charge:
Group of charges forms a charge system. As the charges in a group interact each other, the charge
system posses potential energy. If charges in the charge system, maintain infinite separation from each
other, the PE of system is zero. To remove a charge from the charge system or to add an external charge to
the system, work is necessary to be done.
The potential energy of system of charges is defined as the amount of work done in
bringing all the charges from infinity to the present positions in the system .
Explanation : let q1 and q2 are two charges forming a system are
separated by a distance d as shown in the fig. Let V is the potential
at the point where Q2 is placed. Let q1 and q2 are two charges are
kept at ‘0’ and the other at . As they do not interact potential
energy of either charges is zero. Let charge q2 is brought from  to
a point at a distance d from charge q1. The work done in this case
can be shown as
W
1 q1 q 2

40 d
This work done is stored as PE of system of two charges. If both of the charges are of
same type PE of system is positive and if both charges are of opposite type PE of system
negative.

If number of charge are more than two, then total PE of system of charges is given as
PE 
1
40

qi q j
d ij
The summations is taken over all possible pair of
the number of charges forming the charge system.
****
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QUESTIONS’ BANK
ELECTRIC CHARGES:
Questions carrying one mark each:
1.
How many electrons make one coulomb of electric charge?
[Model QP., Oct. 2002]
2.
The repulsive force between two electrons is F. How will the force change if the electrons are replaced by
protons?
[Model QP.]
3.
What is the practical unit of electric charge?
[Oct. 83, 86, April 85]
4.
Define unit charge.
[April 84, 88]
5.
Define surface density of charge.
[Oct. 84, 2001]
6.
If the shape of an object is non-uniform, where will the surface charge density be maximum? [April 89,90]
7.
In textile mills, the atmosphere is rendered humid. Why?
[Oct. 91]
8.
How many electrons are to e added/ removed to charge a body to +1 coulomb? (Charge of an electron =
1.6x1019 C)
[April 92]
9.
In electrostatics, gravitational forces are not taken into account. Why?
[Oct. 92]
10.
Force between two point charges is 0.5 N. If the distance between them is doubled what will e the force
between them?
[April 95]
11.
Mention one of the methods of charging a body.
[April 99]
12.
How many electrons do constitute a charge of 1 C?
[April 2000,2003]
13.
How does the surface density of charge depend on the curvature of the surface?
[April 2000]
14.
How many electrons make 1 nanocoulomb of electric charge?
[April 2001]
15.
Two spherical droplets have equal surface density of charge. If they are combined to form a single
droplet, what happens to the surface density of charge?
[April 91]
16.
What is the effect of dielectric on a force between two charged spheres separated by a distance? [Oct. 98]
17.
What happens to the magnitude of the force between two charges if glass plate is interposed between
them?
[April 2007]
Questions carrying two marks each:
1.
State and explain Coulomb’s law in electrostatics.
[MQP Apr. 88, 05, July 2006]
2.
Distinguish between conductor and insulator.
[April 85]
3.
What is the effect of dielectric on a force between two charged spheres separated by a distance? [Oct. 98]
4.
Write the expression for Coulomb force between two point charges separated by a distance in vector
form and explain the terms.
[Apr. 2006, M.Q.P]
5.
What is an electric dipole? Define dipole moment.
[July 2006]
Questions carrying five marks each:
1.
Two small spheres each of mass 10 mgm are suspended from a point by threads 0.5 m long. They are
equally charged & repel each other to a distance of 0.28 m. If g=10 m/s 2, what is the charge on each? [Apr. 92]
2.
Two identical pith balls each weighing 30 mg are hung from the same point of a rigid support by two in
extensible threads each of 0.10 m. When equal amount of identical charges are given to the pith balls
they get separated by 0.10 m. Calculate the charge on each of them. (G=10 m/s 2)
[April 95]
3.
Three positive charges 3 nC, 4 nC and 5 nC are placed at the vertices A, B and C respectively of an
equilateral triangle ABC of side 0.2 m. Find only the magnitude of the force on the largest charge. [April. 2002]
4.
Two point charges of +3nC and 3nC are placed at corners A and B of an equilateral triangle ABC of side
0.3 m in air. If a charge +2nC is placed at the point C, what is the force acting on it?
[Oct. 2002]
5.
Two small metal spheres are charged so that they repel each other with a force of 2x10 5 N. The charge
on one sphere is twice that on the other. When they are moved 0.1 m farther apart, the force reduces to
5x106N. What are the charges and what is the initial separation between them?
[April 97]
10
10
10
6.
ABCD is a square of side 1 m. Point charges of +2x10
C, 4x10
C and +8x10 C are placed at
corners A, B, C respectively. Calculate the work done in transferring a charge of 10 C from D to the
point of intersection of diagonals.
[July 2006]
7.
A metal sphere of radius 0.01 m is charged to a potential of 45 kV and kept at the corner B of a rightangled triangle ABC (AB=0.3 m, BC=0.4 m and  ABC=90). Calculate the work required to shift a charge
of 2x106 C from corner C to corner A of the triangle.
[April 2007]
15
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840942174
ELECTROSTATIC FIELD:
Questions carrying one mark each:
1.
A charged spherical conductor is equivalent to a point charge at the centre with respect to a point outside
the conductor, explain.
[Model Q.P.]
2.
What is the potential of the earth?
[Oct. 88, April 95]
3.
Two point charge Q and 4Q are separated by a distance of 12 cm. Find the distance of the neutral point
from Q.
[Oct. 89]
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
o
Find the magnitude of electric field due to a point charge of 1.6x1010 C at a distance of 1 A from it. [Oct. 89]
If 4x104 joules of work is done in moving a charge of 5 microcoulomb from a point A to another point B,
then find the potential difference between A and B.
