notes for std. xiith physics field 2015.pdf · • the electrostatic field at a point due to a...

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 Field Pranjal K. Bharti, B. Tech., IIT Kharagpur

© 2007 P. K. Bharti

All rights reserved.

www.vidyadrishti.org



Prerequisite

• Vectors • Newton’s Laws • Coulomb’s Law

Please revise these topics before studying Electric Field

Topics:

1. Introduction 2. Electric field Intensity

• Definition of electric field intensity • Important points about electric field intensity • Why test charge is small? • Physical significance of electric field

3. Electric field due to a point charge Direction of electric field

4. Force on a charged placed in an electric field Direction of force

5. Electric field lines • Electric field lines to common charges • Electric field lines due to some simple charge

configurations • Properties of electric field lines • Field lines due to surface charges • Field lines close to surface of a conductor

1. Introduction

• We know that electrostatic force (which is an action at a distance force) between two charged particles is given by

1 22

q qF k

r=

• What happens if the distance r between these two particles starts increasing?

• Clearly electrostatic force magnitude will start decreasing and in the limiting case when distance becomes very large (theoretically at infinity), force F will be zero.

• In other words, we can say that electrostatic force magnitude is appreciable up to some distance; after that we can simply ignore this force as the distance becomes large.

• In other words, we can say that there is some kind of electrostatic field near a charge. The magnitude of this field decreases as the distance from this charge is increased. Any other charge will interact with this field (meaning experiences a force) when brought into this field.

• Every charge has its own electrostatic field, where its interaction with another charge is appreciable.

2. Electric field intensity

• Before going to write definition of electric field intensity (or simply electric field) let us see the technique adopted to define electric field.

• Suppose we are interested to find electric field intensity due to a charge Q at a point P. This charge Q is known as source charge Q.

• We need a second kind of small positive charge qo at

point P. This second kind of charge is very small compared to that of source charge Q and is known as test charge q

o.

• In the next step, we find Coulomb’s electrostatic force F

acting on test charge q

o because of source charge Q. In

the figure direction of F

is shown assuming source charge Q to be positive.

• Electric field E

because of source charge Q at point P is nothing but the electrostatic force F

per unit small

positive test charge qo placed at point P.

• Thus,

0 0

0

limq

FEq→

=

(Electric field intensity)

Clearly, the direction of E

is along F

.

• Definition: The electric field intensity or simply electric field E

due to a source charge Q at a point is defined as

the electrostatic force F

per unit small positive test charge q

o acting on q

o due to Q placed at that point.

0 0

0

limq

FEq→

=

(Electric field intensity)

Important points about electric field

1. In the LHS of the expression, E

is the electric field due to source charge Q but in the RHS, positive test charge q

o

appears. Be careful of this. 2. F

is the electrostatic force acting on test charge q

o

because of source charge Q. 3. Electric field intensity is a vector quantity. 4. The direction of E

is along F

.

5. SI unit of electric field: N/C = newton per coulomb Another SI unit: V/m = volt per meter

Q

Source charge

P

F

Positive Test charge

+q0

E



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

FEq

=

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 q

0.

• 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 electromagnetic waves, which then propagate with the speed of light c, reach q2 and cause a force on q2. The notion of field accounts for the time delay. Thus, even though electric and magnetic fields can be detected only by their effects (forces) on charges, they are regarded as physical entities, not merely mathematical constructs. They have an independent dynamics of their own, i.e., they evolve according to laws of their own. They can also transport energy. Thus, a source of time dependent electromagnetic fields, turned on briefly and switched off, leaves behind propagating electromagnetic fields transporting energy.

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 the force is given by Coulomb’s Law:

0 02 2

0 0

1 1= 4 4

qq q qF

r rπε 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

FEq

=

…(2)

Using eqns. (1) & (2) we get:

20

1 4

qE

rπε=

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 towards a 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 charge

– 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)



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 20

1 4.0 109.0 10 4 2.0

9.0N/C

qE

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 the numerator of electric field expression.

9.0N/CE = • 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 = E

1 + 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 qi.

Example 2

• Four point charges qA = 2 μC, q

B = –5 μC, q

C = 2 μC, and

qD = –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 all 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 q

A = 2 μC :

• Since charge qA

is positive in nature therefore, electrostatic field E

A at point O will be away from q

A, i.e.

