Notes on Electrostats for AIEEE

7/29/2019 Notes on Electrostats for AIEEE

1/23

Coulombs Law

ELECTROSTATICS

The branch of physics dealing with charges at rest

Under electrostatic condition, the electric field inside the metallic conductor remains zero.

Application in instruments like microphones, cathode ray tube, capacitors, van de Graaff

generator, photocopier, electrostatic shielder etc.

CHARGE

Benzamin franklin and Milikan

Fundamental property exhibited due to deficiency or sufficiency of electrons

Two types of charges : Positive and negative charge

Unlike charge attract one another and like charges repel

Exhibit properties like Quantization, superposition, invariance, conservation, induction etc

Negatively charged bodies : electron sufficient & slightly mass sufficient also

QUANTISATION

Q = ,n wher n is interger

e = - 1.6 x 10-19C

Exception : Quark particle

Not continuous transfer but in discrete packet

OBJECTIVE 1

How many electrons are there in 1C of charge?i) 1 ii) 6.25 x 10

18iii) 1.6 x 10

-19iv) None

Solution: (ii)

18

191025.6

106.1

1 =

==

qn


2/23

OBJECTIVE 2

Minimum charge that a body can retain is

i) 1.6 x 10-19

C ii) 3.2 x 10-19

iii) 2.4 x 10-19

C iv) None

Solution: (i)

The minimum charge that a body can retain is one electronic charge

SUBJECTIVE 3

What would be the interaction force between two copper spheres, each of mass 1gm, separated by

the distance 1m, if the total electronic charge in them differs from the total charge of the nuclei by

one percent?

Solution:

Atomic weight of 54.635.6329 =C

No. of atoms in 1gm2310023.6

54.63

1 =

No. of electrons 2910023.654.63

1 23 =

Charge C19106.1100

129023.6

54.63

1 =

(Since 1% of electrons has been transferred)

= 4.39 x 102C

2

21

04

1

r

qqF =

( )( )

N152

229

1074.11

1039.4109=

=

Electric force between two charged particle

Inversely proportional to the square of separation between the two particles and directedalong the line joining charges

Proportional to the product of the magnitudes of charges q1 & q2 on the two particles

Attractive if charges are of opposite sign and repulsive if having same sign

Mathematically,2

21

04

1

r

qqF =

2212

0 /1085.8 NmC=


3/23

2

29

0

1094

1

C

Nm=

Where0 is known as permittivity of free space.

Victorial form

Force on 2 due to^

2

21

0

21 .4

11 r

r

qqF ==

r

r

qq.

4

13

21

0

r should be in direction of Colombian force.

Force on 2 due to

== r

r

qqF

3

21

0

214

11

( ) ( ) ( )

( ) ( ) ( )[ ].

4

12/322

^^^

21

0ififif

ififif

zzyyxx

kzzjyyixxqq

++

++

=

OBJECTIVE 4

Calculate the ratio of electrostatic to gravitational interaction between two electrons

Solution:

2

2

2

2

04

1

r

GmFg

r

eFe ==

2

0

2

22

0

22

44 Gm

e

Gmr

re

Fg

Fe

=

=


4/23

( )

( )

42

23111

2199

10

101.91067.6

106.1109

=

Electrostatic force is much stronger compared to gravitational force.

OBJECTIVE 5

A point charge q is situated at a distance d from one end of a thin non conducting rod of length

L having a charge Q (uniformly distributed along its length) as shown in figure. Find the

magnitude of electric force between the two.

Solution:

Consider an element of rod, of length d x, at a distance x from the point charge q as shown in

figure. Treating the element as a point charge, the force between q and charge element will be as:

Charge on elementary strip dxL

Q=

Coulombian force acting on strip

2

04

1

x

qdx

L

Q=

Therefore, +

=Ld

dx

dx

L

qQF

2

04

1

Ld

xL

qQ+

= 1

4

1

0

+=

LddL

qQ 11

4

1

0


5/23

( )LddqQ

+=

04

SUBJECTIVE 6

A thin wire ring of radius R has an electric charge q. What will be the charge of the force

stretching the wire if a point charge q0 is placed at the rings centre?