[April 90]
A potential difference of 300 mV is applied between two plates separated by a distance of 1 mm. What is
the electric intensity in the region between the plates.
[Oct. 90]
What is the magnitude of electric intensity inside a charged hollow sphere?
[Oct. 90, April 97]
If a dielectric is introduced between two charges what happens to the electric field between them?[April 91]
If the electric field at a distance r from a charge q is E, what is the electric potential at that point. [April 91]
Find the electric intensity due to a point charge of 10 mC at a distance of 3 cm.
[April 93]
Three charges +3C, +2C and 5C are placed at the vertices of an equilateral triangle of side ‘d’ m. What
is the potential at the circumcentre of the triangle?
[April 93]
A hollow spherical conductor of radius 0.1 m is charged. At which point is the electric intensity due to this
conductor maximum?
[Oct. 95]
At which point is the electric intensity due to a charged sphere is zero?
[April 96]
State Gauss’ theorem.
[Oct. 97, April 02, 2006]
What is an electric line of force in an electric field?
[April 98]
Is electric potential a scalar or a vector?
[Oct. 2002]
A spherical surface surrounds a point charge Q. What happens to the total flux throught he surface when
the surface is changed to a cube?
[Oct. 2003]
Mention the S.I. unit of electric intensity.
[Oct. 2000]
Can two electric lines of forces intersect?
[April 2004]
The potential at any point inside a hollow charged spherical conductor of radius 0.2 m is 1 V. What is the
potential on the surface of the conductor?
[Oct. 04]
What is an electric dipole?
Mention the S.I unit of electric field.
[Apr. 2005]
What is an equipotential surface?
[July 2006]
Questions carrying two marks each:
1.
Electric intensity inside a charged conductor is zero this follows from Gauss’ theorem. Explain.[Model Q.P.]
2.
Calculate the potential due to a charge of 100 microcoulomb at a distance of 9 metres.
[Oct. 84]
3.
Obtain an expression for the electric intensity at a point due to a point charge.
[Oct. 93]
4.
State and explain Gauss’ theorem in electrostatics.
[April 96, 2007]
5.
Explain why the work done in moving a charge on equipotnetial surface is zero.
[April 99]
6.
State Gauss theorem and mention one of its uses.
[Model QP]
7.
What is electric dipole moment? What is its unit.
8.
Show that a dipole is placed in a electric field experiences a torque.
9.
Give the expression for electric potential due to a system of three point charge.
10.
Obtain the expression for capacity of a spherical conductor.
[Apr. 2006]
Questions carrying four/ five marks each:
1.
State Gauss’ Theorem. Obtain an expression for the electric field at any point near the surface of a
charged conductor of irregular shape by applying Gauss’ theorem.
[Model. Q.P., Oct. 97, Apr. 2003]
2.
Define electric intensity and electric potential. Arrive at the relation E = dV/dx.
[Model Q.P.]
OR
Define electric intensity and electric potential. Deduce the relation connecting them. [Oct. 88, 89, 91, Apr. 04]
3.
State Gauss’ theorem in electrostatic. Hence derive the formula for the electric field at a point outside a
uniformly charged Spherical conductor.
[April 83, Oct. 86]
OR
Obtain an expression for electric intensity at a point outside a charged spherical conductor by using Auss’
theorem.
[April 83, Oct. 99, 2001]
4.
Derive an expression for the electric potential at a point due to a point charge.
[Oct. 83, 87, 88, 2000, 04, April. 97, 98, 2000, 02, July 2006]
16
840942174
5.
6.
SARVAJNYA
Distinguish between electric intensity and electric potential at a point in an electric field. Obtain the
relation between them.
[April 87, 2002]
Write the expression for electric field at any point due to an electric dipole using this expression deduce
the expression for the electric intensity at any point on (i) On the axis of the dipole (ii) the perpendicular
bisector of the axis of the dipole.
CET QUESTIONS
1.
2.
3.
4.
5.
6.
7.
A body
a)
b)
c)
d)
A body
a)
b)
c)
is said to be charged if.
it attracts light objects
it repels light objects
it attracts heavy objects
it repels heavy objects
is said to be positively charged if.
it has no electrons at all
it has only protons
it has more number of electrons than
protons
d) it has less number of electrons than protons
The sure test to know whether a body is charged or
not is.
a) repulsion
b) attraction
c) both 1 and 2
d) neither 1 nor 2
A body is said to e negatively charged if it has.
a) only electrons
b) only protons
c) less number of electrons than protons
d) less number of protons than electrons
Charges are.
a) quantized
b) not quantized
c) always positive integral multiples of the least
unit of charge
d) always negative multiples of the least unit of
charge
The smallest unit of charge is equal to.
a) 1.6x1019C
b) 1.6x1019C
c) 1C
d) 1 C
To charge a body to +1C.
a)
8.
9.
6.25 x 1018 electrons have to be added to
the body
b) 6.25 x 1018 electrons must be removed from
the body
c) one electron must be added to the body
d) one electron must be removed from the
body
To charge a body to 1 C.
a) 6.25 x 1012 electrons must be removed from
the body
b) 106 electrons must be removed from the
body
c) 106 electrons must be added to the body
d) 6.25 x 1012 electrons must be added to the
body.