EA is towards OC.

• Now, electrostatic field at point O due to charge qA when

placed in air is given by:

( )6

92 2

0

6

2 101 9 104 0.1/ 2

3.6 10 N/C (alongOC)

AA

A

qE

r

E

πε

−×= = × ×

⇒ = ×

• Field at point O due to charge qB = -5 μC :

• Since charge qB

is negative in nature therefore, electrostatic field E

B at point O will face towards q

B, i.e.

EB is towards OB.

• Now, electrostatic field at point O due to qB when placed

in air is given by:

( )6

92 2

0

6

5 101 9 104 0.1/ 2

9 10 N/C (alongOB)

AB

B

qE

r

E

πε

−− ×= = × ×

⇒ = × • Similarly,

Field at point O due to charge qC = 2 μC :

63.6 10 N/C (alongOA)CE = ×

• Field at point O due to charge qD = -5 μC :

69 10 N/C (alongOD)DE = × • 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)

qB = –5 μC A B

O

EA C D

qA =2 μC

qB = 2 μC

EC EB

ED

qD = –5 μC

E Positive charge

E

Negative charge



Continuous charge distribution

• Linear charge densityλ:

If a charge Q is uniformly distributed along a line of length L, the linear charge density λ is defined by

QL

λ =

S. I. unit: C/m

• Surface charge density σ :

If a charge Q is uniformly distributed on a surface of area A, the surface charge density σ is defined by

QA

σ =

S. I. unit: C/m2

• Volume charge density ρ :

If a charge Q is uniformly distributed throughout a volume V, the volume charge density ρ is defined by

QV

ρ =

S. I. unit: C/m3

Test your understanding

1. There is a disc of radius R and negligible thickness. A charge Q is distributed uniformly on disc. What kind of charge density we should consider for disc: linear charge density, surface charge density or volume charge density? Also find the appropriate charge density.

2. A charge Q is distributed uniformly over a ring of radius R. Find its linear charge density.

3. A charge Q is distributed uniformly over a square of side a. Find its surface charge density.

4. A charge Q is distributed uniformly over a sphere of radius R. Find its volume charge density.

Ans:

1. Surface charge density

2

QR

σπ

=

2. 2

QR

λπ

=

3. 2

Qa

σ =

4. 3

34

QR

ρπ

=

Example 3 (For Boards)

A ring-shaped conductor with radius a carries a total positive charge Q uniformly distributed around it as shown in figure. Find the electric field at a point P that lies on the axis of the ring at a distance x from its center. Solution: • This is a problem in the superposition of electric fields.

Basic idea is that the charge is distributed continuously around the ring rather than in a number of point charges.

• As shown in Fig., we imagine the ring divided into infinitesimal segments of length dl. Each segment has charge dQ and acts as a point-charge source of electric field. Let dE be the electric field from one such segment; the net electric field at point P is then the vector sum of all contributions dE from all the segments that make up the ring.

• To calculate dE, note that the distance r from a ring

segment to the point is r = 2 2x R+ . Hence the magnitude of this segment's contribution to the electric field at P is

20

2 20

14

1 ...(i)4

dQdE

rdQ

dEx R

πε

πε

=

⇒ =+

• The calculation of net electric field E is greatly simplified because the point P is on the symmetry axis of the ring. Consider two segments at the top and bottom of the ring. The contributions dE to the field at P from these segments have the same x-component i.e., dEcos𝜃𝜃 but opposite y-components i.e, dEsin𝜃𝜃. Hence the total y-component of field due to this pair of segments is zero. So the field at P is described completely by its x-component dEcos𝜃𝜃.

