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2013-2015
Notes for Std. XIIth
Physics
Electric FieldPranjal K. Bharti, B. Tech., IIT Kharagpur
© 2007 P. K. Bharti
All rights reserved.
www.vidyadrishti.org
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Why Test charge is small?
• You must be wondering why test charge is small. There
are two reasons for it.
Reason 1 (For boards)
• If the magnitude of test charge is not very small, the
position of the source charge may change. Expression
0
F E q
=
gives the electric field at the changed position.
Reason 2
• We must assume that the test charge q0 is small enough
that it does not disturb the charge distribution responsible
for the electric field.
• If a vanishingly small test charge q0 is placed near a
uniformly charged metallic sphere, as shown in figure (a),
the charge on the metallic sphere, which produces the
electric field, remains uniformly distributed. If the test
charge is great enough (q 0’ >> q 0) , as shown in figure (b),
the charge on the metallic sphere is redistributed and the
ratio of the force to the test charge is different.
• That is, because of this redistribution of charge on the
metallic sphere, the electric field it sets up is different
from the field it sets up in the presence of the much
smaller q0.
• Hence, test charge must be very small compared to source
charge.
Physical significance of Electric field (NCERT)
Suppose we consider the force between two distant charges q1,q2 in accelerated motion. The field picture is this: the
accelerated motion of charge q1 produces electromagneticwaves, which then propagate with the speed of light c, reachq2 and cause a force on q2. The notion of field accounts for thetime delay. Thus, even though electric and magnetic fields can
be detected only by their effects (forces) on charges, they areregarded as physical entities, not merely mathematicalconstructs. They have an independent dynamics of their own,i.e., they evolve according to laws of their own. They can alsotransport energy. Thus, a source of time dependentelectromagnetic fields, turned on briefly and switched off,leaves behind propagating electromagnetic fields transportingenergy.
3. Electric field due to a point charge
• Suppose we are interested to find out electric field due to
point source charge q at a distance r from it.
• We call the location of the charge the source point, and
we call the point P where we are determining the field the
field point.
• If we place a small test charge qo at the field point P, at a
distance r from the source point, the magnitude F of theforce is given by Coulomb’s Law:
0 0
2 2
0 0
1 1=
4 4
qq q qF
r r πε πε = …(1)
• We have taken qo out of modulus because q
o is positive in
nature.
• Now, from definition of electric field, we have
0
F E
q=
…(2)
Using eqns. (1) & (2) we get:
2
0
1
4
q E
r πε =
Direction of electric field
• Electric field intensity is a vector quantity. The direction
of E
is along F
.
• By definition, the electric field of a point charge always
points away from a positive charge. In general, electric
is radially outward from a point positive charge.
•The electric field of a point charge always points towardsa negative charge. In general, electric field is radially
inward to a point negative charge.
+q
+q0
Source charge
P
E
Test charge
Direction of electric field is away from +ve source charg
– q
+q0
Source charge
P E
Test charge
F
Direction of electric field is towards –ve source charge
r
r
F
(Electric field due to a point charge q)
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Example 1
a) What is the magnitude and direction of the electric field at
a point 2.0 m from a point charge q = 4.0 nC?
b) What about direction if this charge is negative?
Solution:
a) We know that magnitude of electric field due to a point
charge q at a distance r from it is given by:
( )
99
2 2
0
1 4.0 109.0 10
4 2.0
9.0N/C
q E
r
E
πε
−×= = × ×
⇒ =
Since the charge is positive, therefore, electric field points
away from this charge.
b) If this charge is negative electric field magnitude will
remain same at that point because of the term |q| in thenumerator of electric field expression.
9.0N/C E =
• Since the charge is negative, therefore, electric field
points towards this charge.
Principle of Superposition
• The electrostatic field at a point due to a number of
charges is the vector sum of all the electrostatic field at
that point, taken one at a time. The individual fields are
unaffected due to the presence of other charges.
• This is termed as the principle of superposition . E
net
= E1
+ E 2
+ … + E n
•Note: Ei is a vector whose direction is along the line
joining q i & point P and pointing towards or away from
charge q i.
Example 2
• Four point charges q A = 2 μC, q
B = – 5 μC, q
C = 2 μC, and
q D = – 5 μC are located at the corners of a square ABCD
of side 10 cm. What is the net electric field at the centre
of the square?
