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Electrics Review
Electrostatics Review
Electric Potential Review
Electric Flux & Gaussian Theorem Review
Electric Capacitance Review
Akshay
Satagopan
Akshay
Satagopan
2011.09.25
16:30:53 +05'30'
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Electrostatics
Review
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Frictional Electricity:
Frictional electricity is the electricity produced by rubbing two sui
and transfer of electrons from one body to other.
.
+ +
+ +
+ +
+ +
+ +
+ +
 
 
 
 
+ + +
+ + + +
+ + +
Glass
Silk
Electrons in glass are loosely bound in it than the electrons in silglass and silk are rubbed together, the comparatively loosely boufrom glass get transferred to silk.As a result, glass becomes positively charged and silk becomes n
charged.
Electrons in fur are loosely bound in it than the electrons in ebon
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It is very important to note that the electrification of the body (whe
positive or negative) is due to transfer of electrons from one body
i.e. If the electrons are transferred from a body, then the deficiencelectrons makes the body positive.
If the electrons are gained by a body, then the excess of electrons
body negative.
If the two bodies from the following list are rubbed, then the body
early in the list is positively charges whereas the latter is negative
Fur, Glass, Silk, Human body, Cotton, Wood, Sealing wax, Amber
Sulphur, Rubber, Ebonite.
Amber, Ebonite, Rubber, PlasWool, Flannel
SilkGlass
Column II (ve Charge)Column I (+ve Charge)
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Properties of Charges:
1. There exists only two types of charges, namely positive and ne
2. Like charges repel and unlike charges attract each other.
3. Charge is a scalar quantity.
4. Charge is additive in nature. eg. +2 C + 5 C – 3 C = +
5. Charge is quantized.
i.e. Electric charge exists in discrete packets rather than in con
amount.
It can be expressed in integral multiples fundamental electroni
(e = 1.6 x 1019 C)
q = ± ne where n = 1, 2, 3, …………
6. Charge is conserved.
i.e. The algebraic sum of positive and negative charges in an issystem remains constant.
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Note: Recently, the existence of quarks of charge ⅓ e and ⅔ e hapostulated. If the quarks are detected in any experiment with con
practical evidence, then the minimum value of ‘quantum of chargeeither ⅓ e or ⅔ e. However, the law of quantization will hold good
Coulomb’s Law – Force between two point electric ch
The electrostatic force of interaction (attraction or repulsion) betw
electric charges is directly proportional to the product of the charproportional to the square of the distance between them and actsjoining the two charges.
Strictly speaking, Coulomb’s law applies to stationary point charg
r
q1F α q1 q2
F α 1 / r2
or F αq
1
q2
r2F = k
q1
q2
r2or where k is a positiveproportionality called
l i f
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In medium, k =1
4πεwhere ε is the absolute electric pe
the dielectric medium
The dielectric constant or relative permittivity or specific inductiv
dielectric coefficient is given by
F =q1 q2
r2
1
4πε0
In vacuum,
F =
q1 q2
r2
1
4πε0εr
In medium,
ε0 = 8.8542 x 1012 C2 N1 m2
K = εr =εε0
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Coulomb’s Law in Vector Form:
r
+ q1
F12
r12
q1q2 > 0
+ q1 r1
In vacuum, for q1 q2 > 0,
q1 q2
r2
1
4πε0
r21F12 =
q1 q2
r2
1
4πε0
r12F21 =
In vacuum, for q1 q2 < 0,
q1 q21rF
q1 q21rF&
r
 q1
F12
r12
q1q2 > 0
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q1 q2
r3
1
4πε0
r12F12 =
q1 q2
r3
1
4πε0
r2F21 =&
Note: The cube term of the distance is simply because of vector
Otherwise the law is ‘Inverse Square Law’ only.
Units of Charge:
In SI system, the unit of charge is coulomb (C).
One coulomb of charge is that charge which when placed at rest idistance of one metre from an equal and similar stationary chargeis repelled by it with a force of 9 x 109 newton.
In cgs electrostatic system, the unit of charge is ‘statcoulomb’ or
In cgs electrostatic system, k = 1 / K where K is ‘dielectric consta
For vacuum, K = 1.