• Now, 2 2

cos ...(ii)xx R

θ =+

• To find the total x-component dEcos𝜃𝜃 of the field at P, we integrate this expression over all segments of the ring:

2 2 2 20

1cos4

dQ xdEx R x R

θπε

= + +

∫ ∫

(from (i) & (ii))

( )3/22 2

0

1cos4

xdE dQx R

θπε

⇒ =+

∫ ∫

x

R 2 2x R+

dE

dQ dQ

x

R

dE

dE

x

dEsin𝜃𝜃

dEsin𝜃𝜃

dEcos𝜃𝜃

dEcos𝜃𝜃

dQ



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 20 0

3/22 20

1 1cos4 4

14

x xdE dQ Qx R x R

QxEx R

θπε 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

( )3/22 20

14

QxE ix 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 with symmetry 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. If we let λ represent 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

at

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 20 0

1 1 . 4 4

dq dsdE ir r

λπε 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 cancel (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 the differential electric fields set up by all the differential elements of the rod. We can write the x component set up by ds as

( )20

1cos cos 4

dE ds iirλθ θ

πε=

• Equation (ii) has two variables, θ and 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 dθ 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; that will 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 dr

dr r

r

iiir

λθ θ θπε

λ λθ θ θπε πελ

πε

λπε

= =

= =

= − −

=

∫ ∫

∫

• To evaluate λ, we note that the rod has an angle of 120o and so is one-third of a full circle. Its arc length is then

2 ,3rπ and its linear charge density must be

dE

x P θ θ

dEsin𝜃𝜃

y

dEcos𝜃𝜃

ds

ds

dE dEcos𝜃𝜃

dEsin𝜃𝜃

x P

y ds

dθ r



charge 3length 22

3

Q Qr r

λπ π

= = =

• Substituting this into Eq. (iii) and simplifying give us

( )

0 0

2 20

3 3 3 4 4 2

3 3 Answer8

QEr r r

Er

λ λπε πε π

λπ 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 20

3 3 8

E irλ

π ε=

Quick Exercise

The figure here shows three nonconducting rods, one circular and two straight. Each has a uniform charge of magnitude Q along 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

.QdQ dxL

=

The electric, field at P due to this element is

( )20

.4

dQdEAPπε

=

By symmetry, the resultant field at P will be along OP (if the charge is positive). The component of dE along OP is

( ) ( )2 3/22 20 0

cos .4 4

dQ OP a Q dxdEAPAP L a x

θπε πε

= =+

Thus, the resultant field at P is

( )/ 2

3/22 20 /2

cos

....(i)4

L

L

E dE

aQ dxL a x

θ

πε −

=

=+

∫

∫

We have x = a tanθ or dx = a sec2θ dθ

Thus, ( )

2

3/2 3 32 2

secsec

dx a daa x

θ θθ

=+

∫ ∫

( )2 2 2 1/22 2

1 1 1cos sin .xda a a a x

θ θ θ= = =+

∫

From (i),

( )

( )

/ 2

2 1/22 20

/2

2 1/22 20

2 20

4

2 4 4

.2 4

L

L

aQ xELa a x

aQ LLa L a

Qa L a

θε

πε

πε

−

= +

= +

=+

Quick Exercise

1. Two charged particles +5 μC and +10 μC are placed 20 cm 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 of 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 x 0

θ

P

x

P

+

(c)

–

y

x P

–Q

(a)

+Q

y



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 jrπ ε

b) 2 2

0

4

q jrπ ε

−

c) 2 2

0

2

q jrπ ε

−

d) 2 2

0

2

q jrπ ε

8. Let ( ) 4

Qr rR

ρπ

= be the charge density distribution of a

solid sphere of radius R and total charge Q. Find the magnitude of electric field at a point p inside the sphere at a distance r1 from the centre of the sphere.

a) 2 20 1

4

Qrπ ε

b) 2

12 4

0

4

QrRπ ε

c) 2

12 4

0

3

QrRπ ε

d) 0 9. Let there be a spherically symmetric charge distribution

with charge density varying as ( ) 054

rrR

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 rR

ρε

−

b) 0

0

4 5 3 3

r rR

ρε

−

c) 0

0

5 4 3

r rR

ρε

−

d) 0

0

3 5 4 4

r rR

ρε

−

10. Four point charges each of charge +q are rigidly fixed at

the four corners of a square planar soap film of side a. The surface tension of the soap film is S. The system of charges and planar film are in equilibrium, and

1/2 Nqa kS

=

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 krρ = , 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 = 2L

4. (i) 03

qπε

(ii) 05

qπε

5. At a point on the perpendicular bisector of the two charges 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



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 similar charges.

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 positive charges

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



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 would lead 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 closed loops.

