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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
Unit Test Paper No. 6 UT 6/1 – UT 6/3
Unit Test Paper No. 7 UT 7/1 – UT 7/3
Unit Test Paper No. 8 UT 8/1 – UT 8/3
Mock Test Papers 1 – 9
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
13/4 MODERN’S abc OF OBJECTIVE PHYSICS
(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.
13/6 MODERN’S abc OF OBJECTIVE PHYSICS
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
–
13/8 MODERN’S abc OF OBJECTIVE PHYSICS
∴ 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.
13/10 MODERN’S abc OF OBJECTIVE PHYSICS
(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
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Moderns ABC Of Objective Physics JEEMain Part-2
Publisher : MBD GroupPublishers
ISBN : 9789383907588Author : R. P. Arora, O. P.Kakkar, Deepak Chopra,Sameer Arora
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