jee-mainof workbook in physics for of workbook in physics for jee main iit–jee advance abc abc...

Price of Part–I & II : 865.00`

MODERN PUBLISHERS

MBD House, Gulab Bhawan 6, B.S.Z. Marg, New Delhi-110002

PART–IIPART–II

JEE-MainJEE-Main

All Questions

of

IncludedJEE-Main 2016

Joint Entrance Examination (Main)(All India Common Engineering Entrance Test)

THOROUGHLY REVISED AND UPDATED EDITION –2017

(Producers of Quality Text & Competition Books)

We are committed to serve students with best of our knowledge and resources. We have taken utmostcare and attention while editing and printing this book but we would beg to state that Authors andPublishers should not be held responsible for unintentional mistakes that might have crept in. However,errors brought to our notice, shall be gratefully acknowledged and attended to.

© All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted inany form or by any means, electronic, mechanical, photocopying, recording, or otherwise without the prior writtenpermission of the Author and publisher. Any breach will entail legal action and prosecution without further notice.

UR ADDRESSES IN INDIAUR ADDRESSES IN INDIA

MODERN'S abc SERIES OFOBJECTIVE BOOKS

Modern's

Modern's

Modern's

Modern's

Modern's

Modern's

Modern's

Solution of Modern's

Modern's


Modern's


abc JEE Main

abc JEE Main

abc JEE Main

abc NEET

abc NEET

abc NEET

abc JEE Main

abc JEE Main

abc JEE Main

abc JEE Main

abc JEE Main

abc JEE Main

of Objective Physics for

of Objective Chemistry for

of Objective Mathematics for

of Objective Physics for

of Objective Chemistry for

of Objective Biology for

of Workbook in Mathematics for

of Workbook in Mathematics for

of Workbook in Chemistry for

of Workbook in Chemistry for

of Workbook in Physics for

of Workbook in Physics for

JEE Main

IIT–JEE Advance

abcabc

Modern'sModern's

of

of

Sample Papers for

Sample Papers for

Modern's

Modern's

Modern's

Modern's

abc

abc

abc

abc

of Physics

+ of Chemistry

+ of Mathematics

+ of Biology

MODERN'S abc SERIES OFSCIENCE TEXTBOOKS

FOR CLASS XI & XII

New Delhi :

Mumbai :Chennai :Chennai :Kolkata :Jalandhar City :Bengaluru :Hyderabad :Ernakulam :Pune :Nagpur :Ahmedabad :

Cuttack :Guwahati :Lucknow :Patna :Bhopal :Jabalpur :Goa :Jaipur :Raipur :Karnal :Shimla (H.P.) :Jammu (J&K) :Ranchi (Jharkhand) :Sahibabad (U.P.) :Dehradun (Uttarakhand) :

Delhi (Shakarpur) :Delhi (Daryaganj) :Delhi (Patparganj) :

MBD House, Gulab Bhawan, 6, Bahadur Shah Zafar Marg

A-683, T.T.C. Industrial Area, M.I.D.C. Off. Thane-Belapur Road, Navi MumbaiNo. 26 B/2 SIDCO Estate, North Phase, Pataravakkam Ambattur Industrial Estate, Ambattur

Satyam Building, 46-D, Rafi Ahmed Kidwai MargMBD House, Railway Road

124/31, 1st Main, Industrial Town (Near Chowdeshwari Kalyan Mantap), West of Chord Road, Rajajinagar3-4-492, Varun Towers, BarkatpuraSurabhi Building, South Janatha Road, Palarivattom

Survey No. 44, Behind Matoshree Garden, Kondhwa - Khadi Machine - Pisoli Road, At. Post-PisoliNear N.I.T. Swimming Pool, North Ambazari Road, Ambazari Layout

Godown No.10, Vedant Prabha Estate, Opp. ONGC Pumping Station,Sarkhej Sanand Road, Sarkhej