Solution:
• We have to find out net field at centre O. Clearly, we
have to use the principle of superposition. It means, we
have to find out individual fields at point O due to al
charges and then we have to add them using vector
addition rule.
• Clearly, distance of point O from any other charge
= half the length of diagonal
= ½ (√2 side) = (1/√2) (10cm) = (0.1/√2) m.
•Field at point O due to charge qA = 2 μC :
• Since charge q A
is positive in nature therefore
electrostatic field E A
at point O will be away from q A, i.e
E A
is towards OC.
• Now, electrostatic field at point O due to charge q A when
placed in air is given by:
( )
6
9
2 2
0
6
2 101 9 10
4 0.1/ 2
3.6 10 N/C (alongOC)
A
A
A
q E
r
E
πε
−×= = × ×
⇒ = ×
• Field at point O due to charge q
B = -5 μC :
• Since charge q B
is negative in nature therefore
electrostatic field EB at point O will face towards q
B, i.e
EB is towards OB.
• Now, electrostatic field at point O due to q B when placed
in air is given by:
( )
6
9
2 2
0
6
5 101 9 10
4 0.1/ 2
9 10 N/C (alongOB)
A
B
B
q E
r
E
πε
−− ×= = × ×
⇒ = ×
• Similarly,Field at point O due to charge q
C = 2 μC :
63.6 10 N/C (alongOA)C E = ×
• Field at point O due to charge qD = -5 μC :
69 10 N/C (alongOD) D
E = × • Net electric field component along AC = E
A – E
C = 0
• Net electric field component along BD = ED – E
B = 0
• Since, both components of net electric field = 0, therefore
net field at centre O = 0. (Ans)
q B = –5 μC A B
O
E A C
D
q A =2 μC
q B = 2 μC
EC E B
E D
q D = –5 μC
E
Positive charge
E
Negative charge
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x P
60
60
– Q
r
y
• Since x does not vary as we move from point to point
around the ring, all the factors on the right side except dQ
are constant and can be taken outside the integral. The
integral of dQ is just the total charge Q and we finally get
( ) ( )
( )
3/2 3/22 2 2 2
0 0
3/ 22 20
1 1cos
4 4
1
4
x xdE dQ Q
x R x R
Qx E
x R
θ πε πε
πε
= =+ +
∴ =
+
∫ ∫
This is also the net electric field at P. Clearly this field is
directed along x-axis. Hence, net electric field at P in
vector form is given by
( )3/ 2
2 20
1
4
Qx E i
x Rπε =
+
Test your understanding
See last example. Deduce the net electric field at the centre of
the ring using this result. You can also find net electric field at
centre using argument of symmetry. Compare this result withsymmetry argument.
Ans: Zero
Hint: Put x = 0
Example 4
Figure shows a plastic
rod having a
uniformly distributed
charge – Q. The rod
has been bent in a
120o circular arc of
radius r . In terms of
Q and r , what is the
electric field E
due to
the rod at point P?
Solution:
• Consider a differential element having arc length ds and
located at an
angle θ above
the x-axis. Ifwe let λrepresent the
linear charge
density of the
rod, our
element ds has
a differential
charge of
magnitude
dq = λ ds.
• Our element produces a differential electric field d E
a
point P, which is a distance r from the element. Treating
the element as a point charge, we can express the
magnitude of d E
as
( )2 2
0 0
1 1.
4 4
dq dsdE i
r r
λ
πε πε = =
• The direction of d E
is toward ds, because charge dq is
negative.
• Our element has a symmetrically located (mirror image)
element ds in the bottom half of the rod. The electric field
'd E
set up at P by ds also has the magnitude given by
Eq. (i), but the field vector points toward ds as shown in
figure. If we resolve the electric field vectors into x and y
components, we see that their y components cance
(because they have equal magnitudes and are in opposite
directions). We also see that their x components have
equal magnitudes and are in the same direction.
