PHYSICS-MANIA
Chapter -10(Section-2)
Electrostatics
Que 1: Define frictional electricity. Give example of frictional electricity.
Ans: The electricity developed on object, when they are rubbed with each other, is called frictional electricity.
Examples of frictional electricity:
1. Clouds rub against each other and produce lightening.
2. In winter, if a rubber comb is run fast through dry hair, the comb attracts small pieces of paper
Que 2: state the principle of conservation of charges.
Ans :
1. Electrical charge can neither be created nor be destroyed
2. But it is transferred from one part of a system (substance) to other part of the system.
3. so that total charge of an isolated system remains constant
e.g.
I.
When an electron of charge -e and positron of charge +e are brought closed to each other, they annihilated
each other to produce charge less  photons.
II.
In radioactive decay 92U238 is transformed to 90Th234 By emitting an ( 2He4 ) i.e.  - particle
238
→ 90Th234 + 2He4
92U
Thus net charge before decay (92 e) is equal to the net charge after decay (90e +2e) .
Thus charges are remains conserved.
Que 3: state coulomb’s law and explain it in vector form.
Ans: The electrostatic force of attraction or repulsion between two point electric charges at rest is directly
proportion to the product of the magnitude of two charges and inversely proportional to the square of distance
between them. The force acts along the line joining the charges.
Let q1 and q2 are two point electric charges separated by distance r and F is the magnitude of the
electrostatic force of attraction or repulsion between them.
According to coulomb’s law we have
𝐪 𝐪
𝐅 ∝ 𝟏𝟐 𝟐
𝐫
1
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
𝐅=𝐊
𝐪𝟏 𝐪𝟐
𝐫𝟐
…….(i)
Where K is a constant of proportionality which depends upon the nature of medium in which charges are
situated.
The dimensions of K are [M¹ L³ T ⁻⁴ I⁻²]
If the charge is placed in vacuum, then it is expressed as
𝐊=
𝟏
𝟒𝛑𝛜𝐨
Where o is called the permittivity of free space hence in vacuum, coulomb’s law expressed as
𝐅=
𝟏 𝐪𝟏 𝐪𝟐
𝟒𝛑𝛜𝐨 𝐫 𝟐
……(ii)
If q1 and q2 are measured in coulomb, r is measured in metre and F is in newton then
𝟏
= 𝟗 × 𝟏𝟎𝟗
𝟒𝛑𝛜𝐨
𝐍𝐦𝟐
𝐂𝟐
Or o = 8.85 x 10⁻¹² C²/Nm²
If the two point charges q1 and q2 are situated in medium of permittivity  , then Coulomb’s law is expressed
as
𝐅=
𝟏 𝐪𝟏 𝐪𝟐
𝟒𝛑𝛜 𝐫 𝟐
Where  =o k
𝐅=
or k= /o
𝟏 𝐪𝟏 𝐪𝟐
𝟒𝛑𝛜𝐨𝐤 𝐫 𝟐
k is known as dielectric constant of the medium or relative permittivity of the medium with respect to
vacuum.
Que.4 express coulombs law in vector form
2
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
Ans: The force exerted by charge q1 on a like charge
𝐅⃗𝟐𝟏 =
q2 can be represented in vector form as shown in fig.
𝟏 𝐪𝟏 𝐪𝟐 ⃗⃗
𝟒𝛑𝛜𝐨 𝐫𝟐 𝐫𝟏𝟐
Where r12 is unit vector direction from q1 to q2 similarly the force exerted by charge q2 on charge q1 is
𝐅⃗𝟏𝟐 =
𝟏 𝐪𝟏 𝐪𝟐 ⃗⃗
𝟒𝛑𝛜𝐨 𝐫𝟐 𝐫𝟐𝟏
Where r21 is unit vector direction from q2 to q1
As r 12 = - r21
F 12 = - F21
i.e. the force are oppositely direction and have equal magnitude.
Que 4: Define one coulomb
Ans: Force between two charges q1 and q2 separated by distance r in vacuum is given by
𝐅=
𝟏 𝐪𝟏 𝐪𝟐
𝟒𝛑𝛜 𝐫 𝟐
When q1 = q2 = q , r = 1m and F = 9 x 10⁹
Then q² = 1 or q =  1 coulomb
Thus one coulomb is defined as the magnitude of that charge which when placed in vaccum or air at a distance
one meter form an identical charge repel with a force 9 x 10⁹ N
Que 5: Define Dielectric constant (permittivity of medium with respect to vacuum).
Ans : The ratio of permittivity of medium to the permittivity of the vacuum is called dielectric constant.