When a glass rod is rubbed against silk.
a) glass rod loses electrons and becomes
negatively charged
b) silk loses electrons and becomes positively
charged
c) glass rod loses electrons and becomes
positively charged
d) silk loses electrons and becomes negatively
charged
10. When an ebonite rod is rubbed against fur.
a) ebonite rod loses electrons and becomes
negatively charged
b) fur loses electrons and becomes negatively
charged
c) ebonite rod gains electrons and becomes
positvely charged
d) ebonite rod gains electrons and becomes
vely charged
11. When a positively charged body is made to touch an
uncharged conductor.
a) the uncharged conductor also becomes
positvely charged
b) the
uncharged
conductor
becomes
negatively charged
c) neither 1 nor 2
12. When a body is charged by friction.
a) +ve charges are newly created
b) ve charges are newly created
c) charges are conserved. They are simply
transferred from one body to the other
d) a body cannot be charged by friction
13. A charged body induces.
a) an equal amount of the same kind of charge
b) an equal amount of the opposite kind of
charge
c) no charge
d) charge depending on the temperature of the
body
14. When a conductor is charged.
a) all the charges are localised in the central
region
b) all the charges are distributed on the surface
of the conductor
c) more charges are localised at the centre
than on the surface
d) none of the above
15. When an insulator is charged.
a) the charge remains localised
b) the charges are distributed over the surface
c) either 1 of 2 depending on the shape of the
insulator
d) an insulator cannot be charged
16. The distribution of charges on a conductor is such
that.
a) it is always uniform irrespective of the shape
of the body
b) more charges reside at points where the
curvature is less
c) more charges reside at points where the
curvature is more none of the above
17. The surface density of charge is defined as the ratio
of the total charge of the body to its.
a) length
b) surface area
c) volume
d) cross-sectional area
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840942174
18. Repulsion exists between.
a) a positvely charged body and another
positvely charged only
b) a positively charged body and an uncharged
body
c) a negatively charged body and an uncharged
body
d) two uncharged bodies
19. The force ‘F’ between two charged bodies separated
by a distance ‘d’ is such that.
1
a) F 
b) F  d
d
1
c) F  d2
d) F 
d2
20. The force between two charges.
a) depends on the medium between them
b) does not depend on the medium between
them
c) depends on the medium when it is a liquid
only
d) depends on the medium when it is a gas
only
21. The force between two charges in different media
are different because.
a) different media have different densities
b) different media have different viscosities
c) different media have different permittivites
d) different media have different permeabilities
22. Relative permittivity of a medium r is related to the
permittivity of the medium  and the permittivity of
vacuum as follows:

a) r = 0
b) r =  . 0


c) T =
d) 0 = . r
0
23. The force between two charges is.
a) high if the medium in between the two
charges has a high permittivity
b) high if the medium in between the two
charges has a low permittivity
c) independent of the permittivity of the
medium
d) none of the above
24. The permittivity of free space is.
a) 8.85 x 1012 F.m1
b) 8.85 x 1018 F.m1
c) 8.85 x 1018 F.m1
d) 8.85 x 1012 F.m1
25. The relative permittivity of water is 80. This means
that.
a) the force between two charges in water is 80
times less than that in vacuum at the saem
distance
b) the force between two charges in water is 80
times less than that in vacuum at the same
distance
c) relative permittivity has nothing to do with
force between two charges
26. n identical mercury droplets charged to the same
potential V coalesce to form a single bigger drop.
The potential of the new drop will be. (CET 2000)
a) n2/3 V
b) nV2
c) nV
d) V/n
27. Two pith balls carrying identical charges are
suspended from a point in air. Afterwards they are
28.
29.
30.
31.
32.
33.
suspended inside water. The distance between the
two pith balls is now.
a) increased
b) reduced
c) unchanged
d) doubled
When two charges +1C each are separated by a
distance of 1 m in air, they.
a) attract each other with a force of 1 N
b) repel each other with a force of 1 N
c) attract each other with a force of 9x109 N
d) repel each other by a force of 9x109 N
Charge on a body can be detected using.
(CET 1983)
a) an electroscope
b) an electrometer
c) a voltmeter
d) an ammeter
When a glass plate is introduced between two
bodies, the force between them. (CET 1984)
a) increases
b) decreases
c) remains the same
d) becomes zero
Two positive charges are placed with a fixed
separation. A slab of dielectric medium is introduced
between them. As a result, the repulsion between
the charges.
(CET 1985)
a) changes to attraction b) remains constant
c) decreases
d) increases
The intensity of an electric field at a point is defined
as.
a) the force experienced by a charge of +1C
placed at that point
b) the force experienced by a charge of 1
electron placed at the point
c) the force experienced by a proton placed at
that point
d) the force experienced by a neutron placed at
that point
The intensity of an electric field at a point at distance
‘d’ from a point charge q is given by E=
a)
q2
4  0 d
b)
c)
q
4  0 d
d)
4  0 q
d
q
4  0d2
34. The unit of electric intensity is.
a) Nm
b) NC
c) N.C1
d) N.m1
35. An electric field is said to be uniform if.
a) the intensity at all pints in the field is the
same
b) the direction at all points in the field is the
same
c) neither 1 nor 2
d) both 1 and 2
36. A charge q when placed in an electric field of
intensity E experiences a force F given by.
a) F = qE2
b) F = q2E
q
c) F =
d) F = qE
E
37. A proton and an -particle are subjected to the same
electric field. If a1 and a2 denote their respective
accelerations, then a1/a2 is equal to.
a) 1:1
b) 1:2
c) 2:1
d) 1:4
38. If  is the surface density of a charged plane sheet,
the electric intensity at a point t a distance ‘r’ very
close to it is.
(CET 1984, 1985)
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SARVAJNYA
840942174
39.
40.
41.
42.
43.
44.
45.
46.
47.