F =q
1
q2
r2
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1 emu of charge = c esu of charge
1 emu of charge = 3 x 1010 esu of charge
1 coulomb of charge = 3 x 109 statcoulomb
1 abcoulomb = 10 coulomb
Relative Permittivity or Dielectric Constant or SpecificCapacity or Dielectric Coefficient:
The dielectric constant or relative permittivity or specific inductive
dielectric coefficient is given by the ratio of the absolute permittivmedium to the permittivity of free space.
K = εr =εε0
The dielectric constant or relative permittivity or specific inductive
dielectric coefficient can also be defined as the ratio of the electrobetween two charges separated by a certain distance in vacuum to
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Continuous Charge Distribution:Any charge which covers a space with dimensions much less than
away from an observation point can be considered a point chargeA system of closely spaced charges is said to form a continuous c
distribution.
It is useful to consider the density of a charge distribution as we d
of solid, liquid, gas, etc.
(i) Line or Linear Charge Density ( λ ):
If the charge is distributed over a straight line or over the circumfecircle or over the edge of a cuboid, etc, then the distribution is cal
charge distribution’.
Linear charge density is the charge per unit length. Its SI unit is C
qλ =
lλ =
dq
dl
or
+ + + + + + +
dq
dl
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(ii) Surface Charge Density ( σ ):
σ =q
Sσ =
dq
dSor
If the charge is distributed over a surface area, then the distributio
‘surface charge distribution’.
Surface charge density is the charge per unit area. Its SI unit is C
+ + + + + + + +
+ + + + + + + ++ + + + + + + ++ + + + + + + +
dq
dSTotal charge on surface S, q = ∫ σ dS
S
(iii) Volume Charge Density ( ρ ):
ρ =q
ρ =dq
or
If the charge is distributed over a volume, then the distribution is c‘volume charge distribution’.
Volume charge density is the charge per unit volume. Its SI unit is
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Electric Field:Electric field is a region of space around a charge or a system of
within which other charged particles experience electrostatic forcTheoretically, electric field extends upto infinity but practically it i
certain distance.
Electric Field Strength or Electric Field Intensity or El
Electric field strength at a point in an electric field is the electrostunit positive charge acting on a vanishingly small positive test chat that point.
E =∆qFLt
∆q → 0 q0
FE =or or 14πε0
E =
+ q0+ q  q
q – Source charge, q0 – Test charge, F – Force & E
FF
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Note:
1. Since q0 is taken positive, the direction of electric field ( E ) isdirection of electrostatic force ( F ).
2. Electrostatic force on a negatively charged particle will be opdirection of electric field.
3. Electric field is a vector quantity whose magnitude and direcuniquely determined at every point in the field.
4. SI unit of electric field is newton / coulomb ( N C1 ).
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Electric Field due to a Point Charge:
O
Z
Y
+
+ q
r
Force exerted on q0 by q is
q q0
r2
1
4πε0
rF =
q q0
r3
1
4πε0
rF =
Electric field strength isq0
FE =
q
r31
4πε0
E (r) = r
or
orr2
1
4πε0
E (r) =q
r
Th l t i fi ld d t i t h h
E
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Electric field in terms of coordinates is given by
( x2 + y2 + z2 ) 3/2
1
4πε0E (r) =
q
i j k( x + y + z )
Superposition Principle:
 q5
+ q4
+ q1
F12
F15
The electrostatic force experienced by acharge due to other charges is the vector
sum of electrostatic forces due to theseother charges as if they are existingindividually.
F1 = F12 + F13 + F14 + F15
F12
F
qa
qb
1
4πε0
Fa (ra) = ∑b=1b≠a
Nra  rb
ra  rb│ │3
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Superposition principle holds good for electric field also.
Note:
The interactions must be on the charge which is to be studied duecharges.
The charge on which the influence due to other charges is to be f
assumed to be floating charge and others are rigidly fixed.
For eg. 1st charge (floating) is repelled away by q2
and q4
and attra
q3 and q5.
The interactions between the other charges (among themselves)
ignored. i.e. F23, F24, F25, F34, F35 and F45 are ignored.