Quick Exercise

NCERT Questions

1.7 (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not? (b) Explain why two field lines never cross each other at any point? 1.26 Which among the curves shown in Fig. cannot possibly represent electrostatic field lines?

More questions for practice

1. A few electric field lines for a system of two charges Q1 and Q2 fixed at two different points on the x-axis are shown in the figure. These lines suggest that a) |Q1| > |Q2| b) |Q1| < |Q2| c) at a finite distance to

the left of Q1 the electric field is zero

d) at a finite distance to the right of Q2 the electric field is zero

2. Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resulting lines of force should ne sketched as in :

+ +

+ + +

+

+ +

+ +

+ + + +

+ +

conductor



5. Force on a charged placed in an electric field

• If the field E

at a certain point is known, rearranging Eq.

0

FEq

=

, gives the force 0F 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

qEam

=

• 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 electron

due to this field is

( )14 2 1.76 10 m/s downwardFF 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 ×10

14 )× 0.02

u = √ 7.04 ×10 12

u = 2.65 ×10 6 m/s

• Hence, minimum speed = 2.65 ×10 6 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

uHg

θ=

• Replacing g with a we get maximum height as:

( )( )

26 2 02 2

14

2.65 10 sin 30 sin 0.015m=1.5cm2 2 1.76 10

uHa

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



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 is eEam

= 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)ltv

=

The upward component of the velocity of the electron as it emerges from the field region is

0y

eElatmv

υ = =

The horizontal component of the velocity remains vx = v0. The angle θ made by the resultant velocity with the original direction is given by

20

tan y

x

v eElv mv

θ = =

Thus, the electron deviates by an angle

120

tan eElmv

θ −=

Example 10

Positive charge Q is distributed uniformly over the ring of radius 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 force on the particle. Assume x << R, find the time period of oscillation of the particle as it is released from there.

Solution: • From example 3, electric field due to a ring on its axis at a

distance x from its centre is given by expression

( )3/22 20

14

QxEx Rπε

=+

(along x axis)

• Therefore , force on negative charge q is given by

( )3/22 20

14

qQxF qEx Rπε

= =+

(opposite to x axis. Why?)

• For the case, when x << R, we have ( )2 2 2x R R+ ≈

• Hence,

( ) ( )3/2 3/22 2 20 0

1 14 4

qQx qQxFx R Rπε πε

= ≈+

30

14

qQF xRπε

⇒ = (opposite to x axis.)

• As this force is opposite to x axis as well as directly proportional to x, we can write

30

30

3 30

1 4

1 where 4

Time period

16 2

qQF xR

F kxqQkR

mRmTk qQ

πε

πε

π επ

= −

⇒ = −

=

∴

= =

Quick Exercise

NCERT Questions

1.14. Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

1.25. An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 104 NC–1 in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm–3. Estimate the radius of the drop. (g = 9.81 m s–2; e = 1.60 × 10–19 C).

1.34. Suppose that the particle in Example 9 is an electron projected with velocity v0 = 2.0 × 106 m/s . If E between the plates separated by 0.5 cm is 9.1 × 102 N/C, where will the electron strike the upper plate?

Question for practice

Q. A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of strength

581 10 V/m7π

× . When the field is switched off, the drop is

observed to fall with terminal velocity 32 10 m/s−× . Given viscosity of the air = 5 21.8 10 Ns/m−× , and the density of oil =

3900 kg/m . Find the magnitude of q. Ans: 198 10 C−×

v0

θ l

E



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, 𝜏𝜏 is directed into the page).

• Hence the magnitude of the net torque is twice the

magnitude of either individual torque: 𝜏𝜏 = (qE)(2a sin θ)

• Now, since dipole moment p

has a magnitude q2a, therefore torque becomes 𝜏𝜏 = (qE)( 2a sin θ) = (q2a)(E sin θ) = pE sin θ

• We can write this torque in terms of vector product

p Eτ = ×

(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 of

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 dW done by a torque τ during an infinitesimal angular displacement dθ is given by equation dW = τ dθ.