• Thus, to find the electric field set up by the rod, we need
sum (via integration) only the x components of thedifferential electric fields set up by all the differential
elements of the rod. We can write the x component set up
by ds as
( )2
0
1cos cos
4dE ds ii
r
λ θ θ
πε =
• Equation (ii) has
two variables, θand s. Before we
can integrate it, we
must eliminate one
variable. We do so by replacing ds,
using the relation
ds = r dθ,
in which d θ is the angle at P that includes arc length ds .• With this replacement, we can integrate Eq. (ii) over the
angle made by the rod at P, from θ = –60o to θ = 60o; thawill give us the magnitude of the electric field at P due to
the rod:
[ ]
( )
( )
60
2600
60 60
60600 0
0
0
1cos cos
4
cos sin4 4
sin 60 sin 604
3
4
E dE r d r
d r r
r
iiir
λ θ θ θ
πε
λ λ θ θ θ
πε πε
λ
πε
λ
πε
= =
= =
= − −
=
∫ ∫
∫
• To evaluate λ, we note that the rod has an angle of 120and so is one-third of a full circle. Its arc length is then
2 ,3
r π and its linear charge density must be
dE
xP
θ
θ
dE sin
y
dE cos
ds
ds
dE
dE cos
dE sin
x P
y ds
dθ r
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charge 3
length 22
3
Q Q
r r λ
π π = = =
• Substituting this into Eq. (iii) and simplifying give us
( )
0 0
2 20
3 3 3
4 4 2
3 3 Answer
8
Q E
r r r
E r
λ λ
πε πε π
λ
π ε
= = ⋅
⇒ =
• The direction of E
is toward the rod, along the axis of
symmetry of the charge distribution. We can write E
in
unit-vector notation as
2 2
0
3 3
8 E i
r
λ
π ε =
Quick Exercise
The figure here shows three nonconducting rods, one circularand two straight. Each has a uniform charge of magnitude Qalong its top half and another along its bottom half. For each
rod, what is the direction of the net electric field at point P?
Example 5
Find the electric field at a point P on the perpendicular
bisector of a uniformly charged rod. The length of the rod is L,
the charge on it is Q and the distance of P from the centre of
the rod is a.
Solution:Let us take an element of length
dx at a distance x from the centre
of the rod. The charge on this
element is
.Q
dQ dx L
=
The electric, field at P due to this
element is
( )2
0
.4
dQdE
APπε =
By symmetry, the resultant field at P will be along OP (if the
charge is positive). The component of dE along OP is
( ) ( )2 3/2
2 20 0
cos .
4 4
dQ OP a Q dxdE
AP AP L a xθ
πε πε = =
+
Thus, the resultant field at P is
( )
/2
3/ 22 2
0 /2
cos
....(i)4
L
L
E dE
aQ dx L a x
θ
πε −
=
=+
∫
∫
We have x = a tanθ or dx = a sec2θ d θ
Thus,( )
2
3/2 3 32 2
sec
sec
dx a d
aa x
θ θ
θ =
+∫ ∫
( )2 2 2 1/2
2 2
1 1 1cos sin .
xd
a a a a xθ θ θ = = =
+∫
From (i),
( )
( )
/2
2 1/22 2
0/2
2 1/22 2
0
2 2
0
4
2
4 4
.2 4
L
L
aQ x
E La a x
aQ L
La L a
Q
a L a
θε
πε
πε
−
= +
= +
=+
Quick Exercise
1. Two charged particles +5 μC and +10 μC are placed 20cm apart. Find the electric field at the midpoint between
the two charges.
2. Two charged particles A and B have charges +10 μC and
+40 μC are held at a separation of 90 cm from each other
At what distance from A, electric field intensity will be
zero.
3. Two point charges + 8q and – 2q are located at x = 0 and
x = L respectively. Find the location of a point on X-axis
at which the net electric field due to these two charges is
zero.
4. An infinite number of charges, each equal to q are placed
along X-axis at x=1, x=2, x=4, x=8, … and so on.
(i) Find the field at the point x=0 due to this set up o
charges.
(ii) What will be the electric field, if in the above set up
the consecutive charges have opposite charges.
