electric field 2015.pdf

8/16/2019 Electric Field 2015.pdf

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Electric Field Author: Pranjal K. Bharti (B. Tech., IIT Kharagpur) www.vidyadrishti.org

P a g e | 1 Concept: JB 20, Near Jitendra Cinema, Bokaro Mb: 7488044834

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)


4/16



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


9/16



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


10/16



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


11/16



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


12/16



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


13/16



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


14/16


15/16



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


16/16


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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