Badambadi, Link RoadChancellor Commercial, Hem Baruah Road, Paan Bazar173/15, Dr. B. N. Verma Road, Old 30 Kutchery Road

Ist Floor, Annapurna Complex, Naya TolaPlot No. 137, 138, 139, Sector-I, Special Industrial Area,Govindpura

840, Palash Chamber, Malviya ChowkH. No. 932, Plot No. 66, Kranti Nagar (Behind Azad Bhawan), Alto Porvorim, Bardez

C-66A, Malpani Hospital, Road No.1, V.K. Industrial Area, Sikar RoadBehind Kailash Provision Store, Ravi NagarPlot No. 203, Sector-3, HSIDC, Near Namaste Chowk, Opp. New World

C-89, Sector-I, New Shimla-9MBD Office, 48 Gujjar Colony, C/o Gurjar Desh Charitable Trust, N.H. Bye Pass Road

Shivani Complex, 2nd Floor, Jyoti Sangam Lane, Upper BazarB-9 & 10, Site IV, Industrial Area

Plot No. 37, Bhagirathipuram, Niranjanpur, GMS Road

MB 161, Street No. 4MBD House, 4587/15, Opp. Times of IndiaPlot No. 225, Industrial Area

Ph. 30912330, 30912301, 23318301

Ph. 32996410, 27780821, 8691053365Ph. 26359376, 26242350

Ph. 23741471Ph. 22296863, 22161670

Ph. 2458388, 2457160, 2455663Ph. 23103329, 23104667

Ph. 27564788, 9985820001Ph. 2338107, 2347371

Ph. 65271413, 65275071Ph. 2248104, 2248106, 2248649, 2245648

Ph. 26890336, 32986505, 7600024542Ph. 2367277, 2367279, 2313013

Ph. 2131476, 8822857385Ph. 4010992, 4010993

Ph. 2672732, 2686994, 2662472Ph. 2581540, 2601535

Ph. 2405854Ph. 2413982, 2414394Ph. 4050309, 4020168Ph. 4052529, 2445370Ph. 2220006, 2220009Ph. 2670221,2670618

Ph. 2467376, 9419104035Ph. 9431257111

Ph. 3100045, 2896939Ph. 2520360, 2107214

Ph. 22546557, 22518122Ph. 23245676

Ph. 22149691, 22147073

DELHI LOCAL OFFICES :

Plot No. 3018, Old Y Block, 3rd Street, 12th Main Road, Anna Nagar West, Chennai

In front of

MODERN PUBLISHERSMBD House, Railway Road, Jalandhar City.

Printed at :

Published by :

MBD PRINTOGRAPHICS (P) LTD.

Industrial Area Ram Nagar, Gagret,

Tehsil Amb, Distt. Una (H.P.)

13. Electrostatics 13/1 – 13/66

14. Current Electricity 14/1 – 14/47

15. Magnetic Effects of Currents 15/1 – 15/48

16. Magnetostatics 16/1 – 16/24

17. E.M. Induction and A.C. Currents 17/1 – 17/44

18. Ray Optics 18/1 – 18/48

19. Wave Optics 19/1 – 19/36

20. Electromagnetic Waves 20/1 – 20/17

21. Electrons and Photons 21/1 – 21/34

22. Atoms, Molecules and Nuclei 22/1 – 22/45

23. Solids and Semiconductor Devices 23/1 – 23/32

24. Communication System 24/1 – 24/20

Unit Test Paper No. 5 UT 5/1 – UT 5/3




Mock Test Papers 1 – 9

we preserve nature.

and

............write to save nature

IMPORTANT FORMULAE, FACTS AND TERMS

Electric Charges : (i) ELECTRIC CHARGE IS A BASIC property like length, massand time of elementary particles of matter which can explain certain forces of interaction.