𝐤=
𝛜
𝛜𝐨
3
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
𝐪𝟏 𝐪𝟐
𝐤=
𝛜
𝟒𝛑𝛜𝐨𝐤𝐫
=
𝐪𝟏 𝐪𝟐
𝛜𝐨
𝟐
𝟒𝛑𝛜𝐫𝟐
𝐤=
𝐅𝐯𝐚𝐜𝐜𝐮𝐦
𝐅𝐦𝐞𝐝𝐢𝐮𝐦
Dielectric constant of medium can also be defined as the ratio of force between two charges at a certain
distance apart in vacuum to the force between the same two charges separated by the same distance in the
medium
Dielectric constant is the ratio of two similar quantities; hence it has no unit as well as dimension. k has a value
unity (k=1)air , infinity (k= )for metal and for other media k>1
Remark:
1. Charge on electron = 1.602x10⁻¹⁹C
2. Number of electrons carrying a total charge of 1 coulomb is 6.24x10¹⁸ electrons.
3. In CGS system, the unit of charge is electrostatic unit of charge (e.s.u.). It is also called stat coulomb (stat C)
1 C = 3x 10⁹ stat C
4. In CGS system, the unit of charge is electromagnetic unit of charge (e.m.u.)
𝟏
𝟏𝐂 = 𝟏𝟎 𝐞. 𝐦. 𝐮. 𝐨𝐟 𝐜𝐡𝐚𝐫𝐠𝐞
Que 6: What is the principle of superposition of force?
Ans: When a number of charges are interacting, the total force on a given charge is the vector sum of individuals
forces exerted on the given charge by all other charges.
Consider a charge q1, q2, q3,…. qn situated in free space in a discrete manner .if 𝐅⃗𝟏𝟐 , 𝐅⃗𝟏𝟑 , 𝐅⃗𝟏𝟒 , … … . 𝐅⃗𝟏𝐧 be
the forces exerted on charge q1 due to charges q2, q3,…. qn respectively.
4
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
Then according to principle of superposition of force, the resultant force𝐅⃗𝟏 experienced by the charge q1
is given by,
𝐅⃗𝟏 = 𝐅⃗𝟏𝟐 + 𝐅⃗𝟏𝟑 + 𝐅⃗𝟏𝟒 + ⋯ … + 𝐅⃗𝟏𝐧 ..(i)
According to coulomb’s law force acting on charge q1 by other charges is given by,
𝐅⃗𝟏𝟐 =
𝟏 𝐪𝟏 𝐪𝟐 ⃗⃗
𝐫
𝟒𝛑𝛜𝐨 𝐫𝟏𝟐 𝟐 𝟐𝟏
Where
𝐅⃗𝟏𝟐 = force exerted on charge q1 by charge q2.
𝐫⃗𝟐𝟏 = unit vector in the direction from q1 to q2.
𝐫𝟏𝟐 = distance between the charges q1 and q2.
Similarly,
𝐅⃗𝟏𝟑 =
𝟏 𝐪𝟏 𝐪𝟐 ⃗⃗
𝐫
𝟒𝛑𝛜𝐨 𝐫𝟏𝟑 𝟐 𝟑𝟏
𝐅⃗𝟏𝟒 =
𝟏 𝐪𝟏 𝐪𝟐 ⃗⃗
𝐫
𝟒𝛑𝛜𝐨 𝐫𝟏𝟒 𝟐 𝟒𝟏
There for the net force acting on the charge q1 due to all other charges is given by.
5
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
𝐅⃗𝟏 = 𝐅⃗𝟏𝟐 + 𝐅⃗𝟏𝟑 + 𝐅⃗𝟏𝟒
𝐅⃗𝟏 =
𝟏 𝐪𝟏 𝐪𝟐
𝐪𝟏 𝐪𝟐
𝐪𝟏 𝐪𝟐
(
𝐫⃗𝟐𝟏 +
𝐫⃗𝟑𝟏 +
𝐫⃗ )
𝟐
𝟐
𝟒𝛑𝛜𝐨 𝐫𝟏𝟐
𝐫𝟏𝟑
𝐫𝟏𝟒 𝟐 𝟒𝟏
This is an expression for resultant force acting on single charge
Que 7: Explain continuous charge distribution
Ans : In most of the practical situation, charge on a charged body is so large as compare to the magnitude of charge
on an electron or proton , that the quantization of charge may be ignored . So we can assume that the charge on
the charged body of reasonable size has continuous distribution.
Uniform charge distribution:When the charge is not accumulated in some part but is spread uniformly then it is called uniform charge
distribution.
Nonuniform charge distribution:If the shape of conductor is irregular, then charge will be distribution nonuniformly over its surface, more charge
is accumulated at the curves and shape points such distribution of charge is called non uniform charge distribution.
Que 8: explain the concept of charge density:
Ans: The dimensions of geometry of conductor decide the charge density.
Linear charge density (λ):6
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
Charge per unit length of conductor is called liner charge density.
If q is the charge distributed uniformly over a thin conductor of length L, then its liner charge density λ is given
by
𝛌=
𝐪
𝐋
S.I unit of liner charge density is coulomb/metre (C/m) Dimension of λ is [M° L⁻¹ T¹ I¹]
𝐪
Remark : 1. If the charge q is distributed uniformly over a circular ring of radius r, then 𝛌 = 𝟐𝛑𝐫
2. If the charge is not distributed uniformly over a thin conductor then 𝝀 =
𝝀=
𝐂𝐡𝐚𝐫𝐠𝐞 𝐨𝐧 𝐬𝐦𝐚𝐥𝐥 𝐞𝐥𝐞𝐦𝐞𝐧𝐭
𝐋𝐞𝐧𝐠𝐭𝐡 𝐨𝐟 𝐞𝐥𝐞𝐦𝐞𝐧𝐭
𝒅𝒒
𝒅𝒍
Surface charge density (𝝈) :When the charge is distributed uniformly over an area in space, then charge per unit area is called surface
charge density.