4q
a)
b)
r
r
a
1
c)
d)
4  r
4  qr
The electric potential at a distance ‘r’ from a point
charge q is given by V =
(CET 1984)
qr
4q
a)
b)
4 
r
q
1
c)
d)
4  r
4  qr
The unit of electric potential is.
a) the same
b) V/2
c) 1 Volt
The electric potential at a point distant r from a
charge ‘q’ is V. When the charge is replaced by a
charge 4q, the potential at the point will be. (CET
1984)
a) the same
b) V/2
c) 2V
d) 4V
Charges present on the clouds are due to. (CET
1984)
a) motion of water drops
b) earth’s magnetic field
c) lighting
d) motion of the clouds
A deuterium nucleus and a helium nucleus are placed
in the same electric field. The acceleration of helium
is. (CET 1985)
a) greater than that of deuterium
b) less than that of deuterium
c) equal to that of deuterium
d) zero
Inside a hollow charged spherical conductor, the
electric field is found to be.
(CET 1985)
a) zero
b) a function of the area of the sphere
c) proportional to the distance from the centre
d) a function of the charge density of the
sphere
Consider the electric potential due to a point charge
at a distance. The potential due to eight times this
charge at four times the above distance is. (CET
1985)
a) half the original potential
b) twice the original potential
c) the same as the original potential
d) four times the original potential
The accumulation of charge on clouds, which
produces lighting, is caused by.
(CET 1985)
a) ran drops changing into electrons
b) the electric field of the earth
c) ionization of the sun
d) electrification due to motion of water
molecules
State which of the following statements is true. (CET
1985)
a) the number of times electric lines of force
cross depends on the charge distribution
b) no two lines of force intersect each other
c) two lines of force intersect each other at lest
once
d) lines of force in a dielectric medium can
intersect each other
48. A closed surface contains negative charges. Electric
flux across it is.
a) zero
b) directed inwards
c) directed along the normal to the surface
outwards
d) parallel to the surface
49. Four charges of magnitude 1 nC,  4nC, 3 nC and 5
nC are placed at the corners A, B, C and D of a
square ABCD of side 6 m. The potential at the centre
of the square (in volts) is.
a) 7.5
b) 15
2
2
c) 7.5 2
d) 1.5 2
50. Two point charges +2C and +6C repel each other
with a force of 12 N. When a charge q is added to
each of the charges, they attract each other with a
force of 4 N. Value of q is (in C).
a) +4
b) 4
c) +1
d) 1
51. A slab of dielectric is introduced between two equal
negative charges with a fixed separation. As a result.
(CET 1987)
a) the force between the charges decreases
b) the slab gets heated up
c) the two charges attract each other
d) an electric current passes through the slab
52. A hollow spherical conductor carries negative charge.
A positive charge is placed at the centre of the
sphere. Then this positive charge will. (CET 1987)
a) oscillate between the opposite points of the
conductor
b) stick to the conductor
c) move in circle
d) stay at the centre
53. Point charges +50 C, +100C and 75 C are
placed on the circumference of a circle of radius 0.5
m to form an equilateral triangle ABC. The potential
at the centre of the circle is (Cet 1988)
a) 150 C/m
b) 450 C/m
c) 150 C/m
d) none of these
54. The mass of a proton is bout 2000 times the mass of
an electron. An electron and a proton are injected
into a uniform electric field at right angles to the
direction of the field with the same initial kinetic
energy. Then
(CET 1988)
a) the electron trajectory will be less curved
than that of the proton.
b) The proton trajectory will be less curved
than that of the electron
c) Both the trajectories will be equally curved
d) Both the trajectories will be straight
55. Two fixed point charges +4q and q units are
separated by a distance ‘a’. The point where the
resultant field intensity is zero is.
2
a
3
3a
c)
2
a)
b)
a
2
d) none of these
56. To move a unit +charge from one point to another
point on an equipotential surface (CET 1990)
a) work is done b) work is done on the charge
c) no work is done d) work done is a constant
57. Potential at any point inside a charged hollow
sphere.
19
SARVAJNYA
840942174
58.
59.
60.
61.
62.
63.
64.
65.
66.
67.
a) increases with distance b) is a constant
c) no work is done d) work done is a constant
Two point charges +2C and +6C repel each other
with a force of 12 N.If a charge of 2C is given to
each of these chargse, what will be the force now?
(CET 1991)
a) zero
b) 8 N (attraction)
c) 8 N (repulsion)
d) none of these
10 Coulombs of charge are situated at each of the
vertices of a 1 cm cube. Then the electric field at the
centre of the cube is. (CET 1991)
a) 5 N/C
b) 10 N/C
c) 40 N/C
d) zero
P and Q are two points lying on the perpendicular
bisector of the line AB. Work done in taking a charge
of 5 nC from P to Q. (CET 1995)
a) depends only on the charge shifted
b) is zero
c) depends on the distance PQ
d) depends on the distance AB
A conducting sphere of rdius 10 cm is charged with
10 C. Another uncharged sphere of radius 20 cm is
allowed to touch it for enough time. After the two
are separated, the surface density of charge on the
two spheres will be in the ratio (CET 1995)
a) 2:1
b) 1:1
c) 4:1
d) 3:1
n identical mercury droplets charged to the same
potential V coalesce to form a single bigger drop.