Electric Lines of Force:
An electric line of force is an imaginary straight or curved path alounit positive charge is supposed to move when free to do so in an
field.Electric lines of force do not physically exist but they represent re
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1. Electric Lines of Force due to a Point Charge:
q < 0q > 0
a) Reofinfie
Tharre
stel
b) Re
of
infie
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2. Electric Lines of Force due to apair of Equal and Unlike Charges:
(Dipole)
+ q
 q
P
E
3. Electric Lines of Fopair of Equal and Li
.N
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4. Electric Lines of Force due to a Uniform Field:
+
+++
Properties of Electric Lines of Forceor Field Lines:
1. The electric lines of force are imaginary lines.
2. A unit positive charge placed in the electric field tends to followalong the field line if it is free to do so.
3. The electric lines of force emanate from a positive charge and t
a negative charge.
4. The tangent to an electric field line at any point
gives the direction of the electric field at that point.
5. Two electric lines of force can never crosseach other. If they do, then at the point ofintersection, there will be two tangents. It
means there are two values of the electric fieldat that point, which is not possible. E
+1 C
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6. Electric lines of force are closer
(crowded) where the electric field
is stronger and the lines spreadout where the electric field isweaker.
7. Electric lines of force are
perpendicular to the surface of a
positively or negatively chargedbody.
Q > q
Q
8 Electric lines of force contract lengthwise to represent attractio
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10.The number of lines per unit cross – sectional area perpendicu
field lines (i.e. density of lines of force) is directly proportional
magnitude of the intensity of electric field in that region.
11. Electric lines of force do not pass through a conductor. Henceof the conductor is free from the influence of the electric field.
Solid or hollowconductor
No Field
+
++
++
+
+



++++
E E

α E∆N
∆A
(Electrostatic
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Electric Dipole:
Electric dipole is a pair of equal and opposite charges separated small distance.
The electric field produced by a dipole is known as dipole field.
Electric dipole moment is a vector quantity used to measure the s
electric dipole.
 q2 l
p
p = (q x 2l) l
The magnitude of electric dipole moment is the product of magnit
charge and the distance between the two charges.
The direction is from negative to positive charge.
The SI unit of ‘p’ is ‘coulomb metre (C m)’.
Note:
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Electric Field Intensity due to an Electric Dipole:
i) At a point on the axial line:
Resultant electric field intensityat the point P is
EP = EA + EB
The vectors EA and EB are
collinear and opposite.
1
4πε0
iEA =q
(x + l)2
q
(x  l)2
1
4πε0
iEB =
│EP │ = │EB│  │EA│
│EP │ = q
( l)2
q
( l)2
1
4πε ][ 
│EP │ =1
4πε0 (
If l << x, thenEP
ll x
+ q qp
A B E
O
EP =1
4πε0
2
(x2
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θθ
ll
y
θ
θ
+ q qp
A B
EB
EA
EQQ
Resultant electric field intensity
at the point Q is
EQ = EA + EB
The vectors EA and EB are
acting at an angle 2θ.
ii) At a point on the equatorial line:
q
( x2 + l2 )
1
4πε0
EA = i
q1
4πε0EB = i( x2 + l2 )
E
EA co
EB co
The vectors EA sin θ and EB sin θare opposite to each other andhence cancel out.
The vectors EA cos θ and EB cos θ
EQ =q2
4πε0 ( x2 + l2 ) ( x
1E
q . 2l
O
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EQ =
1
4πε0
p
( x2 + l2 )3/2 ( i )
If l << y, then
EQ ≈ p
4πε0 y3
The direction of electric field intensity at a point on the equatorial
dipole is parallel and opposite to the direction of the dipole mome
If the observation point is far away or when the dipole is very sho
electric field intensity at a point on the axial line is double the elecintensity at a point on the equatorial line.
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Torque on an Electric Dipole in a Uniform Electric Fie
The forces of magnitude pE act
opposite to each other and hence netforce acting on the dipole due toexternal uniform electric field is zero.So, there is no translational motion ofthe dipole.
θ
However the forces are along different
lines of action and constitute a couple.Hence the dipole will rotate andexperience torque.
Torque = Electric Force x distance
θ
t = q E (2l sin θ)
= p E sin θ
q E q
2l
t
p
Case i: If θ = 0°, thet = p x E
p
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Work done on an Electric Dipole in Uniform Electric Fi
dW = tdθ
= p E sin θ dθ
W = ∫ p E sin θ dθ
W = p E (cosθ1  cos θ2)
θ1
θ2
If Potential Energy is arbitrarily taken zero when the dipole is at 9then P.E in rotating the dipole and inclining it at an angle θ is
Potential Energy U =  p E cos θ
Note: Potential Energy can be taken zero arbitrarily at any positidipole.
 q
2l θ
q E
q E
When an electric dipole is placed in a uniform electric field, it exp
torque and tends to allign in such a way to attain stable equilibriu
d
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Line Integral of Electric Field (Work Done by Electric F
Negative Line Integral of Electric Field represents the work done b
field on a unit positive charge in moving it from one point to anotelectric field.
