Chapter 4 Electric Charges and Fields (Electrostatics Part 4) – Physics free study material by TEACHING CARE online tuition and coaching classes
Electric Dipole.
 General information : System of two equal and opposite charges separated by a small fixed distance is called a
 Dipole axis : Line joining negative charge to positive charge of a dipole is called its It may also be termed as its longitudinal axis.
 Equatorial axis : Perpendicular bisector of the dipole is called its equatorial or transverse axis as it is perpendicular to
 Dipole length : The distance between two charges is known as dipole length (L = 2l)
 Dipole moment : It is a quantity which gives information about the strength of dipole. It is a vector quantity and is directed from negative charge to positive charge along the axis. It is denoted as p and is defined as the product of the magnitude of either of the charge and the dipole
i.e.
r
p = q(2l )
Its S.I. unit is coulombmetre or Debye (1 Debye = 3.3 × 10^{–30} C ´ m) and its dimensions are
M^{0}L^{1}T^{1}A^{1}.
Note : @A region surrounding a stationary electric dipole has electric field only.
@ When a dielectric is placed in an electric field, its atoms or molecules are considered as tiny dipoles.
@ Water (H_{2}O), Chloroform (CHCl_{3}), Ammonia (NH_{3}), HCl, CO molecules are some example of permanent electric dipole.
 Electric field and potential due to an electric dipole : It is better to understand electric dipole with magnetic
S.No. Electric dipole Magnetic dipole
 System of two equal and opposite charges separated by a small fixed
System of two equal and opposite magnetic poles (Bar magnet) separated by a small fixed distance.
A B – m + m

–q +q
2l
 Electric dipole moment :
r
p



r = r , directed from
Magnetic dipole moment :
M


r r
= m(2 ) , directed
 q to +q. It’s I. unit is coulomb × meter or from S to N. It’s S.I. unit is ampere × meter^{2}.
Debye.
 Intensity of electric field
r
e E
r
Ee g
Equatorial line
q
– q +q
2l
q + a

r
Ea
Intensity of magnetic field
r e
Be
Equatorial line
q
2l
r
B
g
q + a

a r
Ba
r Axial line
p
Axial line
M
If a, e and g are three points on axial, equatorial and general position at a distance r from the centre of dipole
If a, e and g are three points on axial, equatorial and general position at a distance r from the centre of dipole
on axial point
on equatorial point
(directed from – q to +q)
(directed from +q to –q)
on axial point
on equatorial point
(directed from S to N) (directed from N to S)
on general point Ea = 1 . p
(3 cos ^{2} q + 1)
on general point
4pe _{0} r 3
Angle between – r and r is 0^{o}, r and r is 180^{o}, Angle between – r and r is 0^{o}, r and r is
Ea p
Ee p
Ba M
Be M


r and r is (q + a) (where
r r


) 180^{o}, and is (q + a) (where )
Electric Potential – At a Va = 1 . p , At e V = 0
At g
4pe _{0} r 2
(3) Dipole (electric/magnetic) in uniform field (electric/magnetic)
 Torque : If a dipole is placed in an uniform field such that dipole (e. p or M ) makes an angle q with direction of field then two equal and opposite force acting on dipole constitute a couple whose tendency is to rotate the dipole hence a torque is developed in it and dipole tries to align it self in the direction of field.
 Work : From the above discussion it is clear that in an uniform electric/magnetic field dipole tries to align itself in the direction of electric field (i.e. equilibrium position). To change it’s angular position some work has to be
Suppose an electric/magnetic dipole is kept in an uniform electric/magnetic field by making an angle q_{1} with the field, if it is again turn so that it makes an angle q_{2} with the field, work done in this process is given by the formula
 Potential energy : In case of a dipole (in a uniform field), potential energy of dipole is defined as work done in rotating a dipole from a direction perpendicular to the field to the given direction e. if q_{1} = 90^{o} and q_{2} = q then –
 Equilibrium of dipole : We know that, for any equilibrium net torque and net force on a particle (or system) should be
We already discussed when a dipole is placed in an uniform electric/magnetic field net force on dipole is always zero. But net torque will be zero only when q = 0^{o} or 180^{o}
When q = 0^{o} i.e. dipole is placed along the electric field it is said to be in stable equilibrium, because after turning it through a small angle, dipole tries to align itself again in the direction of electric field.
When q = 180^{o} i.e. dipole is placed opposite to electric field, it is said to be in unstable equilibrium.
r
p
E E E
r
p
M
B B B M M
q = 0^{o} q = 90^{O} q = 180^{o}
Stable equilibrium Unstable equilibrium
t = 0 t_{max} = pE t = 0
W = 0 W = pE W_{max} = 2pE U_{min} = – pE U = 0 U_{max} = pE
q = 0^{o} q = 90^{O} q = 180^{o}
Stable equilibrium Unstable equilibrium
t = 0 t_{max} = MB t = 0
W = 0 W = MB W_{max} = 2MB
U_{min} = – MB U = 0 U_{max} = MB
 Angular SHM : In a uniform electric/magnetic field (intensity E/B) if a dipole (electric/magnetic) is slightly displaced from it’s stable equilibrium position it executes angular SHM having period of If I
= moment of inertia of dipole about the axis passing through it’s centre and perpendicular to it’s length.
For electric dipole :
T = 2p
and For Magnetic dipole :
T = 2p
 Dipolepoint charge interaction : If a point charge/isolated magnetic pole is placed in dipole field at a distance r from the mid point of dipole then force experienced by point charge/pole varies
according to the relation
F µ 1
r 3
 Dipoledipole interaction : When two dipoles placed closed to each other, they experiences a force due to each If suppose two dipoles (1) and (2) are placed as shown in figure then
Both the dipoles are placed in the field of one another hence potential energy dipole (2) is
U = – p E
cos 0 = – p E = – p
´ 1 . 2p_{1}
2 2 1
2 1 2
4pe _{0} r ^{3}
then by using
F = – dU , Force on dipole (2) is
dr
F = – dU2
^{2} dr