5. Two point charges +q and –q are placed distance d apart
What are the points at which the resultant electric field is
parallel to the line joining the two charges?
x P
+Q
(b)
+Q
y
θ
dE
a
L
A x0
θ
P
x
P
+
(c)
–
y
x P
–Q
(a)
+Q
y
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Q
R
S
U O
P
T
6. Six charges, three positive and three negative of equal
magnitude are to be placed at the vertices of a regular
hexagon, such that the electric field, at the centre O is
double the electric field when only positive charge of
same magnitude is placed at R. Which of the following
arrangements of charges is possible for P, Q, R, S, T and
U respectively?
a)
+, –, +, –, –, + b) +, –, +, –, +, –
c) +, +, –, +, –, –
d) –, +, +, –, +, –
7. A thin semicircular ring of radius r has a positive charge q
distributed uniformly over it. The net field at the centre is
a) 2 2
0
4
q j
r π ε
b) 2 2
0
4
q j
r π ε −
c) 2 2
0
2
q j
r π ε −
d) 2 2
0
2
q j
r π ε
8. Let ( )4
Qr r
R ρ
π = be the charge density distribution of a
solid sphere of radius R and total charge Q. Find themagnitude of electric field at a point p inside the sphereat a distance r 1 from the centre of the sphere.
a) 2 2
0 1
4
Q
r π ε
b) 2
1
2 4
0
4
Qr
Rπ ε
c) 2
1
2 4
0
3
Qr
Rπ ε
d) 0
9. Let there be a spherically symmetric charge distribution
with charge density varying as ( ) 05
4
r r
R ρ ρ
= −
upto r
= R, and ρ (r )= 0 for r > R, where r is the distance from
the origin. Find the electric field at a distance r (r < R)from the origin.
a) 0
0
3 5
4 3
r r
R
ρ
ε
−
b) 0
0
4 5
3 3
r r
R
ρ
ε
−
c) 0
0
5
4 3
r r
R
ρ
ε
−
d) 0
0
3 5
4 4
r r
R
ρ
ε
−
10. Four point charges each of charge +q are rigidly fixed athe four corners of a square planar soap film of side aThe surface tension of the soap film is S . The system ocharges and planar film are in equilibrium, and
1/2
N
qa k
S
=
where k is a constant. Find the value of N .
11. A solid sphere of radius R has a charge Q distributed in its
volume with a charge density ( ) ar kr ρ = , where k and a
are constants and r is the distance from its centre. If the
electric field at r = R/2 is 1/8 times greater that at r = R
find the value of a.
12. Two identical point charges are placed at a separation of l
P is a point on the line joining the charges, at a distance x
from any one charge. The field at P is E. E is plotted
against x for values of x from close to zero to slightly less
than l. Which of the following best represents the
resulting curve?
Answers:
1. 4.5 × 106 N/C
2. 30 cm
3. x = 2 L
4. (i)0
3
q
πε (ii)
05
q
πε
5.
At a point on the perpendicular bisector of the twocharges and on the either side of the line joining the
two charges.
6. d
7. c
8. c
9. c
10. 3
11. 2
12. c
x O
E
l x O
E
l
x
O
E l x
O
E l
(a) (b)
(d)(c)
y
x
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4. Electric Field Lines
• An electric field line is an imaginary continuous curve
drawn through a region of space so that its tangent at
any point is in the direction of the electric-field vector at
that point.
• Electric lines of force exist throughout the region of an
electric field. The electric field intensity of a charge
decreases gradually with increasing distance from it and becomes zero at infinity i.e., electric field cannot vanishes
abruptly. So a line of force cannot have sudden breaks, it
must be a continuous curve. Figure shows the basic idea.
• Electric field lines show the direction of E
at each point
and their spacing gives a general idea of the magnitude of
E
at each point. Electric field is stronger where field
lines are closer and field is weaker where field lines are
farther.
• At any particular point. the electric field has a unique
direction, so only one field line can pass through each
point of the field. In other words, field lines never
intersect .
Electric field lines due to some simple charge configuration
Properties of electrostatic field lines
1. Every electric field line begins either at a positive source
charge or at infinity .
2. Every electric field line ends either at a negative source
charge or at infinity.
3. A line of force cannot have sudden breaks, it must be a
continuous curve.
4. The direction of electric field at any point is given by
tangent to the electric field line at that point.
5. Electric field lines never cross each other or themselves.
6. The regions, where field lines are closer, the electric field
is strong. Similarly, the regions where the field lines are
farther apart, the field is weak.
7. Electrostatic field lines do not form any closed loops. This
follows from the conservative nature of electric field.
8. The lines of force have a tendency to contract lengthwise
This explains attraction between two unlike charges.
9. The lines of force have a tendency to expand laterally so
as to exert a lateral pressure on neighbouring lines of
force. This explains repulsion between two similarcharges.
10. The lines of force do not pass through a conductor
because the electric field inside a conductor is zero in
electrostatic.