Electric charges are of two types—positive and negative.(ii) Relation Between Units of Charges : S.I. Units — Coulomb (C)

1 Stat C or 1 e.s.u. of chargeC.G.S. System

1 a b C or 1 e.m.u. of charge

1 C = 3 × 109 stat C = 1

10 a b C

(iii) COULOMB’S LAW- ACTION AT A DISTANCE LAWThe force of interaction (attraction or repulsion) between two point charges q1 and

q2 in free space or air or vacuum (i) is directly proportional to product of two point charges.(ii) is inversely proportional to square of distance between their centres.

i.e., 1 22

F∝q q

r ⇒ 1 2

2F=k q q

r

where k is proportionality constant called ELECTROSTATIC CONSTANT ORCOULOMB CONSTANT whose value depends on two factors.

(a) System of units (b) Medium in which charges are placed.

In S.I. System : 0

1

4=

π∈

k = 9 × 109 Nm2 C–2

In C.G.S. System : k = 1where ∈0 = Permittivity of free space = 8·854 × 10–12 C2 N–1 m–2.(Experimentally determined).

(iv)RELATIVE PERMITTIVITY of medium OR DIELECTRIC CONSTANT(K) OR SPECIFIC INDUCTIVE CAPACITY

Dielectric constant of medium (K = ∈r) = (0

(permittivity of the medium)Absolute permittivity)

∈

∈

(v) COULOMB’S LAW IN TERMS OF DIELECTRIC CONSTANTForce between two point charges when they are placed in a medium

1 2

2

1F

4=

π∈

mq q

r

∵ 0

K∈

=

∈

⇒ ∈ = K∈0

∴ 1 2

20

1F

4 K=

π ∈

mq q

r ⇒ Fm =

F

K ⇒ K =

F

Fm

13/1

Pair production can not take placein vacuum.

There is also some change in massof charged body. A positivelycharged body loses mass whilenegatively charged body gain thatmass.

Key Point

The minimum energy required toproduce e– – e+ pair is 1·02 MeV.In annihilation of matter, twoγ-ray photons are producedinstead of one. To conserve thelaw of conservation of linearmomentum, two γ-ray photons areproduced.

Key Point

For vacuum, K is min. i.e., 1For air, K ≈ 1. The exact valueof K for air is 1·0006For conductors, K is max. i.e., ∞K is a unitless, dimensionless andcharacteristic constant.

ELECTROSTATICS13

13/2 MODERN’S abc OF OBJECTIVE PHYSICS

(vi) Electrostatic forces are central forces in nature

(vii) Electrostatic forces ( 12F→

and 21F→

) obey NEWTON’s III RD LAW OF MOTION.

i.e., 12 21F F→ →

= −

1 2(ATTRACTION)

0q q < 1 2

(REPULSION)0q q >

(viii) In vector form, Coulomb’s law :

F→

=^1 2

20

1.

4π ∈q q

rr

(ix) METHOD OF ELECTRICAL IMAGES : (a) Force between point charge + q

and an earthed conductor = Force between charges + q and – q separated by distance ‘2 ’r

=2

2(2 )

kq

r

(b) Pot. energy of the above system 2

(U) =2

q k

r

− .

(x) Coulomb’s force is independent of presence of other charges between two givenpoint charges.

(xi)If two point charges q1 and q2 are separated by a number of media of thicknesst1, t2 t3 ....... tn having dielectric constants K1, K2, K3.......... Kn respectively, then electrostaticforce between given two point charges is

1 2

2

01

F

4 Kn

i ii

q q

t=

=

⎡ ⎤

⎢ ⎥π∈

⎢ ⎥⎣ ⎦

∑

...(1)

(xii) Equivalent Dielectric Constant : In above case, if we suppose that the distancebetween two point charges q1 and q2 is t, then

2

1

1

K

K

n

i ii

n

ii

t

t

=

=

⎡ ⎤

⎢ ⎥

⎢ ⎥

⎢ ⎥=

⎢ ⎥

⎢ ⎥

⎢ ⎥⎣ ⎦

∑

∑

(xiii) The magnitude of induced charge qi = ± q 1

1K

⎛ ⎞

−⎜ ⎟

⎝ ⎠

where K = Dielectric

constant of an object on which charge is induced.(xiv) CONDITIONS FOR EQUILIBRIUMCase–I. These three charges are in equilibrium, when net force on every charge is zero.