If q is the charge distributed uniformly over a area A , then surface charge density.
𝛔=
𝐪
𝐀
SI unit of surface charge density is coulomb/square metre(C/m²).
Dimension of  is [M° L⁻² T⁻¹I¹]
If q is the charge distributed uniformly over a area of a spherical conductor of radius r, then
𝛔=
𝐪
𝟒𝛑𝐫 𝟐
If the metallic conductor is irregular shape then
𝛔=
𝐂𝐡𝐚𝐫𝐠𝐞 𝐨𝐧 𝐬𝐦𝐚𝐥𝐥 𝐞𝐥𝐞𝐦𝐞𝐧𝐭
𝐝𝐪
=
𝐬𝐮𝐫𝐟𝐚𝐜𝐞 𝐚𝐫𝐞𝐚 𝐨𝐟 𝐞𝐥𝐞𝐦𝐞𝐧𝐭
𝐝𝐀
Volume charge density ():-
7
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
When the charge is distributed uniformly over a volume in space, then charge per unit volume is called
volume charge density.
If q is the charge distributed uniformly over a volume V, then volume charge density.
𝛒=
𝐪
𝐯
SI unit of volume charge density is coulomb/cubic metre (C/m³)
Dimension of  is [M° L⁻³ T¹I¹]
If the charge is not distributed uniformly
𝛒=
𝐂𝐡𝐚𝐫𝐠𝐞 𝐨𝐧 𝐬𝐦𝐚𝐥𝐥 𝐞𝐥𝐞𝐦𝐞𝐧𝐭
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐞𝐥𝐞𝐦𝐞𝐧𝐭
𝐝𝐪
𝛒 = 𝐝𝐯
⃗⃗).
Que 9:Define electric field and electric field intensity(𝐄
Ans : Electric field :The space surrounding an electric charge, in which any other electric charge experiences a force is called
electric field of the charge.
⃗⃗) :Electric field intensity (𝐄
The electric field intensity due to point charge at any point in the electric field is defined as the electrostatic
force exerted by it on a unit positive charge placed at that point.
⃗⃗ =
𝐄
𝐅⃗
𝐪𝐨
The electric intensity at point is a vector
quantity. The direction of field intensity is same as that of
electrostatic force acting on the positive test charge.
S.I. unit of electric intensity is newton /coulomb (N/C) or volt /metre (V/m)
⃗⃗ is [M¹ L¹ T⁻³ I⁻¹]
Dimension of 𝐄
⃗⃗ is force per unit charge then force on a charge q is given by 𝐅⃗ = q𝐄
⃗⃗
Note : If E
8
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
Que 10: obtain an expression for electric field intensity due to point charge.
Ans :
Consider a point charge at a point o in a medium of dielectric constant K as shown in fig. consider the point charge
qo is situated at point A , at a distance r from q. force exerted by charge q on qo is
𝐅⃗ =
𝟏 𝐪𝐪𝐨
⃗⃗⃗⃗
𝐫
𝟒𝛑𝛜𝐨 𝐫 𝟐 𝐨
Where 𝐫⃗𝐨 is unit vector in the direction of force i.e. OA
𝟏 𝐪𝐪𝐨 ⃗⃗⃗⃗⃗⃗
𝐫
𝐅⃗
𝟒𝛑𝛜
𝐤
𝐫𝟐 𝐨
𝐨
⃗𝐄⃗ =
=
𝐪𝐨
⃗⃗ =
𝐄
𝐪𝐨
𝟏 𝐪 ⃗⃗⃗⃗⃗⃗
𝐫
𝟒𝛑𝛜𝐨𝐤 𝐫𝟐 𝐨
The magnitude of electric field intensity is given by for air or vacuum medium k =1 then
𝐄=
𝟏 𝐪
𝟒𝛑𝛜𝐨 𝐫𝟐
Physical significance of equation
i.
E is large if point is near to charge and E goes on decreasing as the distance from the charge increases.
ii.
The electric intensity due to point charge is not same at all the point in the field.
Que 11 : Define the term
i.
Uniform electric field.
ii.
Non uniform electric field.
9
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
iii.
Radial electric field.
Ans :i. Uniform electric field : when the magnitude as well as direction of electric field intensity E is same at all
the point in the electric field .it is represented by the set of equidistant ,parallel straight line (fig a )
fig a
fig b
fig c
ii. Non uniform electric field: when the magnitude and direction of electric field intensity E or both are different
at different point in the electric field then it is called non uniform electric field.(fig b)
iii. Radial electric field: when electric field intensity E at any point in the electric field is directed toward or
away from the same fixed point then the field is called Radial electric field. (fig c)
Que 12: what are electric lines of force? State the properties of line of force.