O
Z
Y
+ q
ArA
WAB = dW =  E . dl
A
B
Let q0 be the test charge in place of the unitpositive charge.
The force F = +q0E acts on the test charge
due to the source charge +q.
It is radially outward and tends to acceleratethe test charge. To prevent thisacceleration, equal and opposite force –q0Ehas to be applied on the test charge.
Total work done by the electric field on the test charge in movingin the electric field is
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Electric Potential
Review
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Electric potential is a physical quantity which determines the flow
from one body to another.
It is a physical quantity that determines the degree of electrificatio
Electric Potential at a point in the electric field is defined as the w
moving (without any acceleration) a unit positive charge from infi
point against the electrostatic force irrespective of the path follow
Electric Potential:
=11qq0
4πε0
][ rB rA
WAB =  E . dl
A
B
=q
4πε0
WAB
q0
According to definition, rA = ∞ and rB = r
(where r is the distance from the sourcand the point of consideration)
=q
4
W∞B
q= V V =
W∞B
q
or
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Electric Potential due to a Single Point Charge:
r
+ q B
x
Q
dxE +q0
Let +q0 be the test charge
placed at P at a distance xfrom the source charge +q.
To prevent this acceleration, equal and opposite force –q0E has to
on the test charge.
The force F = +q0E is
radially outward and tends
to accelerate the test charge.
Work done to move q0 from P to Q through ‘dx’ against q0E is
dW = F . dx = q0E . dx dW = q0E dx cos 180° =  q 0E
dW =  dxq q0
4πε0 x2E =
q
4πε0 x2
or
Total work done to move q0 from A to B (from ∞ to r ) is W∞q0
P
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Electric Potential due to a Group of Point Charges:
+qn
q4
r4
rnVP = V1 + V2 + V3 + V4 + …………+ Vn
│ │
1
4πε0V = ∑
i=1
n qi
r  ri
( in terms of
position vector )
The net electrostatic potential at a point in the
electric field due to a group of charges is thealgebraic sum of their individual potentials at thatpoint.
1. Electric potential at a point due to a charge is not affected by tof other charges.
2. Potential, V α 1 / r whereas Coulomb’s force F α 1 / r2.
1
4πε0
V = ∑i=1
n qi
ri
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Electric Potential due to an Electric Dipole:
llx
+ q qp
A B
qVP q+
=4πε0 (x – l)
1
VP = VP q++ VP q
VP q=
4πε0 (x + l)1
 q
VP =4πε0
q
(x – l)
1[ 
(x + l)
1
]
VP =1
4πε0
q . 2l
(x2 – l2)
i) At a point on the axial line:
O
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 qp
A θθ
y
O
ii) At a point on the equatorial line:
qVQ q+=
4πε0 BQ1
VQ = VP q++ VP q
VQ q=
4
πε0AQ
1  q
VQ =4πε0
q
BQ
1[ AQ
1 ]
VQ = 0 BQ = AQ
Q
ll
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Equipotential Surfaces:
A surface at every point of which the potential due to charge distr
the same is called equipotential surface.
i) For a uniform electric field:
E
V1 V2V3
Plane Equipotential Surfaces +
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Properties of Equipotential Surfaces:
2. The electric field is always perpendicular to the element dl of th
equipotential surface.
1. No work is done in moving a test charge from one point to anot
equipotential surface.
WAB
q0
VB  VA = ∆V =
If A and B are two points on the equipotential surface, then VB
WAB
q0= 0 or WAB
= 0
WAB
= E . dl
A
B
= 0
Since no work is done on equipotential surface,
i.e.E dl cos θ =
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ii) Electrostatic Potential Energyof a Three Charges System:
OZ
Y
A (q1)
r1
or
r
U =q1q2
4πε0 │ │ r1r2
1+
q1q3
4πε0 │ │ r1r3
1
+
q2q3
4πε0 │ │ r2r3
1
U =
q1q2
4πε0 r12
1
[
q1q3
r31
q2q3
r32+ + ]iii) Electrostatic Potential Energy of an n  Charges Syste
]
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Electric Flux &
Gaussian Theorem
Review
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Area Vector:
Small area of a surface can be represented by a vector.