∴ Force on charge 1 21 2 2

0 1 12

Q

4 ( )

q qq

r r r

⎡ ⎤

⎢ ⎥= −

π∈ ⎢ ⎥+⎣ ⎦

= 0

⇒ 2

1 222

1

( )

Q

r rq

r

+=

Key Point

0FF ,

K= the force decreases when

K increases

ELECTROSTATICS 13/3

Force on charge 2 12 2 2

0 1 22

Q

4 ( )

q qq

r r r

⎡ ⎤

= −⎢ ⎥

π ∈ +⎢ ⎥⎣ ⎦

= 0

⇒

21 21

22

( )

Q

r rq

r

+=

Force of charge 1 22 2

0 1 2

QQ 0

4

q q

r r

⎡ ⎤−

= + =⎢ ⎥

π ∈⎢ ⎥⎣ ⎦

⇒ 2

222

1 1

rq

q r=

Case II. If Q is in equilibrium, then force on Q = 0

i.e., 1 22 2

0 1 2

Q0

4

q q

r r

⎡ ⎤

− =⎢ ⎥

π ∈⎢ ⎥⎣ ⎦

2

222

1 1

rq

q r=

(xv) When three point charges are placed at the corners of equilateral triangle asshown in figure.

| F |→

2 2 2 2331 32 1 2 1 22

0

F F4

qq q q q

r= + = + +

π ∈

if q1 = q2 = q3 = Q (say)

Then 2–

20

3 Q| F |

4 r

→=

π∈

(xvi) To have same magnitude of force both in vacuum and in medium

1 2 1 22 2

0 04 4 K

q q q q

r r=

π∈ π∈ ′ where K = dielectric constant of the medium.

⇒ K

rr′ =

(xvii) When the force between two point charges q1 and q2 is maximum. Then charge

on each should be 1 2 .2

q q+

(xviii) Let F = Force between two identical bodies having charges Q and – 2Qseparated by a distance ‘r’.

When these two bodies are kept in contact and then kept at same distance again. Then

repulsive force between them becomes 2

Q Q– –

2 2F = k

r

⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

′F

8= .

Key Point

1 e.s.u. of charge = 1 Frankline


(xix) If two point charges q1 and q2 are separated by a distance ‘r’ then the distance (d) between charge q1 and null

point is given by 2

11

rd

q

q

=±

if q2 > q1 :

If we have two like charges then we use d = 2

1

1

r

q

q+

If we have two unlike charges then we use d = 2

1

1−

r

q

q

(xx) When two identical balls each of mass m are charged with a charge q. Thenthese two balls are suspended by silk threads of length l. Then distance between balls are

given by Ftan

2

x

mg l

⎛ ⎞

= θ=⎜ ⎟

⎝ ⎠

∴ x = 2l F

mg

⎛ ⎞

⎜ ⎟⎝ ⎠

Or x =

1/ 32

0

2

4

⎛ ⎞

⎜ ⎟⎜ ⎟

π∈⎝ ⎠

q l

mg

(xxi) In above case, if whole set-up placed in a artificial satellite (where gravity iszero). Then, angle between two strings becomes 180° and tension (T) in each string is equal

to 2

20

.4 4

q

lπ∈

(xxii) In above case, if balls are suspended in a liquid of density ρ and distance between

balls remains same. Then, dielectric constant of liquid is given by Kρ′

=

ρ′−ρ

where

ρ′ = density of material of ball

∵ F

FKm = ⇒ F = Fm K

⇒ FF

tanm

mg mg

⎛ ⎞ρ′

= = θ⎜ ⎟

ρ′− ρ⎝ ⎠

ELECTROSTATIC FIELD

ELECTRIC FIELD INTENSITY OR ELECTRIC FIELD STRENGTH : (i) Theelectric field strength at a point is the force experienced per unit test charge at that point.