Ans: Electric line of force defined as a curve such that the tangent at any point to this curve gives the direction of
non uniform electric field at that point.
OR
The path along which the unit positive charge moves is called line of force.
Properties of electric line of force:
i.
The line of originated from a positive charged conductor and end on negative charged conductor.
10
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
ii.
The line of force leaves or ends on conductor normally.
iii.
The line of force neither intersects nor meets each other.
iv.
The line of force do not pass through conductor i.e. electric field inside the conductor is always zero but it
passes through insulator.
v.
The line of force have tendency to contract along their lengths .this property explain the attraction
between unlike charges.
vi.
The lines of force have a tendency to exert a lateral pressure on each other. This property explains the
repulsion between like charges.
11
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
vii.
Magnitude of electric field intensity is proportional to the number of line of force per unit area of the
surface held perpendicular to the field.
viii.
Electric lines of force are crowded in region where electric intensity is large and electric lines of force are
widely separately from each other in a region where electric intensity is small.
ix.
In uniform electric field electric line of force are parallel to each other and equally spaced.
Que 13: Define electric dipole and electric dipole moment.
Ans: Two equal and opposite electric point charge separated by a finite distance from an electric dipole.
The line passing through the center of two charges forming electric dipole is called axis of dipole.
A straight line drawn perpendicular to the axis and passing through center ‘O’of electric dipole is called
equatorial of dipole.
Electric dipole moment:
The product of magnitude of one charge and the distance between the two charge is called dipole moment
(p)of the electric dipole.
p =q x 2l
⃗⃗) and directed from negative charge to positive charge.
Electric dipole moment is vector (𝐩
The S.I. unit of dipole moment is coulomb metre (C.m.)
12
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
In vector notation
⃗⃗ = 𝒒 × 𝟐𝒍⃗
𝐩
Que 14: Obtain an expression for the torque acting on the dipole in uniform electric field and hence define dipole
moment of an electric dipole.
Ans:
1. Consider an electric dipole placed in uniform electric field E.
2.  Be the angle between axis of the dipole and direction of electric intensity.
3. The force acting on charge +q at B is +qE in the direction of E and the force acting on charge –q at A is –qE
in opposite direction of the field.
4. These two equal and opposite and parallel non collinear forces separated by perpendicular distance (BP)
acting on the electric dipole form a couple.
5. Expression for torque:
Magnitude of moment of couple (Torque) = {Magnitude of one of the force forming
{perpendicular distance between two parallel forces} .
the couple} x
 = qe x BP
 = qe x 2l sin
But p = q x 2l
 = p E sin
13
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
In vector notation  = p xE.
6. If  =90°, sin = 1 then  = pE
If E =1 then  = p
Thus dipole moment of an electric dipole is equal to the moment of couple acting on the dipole which held with its
axis perpendicular to a uniform electric field of unit intensity.
Que 15: Derive an expression for potential energy of a system of charges
Ans: - The electrostatic potential energy of a system of point charges is defined as the work required to assemble
the system of charges by bringing them from infinity to their present locations.
Consider two charges q1 and q2 initially at infinity and determine the work done by external agency to bring the
charges to the given location A and B
Let r1 and r2 be the position vector of point A and point B .
Suppose first the charge q1 is brought from infinity to the point A . There is no external field, against which work
needs to be done, so that work done to bring charge q1 from infinity to point A is zero
Now charge q2 move from infinity to point B against the electric field due to charge q1 it work done is given by,
W = ( Electric potential at B due to charge q1 ) x q2
14
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
𝐖= (
𝟏 𝐪𝟏
).𝐪
𝟒𝛑𝛜𝐨 𝐀𝐁 𝟐
⃗⃗𝟏 − 𝒓
⃗⃗𝟐 | = 𝒓𝟏𝟐
since 𝐀𝐁 = |𝒓
𝐖=
𝟏 𝐪𝟏 𝐪𝟐
𝟒𝛑𝛜𝐨 𝐫𝟏𝟐
This work done in bringing the two charges to their respective location is stored as the potential energy of the
configuration of two charges.
∴𝐔=
𝟏 𝐪𝟏 𝐪𝟐
𝟒𝛑𝛜𝐨 𝐫𝟏𝟐
now for three charges :-Let’s consider system of three charges q1 , q2 and q3 having position r1, r2 and r3 relative to
same origin.