Þ F = – d ì 1 . 2p_{1} p_{2} ü = – 1 . 6 p_{1} p_{2}
^{2} dr î 4pe
ý

r 3 þ
4pe _{0} r ^{4}


Similarly force experienced by dipole (1)
F = –
1 . 6 p_{1} p_{2} so
F = F
= – 1 . 6 p_{1} p_{2}
1 4pe r 4 1 2 4pe r 4
Negative sign indicates that force is attractive.  F = 1 . 6 p1 p2
and
F µ 1
4pe _{0} r ^{4} r ^{4}
Note : @ Same result can also be obtained for magnetic dipole.
 Electric dipole in nonuniform electric field : When an electric dipole is placed in a non uniform field, the two charges of dipole experiences unequal forces, therefore the net force on the dipole is not equal to The magnitude of the force is given by the negative derivative of the potential energy w.r.t.
distance along the axis of the dipole i.e. F
= – dU
dr
= – p . d E .
dr
Due to two unequal forces, a torque is produced which rotate the dipole so as to align it in the direction of field. When the dipole gets aligned with the field, the torque becomes zero
and then the unbalanced force acts on the dipole and the dipole then moves linearly along the direction of field from weaker portion of the field to the stronger portion of the field. So in nonuniform electric field
 Motion of the dipole is translatory and rotatory
 Torque on it may be zero.
Example: 84 If the magnitude of intensity of electric field at a distance x on axial line and at a distance y on equatorial
line on a given dipole are equal, then x : y is [EAMCET 1994]
(a) 1 : 1 (b) 1 : 2 (c) 1 : 2 (d) 1
Solution: (d) According to the question
1
4pe _{0}
. 2p =
x 3
1 . p
4pe _{0} y 3
Þ x = (2)^{1/} ^{3} : 1
y
Example: 85 Three charges of (+2q), (– q) and (– q) are placed at the corners A, B and C of an equilateral triangle of
side a as shown in the adjoining figure. Then the dipole moment of this combination is [MP PMT 1994; CPMT 1994]
 qa
 Zero
q a
2 qa
Solution: (c) The charge +2q can be broken in +q, +q. Now as shown in figure we have two equal dipoles inclined at an angle of 60^{o}. Therefore resultant dipole moment will be
pnet =
Example: 86 An electric dipole is placed along the xaxis at the origin O. A point P is at a distance of 20 cm from this
origin such that OP makes an angle
p with the xaxis. If the electric field at P makes an angle q with x–
3
axis, the value of q would be [MP PMT 1997]
p p _{–}_{1} æ 3 ö 2p
1 æ 3 ö
(a)
(b)
 tan
ç ÷
(c)
(d)
tan
ç ÷