• Because the torque is in the direction of decreasing θ (as θ is anti-clockwise, whereas τ is clockwise), we have dW = τ dθ = – pE sin θ dθ

• In a finite angular displacement from θ1 to θ2 the total work done on the dipole is

2

1

2 1sin cos cosW pE d pE pEθ

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 potential energy. W = – (U1 – U2 )

• Thus, we see that a suitable definition of potential energy U for this system is U (θ) = – pEcos 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 ,2πθ = where P

is perpendicular to ,E

U is zero.

2a

p

F qE+ =

F qE− = −

- q

+ q

θ

E

d

– q

p

+ q



The field of an electric dipole at a point on its axis

• The electric field of the pair of charges (–q and +q) at any point in space can be found out from the superposition principle.

• Let the point P be at distance r from the centre of the dipole on the side of the charge +q, as shown in Fig.

• As point P is at a distance (r – a ) from charge +q,

therefore, magnitude of electric field E+

due to +q at P is

( )( )2

0

leftward4

qEr aπε

+ =−

• As point P is at a distance (r + a ) from charge –q, therefore, magnitude of electric field E−

due to –q at P is

( )( )2

0

rightward4

qEr aπε

− =+

• Now we know that dipole moment is a vector quantity whose magnitude is p = q(2a) and direction is from negative to positive charge.

• Therefore, net electric along the direction of dipole axis is given by

( ) ( )

( )( )

( ) ( )

2 20

2 2 22 2 2 2 2 20 0 0

1 14

2 24 1 1 2= = 4 4 4

(rearranging and using =2 )

qE E Er a r a

r q aq ar rpEr a r a r a

p qa

πε

πε πε πε

+ −

= − = −

− +

=− − −

• Since net electric field is taken along dipole axis, we can write net electric field at point P in vector form as,

( )22 20

1 24

rE pr aπε

=−

• Electric field due to a dipole on its axis at a very large distance from its centre

• In this case, we can write r > > a

•

2 2 2r a r∴ − ≈

• Hence, net electric field

( )30

2 4

E p r arπε

= >>

• Clearly,

3

1 Er

∝

The electric field due to dipole at a point on

Equatorial Plane

The equatorial plane of the dipole is a plane perpendicular to the dipole axis through its centre.

The magnitudes of the electric fields due to the two charges +q and –q are given by

2 2 2 20 0

1 1 and 4 4q q

q qE Er a r aπε πε+ −= =

+ +

Clearly qE+

and qE−

have equal magnitudes. The directions

of qE+

and qE−

are as shown in figure. Clearly, the normal

components qE+ sinθ and qE− sinθ cancel away. The

components qE+ cos𝜽𝜽 and qE− cos𝜽𝜽 along the dipole axis add

up. Therefore, net electric field at P is in rightward direction and is given by

( )cos cos cos q q q qE E E E Eθ θ θ+ − + −= + = +

Also using little Trigonometry, we have

2 2cos a

r aθ =

+

Therefore, net electric field at P is in rightward direction and is given by

( ) ( )( )

3/2 3/22 2 2 20 0

2 E = (rightward)4 4

2

qa p

r a r a

p qa

πε πε=

+ +

=

Since, net electric is along rightward direction and direction of dipole moment is leftward, we can write in vector form as

( )3/22 20

4

pEr aπε

= −+

At large distances (r >> a), this reduces to

30

4

pErπε

= −

• Clearly,

3

1 Er

∝

E−

+q -q

E+

E

P

P

r

2a

2 2r a+

𝜽𝜽 𝜽𝜽

𝜽𝜽 𝜽𝜽 E+ cos𝜽𝜽

E− cos𝜽𝜽

E+ sin𝜽𝜽

E− sin𝜽𝜽 P

E−

+q

E+

r

-q

2a

p



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 Eτ = ×

is out of the page. This

corresponds to a counterclockwise torque that tends to align P

with .E

(d) The potential energy is U = – 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 the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 105 NC–1 per metre. What are the force and torque experienced by a system having a total 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 far away from the dipole and at equal distances from it. The

fields at A and B are AE

and .BE

(a) A BE E=

(b) 2A BE E=

(c) 2A BE E= −

(d) 1 ,2B BE E= and BE

is perpendicular to AE

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) 1tan2

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) pEsin θ (b) pEcos θ (c) pE(1 – cos θ) (d) pE(1 – sin θ)

Answers: 1. c 2. c 3. c

- q

+ q

035

E

0145



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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