SElectric Flux:
Electric flux linked with any surface is defined as the total numbe
lines of force that normally pass through that surface.
90°
θ
dS
S
Electric flux dΦ through a small area
element dS due to an electric field E at an
angle θ with dS is
= E dS cos θE . dSdΦ =
Total electric flux Φ over the wholesurface S due to an electric field E is
Φ = E . dS = E S cos θ = E . S
dS = dS n
dS
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Solid Angle:
Solid angle is the threedimensional equivalent of an ordinary two
dimensional plane angle.
SI unit of solid angle is steradian.
Solid angle subtended by area element dS at thecentre O of a sphere of radius r is
d =dS cos θ
r2
dS θ
1. For 0°< θ < 90°, Φ is positive.
2. For θ = 90°, Φ is zero.
3. For 90°< θ < 180°, Φ is negative.
Special Cases:
r
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Gauss’s Theorem:
The surface integral of the electric field intensity over any closed
surface (called Gaussian surface) in free space is equal to 1 / ε0 ticharge enclosed within the surface.
E . dS =
S
ΦE =1
ε0
∑i=1
n
qi
Proof of Gauss’s Theorem for Spherically Symmetric
E . dSdΦ =r2
1
4πε0
=q
r . dS n
dΦ =r2
1
4πε0
q dSr n.
Here, = 1 x 1 cos 0°= 1r n.
dΦ1 q dS
O •
r
r+q
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Proof of Gauss’s Theorem for a Closed Surface of any
E . dSdΦ =r2
14πε0
= q r . dS n
dΦ =
r2
1
4πε0
q dSr n.
Here, = 1 x 1 cos θ= cos θ
r n.
dΦ =r2
q
4πε0
dS cos θ
Φ dΦq
4q
dq
d
r
+q •
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O •
r
r+q
Deduction of Coulomb’s Law from Gauss’s Theorem:
From Gauss’s law,
E . dS =
S
ΦE =q
ε0
E dS =
S
ΦE =q
ε0
or dS =
S
ΦE =
q
ε0E
E =q
4πε0 r2E x 4π r2
q
ε0
=
If a charge q0 is placed at a point where E
Since E and dS are in the same direction,
or
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Applications of Gauss’s Theorem:
1. Electric Field Intensity due to an Infinitely Long Straight
Wire:
Gaussia
closed saround distribu
that theintensit
fixed vapoint on
From Gauss’s law,
E . dS =
S
ΦE =q
ε0
E . dS = E . dS + E . dS + E . dS
 ∞ B A
C
E
E E
dSdS
dS
l
r
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q
ε0
=λ l
ε0(where λ is the liner charge density)
E x 2 π r l =λ l
ε0
or E = 2 πε0
1 λ
r
or E =4 πε0
1 2λr
In vector form, E (r) =4 πε0
1 2λr
r
The direction of the electric field intensity is radially outward from
line charge. For negative line charge, it will be radially inward.
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dS
C
BA
E dS r
l
2. Electric Field Intensity due to an Infinitely Long, Thin PlaCharge:
From Gauss’s law,
σ
E . dS =
S
ΦE =q
ε0
E . dS = E . dS + E . dS + E . dS
TIP:
The fieldstraight,
uniforml
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(where σ is the surface charge density)
or E =2 ε0
σ
In vector form, E (l) = 2 ε0
σ
l
The direction of the electric field intensity is normal to the plane a
from the positive charge distribution. For negative charge distrib
be towards the plane.Note:
The electric field intensity is independent of the size of the Gauss
constructed. It neither depends on the distance of point of considthe radius of the cylindrical surface.