Solution:

When the wire is given charge q, it comes in tension, called hoop tension. When another charge

q0 is placed at the centre of the ring, the tension in the wire increases to keep this segment into

equilibrium again.

Consider an elementary strip subtending angle d at its centre.

The string is in equilibrium in X-direction. For equilibrium in y-direction,

2sin2

dTFe= (i)

Charge on elementary strip

dq =2

2

0

0 24

1

R

qdqF

=

.(ii)

Substituing (ii) in (i)

2

0

0

.

24

1

22

R

dqqdT

=


6/23

2

0

2

0

8 R

qqT

=

SUBJECTIVE 7

The identical beads each having a mass m and charge q, when placed in hemi spherical bowl of

radius R with frictionless, non-conducting walls, the beads move, and at equilibrium they are at a

distance R apart. Determine the charge on each bead.

Solution:

The bead is under electrostatic equilibrium and force acting are its weight, electrostatic force and

Normal Reaction.

For equilibrium

N sin 600 = mg ..(i)

2

2

0

0

4

160cos

R

qN =

..(ii)

Dividing (i) by (ii),

2

2

00 460tanq

Rmg =


7/23

3

42

02 Rmgq

=

2/1

2

02

3

4

=

Rmgq

SUPERPOSITION

The interaction between any two charges completely unaffected by the presence of other charges

Net resultant force on any charge equals vectorial sum o coulombian force contributed by

individual particle in surroundings

Net force on q1 = 432

++ FFF

All vectorial technique to be applied for calculation of net force.

For three dimensional distribution of charges, the coulombs law in vectorial form should be

used.

OBJECTIVE 8

Three charges q are placed at the vertices of equilateral triangle of side a. Find the electrostatic

force on the charge Q kept at the centroid of triangle.

Solution :

2

04

1

r

qQF =

Since three equal forces are acting at a point making angle of 1200, net resultant force is zero.

Concept : The net force acting on the charge kept at centre of regular polygon is zero due to

charges kept at the vertices provided they are of same magnitude & sign.


8/23

OBJECTIVE 9

Five point charges, each of value +q coulomb, are placed on five vertices of a regular hexagon of

side Lm. The magnitude of the force on the point charge of q coulomb placed at the centre of

hexagon will be:

i) Zero ii)2

0

2 4/ La iii) 202

4/2 Lq iv) None of these

Solution :

If there would have been charge at the left vertices also, net force at the centre will be zero.

The coulombian force due to five charge should balance the coulombian force exerted by the

charge dept at left vertices. Force due to all charges should be along OA.

2

2

0

.

4

1

L

qF

=

OBJECTIVE 10

There equal charges, each equal to q, are placed at the vertices of an equilateral triangle of side a

find the force on the left most corner of charge q.

Solution:


9/23

2

2

0

.4

1

a

qF

=

The net resultant force

= F1

FFFFF 360cos2 022 =++=

2

0

2

4,3

a

q

=

OBJECTIVE 11

A charge Q is placed at each of the opposite corner of a square. A charge q is placed at each at

each of the two other two corners. If the resultant force on Q is zero, then:

a) qQ 2= b) qQ 2= c) qQ 22= d) qQ 22=

Solution:

2

04

1

a

QqF =

2

2

0 24

1

a

QF =

Net resultant = FF 2+

0224

1 2

2

0

=

+= qQQ

a

qQQ

22

2

=

qQ 22=

OBJECTIVE 12


10/23

A cube of edge a carries a point charge q at each corner. Show that the magnitude of the

resultant force on any one of the charge is

2

0

2261.0

a

qF

=

Solution:

Refer to coulombs law in vector form

34

.

0

2

=r

rqF

3

^

0

2

41

4 a

iaqF

=

^

2

0

2

4

i

a

q

;4

;4

^

2

0

2

21

^

2

0

2

21

=

=

k

a

qFj

a

qF

( )

=

= ^^

2

0

2

3

^^

0

2

31

22424ji

a

q

a

jaiaq

F

=

=

^^

20

2

7120

^^2

51 224;224 kia

q

Fa

kiq

F

( ) 20

^^^2

^^^

3

0

2

31

334.