The potential of the new drop will be (CET 1995)
a) nV
b) V/n
c) n2/3V
d) nV2
The electric field intensity due to a hollow spherical
conductor is maximum (CET 1996)
a) outside the sphere
b) on the surface of the sphere
c) at any point inside the sphere
d) only at the centre of the sphere
A hollow metal sphere of radius 5 cm is charged such
that the potential on its surface is 10V. The potential
at the centre of the sphere is.
a) Zero
b) 10 V
c) the same as that at a point 5 cm away from
the surface
d) the same as that at a point 25 cm away from
the surface
Two equal negative charges q are fixed at points
(O, q) and (O, a) on the y-axis. A positive charge Q
is released from rest at the point (2a, O) on the Xaxis. The charge Q will
a) execute simple harmonic motion about the
origin
b) move to the origin and remain at rest there
c) move to infinity
d) execute oscillatory but not simple harmonic
motion
Three point charges 4q, a and Q are placed in a
straight line of length 1 at points ½ and zero. The
net force on the charge q is zero. The value of Q is.
a) q
b) 2q
c)  q/2
d) 4q
Two point charges 12 C and 8 C respectively are
placed 10 cm apart in air. The work done to bring
them 4 cm closer is.
a) zero
b) 3.8 J
c) 4.8 J
d) 5.8 J
68. The work done in carrying a charge q once round a
circle of radius r with a charge Q at the centre is.
a) qQ/ 40r
b) qQ/ 40r
c) qQ/ 40 (1/2r)
d) zero
69. A charge q is placed at the centre of the line joining
two equal charges Q. The system of the three
charges will be in equilibrium if q=
a) Q/2
b)  Q/2
c) + Q/2
d) + Q/4
70. If dV is the potential difference between two points
separated by a distance, then electric intensity E is
given by.
dV
dx
dx
c) E =
dV
a) E =
dx
dV
dV
d) E = 
dx
b) E =
71. The electric potential in a region along along the Xaxis varies with x according to the relation V (x) =
5+4x2. Then
a) p.d. between the points x=1 and x=3 is
32V
b) force exerted by a charge of 1 C at x=1 m
is 8 N
c) the force exerted by the above charge is
along the +ve X-axis.
d) a uniform electric field exists in this region
along the x-axis
72. Inside a uniform charged spherical conductor, the
electric
a) field is zero every where
b) potential is zero everywhere
c) potential is same as at any point outside
d) field has the same magnitude everywhere
but it is not zero
73. The distance of closet approach between two
protons in vacuum is 1014 metre and the force
between them is 2.31N. The charge of a proton is.
a) 1.6x1017C
b) 1.6x1018C
c) 1.6x1019C
d) 2.31x1014C
74. A charge A of 25 C is placed on the line between
two charges B of 5 C and C of 30C. The charge is
0.05 m from B and 0.1 m from C. What is the force
on A?
a) 1.125 N
b) 11.25 N
c) 112.5 N
d) 1125 N
75. What is the strength of the electric field such that an
electron placed in the field would experience an
electrical force equal to its weight? [Charge of the
electron = 1.6 x 1019 C, mass of the electron = 9.1
x 1031 kg]
a) 5.57x1010 N/C
b) 5.57 x 109 N/C
8
c) 5.57 x 10 N/C
d) 5.57 x 1011 N/C
76. A and B are two small spheres with charges 9 C and
16 C. The distance between them is 0.28 m. How
far from A along the line AB, will the intensities due
to the two charges be equal?
a) 0.12 m from A in between A and B
b) 0.84 m from A outside AB
c) both 1 and 2
d) neither 1 nor 2
77. Two equal charges repel each other with a force of
0.1 N, when situated 0.45 apart. The medium in
20
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between the charges has dielectric constant 9. The
magnitude of each charge is.
a) 4.5 C
b) 4.5 mC
c) 4.5 C
d) 4.5 nC
78. Three small spheres each carrying a charge q are
placed on the circumference of an equilateral triangle
of side ‘a’. The potential at the centre of the triangle
is.
a) zero
c)
3 3q
 0 a
b)
3 3q
2 0 a
d)
3 3q
4  0 a
79. A charged oil drop remains stationary when sitauted
between two parallel horizontal metal plates 25 mm
apart and a p.d. of 1000 V is applied to the plates.
Find the charge on the drop if it has a mass of
5x1015 kg (g=10 m.s2)
a) 1.25x1017C
b) 1.25x1016C
c) 1.25x1019C
d) 1.25x1018C
80. An isolated sphere of radius 0.1 m has lost 1012
electrons. The intensity of the electric field on the
surface of the sphere is.
a) 1.44x1016 N/C
b) 1.44x105 N/C
4
c) 1.44x10 N/C
d) 1.44 N/C
81. The radius of a spherical conductor which will have a
potential of 6000 V when surrounded by an oil of
dielectric constant 1.5 and charged with 5x108C is.
a) 0.5 m
b) 0.005 m
c) 5 m
d) 0.05 m
82. A hollow spherical conductor is charged to 2x108C.
The potential inside the sphere is.
a) 0
b) 900 V
c) 1800 V
d) 600 V
83. To what potential must an insulated sphere be
charged so that its surface density of charge is 1012
C/cm2? [Radius of the sphere = 7 cm]
a) 7920 V
b) 792 V
c) 79.2 V
d) 7.92 V
84. Two metal balls of radii 0.05 m and 0.04 m are
charged to the same potential. The surface densities
of charge are in the ratio.
a) 5:4
b) 4:5
c) 25:16
d) 16:25
85. Two point charges, one ten times as strong as the
other, repel each other with a force of 81x104N
when separated by a distance of 0.02 m in air. The
charges are.
a) 9 nC, 90 nC
b) 6 C, 60 C
c) 6 nC, 60nC
d) 9 C, 90 C
86. Two charged spheres having the same radius and
charges 9 nC and +5 nC are separated by a certain
distance in air. They are brought into contact and
then replaced in their original positions. The ratio of
the forces between them before and after contact is.
a) 9:5
b) 5: 9
c) 45:4
d) 45:4
87. The separation between two pints in order that the
electrostatic force of repulsion acting on either of
them may be equal to its weight is.