Field lines due to some surface charges
Positive charge
(Radially outward)
Negative charge
(Radially inward)
System of two
identical positivecharges
System of identical
positive and negative
charges
Two infinite plane
sheet of charges
(3-D view)
Infinite plane sheet of
positive charge
Infinite plane sheet of
negative charge
Two infinite plane
sheet of charges
Infinite plane
sheet of negative charges
and point positive charge
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Field lines close to the surface of a conductor
• The electric force, and thus the electrostatic field, is
always directed perpendicular to the surface of a
conducting object. This is true when we are observing
electric field very close to the surface.
• If there were ever any component of force parallel to the
surface, then any excess charge residing upon the surface
of a source charge would begin to accelerate. This wouldlead to the occurrence of an electric current within the
object; this is never observed in electrostatic.
• Please note that, electric field inside a conductor is zero.
Example 6
Field lines due to three charges are shown in figure. State
which is the largest and which is smallest charge in
magnitude.
• Solution:
• By convention, objects having greater amount of
charge are surrounded by more field lines.
• Bodies having greater charge create stronger electric
fields.
• As the density of electric field lines near charge A is least
& near charge C is greatest, therefore, A has least amount
of charge & C has largest amount of charge.
Example 7
• State whether given figure represents correct field lines in
electrostatics? Why?
Solution:
• No, because electrostatic field lines cannot form closedloops.
Quick Exercise
NCERT Questions
1.7 (a) An electrostatic field line is a continuous curveThat is, a field line cannot have sudden breaks. Why not?(b) Explain why two field lines never cross each other aany point?1.26 Which among the curves shown in Fig.cannot possibly represent electrostatic fieldlines?
More questions for practice
1. A few electric field lines for a system of two charges Qand Q2 fixed at two different points on the x-axis areshown in the figure. These lines suggest thata) |Q1| > |Q2|
b) |Q1| < |Q2|
c) at a finite distance tothe left of Q1 theelectric field is zero
d) at a finite distance to the right of Q2 the electric fieldis zero
2. Three positive charges of equal value q are placed at thevertices of an equilateral triangle. The resulting lines of
force should ne sketched as in :
++
++
+
+
+
+
++
+++
+
++
conductor
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5. Force on a charged placed in an electric field
• If the field E
at a certain point is known, rearranging Eq.
0
F E
q=
, gives the force0F q E =
experienced by a point
charge qo placed at that point.
• In general, force experienced by a charge q at a certain
point in an electric field E
is given by
F qE =
Direction of force
• If q is positive, the force F qE =
experienced by the
charge is in the same direction as that of E
.
• If q is negative, the force F qE =
experienced by the
charge is in the opposite direction of E
.
Motion of charged particles in a uniform electric field
• When a charged particle of charge q and mass m is placed
in an electric field E
, the electric force exerted on the
charge is q E
.• If this is the only force exerted on the particle, it must be
the net force and so must cause the particle to accelerate.
• In this case, Newton’s second law applied to the particle
gives
F ma qE ma= ⇒ =
• The acceleration of the particle is therefore
qE a
m=
• If E
is uniform (that is, constant in magnitude and
direction), then the acceleration is constant.
Example 8
• An electric field of magnitude 1000 N/C is produced
between two parallel plates having a separation of 2.0 cm.
(a) With what minimum speed should an electron be
projected in the direction of field so that it may reach the
upper plate? (b) Suppose the electron is projected from
the lower plate with the speed calculated in part (a). The
direction of projection makes an angle of 600 with the
field. Find the maximum height reached by the electron.
Solution:
(a) Let the required minimum speed be u.
• Force on electron,
F = q E = 1.6 ×10 – 19 × 1000
= 1.6 ×10 – 16 N
• This force is directed downward because electron has
negative charge.
• Hence, from Newton’s 2nd law, acceleration of electrondue to this field is
( )14 2 1.76 10 m/s downward F
F ma am
= ⇒ = = ×
• Clearly, this acceleration is much larger than acceleration
due to gravity. Hence, we can neglect acceleration due to
gravity here.
• For minimum velocity electron just reaches the upper
plate; meaning final velocity at upper plate becomes zero.
• Let us consider all quantities in the upward direction to be
positive. Thus, we have
Initial velocity, u
Final velocity, v = 0
Acceleration, a = – 1.76 ×1014 m s – 2
.