34 a

kjiq

kajaiaa

qF

=

=

According to superposition principle,

++++= 8161514121 FFFFFF


11/23

++

++=2

1

33

11

4

^^^

2

0

2

kjia

q

2

0

2

2

0

2 261.0

23

33263.3

4 a

q

a

qF

=

++=

SUBJECTIVE 13

Two point charges A and B have charges CC 2&2

1respectively and position vectors

++

^^^

kji and

^^^

kji respectively. Find the force on the charge at A due to B.

Solution :

Force on A and to B

= ABF

= BA

AB

qq BA ..

4

13

0

3

9 22

1109

=BA

rr BA

( )3

^^^9

32

222109

+

=kji

=

^^^9

222324

109kji


12/23

STRING RELATED PROBLEMS

Draw the free body diagram of force

Write down the condition for equilibrium

Solve the equation after co-relating situation.

SUBJECTIVE 14

Two small charged spheres, each of charge q and each of mass m, are suspended form the same

point by silk threads of length 1

(a) Find the distance between the spheres x, if x < < < 1, for the equilibrium of charges.

(b) If the charge starts leaking form both the spheres at the constant ratetd

Qd, find the velocity of

approach of the spheres.

Solution:

The force acting on the charged spheres are shown in the equilibrium.

FT =sin ..(i)

mgT =cos ..(ii)

Dividing (i) by (ii),

mg

F=tan

If x < < < 1, then sintan

mgx

q

l

x

=

2

2

04

1

2

mg

lqx

0

23

2=


13/23

Or 3/23/1

0

.

2

q

mg

lx

=

ii)2/30 .

2x

l

mgq

=

Differentiating both side with respect to t

dt

xdx

l

mg

td

Qd= 12/30 .

2

3.

2

Vxl

mg=

2

32 0

332

/2

0

=mg

atdlV

SUBJECTIVE 15

If the two balls in the above mentioned problem contains different charges q 1 > q2. The deflection

in the cases are21 and as shown in figure. The two balls are of same mass, then

i)21 > ii) 12 < iii) 21 = iv) Cannot be

predicated

Solution: (iii)

;tanmg

F= since two balls are of same mass and electrostatic force is also same.

SUBJECTIVE 16

Three small balls, each of mass m are suspended separately from a common point by silk threads,

each of length 1. The balls are identically charged and hung at the corners of equilateral triangle

of side a (a < < < 1). What is the charge on each ball?

Solution:


14/23

The lengths of the threads are same, the mass of each ball is same, the balls are identically

charged, and the balls are suspended form the same point. Under the equilibrium, the balls will

make an equilateral triangle in a horizontal Plane.

Let ABC be the triangle formed by them.

32

3

2

aP

P

a==

The force on one of the balls at c are

(i) Coulombic repulsion by charges at A & B

FFFFFF 360cos2 0221 =++=

2

0

2

4

3

a

q

=

ii) mg, acting down ward

iii) Tension in thread

FT =sin (1)

mgT =cos (2)

From (1) and (2)

2

0

2

4

3tan

'tan

amg

q

mg

F

== (3)

If tan is small, then sintan


15/23

l

a

3sin = (4)

Substituting (4) in equal no. (3),

2

0

2

4

3

31 amg

qa

=

3

4 302 mgaq

=

3

4 30 amgq

=

SUBJECTIVE 17

A particle A having a charge of 5 x 10 -7C is fixed in a vertical wall. A second particle B of mass

100g and having equal charge is suspended by a silk thread of length 30cm form the wall. The

point of suspension is 30cm above particle A. Find the angle of the thread with the vertical when

it stays in equilibrium?

Solution:

2sin2

lAB =

So the electrostatic force between the charge will

Be2

2

0

2sin2

4

1

=

l

qF .(i)

Lamis theorem can be applied in order to solve this question

( ) ( ) ( )2/90sin180sin2/90sin 000 +=

=

+mgFT


16/23

Or,2/cossin

2cos

mgFT

==

2sin2

mgF = .(ii)

Equating (i) and (ii),

2sin16

2sin2

2

0

2

2

=l

qmg

20

23

322sin

lmg

q

=

In the given problem,

q = 5 x 10-7C

mcmkggm 3.0301,10

1100 ====

914914

3 109108

25

3.03.018

1091025

2sin =

=

Or

033

1715.02sin0032.02sin ===

or

CONSERVATION

Total charge of isolated system remains constant,

Appearance of charge is due to slight disturbance of electrical neutrality of each.