3 3 ç ÷ 3 ç ÷
Solution: (b) According to question we can draw following figure.
As we have discussed earlier in theory q = p + a
3
tana = 1 tan p Þ a = tan^{–}^{1} 3
So,
q = p
3
2 3 2
x
 tan^{–}^{1}
2
Example: 87 An electric dipole in a uniform electric field experiences [RPET 2000]
(a) Force and torque both (b) Force but no torque (c) Torque but no force (d) No force and no torque
Solution: (c) In uniform electric field F_{net} = 0, t_{net} ¹ 0
Example: 89 Two opposite and equal charges 4 × 10^{–8} coulomb when placed 2 × 10^{–2} cm away, form a dipole. If this dipole is placed in an external electric field 4 × 10^{8} newton/coulomb, the value of maximum torque and
the work done in rotating it through 180^{o} will be [MP PET 1996 Similar to MP PMT 1987]
(a) 64 × 10^{–4} Nm and 64 × 10^{–4} J (b) 32 × 10^{–4} Nm and 32 × 10^{–4} J
(c) 64 × 10^{–4} Nm and 32 × 10^{–4} J (d) 32 × 10^{–4} Nm and 64 × 10^{–4} J
Solution: (d) t_{max} = pE and W_{max} = 2pE Q p = Q × 2l = 4 × 10^{–8} × 2 × 10^{–2} × 10^{–2} = 8 × 10^{–12} Cm
So, t_{max} = 8 × 10^{–12} × 4 × 10^{8} = 32 × 10^{–4} N–m and W_{max} = 2 × 32 × 10^{–4} = 64 × 10^{–4} J
Example: 90 A point charge placed at any point on the axis of an electric dipole at some large distance experiences a force F. The force acting on the point charge when it’s distance from the dipole is doubled is
[CPMT 1991; MNR 1986]
(a) F (b)
F (c)
2
F (d) F
4 8
Solution: (d) Force acting on a point charge in dipole field varies as
F µ 1
r 3
where r is the distance of point charge
from the centre of dipole. Hence if r makes double so new force F‘ = F .
8
Example: 91 A point particle of mass M is attached to one end of a massless rigid nonconducting rod of length L. Another point particle of the same mass is attached to other end of the rod. The two particles carry charges +q and – q respectively. This arrangement is held in a region of a uniform electric field E such that the rod makes a small angle q (say of about 5 degrees) with the field direction (see figure). Will be minimum time, needed for the rod to become parallel to the field after it is set free
E
(a)
t = 2p
(b)
t = p
2
(c)
t = 3p
2
(d)
t = p
Solution: (b) In the given situation system oscillate in electric field with maximum angular displacement q.
It’s time period of oscillation (similar to dipole)
T = 2p
where I = moment of inertia of the system and
p = qL

Hence the minimum time needed for the rod becomes parallel to the field is t = T = p
Here
4 2
I = M
è ø è ø
Electric Flux.
 Area vector : In many cases, it is convenient to treat area of a surface as a The length of the vector represents the magnitude of the area and its direction is along the outward drawn normal to the area.
 Electric flux : The electric flux linked with any surface in an electric field is basically a measure of total number of lines of forces passing normally through the or
Electric flux through an elementary area ds is defined as the scalar product of area of field i.e.
df = E × ds = Eds cosq
Hence flux from complete area (S) f = ò E ds cosq
= ES cosq
If q = 0^{o}, i.e. surface area is perpendicular to the electric field, so flux linked with it will be max.
i.e. f_{max} = E ds and if q = 90^{o}, f_{min} = 0
(3) Unit and Dimensional Formula
S.I. unit – (volt × m) or
N – C m2
It’s Dimensional formula – (ML^{3}T^{–3}A^{–} ^{1})
 Types : For a closed body outward flux is taken to be positive, while inward flux is to be negative
Gauss’s Law.
 Definition : According to this law, total electric flux through a closed surface enclosing a charge is
1 times the magnitude of the charge enclosed i.e. f = 1 (Q
e _{0} e _{0}
enc. )
 Gaussian Surface : Gauss’s law is valid for symmetrical charge distribution. Gauss’s law is very helpful in calculating electric field in those cases where electric field is symmetrical around the source producing Electric field can be calculated very easily by the clever choice of a closed surface that encloses the source charges. Such a surface is called “Gaussian surface”. This surface should pass through the point where electric field is to be calculated and must have a shape according to the symmetry of source.
e.g. If suppose a charge Q is placed at the centre of a hemisphere, then to calculate the flux through this body, to encloses the first charge we will have to imagine a Gaussian surface. This imaginary Gaussian surface will be a hemisphere as shown.
Net flux through this closed body f = Q
e _{0}
Hence flux coming out from given hemisphere is f =
Q .
2e _{0}