a) 12 cm
b) 12 mm
c) 1.2 m
d) 12 m
Charge of a proton = 1.6 x 1019C, mass of a proton =
1.67 x 1027 kg, g =9.8 m/s2]
88. Charges equal to q1 are palced at two diagonally
opposite corners of a square and charges equal to q2
are placed at the remaining corners. If the resultant
force on q2 is zero, then q2=
a) + 2 q1
b)  2 q1
c) +2 2 q1
d) 2 2 q1
89. Two point charges +9x108C and 9x108C are
placed 0.5 m apart in air. The magnitude of resultant
intensity at a point located 0.5 m from either charge
is
a) 3.24x104 N/C
b) 3.24x103 N/C
c) 32.4 N/C
d) zero
90. Point charges 2nC and 2nC are placed at the
points (3, 4) and (3, 4) in the X-Y plane. Find the
electric intensity at the origin. All the co-ordinate
distances are in metres.
a) 0.12 NC1
b) 12 NC1
c) 1.2 NC1
d) 120 NC1
91. ABC is an equilateral triangle of side 3 cm. Charges
of +5 nC and 5nC are placed at the corners A and
B. The magnitude of the electric intensity at C is
a) 5 NC1
b) 50 NC1
1
c) 500 NC
d) 5x104 NC1
92. ABC is a triangle with sides AB=3m, BC=4m and
B=90º. Charges of magnitude 9x105C and
16x105C are placed at the corners A and C
respectively. the magnitude of the electric intensity
at B is.
a) 1.273x104 NC1
b) 1.273x105 NC1
3
1
c) 1.273 x 10 NC
d) 1.273 NC1
93. An infinite number of charges each equal to 1 nC are
placed along the X-axis at x=1 m, x=2 m, x=4 m,
x=8 m…. The electric intensity at x=0 due to this set
of charges is.
a) 
b) 0
c) 12 NC1
d) 120 NC1
94. Volt/ metre is a unit of.
a) potential
b) potential difference
c) distance
d) electric intensity
95. A metal sphere of radius 0.05 m has a charge of +1
nC. The surface density of charge is.
a) 3.18x108 C.m2
b) 3.18x107 C.m2
c) 3.18x107 C.m2
d) 3.18x108 C.m2
96. A point charge of +10 C is at the centre of a cubical
gaussian surface of side 0.1 m. The flux of the
electric field from the surface of the cube is.
a) 1.13 x 105 NC1
b) 1.31x106 NC1
c) 1.13x106 NC1
d) 1.31 NC1
97. ABC is a right angled triangle with AB = 0.2 m, BC =
0.6 m and B = 90º. A metal sphere of radius 2 cm
is charged to a potential of 9x104V and is placed at
B. The work done is carrying a charge of 1 C from C
to A is.
a) 60 J
b) 6 J
c) 600 J
d) 6000 J
98. ABCD is a square of side
2 m. Charges of 5 2
nC, +10 nC and 5 2 nC are placed at the corners
A, B and C respectively. The work done in
transferring a charge of 5 C from D to 0 (the centre
of the square) is.
a) 2.25x103 J
b) 225 J
c) 2.25 x 104J
d) 0.225 J
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99. An infinite number of charges each equal to 1 nC are
placed along the X-axis at x=1, x=2, x=4…. The
distances are in m. The electric potential at the origin
due to this set up of charge is.
a) 6 V
b) 12 V
c) 18 V
d) 24 V
100. A work of 16x1016 J is done to transfer an electron
between two points in an electric field. The potential
difference between the two points is.
a) 1 KV
b) 10 KV
c) 100 KV
d) 1000 KV
101. An electron is accelerated from rest in a uniform
electric field of intensity 10 KV.m1. Its velocity at the
end of 1 ns is.
a) 1.76x104 ms1
b) 1.76x105 ms1
6
1
c) 1.76x10 ms
d) 1.76x108 ms1
102. The potential energy of a charge 1 C in an electric
field is 3x104 J. The potential of the field is.
a) 3 V
b) 30 V
c) 300 V
d) 3000 V
103. A spherical conductor of radius R, placed in air, is
given a charge Q. then the potential at a point inside
the conductor and at a distance R/2 from its centre is
(CET 1997)
a) V =40R
b) V=4R
c) V =
1 Q
4  0 2R
d) V=
1Q
4  0 R
104. Two spherical conductors of radii 4 m and 5 m are
charged to the same potential. If 1 and 2 be the
respective values of the surface density of charge on
the two conductors, then the ratio
1
is(CET 1998)
2
a) 1.5
b) 0.5
c) 1
d) 2
105. A point charge A of charge +4 C and another point
charge B of charge 1 C are placed in air at a
distance 1 metre apart. Then the distance of the
point on the line joining the charges and from the
charge B, where the resultant electric field is zero, is
(metre) (CET 1998)
a) 1.5
b) 0.5
c) 1
d) 2
106. When a positvely charged conductor is earth
connected (CET 1998)
a) electrons flow from the earth to the
conductor
b) protons flow from the conductor to the earth
c) no charge flow occurs
d) electrons flow from the conductors to the
earth
107. Electric charges +10 C, +5C, 3C and +8C are
placed at the corners of a square of side
potential at the centre of the square is
1999)
a) 18x105
b) 1.8x106
c) 1.8
d) 1.8x105
108. When 1019 electrons are removed from
metal plate, the electric charge on it is
1999)
a) 10+19
b) +1.6
2 m. The
(V) (CET
a neutral
(C) (CET
c) 1.6
d) 1019
109. Two metal spheres of radii R1 and R2 are charged to
the same potential. The ratio of the charge on the
two spheres is. (CET 1999)
a) R1/R2
b) 1
c) ½
d) R1 – R2
110. A cube of side b has charge q at each of its vertices.
The electric field at the centre of the cube (CET
2000)
a) q/b2
b) q/2b2
c) 32q/b2
d) zero
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ELECTRIC CHARGE AND ELECTRIC FIELD & ELECTRIC POTENTIAL
QUESTION BANK:
One marks type questions:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
Define charge.