To bring q1 first from infinity to its location 𝑟⃗1 no work is required. When q2 bring from infinity to its location 𝑟⃗2
. then work done is given by,
𝐖𝟏𝟐 =
𝟏 𝐪𝟏 𝐪𝟐
𝟒𝛑𝛜𝐨 𝐫𝟏𝟐
⃗⃗𝟏 − 𝒓
⃗⃗𝟐 |
Where 𝑟12 = |𝒓
⃗⃗𝟑 the work done say
Now move the charge q3 from infinity to its location 𝒓
𝐖𝟏𝟑 for moving charge q3 in the
electric field of q1, similarly work done 𝐖𝟐𝟑 has to be done for moving charge q3 in electric field of q2
𝐖𝟏𝟑 =
𝟏 𝐪𝟏𝐪𝟑
|𝒓
⃗⃗
⃗⃗ |
𝟒𝛑𝛜𝐨 𝐫𝟏𝟑 Where 𝒓𝟏𝟑 = 𝟏 − 𝒓𝟑
𝐖𝟐𝟑 =
𝟏 𝐪𝟐𝐪𝟑
|𝒓
⃗⃗
⃗⃗ |
𝟒𝛑𝛜𝐨 𝐫𝟐𝟑 Where 𝒓𝟐𝟑 = 𝟐 − 𝒓𝟑
The total work done in assembling the charge at given location is
𝐔 = 𝐖𝟏𝟐 + 𝐖𝟏𝟑 + 𝐖𝟐𝟑
𝐔=
𝟏 𝐪𝟏 𝐪𝟐 𝐪 𝟏 𝐪𝟑 𝐪𝟐 𝐪𝟑
+ 𝐫 + 𝐫 )
(
𝟒𝛑𝛜𝐨 𝐫𝟏𝟐
𝟏𝟑
𝟐𝟑
15
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
Que 16: What do you understand by electric potential difference?.
Ans: Let the work done in moving the charge qo from point A to point B be WBA. The energy stored is grater at
point B than at point A .it observed that WBA is proportional to qo
Electric potential energy difference between point A and B = WBA
𝐖𝐁𝐀
= 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝒒𝒐
This constant is defined as electric potential difference VAB between the point A and B
𝐕𝐀𝐁 =
𝐖𝐁𝐀
𝒒𝒐
Let VB be the electric potential at B,
VA be the electric potential at A
𝐕𝐀 − 𝐕𝐁 =
𝐖𝐁𝐀
𝐪𝐨
Que 17: Define electric potential due to a point charge
Ans : the electric potential at point at distance r in electric field is
𝐰𝐨𝐫𝐤 𝐝𝐨𝐧𝐞 𝐭𝐨 𝐛𝐫𝐢𝐧𝐠 𝐭𝐡𝐞 𝐜𝐡𝐚𝐫𝐠𝐞 𝐪𝐨
𝐟𝐫𝐨𝐦 𝐚 𝐫𝐞𝐟𝐞𝐫𝐞𝐧𝐜𝐞 𝐩𝐨𝐢𝐧𝐭 𝐭𝐨 𝐩𝐨𝐢𝐧𝐭 𝐫
𝐕(𝐫) − 𝐕𝐫𝐞𝐟 =
𝐪𝐨
Electric potential at a point in electric field is defined as the amount of work done to bring a unit positive charge
from infinity to that point against the direction of electric intensity
It is scalar quantity and is denoted by V.
Let W is work done to bring positive test charge from infinity to point in electric field of charge.
𝐕=
𝐖
𝐪𝐨
S.I unit of electric potential is joule / coulomb (J/C)
16
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
It is named as volt.
Dimension of electric potential is [M¹ L² T⁻³ I⁻¹]
𝟏𝐯𝐨𝐥𝐭 =
𝟏 𝐣𝐨𝐮𝐥𝐞
𝟏 𝐜𝐨𝐮𝐥𝐨𝐦𝐛
Que.18: Define one volt
Ans: One volt:: Electric potential at a point in electric field is said to be one volt if 1 joule amount of work done to bring a 1
coulomb charge from infinity to that point.
𝟏𝐯𝐨𝐥𝐭 =
𝟏 𝐣𝐨𝐮𝐥𝐞
𝟏 𝐜𝐨𝐮𝐥𝐨𝐦𝐛
Que 19: what is electron volt?
Ans : Electron volt (eV): An electron volt is defined as the amount of energy gained by an electron when it is
displaced from one point to another point whose potential is higher by one volt.
Work must be done to displace a charge from a point of lower potential to the point of higher potential .This
work increases, the potential energy of charge.
Potential energy gained by charge
= work done
= Charge x potential difference
W = q V
“The electron volt is the kinetic energy gained by an electron when it is accelerated through a potential
difference of one volt.”
1 electron volt = 1.602 x 10⁻¹⁹coulomb x 1 volt
1 eV = 1.602 x 10⁻¹⁹ J
Where -1.602 x 10⁻¹⁹ C is charge on electron.
17
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
Que 18: obtain an expression for the electric potential due to point charge.
1. Consider a point charge +q is situated at point O in a dielectric medium of constant k.
2. We have to find out electric potential at point A is situated at distance r from O.
3. consider test charge qo at infinity from charge q when the test charge is at M it experience the force F which is
given by
𝐅=
𝟏 𝐪𝐪𝐨
𝐚𝐥𝐨𝐧𝐠 𝐎𝐀
𝟒𝛑𝛜𝐨 𝐤 𝒙𝟐
4. If the test charge qo is displaced from point M to N through the infinitesimal distance dx against the force then
work done is,
dw = - F dx
-ve sing indicate the force and displacement have opposite directions.