q
ε0
=σ π r2
ε0
2 E x π r2 =σ π r2
ε0
If the plane sheet is thick, then the charge distribution will be ava
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3. Electric Field Intensity due to Two Parallel, Infinitely LonPlane Sheet of Charge:
σ1σ2
E1E1
E2E2E2
E E E
Region I Region II Reg
σ1 > σ2( )
Case 1: σ1 > σ2
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Case 2:
σ1 σ2
E1E1
E2E2E2
E E E
Region I Region II Reg
(σ1 > σ2
( )
+ σ1 &  σ2
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Case 3:
σ1 σ2
E1E1
E2E2E2
E = 0 E ≠ 0
Region I Region II Reg
E
E E
+ σ &  σ
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4. Electric Field Intensity due to a Uniformed Charged This Shell:
q
r
R
From Gauss’s law,
E . dS =
S
ΦE =q
ε0
E dS =
S
ΦE =q
ε0
or dS =
S
ΦE =q
ε0
E
E x 4π r2 q=
Since E and dS are in the same direction,
or E = q
4 2
i) At a point P outside the shell:
Electric field due charged thin sphe
HOLLOW
……… Gaussia
O •
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From Gauss’s law,
E . dS =
S
ΦE =q
ε0
E dS =
S
ΦE =q
ε0
or dS =
S
ΦE =q
ε0
E
E x 4π R2q
ε0
=
Since E and dS are in the same direction,
or E =q
4πε0 R2
ii) At a point A on the surface of the shell:
Electric field due Since q = σ x 4π R2
q
HOLLOW
O •
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r’
Oq
HOLLO
•
From Gauss’s law,
E . dS =
S
ΦE =q
ε0
E dS =
S
ΦE =q
ε0
or dS =
S
ΦE =q
ε0
E
E x 4π r’2q
ε0=
Since E and dS are in the same direction,
or E =0
4πε0 r’2
iii) At a point B inside the shell:
(since q = 0 inside the Gaussian surface)
E
Emax
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Capacitance
Review
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Behaviour of Conductors in the Electrostatic Field:
1. Net electric field intensity in the interior of a
conductor is zero.When a conductor is placed in an
electrostatic field, the charges (freeelectrons) drift towards the positive plateleaving the + ve core behind. At an
equilibrium, the electric field due to thepolarisation becomes equal to the appliedfield. So, the net electrostatic field insidethe conductor is zero.
E
Ene
2. Electric field just outside the chargedconductor is perpendicular to the surface ofthe conductor.
Suppose the electric field is acting at an
angle other than 90°, then there will be a
component E cos θ acting along the tangentat that point to the surface which will tend to
E
T P
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3. Net charge in the interior of a conductor is zero.
The charges are temporarily separated. The total
charge of the system is zero.
E . dS =
S
ΦE =q
ε0
Since E = 0 in the interior of the conductor,
therefore q = 0.
4. Charge always resides on the surface of a
conductor.
Suppose a conductor is given some excess
charge q. Construct a Gaussian surface justinside the conductor.
Since E = 0 in the interior of the conductor,therefore q = 0 inside the conductor.
q
5. Electric potential is constant for the entireconductor
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6. Surface charge distribution may be different
at different points.
σ=
q
S σmin
Every conductor is an equipotential volume
(three dimensional) rather than just anequipotential surface (two dimensional).
Electrical Capacitance:
The measure of the ability of a conductor to store charges is kn
capacitance or capacity (old name).
q α V or q = C V or C =
q
VIf V = 1 volt, then C = q
Capacitance of a conductor is defined as the charge required to
potential through one unit.
SI Unit of capacitance is ‘farad’ (F). Symbol of capacitance:
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Capacitance of an Isolated Spherical Conductor:
Let a charge q be given to the sphere which
is assumed to be concentrated at the centre.
Potential at any point on the surface is
V =q
4πε0 r
C =q
V
C = 4πε0 r
1. Capacitance of a spherical conductor is directly proportional t
2. The above equation is true for conducting spheres, hollow or s
3. IF the sphere is in a medium, then C = 4πε0εr r.
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Principle of Capacitance:A
A
Step 1: Plate A is positively charged and B is neutral.
Step 2: When a neutral plate B is brought near A,charges are induced on B such that the side near A is
negative and the other side is positive.
The potential of the system of A and B in step 1 and 2
remains the same because the potential due to positiveand negative charges on B cancel out.
Step 3: When the farther side of B is earthed thepositive charges on B get neutralised and B is left onlywith negative charges.
Now, the net potential of the system decreases due tothe sum of positive potential on A and negativepotential on B.
To increase the potential to the same value as was in
step 2, an additional amount of charges can be given tol t A
P
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Capacitance of Parallel Plate Capacitor:
Parallel plate capacitor is an arrangement of two
parallel conducting plates of equal areaseparated by air medium or any other insulating
medium such as paper, mica, glass, wood,ceramic, etc.