What is electrification?
How many electrons constitute a charge of 1 coulomb?
A body is at positive potential. What does it mean?
A body is at negative potential. What does it mean?
What is the relation between curvature of the surface and the concentration of charge on it?
Define the term linear density of charge.
What is meant by electric permittivity of the medium?
Write the SI unit of electric field strength.
Define the term, dielectric strength of a medium.
What is the difference in charging a conductor & an insulator?
What is the minimum amount of charge that can be given to a body?
Define an electric line of a force.
What is the field inside the charged conductor?
What is the direction of an electric line of force on an equipotential surface?
What is electrostatic shielding?
Write the SI unit of electric permittivity.
Define an electron volt.
Write an example of equipotential line.
A charge is moved on equipotential surface, what is the net work done?
What is electric dipole?
Define electric dipole moment.
An electron is kept in a field of strength (1/1.6x10-19) N/c. What is the force experienced by it?
An electron with KE of 2J enters a region of Pd ( 2/e) Volts. What will be its new KE?
A charge of 2C is enclosed by a cubical Gaussian surface, is transferred to the spherical box.
What is the change electric flux through the surface?
What is the potential of earth?
What is the direction of field due to a positive charge?
What is the direction of field due to a negative charge?
What is the direction of electric dipole moment?
If a slab of dielectric material is introduced between two charged bodies, the force between them
reduces to half. What is the dielectric constant of the slab?
Write the dimensional formula for electric flux.
Who classified the charges as positive and negative?
Which of these two cannot be charged by induction? A glass rod, a metal sphere.
Charged bodies can attract light uncharged bodies. State true/ false.
What is meant by discharging action of points?
How is the lighting caused?
Write a limitation of Coulomb’s law.
Write an example of polar molecule.
Write an example of non-polar molecule.
What is meant by induced dipole moment?
41.
42.
43.
44.
45.
46.
47.
48.
49.
50.
51.
52.
Define strength of electric field. Is it a scalar or a vector?
Define the uniform and non-uniform electric field.
Obtain an expression for electric field at a point due to a point charge.
Write any four properties of electric lines of force.
Define electric flux. Write its SI unit.
Derive the relation between electric field and electric potential.
Define the terms electric potential and potential difference.
Write any four properties of electric charge.
Write a note on charging by friction.
Write a note on charging by conduction.
Write the formula for electric field on the axis of electric dipole and define the terms.
Write the formula for electric field on the equatorial line of electric dipole and define the terms.
26.
27.
28.
29.
30.
Two marks type questions:
23
840942174
53.
54.
55.
SARVAJNYA
Write the formula for the potential energy of system of charge, containing n number of charge.
Define the terms used in the formula.
Write the note on the potential energy of the system of charges.
Write the formula for electric field near charged plate & define the terms.
Five marks type questions:
56.
57.
58.
59.
60.
61.
What is charging? Explain charging by induction.
State and explain Coulomb’s law in electrostatics.
Explain the distribution of charge on a charged body. Define surface density of charge.
State Gauss theorem. Show that electric field at a point near a charged conductor is directly
proportional to the charge density below the point.
Define electric potential. Derive an expression for electric potential at a point due to an isolated point
charge.
What is a dielectric material? Explain the phenomena of dielectric polarization.
NUMERICAL BANK:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
Two point charges +3C are and 6C located 1 m apart in air. Find the resultant electric intensity
midway between them.
[3.24x105 N/C]
Two point charges 9 nC and +18 nC are separated by a distance of 0. 1m in air. Find the resultant
electric intensity at a point between them and 0.01 m from 9nC.
[83x104 N/C]
Two point charges of +6x108 C and 6x108C are 0.4 m apart in air. Find the resultant electric intensity
at a point 0.4 m from either charge.
[3.375x103 N/C]
Two point charges of +1 nC and +4 nC are located 0.1 m in air. Find the position between them and
along the line joining them at which resultant electric intensity is zero.
[0.0333 m]
Two point changes +2nC and 18 nC are located 1 m apart in air. Find the position along the line joining
the two charges at which resultant electric intensity is zero.
[0.5 m]
ABC is an equilateral triangle of side 0.1 m. Point charges of +3 nC and 3 nC are placed at corners A
and B respectively. Calculate the resultant electric intensity at C.
[27x102 N/C]
ABC is a right angled triangle with sides AB=0.3 m, BC= 0.4 m and B=90º. Point charges of +18 nC and
+32 nC are placed at corners A and C respectively. Find the resultant electric intensity at B. [2.546 x 103
N/C]
ABC is triangle with sides AB = 3m, BC = 4 m and  ABC = 90º. Charges of +9x1010 C and 16x1010C
are placed at corners A and C respectively. Find the resultant at B.
[0.9 N/C]
ABC is right angled triangle with sides AB = 0.2 m, BC = 0.4 m and  ABC = 90º. Point charges of +4
x1012 C and +32x1012 C are placed at corners A and B respectively. Calculate the resultant electric
intensity at B. Suppose a point charge of 5C is placed at B, what force would it experience?
[2.102
N/C, 10.06x106 N]
ABCD is a square of side 1 m. Point charges of +4C, +8C and +4C are placed at corner A, B and C
respectively. Find the resultant electric intensity at D.