5. Consider the test charge moved from infinity to point A. The total work done is given by,
𝐫
𝐫
𝐖 = ∫ 𝐝𝐰 = ∫ −𝐅𝐝𝐱
𝐱=∞
𝐫
𝐖=
∫−
𝐱=∞
𝐱=∞
𝟏 𝐪𝐪𝐨
𝐝𝐱
𝟒𝛑𝛜𝐨 𝐤 𝐱 𝟐
𝐫
𝐪𝐪𝐨
𝐝𝐱
𝐖 = −(
) ∫ 𝟐
𝟒𝛑𝛜𝐨 𝐤
𝐱
𝐱=∞
𝐫
𝐪𝐪𝐨
𝐖 = −(
) ∫ 𝐱 −𝟐 𝐝𝐱
𝟒𝛑𝛜𝐨 𝐤
𝐱=∞
18
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
𝐪𝐪𝐨
𝐖 = −(
𝐱 −𝟐+𝟏
) [ −𝟐+𝟏 ]
𝟒𝛑𝛜 𝐤
𝐨
𝐫
𝐱=∞
𝐪𝐪𝐨
𝐫
𝐖= (
) [𝐱 −𝟏 ]𝐱=∞
𝟒𝛑𝛜𝐨 𝐤
𝐪𝐪𝐨
𝟏𝐫
𝐖= (
)[ ]
𝟒𝛑𝛜𝐨 𝐤 𝐱 𝐱=∞
𝐖=
𝟏
𝐪𝐪𝐨
𝟒𝛑𝛜𝐨 𝐤 𝐫
By the definition of electric potential
𝐕=
𝐖
𝟏
𝐪
=
𝐪𝐨 𝟒𝛑𝛜𝐨 𝐤 𝐫
For air or vacuum k=1
𝐕=
𝐖
𝟏
𝐪
=
𝐪𝐨 𝟒𝛑𝛜𝐨 𝐫
This is an expression for electric potential
Remark: Physical significance of equation:It clear that potential is inversely proportion to its distance. The potential due to positive charge is positive while
potential due to negative charge is negative
Que 19: Define electric potential gradient. Obtain the relation between electric field intensity and potential
gradient at a point in the electric field.
Ans: consider a point charge +q is situated at o in a medium of dielectric constant k as shown in fig.
19
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
Let A and B are the two point separated by the infinitesimal distance dx in non uniform electric field of
charge +q. Let E be the electric field near the point B. if small test charge qo is placed at B it experiences a repulsive
force which is
F =E qo
If test charge move from B to A through small distance against electric field. The work done on the charge qo is
dw =-Fdx = -E qodx
𝐝𝐰
= −𝐄𝐝𝒙
𝐪𝐨
The work done per unit charge displaced is the increase in potential dv.
dv = -Edx
𝐝𝐯
𝐝𝐱
= −𝐄
The quantity dv/dx represents the rate of change of potential. It is called potential gradient.
The electric field intensity at any point in the electric field is equal to negative rate of change of potential with
respect to distance at that point measured in the direction of electric intensity.
SI unit of potential gradient is volt / meter (V/m)
Que.20 : Derive an expression for electric potential due to electric dipole.
Ans:Consider an electric dipole AB Having -q charge at A and +q Charge at B. Let O be the
center of dipole and P be any point at a distance r from the center of dipole, where
electric potential due to an electric dipole is to be determined.
Let
∠𝑷𝑶𝑩 = 𝜽 𝒂𝒏𝒅 𝒓 ≫ 𝒍
Let AA’ be the perpendicular from A to PO and
BB’ be the perpendicular from B to PO
As 𝒍 is very small as compare to r
𝐀𝐏 ≈ 𝐀𝐏 ′ = 𝐎𝐏 + 𝐎𝐀′ = 𝐎𝐏 + 𝐀𝐎 𝐜𝐨𝐬 𝛉
20
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
= 𝐫 + 𝒍 𝐜𝐨𝐬 𝛉
Similarly
𝐁𝐏 ≈ 𝐁𝐏 ′ = 𝐎𝐏 − 𝐎𝐁′ = 𝐎𝐏 − 𝐎𝐁 𝐜𝐨𝐬 𝛉
= 𝐫 − 𝒍 𝐜𝐨𝐬 𝛉
The potential at point P due to charge -q is
𝐕𝟏 = −
𝐕𝟏 = −
𝟏
𝐪
𝟒𝛑𝛜𝐨 𝐀𝐏
𝟏
𝐪
×
𝟒𝛑𝛜𝐨 (𝐫 + 𝒍 𝐜𝐨𝐬 𝛉)
The potential at point P due to charge +q is
𝐕𝟐 =
𝐕𝟐 =
𝟏
𝐪
𝟒𝛑𝛜𝐨 𝐁𝐏
𝟏
𝐪
×
𝟒𝛑𝛜𝐨 (𝐫 − 𝒍 𝐜𝐨𝐬 𝛉)
The net potential at P due to dipole is
∴ 𝐕 = 𝐕𝟏 + 𝐕𝟐
∴𝐕=−
𝟏
𝐪
𝟏
𝐪
.