Aσ
V = E dσε0
= d
orq d
V =A ε0
But dC =
A ε0
C =
q
V
If the space between the plates is filled with dielectric medium of permittivity εr, then
dC =
A ε0 εr
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Series Combination of Capacitors:
V1 V2
V
C1 C2In series combination,
i) Charge is same in each capacitor
ii) Potential is distributed in inverse
proportion to capacitances
i.e. V = V1
+ V2
+ V3
Butq
V1 =C1
V2 =C2
qV3 =
C3
q, and
qV =
C
,
(where C is the equivalent cap
effective capacitance or net catotal capacitance)
q
= C1+ C2
q
+ C3
C
or
q q
1
C=
1=
C1
+C2
1+
C3
11
C
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Parallel Combination of Capacitors:
In parallel combination,
i) Potential is same across each capacitor
ii) Charge is distributed in direct proportion to
capacitances
i.e. q = q1 + q2 + q3
But , and,
(where C is the equivalent
capacitance)
or
Th ff ti it i th f th i di id l it
q1 = C1 V q2 = C2 V q3 = C3 V q = C V
C V = C1V + C2 V + C3 V
∑i=1
n
CiC =C = C1 + C2 + C3
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Energy Stored in a Capacitor:
The process of charging a capacitor is
equivalent to transferring charges from oneplate to the other of the capacitor.
The moment charging starts, there is a potentialdifference between the plates. Therefore, totransfer charges against the potential difference
some work is to be done. This work is stored aselectrostatic potential energy in the capacitor.
If dq be the charge transferred against the
potential difference V, then work done is
dU = dW = V dq
q=
Cdq
The total work done ( energy) to transfer charge q is
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Energy Density:
U =
1
2C V2
dC =
A ε0 V = E dand
U =1
2ε0 Ad E2
1
2ε0 E2=
U
Ad
1
2ε0 E2=U
But
or or
SI unit of energy density is J m3.
Energy density is generalised as energy per unit volume of the fi
Energy Stored in a Series Combination of Capacitors:
1=
C1
+C2
1+
C3
11
C+
Cn
1………. +
U =q21 U = 1
2q2 1[ C
+C1 +
C1 +
C1………. +
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Energy Stored in a Parallel Combination of Capacitors
U = U1 + U2 + U3 + ………. + Un
The total energy stored in the system is the sum of energy stored
individual capacitors.
C = C1 + C2 + C3 + ……….. + Cn
U =1
2C V2 U =
1
2V2 ( C1 + C2 + C3 + ……….. + Cn )
Loss of Energy on Sharing of Charges between the Ca
Parallel:Consider two capacitors of capacitances C1, C2, charges q1, q2 anpotentials V1,V2.
Total charge after sharing = Total charge before sharing
(C + C ) V = C V + C V
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The total energy before sharing is
Ui =
1
2C1 V12
1
2C2 V22+
The total energy after sharing is
Uf =
1
2(C1 + C2) V2
Ui – Uf =C1 C2 (V1 – V2)2
2 (C1 + C2)
Ui – Uf > 0 or Ui > Uf
Therefore, there is some loss of energy when two charged capacit
connected together.
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Polar Molecules:A molecule in which the centre of positive charges does
not coincide with the centre of negative charges is calleda polar molecule.
Polar molecule does not have symmetrical shape.
Eg. H Cl, H2 O, N H3, C O2, alcohol, etc.H
Effect of Electric Field on Polar Molecules:E = 0 E
p = 0 p
In the absence of external electricfield the permanent dipoles of the
When electric field is adipoles orient themse
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Non  polar Molecules:
A molecule in which the centre of positive charges coincides wit
negative charges is called a nonpolar molecule.Nonpolar molecule has symmetrical shape.
Eg. N2 , C H4, O2, C6 H6, etc.
Effect of Electric Field on Nonpolar Molecules:E = 0 E
p = 0 p
In the absence of externalelectric field the effective
When electric field is applied, thecharges are pushed in the direct
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Dielectrics:Generally, a nonconducting medium or insulator is called a ‘diele
Precisely, the nonconducting materials in which induced chargeson their faces on the application of electric fields are called dielect
Eg. Air, H2, glass, mica, paraffin wax, transformer oil, etc.