0 0 0

FE Lt

q q

→→

→=

(ii) Electric field due to a point charge q at a distance ‘r’ from it is (OR when pointcharge q is at origin)

^

2 30 0

E4 4

q qr r

r r

→

→

= =

π∈ π∈

(iii) When charge is situated any where in above case

∴ E→

(at point P) 3

0

( )

4 | |

q r r

r r

→ →

→ →

− ′

=

π∈ − ′

.

ELECTROSTATICS 13/5

(iv) The resultant electric field at any point is equal to vector sum of all electric fieldsdue to ‘n’ charges. According to the principle of superposition.

1 21

E E E ......... E En

n ii

→ → → → →

=

= + + + = ∑

(v) The magnitude of resultant field of two electric fields is

2 21 2 1 2| E | E E 2E E cos

→= + + θ

(vi) ELECTRIC LINES OF FORCE FOR A CHARGE + Q is fixed at a distance‘d’ in front of infinite metal plate

(vii) ELECTRIC DIPOLE : An arrangement of two equal and opposite point charges

infinitesimal distance apart constitutes an electric dipole.

(a) Ideal Electric Dipole : An ideal electric dipole is that dipole whose magnitude

of the charge tends to infinity while separation between them tends to zero.

(b) Electric Dipole Moment : It is a vector quantity (→p ) whose magnitude is equal

to the magnitude of one charge and distance between two charges.

| |→p = (q) (2a)

Its direction is from negative charge to positive charge.(c) Field Intensity on Axial Line of Dipole : An axial line is line joining the centres

of two charges forming an electric dipole.

2 2 2(at pt.P) 0

2E

4 ( )

p r

r a

→→ +=

π ∈ −.

For a short dipole (i.e., 2a < < r)

3

0

2E

4

p

r

→→

=π ∈

.

(d) Field Intensity on Equatorial or Equitorial Line of Dipole : A line passingthrough mid-point of an electric dipole and ⊥ is axial line is called equitorial line.

2 2 3 / 2

0

E (at pt. P)4 ( )

p

r a

→→ −

=π ∈ +

For a short dipole ; 3

0

E4

→→ −=

π ∈p

r

For a short dipole, axial

equitorial

E 2

E 1= .

(e) Electric Field at Intensity Any Point Due to Short Electric Dipole :

23

0

| E | (at pt. P) 1 3 cos4

p

r

→= + θ

π ∈and tan α =

1

2 tan θ.

(f) When two short dipoles of dipole moments p1 and p2 separated by distance ‘r’ asshown in figure.

Then electrostatic force between two short dipoles 1 24

0

61

4

p p

r=

π ∈.

Field intensity due to dipole variesinversely as cube of the distanceof the point, where as fieldintensity due to single chargevaries inversely as the square ofthe distance of the point from thecharge.


Key Point

Electric field due to a singlecharge is spherically symmetric,while that due to dipole issymmetric cylinderically.

(g) When two short dipoles whose dipole moment are 1→p and 2

→p placed parallel to

each other. Then, electrostatic force between two short dipoles 1 24

0

31

4=

π∈

p p

r. In both

above cases, couple is zero.

(h) When two short dipoles of dipole moments 1→p and 2

→p are placed mutually

perpendicular to each other. Then electrostatic force between these two

1 24

0

21

4=±

π∈

p p

r . In this case couple on p1 is 1 2

30

2

4π∈

p p

r and couple on p2 is 1 2

304π∈

p p

r.

(viii) ELECTRIC FIELD INTENSITY due to uniformly charged ring or circularloop :

3(at pt.P) 2 2 2

0

| E | modulus

4 ( )

→

=

π∈ +

qx

x a.