+
.
𝟒𝛑𝛜𝐨 (𝐫 + 𝒍 𝐜𝐨𝐬 𝛉) 𝟒𝛑𝛜𝐨 (𝐫 − 𝒍 𝐜𝐨𝐬 𝛉)
∴𝐕=
𝐪
𝟏
𝟏
(
−
)
𝟒𝛑𝛜𝐨 (𝐫 − 𝒍 𝐜𝐨𝐬 𝛉) (𝐫 + 𝒍 𝐜𝐨𝐬 𝛉)
∴𝐕=
𝐪 𝐫 + 𝒍 𝐜𝐨𝐬 𝜽 − 𝐫 + 𝒍 𝐜𝐨𝐬 𝜽
[
]
𝟒𝛑𝛜𝐨
𝐫 𝟐 − 𝒍𝟐 𝐜𝐨𝐬 𝟐 𝛉
∴𝐕=
𝐪
𝟐𝒍 𝐜𝐨𝐬 𝜽
[ 𝟐
]
𝟒𝛑𝛜𝐨 𝐫 − 𝒍𝟐 𝐜𝐨𝐬 𝟐 𝛉
But 𝒒 × 𝟐𝒍 = 𝒑
21
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
∴𝐕=
𝐩 𝐜𝐨𝐬 𝜽
𝟒𝛑𝛜𝐨 (𝐫 𝟐 − 𝒍𝟐 𝐜𝐨𝐬 𝟐 𝛉)
∴ 𝐫 ≫ 𝒍, 𝐫𝟐 − 𝒍𝟐 𝐜𝐨𝐬𝟐 𝛉 = 𝐫 𝟐
∴𝐕=
𝟏 𝐩 𝐜𝐨𝐬 𝜽
𝟒𝛑𝛜𝐨 𝐫 𝟐
Case I: When point P lies on the axial line nearer the +q .
∴𝐕=
𝜽 = 𝟎° 𝒂𝒏𝒅 𝐜𝐨𝐬 𝜽 = 𝟏
𝟏 𝐩
𝟒𝛑𝛜𝐨 𝐫 𝟐
When point p lies on the axial line nearer -q
∴𝐕=−
𝟏
𝐩
𝟒𝛑𝛜𝐨 𝐫 𝟐
Case II : When point P lies on the equatorial line
i.e. 𝜽 = 𝟗𝟎° 𝒐𝒓 𝟐𝟕𝟎° 𝒂𝒏𝒅 𝐜𝐨𝐬 𝜽 = 𝟎
Then V = 0
Que.21 : Derive an expression for electric potential due to a system of charges.
Ans : Consider a system of charge q1 ,q2 , q3 ,…..qn at a distance r1,r2,r3,…….rn from point P.
22
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
The potential V1 at P due to charge q1 is,
∴ 𝐕𝟏 =
𝟏
𝐪𝟏
𝟒𝛑𝛜𝐨 𝐫𝟏
similarly the potential V2 , V3 …..Vn at point P due to charges q2 , q3 ,…..qn are given by
𝟏
𝐕𝟐 = 𝟒𝛑𝛜
𝐪𝟐
𝐨
𝐫𝟐
𝟏
, 𝐕𝟑 = 𝟒𝛑𝛜
𝐪𝟑
𝐨
𝟏
… … 𝐕𝐧 = 𝟒𝛑𝛜
𝐫
𝟑
𝐪𝐧
𝐨
𝐫𝐧
Net potential is algebraic sum of the potential due to the individual charges
V = V1 + V2 + V3 +…..+Vn
𝐕=
𝟏 𝐪𝟏
𝟏 𝐪𝟐
𝟏 𝐪𝟑
𝟏 𝐪𝐧
+
+
+⋯+
𝟒𝛑𝛜𝐨 𝐫𝟏 𝟒𝛑𝛜𝐨 𝐫𝟐 𝟒𝛑𝛜𝐨 𝐫𝟑
𝟒𝛑𝛜𝐨 𝐫𝐧
𝐕=
𝐕=
𝟏
𝟒𝛑𝛜𝐨
𝟏
𝐪𝟏 𝐪𝟐 𝐪𝟑
𝐪𝐧
( +
+
+ ⋯+
)
𝟒𝛑𝛜𝐨 𝐫𝟏
𝐫𝟐
𝐫𝟑
𝐫𝐧
𝐪
∑𝐧𝐢=𝐫 𝐢
𝐫𝐢
Que.22 : Explain the concept of equipotential surface.
Ans: An equipotential surface is a surface with constant value of potential at all point on the surface.
For a single charge q , the potential is given by equation. 𝐕 =
𝟏
𝐪
𝟒𝛑𝛜𝐨 𝐫
This show that if r is constant then V is also
constant
Shape of equipotential surface:I) Spherical equipotential surface
23
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
If a single charge is situated at center of sphere then it electric field at every point of spherical is normal to
equipotential surface. Therefore equipotential surface is in the form of concentric sphere of different radius.