(negative sign, because it is downward)
Distance, s = 2 cm = 0.02 m.
• Now, using v 2
= u 2
+ 2as, we have
0 2
= u 2
+ 2 × (– 1.76 ×1014
)× 0.02
u = √ 7.04 ×1012
u = 2.65 ×106
m/s
• Hence, minimum speed = 2.65 ×106
m/s (Ans)
(b) Here electron will move like a projectile, only difference
will be that we have to use ‘a’ in place of ‘g’.• For a projectile, maximum height is given by:
2 2 sin
2
u H
g
θ =
• Replacing g with a we get maximum height as:
( )( )
26 2 02 2
14
2.65 10 sin 30sin0.015m=1.5cm
2 2 1.76 10
u H
a
θ ×= = =
× ×
• Hence, maximum height = 1.5 cm (Ans)
E F a 2 cm
a Upward+ ve
- ve
u
+ ve
(Force on point charge q in electric field E
)
+q
E
Positive charge
F is along E
F
- q
E
Negative charge
F is opposite to E
F
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Example 9
A uniform electric field E is created between two parallel,
charged plates as shown in figure. An electron enters the field
symmetrically between the plates with a speed v0. The length
of each plate is l. Find the angle of deviation of the path of the
electron as it comes out of the field.
Solution:
The acceleration of the electron iseE
am
= in the upward
direction. The horizontal velocity remains v0 as there is no
acceleration in this direction. Thus, the time taken in crossing
the field is
0
...(i)l
t v
=
The upward component of the velocity of the electron as it
emerges from the field region is
0
y
eElat
mvυ = =
The horizontal component of the velocity remains
v x = v0.
The angle θ made by the resultant velocity with the originaldirection is given by
2
0
tan y
x
v eElv mv
θ = =
Thus, the electron deviates by an angle
1
2
0
tan eEl
mvθ
−=
Example 10
Positive charge Q is distributed uniformly over the ring ofradius R. A particle having a mass m and negative charge q, is
placed on its axis at a distance x from its centre. Find the forceon the particle. Assume x
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Dipole
• Definition: A rigid combination of two charges with
equal magnitude and opposite sign separated by a
small distance is called an electric dipole.
• Such combinations
occur frequently in
nature. Examples are
H2O, HCl, C2H5OH,CH3COOH, etc.
• Dipole Moment ( p
) :
The dipole moment of an electric dipole is a vector whose
magnitude is the product of the either charge q and the
separation d (= 2a) and the direction along the dipole axis
from the negative to the positive charge.
p = q d = q2a (magnitude of electric dipole moment)
• Dipole moment is a vector quantity, which can be
expressed in vector form as:
p qd =
(dipole moment)
• Direction of d
and hence, p
is from negative to
positive charge.
Force and Torque on a Dipole in a
Uniform Electric Field
• To start with, let's place an electric dipole in a uniform
external electric field E
, as shown in Fig. The forces F +
and F −
on the two charges both have magnitude qE , but
their directions are opposite, and they add to zero. The net
force on an electric dipole in a uniform external electric
field is zero.
• However, the two forces don't act along the same line, so
their torques do not add to zero.
• We calculate torques with respect to the center of the
dipole. Let the angel between the electric field E and the
dipole axis be θ ; then the lever arm for both F +
and F −
is
sin .a θ
• The torque of F +
and the torque of F −
both have the
same magnitude of ( ) sin ,qE a θ and both torques tend to
rotate the dipole clockwise (that is, is directed into the
page).
• Hence the magnitude of the net torque is twice the
magnitude of either individual torque:
= (qE )(2a sin θ)
• Now, since dipole moment p
has a magnitude q2a
therefore torque becomes
= (qE )( 2a sin θ) = (q2a)( E sin θ) = pE sin θ
• We can write this torque in terms of vector product
p E τ = ×
(torque on dipole in a uniform field)
• The torque is greatest when p
and E
are perpendicular
and is zero when they are parallel or anti-parallel.
• The torque always tends to turn p
to line it up with E
.
• The position θ = 0, with p
parallel to E
, is a position o
stable equilibrium, and the position θ = π, with p
and
E
anti- parallel, is a position of unstable equilibrium.
Potential Energy of a dipole in a
Uniform Electric Field
• When a dipole changes direction in an electric field, the
electric-field torque does work on it with a corresponding
change in potential energy. The work d W done by a
torque τ during an infinitesimal angular displacement d θ
is given by equation dW = τ d θ .