OBJECTIVE 18

A certain charge Q is to be divided into two parts, q and (Q-q). What is the relationship of Q to q

of the two parts, placed at a given distance apart are to have maximum coulomb repulsion?

Solution:

( )2

04

1

r

qQqF

=

For F to be maximum, 0=dq

dF,


17/23

[ ] 02.4

12

0

== qQrdq

Fd

2

02Q

qqQ ==

SUBJECTIVE 19

Two identical oppositely charged metallic sphere placed 0.5m apart attract each other with a force

of 0.108 N. When they are connected by a Cu wire, they begin to repel each other with a force of

0.036 N. What are the initial charges on them?

Solution:

Let the charge be q1 and q2 before connecting by cu-wire. After connection, charges on each will

be

2

21 qq

( )108.0

5.04

12

21

0

=qq

..(i)

( )

036.0

5.0

2

4

12

2

21

0

=

qq

(ii)

After solving these two equations,

CqCq 626

1 10103 ==

OSCILLATIONS

Under equilibrium condition, net force on particle = 0

By displacing slightly, the net force should be restoring in nature.

If net force displacement, it will be making SHM

Exampls


18/23

SUBJECITIVE 20

Two particles A and B, each carrying a charge Q, are held fixed with O separation 2l between

them. A particle C having mass m and charge q is kept at the middle points of the line AB.

If the particle c is displaced through a distance x and x


19/23

( ) ( ) 2/12222042

lx

x

lx

qQ

+

+=

( ) 2/32204 lxxqQ

+=

If x


20/23

2/32

2

^^^

0

1

3

322

4

+

=

ax

kxja

ia

qQF

2/32

2

^^^

0

2

3

322

4

+

=

ax

kxja

ia

qQF

2/32

2

^^

0

3

3

34

+

=a

x

kxja

qQF

2/32

2

0

^

321

34

3

+

=++=

ax

kxqQFFFF

If x


21/23

Third charge can be of any sign for its equilibrium. But it has to be negative in order to make

force on q and 4q equal to zero. Then only the entire system will be in equilibrium. For net force

on Q1

to be zero.

( )2

4

4

1

4

12

0

2

0

=

=x

xL

xL

qQ

x

qQ

3/2 LxxxL ==

For net force on q to be zer,

( ) 9

4

3

2

4

4

1

3/4

12

2

0

2

0

qQor

L

q

L

qQ=

=

EFFECT OF MEDIUM

In medium the electrostatic force reduces by factor of K (dielectric constant)

If charge is kept in interfacing media, then effective distance has to be calculated for calculation

of force

Effective distance2211 rkrk +=

Force in interfacing media

[ ]22211021

4 rkrk

qq

+=

OBJECTIVE 23If charge q1 and q2 kept in interfacing media of three different dielectric constant 9, 81 and air.

The thickness of each median in interfacing system is equal. If the distance between the charge is

r. Then the coulomb force acting between than.

Solution:

The effective distance in medium rk=

Net effective distance


22/23

( )3

13193

3381

39

3

rrrrr=++=++=

Colombian force

( ) 20

21

2

21

0 1694

9

3/134

1

r

qq

r

qq

==

SUBJECTIVE 24

Two balls each having a density d are suspended form Q common point by two insulated string of

equal length. Both the balls have same mass and charge. In equilibrium, each string makes

angle with the vertical. Both the balls are immersed in a liquid. As a result, angle does not

change. The density of liquid is . Find the dielectric strength?

Solution:

Refer to concept section,

vdg

F

=tan .. (i)

When the system is immersed in liquid medium, then the force (electrostatic) becomes F/K and

upward thrust acts on the system.

( )gdvkF

=

.tan .. (ii)


23/23

Equating (i) and (ii)

( ) =

=

d

dk

dkd

11

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Notes on Electrostats for AIEEE