 Zero flux : The value of flux is zero in the following circumstances
 Flux emergence : Flux linked with a closed body is independent of the shape and size of the body and position of charge inside it
Example: 91 Electric charge is uniformly distributed along a long straight wire of radius 1 mm. The charge per cm length of the wire is Q coulomb. Another cylindrical surface of radius 50 cm and length 1 m symmetrically encloses the wire as shown in the figure. The total electric flux passing through the cylindrical surface is
[MP PET 2001]
 Q
e
0
100 Q
e _{0}
10 Q
(pe _{0} )
100 Q
(pe _{0} )
Solution: (b) Given that charge per cm length of the wire is Q. Since 100 cm length of the wire is enclosed so
Qenc = 100 Q
Þ Electric flux emerging through cylindrical surface f = 100 Q .
e _{0}
Example: 92 A charge Q is situated at the corner A of a cube, the electric flux through the one face of the cube is
[CPMT 2000]
(a)
Q (b)
6e _{0}
Q (c)
8e _{0}
Q (d)
24e _{0}
Q
2e _{0}
Solution: (c) For the charge at the corner, we require eight cube to symmetrically enclose it in a Gaussian surface. The total flux f_{T} = Q . Therefore the flux through one cube will be f_{cube} = Q . The cube has six faces and
e _{0} 8e _{0}
flux linked with three faces (through A) is zero, so flux linked with remaining three faces will
f . Now
8e _{0}
as the remaining three are identical so flux linked with each of the three faces will be
1 é 1 æ Q öù 1 Q
= 3 ´ ê 8 ç e
÷ú = 24 e .
ëê è 0 øúû 0
Example: 93 A square of side 20 cm is enclosed by a surface of sphere of 80 cm radius. Square and sphere have the same centre. Four charges + 2 × 10^{–6} C, – 5 × 10^{–} ^{6} C, – 3 × 10^{–} ^{6} C, +6 × 10^{–} ^{6} C are located at the four corners of a square, then out going total flux from spherical surface in N–m^{2}/C will be
(a) Zero (b) (16 p) × 10^{–} ^{6} (c) (8p) × 10^{–6} (d) 36p × 10^{–6}
Solution: (a) Since charge enclosed by Gaussian surface is
f_{enc}_{.} = (2 ´ 10^{–}^{6} – 5 ´ 10^{–}^{6} – 3 ´ 10^{–}^{6} + 6 ´ 10^{–}^{6} ) = 0
so f = 0
Example: 94 In a region of space, the electric field is in the xdirection and proportional to x, i.e., E
= E_{0} xˆi . Consider
an imaginary cubical volume of edge a, with its edges parallel to the axes of coordinates. The charge inside this cube is
 Zero (b)
e _{0} E0 a ^{3}
(c)
1 E0 a 3
e _{0}
(d)
1 e _{0} E0 a ^{2}
6
Solution: (b) The field at the face ABCD =
E x ˆi.

\ Flux over the face ABCD = – (E_{0}x_{0})a^{2}
The negative sign arises as the field is directed into the cube. The field at the face EFGH = E0 (x0 + a)ˆi.
\ Flux over the face EFGH = E_{0}
(x0
 a) a ^{2} F
The flux over the other four faces is zero as the field is parallel to the X
surfaces.
\ Total flux over the cube = E0 a ^{2} = 1 q
2
where q is the total charge inside the cube. \ q = e _{0} E0 a ^{3} .
Application of Gauss’s Law.
Gauss’s law is a powerful tool for calculating electric field in case of symmetrical charge distribution by choosing a Gaussian surface in such away that E is either parallel or perpendicular to it’s various faces.
e.g. Electric field due to infinitely long line of charge : Let us consider a uniformly charged wire of infinite length having a constant linear charge density is

l ç l =
è
charge ö
÷. Let P be a point distant r from the wire at which the electric field is to be
length ø
calculated.
Draw a cylinder (Gaussian surface) of radius r and length l around the line charge which encloses the charge Q ( Q = l . l ). Cylindrical Gaussian surface has three surfaces;
two circular and one curved for surfaces (1) and (2) angle between electric field and normal to the surface is 90^{o} i.e., q = 90^{o}.
So flux linked with these surfaces will be zero. Hence total flux will pass through curved surface and it is
f = ò E ds cos q
According to Gauss’s law
f = Q
e_{0}

Equating equation (i) and (ii) E ds = Q
e_{0}
…. (i)
…. (ii)
Þ Eò ds = Q Þ Ex2prl = Q
e_{0} e_{0}

Þ E = Q = l = 2kl ìK = 1 ü

ý
2pe_{0}rl 2pe_{0}r r î 4pe