At the centre of ring, x = 0

Then | E→

| = 0.(ix) Electric field will be maximum at a distance ‘x’ from centre of ring (O) at pt. P

due to charged ring when 2

=

ax

∴ max. 1 23 0

2 1E

4(3)

q

aπ⎛ ⎞

= ⋅⎜ ⎟

∈⎝ ⎠

(x) When a charged rod of length ‘L’ having charge Q is bent in the form of

semi-circle. Then, electric field at centre (C) of semi-circle (E) = 20

Q

2 L∈

(xi) Torque Acting on electric dipole in Uniform Electric Field :

E→ → →τ = ×p

|→τ | = p E sin θ.

Special cases : (i) When θ = 0° or 180°, τ = 0 (ii) When θ = 90° , τ max. = + pE (iii) When θ = 270°, τmin. = – pE

(a) Graph between τ and θ :

(b) In a uniform electric field an electric dipole experiences no force but only torque,but in non-uniform field it experiences both force and torque.

(c) When electric field is non-uniform, some net force and torque do act on the dipole.However, when the dipole sets itself parallel to the field, torque becomes zero. But notforce persists.

ELECTROSTATICS 13/7

(d) The torque on dipole in uniform electric field is zero both in stable as well as theunstable equilibrium.

(xii) WORK DONE in rotating an electric dipole in uniform electric field :If dipole is rotated through a very small angle dθ, then small amount of work done

(dW) = τ dθ

= p E sin θ dθ

∴ Total work done W = 2

1

W = E sind p dθ

θ

θ θ∫ ∫

= – p E (cos θ2 – cos θ1)

If θ1 = 90° and θ2= θ

Then W = – pE cos θ, which is stored in the dipole in the form of potential energy.

∴ U( Pot. energy stored in dipole) = – pE cos θ = – →p . E

→.

Special Cases : (i) When θ = 0°, then U (min.) = – pE. (Stable equilibrium)

(ii) When θ = 180°, then U (max.).= + pE. (Unstable equilibrium)

(iii) When θ = 90° , then U = 0.

Graph between U and θ :

(xiii) When a charged particle q (at rest) placed in uniform electric field ( )E→

:

1

V2

2 =m qυ ...(1) ⇒

1E

2m q d2 =υ (∵V = Ed)

Where m = mass of given charged particle V = Pot. difference applied d = distance travelled by particle.

Then from eqn. (1) 2 V= q

mυ

(a) Velocity acquired by a particle when moves distance ‘x’ in time t υ = u + at

E

0= + qt

m.

(b) K.E. acquired by particle 21 E

2

2 2= q

tm

.

(xiv) Time Period of Oscillation of charged body in electric field :

(a) If charge q is given to bob and electric field E→

is applied as shownin Fig. then its equilibrium position changes from O to O′.

∴ m W Fg→ →

→

′= +

∴ T = 2π′

l

g

–


∴ g′ = 2 2+g a =

22 E⎛ ⎞

+⎜ ⎟

⎝ ⎠

qg

m

(b) If electric field is applied in downward direction then

g′ = E

effective accelerationq

gm

+ =

∴ T = 2E

π

+

lq

gm

= Time period of simple pendulum.

(c) If electric field is applied in upward direction then effective acceleration

g′ = Eq

gm

−

T = 2 2E

π = π

−

−

l lqg a gm

Gauss’s Law

(i) Solid Angle : Let dS = small area element of spherical surface.

The solid angle (dΩ) subtended by the area element dS at the centre of sphere is

defined as 2

SΩ = dd

r. ⇒ Ω = dΩ∫ = 2

1Sd

r ∫ = 4π sr = Solid angle subtended by surface

area of solid sphere.(a) The unit of solid angle is steradian (sr) and it is a dimensionless quantity.