[8.691x104 N/C]
ABCD is a square of side 2 m. Point charges of 2x109 C, +3x109 C, +4x109 C and +5x109 C are
placed at corners A, B, C and D respectively. Calculate the magnitude of resultant electric intensity at O,
the point of intersection of diagonals.
[25.45 N/C]
An electric dipole with a dipole moment 5x109 C-m is placed in a uniform electric field of magnitude
4x104 N/C at (i) 30º and (ii) 45º. Calculate the magnitude of the torque acting on the dipole in each case.
[10x105 N-m, 1.414x104 N-m]
o
An electric dipole consisting of an electron and a proton separated by a distance of 5 A is located in an
electric field of intensity 4x105 N/C at an angle of 30º with the field. Calculate the dipole moment and the
torque acting on it. Given: Charge on electron=1.602x1019 C.
[8.01x1029 C-m, 1.602x1023 N-m]
An electric dipole consists of two opposite point charges each of 2C separated by 1 cm. The dipole is
placed in an external uniform electric field of 3x10 5 N/C. Find (i) maximum torque exerted by the field on
the dipole and (ii) the work done in rotating the dipole through 180º starting from the position =0º.
[6x103 N-m, 3600 J]
12
12
Two point charges of 0.1x10 C and 0.1x10 C are separated by a distance of 108 m. Determine the
electric field (i) at an axial point distant 0.1 m from the mid point of the dipole and (ii) at an equatrial
point distant 0.15 m from the dipole.
[1.8x108 N/C, 2.67x109 N/C]
A cubical surface encloses a system of 3 charges +4nC, 1 nC and +2nC. What is the electric flux over
the cubical surface?
[564.7 N-m2/C]
A cube of side 1 m enclose a point charge of +2nC at its centre.
24
SARVAJNYA
840942174
(i)
Find the total flux emanating from the cubical surface.
(ii)
What is the flux through each surface of the cube?
[225.9 N-m2/C, 37.65 N-m2/C]
8
A spherical shell of radius 0.1 m is charged with 10 C of electricity. Find the potential at (i) the surface
of the shell and (ii) at a distance of 0.5 m from the surface of the shell.
[900 V, 150 V]
The following figure shows variation of electric potential with distance due to an isolated charge. From
the figure calculate the electric intensity at (i) x=1 m, (ii) x=3 m and (iii) x = 5 m.[2 V/m, 1.33 V/m, +2
V/m]
Electric potential due to a point charge varies according to V = 3x 2+2. Calculate electric intensity at (i) x
= 0.1 m and (ii) x=2 m from the charge.
[0.6 V/m, 12 V/m]
Two point charges of +2 C and 8 C are placed 2 m apart in air. Find the positions along the line
joining the two charges at which resultant electric potential is zero.
[0.4 m, 0.67 m]
Two point charges of +1 C and 9C are placed 1 m apart in air. Find the positions along the line
joining the two charges at which the resultant electric potential is zero.
[0.1 m, 0.125 m]
ABC is a right angled triangle with AB = 0.3m, BC = 0.6 m and  ABC = 90º. A metal sphere of radius 1
cm is charged to a potential of 5x104 V and is placed at B. What is the work done in moving a charge of 1
C from C to A?
[832.5 J]
ABCD is a square of side 1 m. Point charges of +1x1010 C, 2x1010 C and +5x1010 C are placed at
corners A, C and D respectively. Find the resultant electric potential at B.
[2.2815 J/C]
ABCD is a square of side 2 m. Point charges of +1 nC, +2 nC, 3 nC and +4 nC are placed at corners
A, B, C and respectively. Find the resultant electric potential at the point of intersection of the diagonals.
[36 J/C]
8
8
8
ABCD is a square of side 2 m. Point charges of 2x10 C, 4 x10 C and 5x10 C are placed at corners
A, B, C and D respectively. Calculate the work done in moving a point charge of 2C from B to be point
of intersection of diagonals.
[7.664x104 J]
n identical drops, each charged to a potential of V, combine to form a bigger drop. What is the potential
of the bigger drop?
[n2/3 V]
Three point charges are arranged at the three vertices of an equilateral triangle of side 0.1 m as shown.
Calculate the electrostatic potential energy of the system. Given: q=10 9 C.
[9x107 J]
Three point charges +2 nC, +4nC and +8nC are placed at the corners of an equilateral triangle of side
0.2 m. What is the potential energy of the system? What is the work done to remove 8 nC from the
triangle to infinity?
[2.52x105 J, 3.6x106 J]
Two positive charges 8C and 4C are 0.1 m apart in free space. Calculate the work done in bringing
them 0.02 m closer.
[0.72 J]
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
Electric potential energy of electric dipole in uniform electric field or work done in rotating and
electric dipole in uniform electric field:


Consider an electric dipole of dipole moment p placed in a uniform electric field E . Let the dipole
moment of the dipole make an angle  to the direction of electric field. Then torque acting on the dipole is given
by;
 = pE sin
If the dipole is rotated through a very small angle d against this torque, then small amount of work
done is.
dW = d = pE sin d
Total work done in rotating the dipole from 1 to 2 is
W
2

pE sin  d
1
Since p and E are constant, they can be taken out of the integral.
 W = pE
2

1
sin  d = pE  cos 2 = pE (cos 2 - cos2)

1
Special cases:
(i) When =0º; U = pE cos 0º =  pE
In this position, the dipole has minimum potential energy
and hence it is in stable equilibrium. The dipole has more potential energy in all other positions.
(ii) When =90º;
U = pE cos 90º = 0
In this position, the potential energy of the dipole is zero.
(iii)
When =180ºl U=pE cos 180º = +pE
In this position, the dipole has maximum potential energy and is in unstable equilibrium.
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