Polarization of Dielectrics:
E0
When a nonpolar dielectric slab issubjected to an electric field, dipoles
are induced due to separation ofeffective positive and negative centres.
E0 is the applied field and Ep is theinduced field in the dielectric.
The net field is EN = E0 – Ep
i.e. the field is reduced when a
dielectric slab is introduced.
E = 0
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Capacitance of Parallel Plate Capacitor with Dielectric
EpE0 EN = E0 
V = E0
(d – t) + EN
t
EN
K =E0 or EN =
K
E0
V = E0 (d – t) +KE0 t
V = E0 [ (d – t) +K
t]
But E0 =ε0
σ=
ε0
qA
and C =q
VA ε
or C =
A ε0
d [1 –K
t
d
t(1 
or C =
C0
[ tt(
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If the dielectric slab occupies the whole space between the plates,
then
WITH DIELECTRIC SLAB
Remains the sDecreases
EN = E0 – Ep
Electric Field
Increases (K Increases (K C0)Capacitance
Increases (K Remains the sameCharge
With Batteryconnected
With Batterydisconnected
Physcial Quantity
C0
K =C
C = K C0
Dielectric Constant
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Van de Graaff Generator:
C2
C1
P2
S
S – Large Copper sphere
C1, C2 – Combs with sharp points
P1, P2 – Pulleys to run belt
HVR – High Voltage Rectifier
M – Motor
IS – Insulating Stand
D – Gas Discharge Tube
T  Target
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Principle:
Consider two charged conducting spherical shells such that osmaller and the other is larger. When the smaller one is kept inlarger one and connected together, charge from the smaller ontransferred to larger shell irrespective of the higher potential o
shell. i.e. The charge resides on the outer surface of the outer
the potential of the outer shell increases considerably.
Sharp pointed surfaces of a conductor have large surface char
densities and hence the electric field created by them is very hcompared to the dielectric strength of the dielectric (air).
Therefore air surrounding these conductors get ionized and th
charges are repelled by the charged pointed conductors causi
discharging action known as Corona Discharge or Action of Posprayed charges moving with high speed cause electric wind.
Opposite charges are induced on the teeth of collecting comb
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Construction:
Van de Graaff Generator consists of a large (about a few mradius) copper spherical shell (S) supported on an insulating s
which is of several metres high above the ground.
A belt made of insulating fabric (silk, rubber, etc.) is made tthe pulleys (P1, P2 ) operated by an electric motor (M) such that
on the side of the combs.
Comb (C1) near the lower pulley is connected to High Volta
(HVR) whose other end is earthed. Comb (C2) near the upper pconnected to the sphere S through a conducting rod.
A tube (T) with the charged particles to be accelerated at it
the target at the bottom is placed as shown in the figure. The b
of the tube is earthed for maintaining lower potential.
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Working:
Let the positive terminal of the High Voltage Rectifier (HVR) iconnected to the comb (C1). Due to action of points, electriccaused and the positive charges are sprayed on to the belt (srubber). The belt made ascending by electric motor (EM) and
(P1) carries these charges in the upward direction.
The comb (C2) is induced with the negative charges which arcarried by conduction to inner surface of the collecting sphe(dome) S through a metallic wire which in turn induces posi
charges on the outer surface of the dome.
The comb (C2) being negatively charged causes electric windspraying negative charges due to action of points which neuthe positive charges on the belt. Therefore the belt does not
any charge back while descending. (Thus the principle ofconservation of charge is obeyed.)
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The process continues for a longer time to store more and mcharges on the sphere and the potential of the sphere increa
considerably. When the charge on the sphere is very high, t
leakage of charges due to ionization of surrounding air alsoincreases.
Maximum potential occurs when the rate of charge carried in
the belt is equal to the rate at which charge leaks from the shdue to ionization of air.
Now, if the positively charged particles which are to be
accelerated are kept at the top of the tube T, they get accelerdue to difference in potential (the lower end of the tube isconnected to the earth and hence at the lower potential) and
made to hit the target for causing nuclear reactions, etc.
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Uses:
Van de Graaff Generator is used to produce very highpotential difference (of the order of several million volts) fo
accelerating charged particles.
The beam of accelerated charged particles are used totrigger nuclear reactions.
The beam is used to break atoms for various experimein Physics.
In medicine, such beams are used to treat cancer.
It is used for research purposes.
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