(b) Solid angle subtended at pt. O by whole surface of sphere 2

2

44

πΩ = = πr

rsteradian.

(c) Solid angle subtended by dS at point O. When area vector S−→d makes some angle

θ with unit vector n^

2

S cos θΩ = dd

r

(d) Solid angle subtended by closed surface of any arbitrary shape point inside it.

2

2 2S S

S cos 44 .

d rd

r r

θ π

Ω = Ω = = = π∫ ∫

(ii) Electric Flux : The product of electric field strength ‘E’ and area ‘A’ ⊥ to fieldis called ‘ELECTRIC FLUX.’

E EAφ =

∴ Electric flux through area element Sd→

is given as E( )dφ = E . S E S cosd d→ →

= θ

∴ Total electric flux through closed surface E SE . Sd

→ →

φ = ∫

Positive electric flux : When θ < 90º

Where θ = Angle between E and Sd→ →

.

∫

ELECTROSTATICS 13/9

Then cos θ and hence Eφ is +ve means electric lines leaves the surface at that point.

Negative electric flux : When θ > 90º

Then cos θ and hence Eφ is –ve means electric lines enter the surface at that point.

Zero electric flux : When θ = 90°. Then cos θ and hence Eφ is zero.

(iii) GAUSS’s law : The total electric flux through closed surface (Gaussian surface)

is equal to 0

1

∈times the charge enclosed by closed surface.

i.e., Electric flux (φE) = S

E . Sq

d→ →

0

=

∈

∫

(a) Gaussian surface : An imaginary closed surface enclosing a charge chosen tocalculate the surface integral of electric field.

(iv) Electric Flux for different charge arrangements :

CHARGE ARRANGEMENTS ELECTRIC FLUX φ E

(a) φE (Electric flux for given closed surface S = 0) becomes

the net charge inside the closed surface is zero.

(b)

S

+qφE

= 0

(c) φE = 0

(d) φE = 0∈

q

(e)

+q

φE = 08∈

q Because

1

8 of given charge contributes to cube.


(f)

+q

E02

φ =∈q

(g)+q

φE = πr2E. Where r = radius of hemisphere.

and E = electric field due to point charge +q.

(h)

Up down

down φE = 0 Because the charge on up quark 2

3

+= e and charge on

down quark = 3

− e

∴

2 20.

3 3

e eq

+

= − =∑

(v) Applications of Gauss’s Law : (a) Electric field intensity due to line charge.Electric field intensity due to thin charged rod of uniform.

Charge density ^

0( E )

2

→− λλ =π ∈

rr

(b) Electric field intensity due to uniformly charged spherical shell or chargedspherical conductor :

(i) Outside the shell : When observation point P lies outside the shell of radius Rand centre O having uniform charge density σ.

∴ The magnitude of electric field 20

| (E) |4

=π ∈

q

r

(ii) On the surface of shell : When observation pt. P lies on the surface of shell,then r = R

200

σ

E4 R

= =∈π∈

q

where σ = surface charge density = 24 Rπq

(iii) Inside the shell : When observation pt. P inside the shell.

E = 0 (∴ q∑ = net charge inside Gaussian surface is zero).

(c) Electric field intensity due to solid sphere of charge : Inside the sphere, E ∝ r

E3

rρ=∈

where ρ = volume charge density

P

r

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

Moderns ABC Of Objective Physics JEEMain Part-2

Publisher : MBD GroupPublishers

ISBN : 9789383907588Author : R. P. Arora, O. P.Kakkar, Deepak Chopra,Sameer Arora

Type the URL : http://www.kopykitab.com/product/12171

Get this eBook

40%OFF

https://www.kopykitab.com/MOD-ABC-Of-Objective-Physics-JEE-Main-Part-2-by-R-P-Arora-O-P-Kakkar-Deepak-Chopra-Sameer-Arora

jee-mainof workbook in physics for of workbook in physics for jee main iit–jee advance abc abc...

Documents