• Because the torque is in the direction of decreasing θ (as θis anti-clockwise, whereas τ is clockwise), we have
dW = τ d θ = – pE sin θ d θ
• In a finite angular displacement from θ1 to θ2 the tota
work done on the dipole is2
1
2 1sin cos cosW pE d pE pE
θ
θ
θ θ θ θ = − = −∫
• As we know that electric field is a conservative field
hence, the work is the negative of the change of potentia
energy.
W = – (U 1 – U 2 )
• Thus, we see that a suitable definition of potential energy
U for this system is
U (θ) = – pE cos θ
.U p E = −
(Potential energy on dipole in a uniform field)
• The potential energy has its minimum value U = – pE
(i.e., its most negative value) at the stable equilibrium
position, where θ = 0 and P
is parallel to E
.
• The potential energy is maximum when ø = π and P
is
anti-parallel to ; E
then U = + pE .
• At ,2
π θ = where P
is perpendicular to , E
U is zero.
2a
p
F qE + =
F qE − = −
- q
+ q
θ
E
d
– q
p
+ q
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Example 11
• Figure shows an electric dipole in a uniform electric field
with magnitude 5 ×10 –5
N/C directed parallel to the plane
of the figure. The charges are ± 1.6 × 10 –
19
C ; both lie in
the plane and are separated by 0.125 nm = 0.125 × 10 –
9
m. Find (a) the net force exerted by the field on the
dipole; (b) the magnitude and direction of the electric
dipole moment; (c) the magnitude and direction of the
torque; (d) the potential energy of the system in the
position shown.
Solution:
(a) Since the field is uniform, the forces on the two charges
are equal and opposite. and the total force is zero.
(b) The magnitude p of the electric dipole moment P
is
p = qd = (1.6 × 10 –
19
C) (0.125 × 10 –
9
m)
292.0 10 C mP −⇔ = × −
The direction of p
is from the negative to the positive
charge, 145° clockwise from the electric-field direction
(Fig.)
(c) The magnitude of the torque is
τ = pE sin θ = (2.0 ×10 –
29
) (5×10 –5
) (sin 145°)
τ = 5.7 × 10 –
24
Nm
From the right-hand rule for vector products, the direction
of the torque p E τ = ×
is out of the page. This
corresponds to a counterclockwise torque that tends to
align P
with . E
(d) The potential energy isU = – pE cos θ
= –(2.0 ×10 –
29
) (5×10 –5
) ( cos145°) = 8.2 × 10 –
29
J
Quick Exercise
NCERT Questions
1.10 An electric dipole with dipole moment 4 × 10 –9 C m is
aligned at 30° with the direction of a uniform electric field of
magnitude 5×104 NC –1. Calculate the magnitude of the torque
acting on the dipole.
1.27 In a certain region of space, electric field is along thez-direction throughout. The magnitude of electric field is
however, not constant but increases uniformly along th
positive z-direction, at the rate of 105 NC –1 per metre. What ar
the force and torque experienced by a system having a tota
dipole moment equal to 10 –7 Cm in the negative z-direction?
Question for practice
1. A and B are two points on the axis and the perpendicular
bisector respectively of an electric dipole. A and B are fa
away from the dipole and at equal distances from it. The
fields at A and B are A E
and . B E
(a) A B E E =
(b) 2 A B E E =
(c) 2 A B E E = −
(d) 1
,2
B B E E = and B E
is perpendicular to A E
2. An electric dipole is placed at the origin and is directed
along the x-axis. At a point P, far away from the dipole
the electric field is parallel to the y-axis. OP makes an
angle θ with the x-axis.
(a) tan 3θ =
(b) tan 2θ = (c) θ = 45o
(d) 1
tan2
θ =
3. An electric dipole of moment p
is placed in a uniform
electric field , E
with p
parallel to . E
It is then rotated
by an angle θ. The work done is
(a)
pE sin θ (b) pE cos θ (c) pE (1 – cos θ) (d) pE (1 – sin θ)
Answers:
1. c
2. c
3. c
- q
+ q
035
E
0145
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Final Exercise
NCERT:
Solve all problems on Electric field and dipole.
H C Verma:
Page No: 122-123
Qs No: 34 to 52.
Qs No: 70 to 75
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