II) Plane equipotential surface
Let uniform electric field E along X- axis ,the equipotential surface are plane normal to X-axis I.e. planes parallel to
the Y-Z plane.
Remark:
Potential energy due to single charge:-According to the definition of potential energy require to bring unit
positive charge from infinity to the point in an electric field is the potential at that point ,thus
V= w/q , w= Vq
Thus energy is E= qv
Que.23 Derive an expression for potential energy of system of two charge in an external field
⃗⃗𝟏 & 𝒓
⃗⃗𝟐 w.r.to origin
Ans: Consider the two charges q1 & q2 lying at point A & B with position vector 𝒓
24
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
The work done in bringing the charge q1 from infinity to the point A is given by,
Work done on q1 against external field = q1 V1
In bringing the charge q2 work is done against the External field also field due to charge q1.
Work done in bringing q2 to B
= 𝐪𝟐 𝐕𝟐 +
𝟏 𝐪𝟏 𝐪𝟐
𝟒𝛑𝛜𝐨 𝐫𝟏𝟐
Thus the potential energy of the system
= q1 V1 + 𝐪𝟐 𝐕𝟐
+
𝟏
𝐪𝟏 𝐪𝟐
𝟒𝛑𝛜𝐨 𝐫𝟏𝟐
Que.24 Derive an expression for potential energy of Dipole in an external field.
⃗⃗ , a torque acting on a dipole is 𝝉
⃗⃗
⃗⃗ × 𝑬
⃗⃗ = 𝒑
Ans : -Consider a electric dipole is placed in an electric field 𝑬
Which produces rotation of dipole (except p is parallel to E )
⃗⃗𝒆𝒙𝒕 ) is applied in such manner that previous torque (𝝉
⃗⃗ ) is neutralizes and rotate the
Suppose an external torque (𝝉
dipole in the plane of paper from angle 𝜽𝒐 𝒕𝒐 𝒂𝒏 𝒂𝒏𝒈𝒍𝒆 𝜽
The amount of work done by the external torque is
25
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
𝛉
𝐖 = ∫ 𝛕𝐞𝐱𝐭 (𝛉)𝐝𝛉
𝛉𝐨
𝛉
𝐖 = ∫ 𝐩𝐄 𝐬𝐢𝐧 𝛉 𝐝𝛉
𝛉𝐨
∴ 𝐖 = 𝐩𝐄[− 𝐜𝐨𝐬 𝛉]𝛉𝛉𝐨 = 𝐩𝐄 [𝐜𝐨𝐬 𝛉𝐨 − 𝐜𝐨𝐬 𝛉]
This work is stored as the potential energy of the system with an inclination θ of the dipole.
∴ 𝐔(𝛉) = 𝐩𝐄 [𝐜𝐨𝐬 𝛉𝐨 − 𝐜𝐨𝐬 𝛉]
Case I :
If initially dipole is perpendicular to the field E 𝜽𝒐 =
𝝅
𝛑
, then ∴ 𝐔(𝛉) = 𝐩𝐄 [𝐜𝐨𝐬 − 𝐜𝐨𝐬 𝛉]
𝟐
𝟐
𝐔(𝛉) = −𝐩𝐄 𝐜𝐨𝐬 𝛉
⃗⃗
⃗⃗ ∙ 𝐄
𝐔(𝛉) = −𝐩
Case II :
If initially dipole is parallel to the field E 𝜽𝒐 = 𝟎 , then
∴ 𝐔(𝛉) = 𝐩𝐄 [𝐜𝐨𝐬 𝟎 − 𝐜𝐨𝐬 𝛉]
𝐔(𝛉) = −𝐩𝐄(𝟏 − 𝐜𝐨𝐬 𝛉)
Que 20: explain electric flux and flux density.
Ans:
26
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com
PHYSICS-MANIA
⃗⃗⃗⃗⃗ in an electric field and represented by vector draw perpendicular to it. Let  angle
Consider small surface area 𝒅𝒔
between electric field ⃗𝑬⃗ and area vector ⃗⃗⃗⃗⃗
𝒅𝒔 .then electric flux d passing through area ⃗⃗⃗⃗⃗
𝒅𝒔 are given by.
d= (component of ⃗𝑬⃗ along ⃗⃗⃗⃗⃗
𝒅𝒔)(area ⃗⃗⃗⃗⃗
𝒅𝒔)
d= (Ecos )( ds)
⃗⃗⃗⃗⃗
⃗⃗ ∙ 𝒅𝒔
d= 𝑬
Total electric flux through whole surface area is given by
⃗⃗⃗⃗⃗
⃗⃗ ∙ 𝒅𝒔
=  𝑬
SI unit of electric flux is volt-metre (Vm)
Electric flux density:-It is defined as the electric flux per unit area.
SI unit of electric flux density is volt per metre (V/m) or weber per square metre (Wb/m²).
It is equivalent with to electric field intensity
27
© 2016 PhYsIcS-MaNiA. All rights reserved.
www.physics-mania.in\
E-mail: pm.contactus@gmail.com