ELECTROSTATICS 1(a)

Charge
Charge is that property of an object by virtue of which it applies electrostatic
force of interaction on other objects.
Charges are of two types
(i) Positive charge
(ii) Negative charge
Like charges repel and unlike charges attract each other.

Properties of charges:
Same charges repel each other and opposite charges attract each other.

1. Quantization of Charge
Charge on any object can be an integer multiple of a smallest charge (e).
Q = ± ne
where, n = 1, 2, 3,……. and e = 1.6 * 10-19 C.

2. Conservation of Charge
Charge can neither be created nor be destroyed. but can be transferal from one
object to another object.
Recently a new particle has been discovered called ‘Quark’. It contains charge ±
e / 3, ± 2e / 3.

Coulomb’s Law of Electrostatics:

The force between two charges is directly proportional to product of both
charges and inversely proportional to square of the distance between them.
q1 q2
Fα 2
r
q1 q2
OR, F = K 2
r
Where K is the constant of proportionality.
1
K = 4πε
0

Where, ε0 = 8.85 x 10 -12CN -1

m -2
is the permittivity of free space.
And K= 9 x 109Nm2/c2

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Vector form of coulombs law:
Consider a XY Cartesian frame with origin O. Let q 1 and q2 are the two charges
with position vector r1 and r2.
F12 = force on charge q2 by q1.
F21 = force on charge q1 by q2.
r12 = displacement vector from charge q1 to q2.
r21 = displacement vector from charge q2 to q1.

Therefore, we can say F12 = -F21

i.e. we can say force applied by both charges on each other will be equal and
opposite to each other.

NOTE:
1. If coulomb force is positive then forces will be repulsive in nature. Whereas If
coulomb force is negative then forces will be attractive in nature.
2. Effect of medium: if an insulator i.e dielectric is placed between the both
charges , then force between them will decrease by a certain factor “E r”.
where r stands for relative permittivity.
3. For vacuum: Er = 1 , for air Er = 1.006
For other mediums Er >1 , for metal Er =∞
Superposition principle:

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Total Coulomb force on a test charge due to a group of charges is equal to the
vector sum of all the Coulomb forces between the test charge and other
individual charges. The superposition of forces is not limited to Coulomb forces.
It applies to any types (or combinations) of forces.
i.e F = F1 + F2 ………………….+ Fn
Derivation:
Consider a XY Cartesian frame with origin O. Let q 1 , q2………….qn are the n
charges with position vector r1, r2 and rn. q1 is the test charge on which we
want to calculate the net force by all the charges.
F12 = force on charge q1 by q2.
F13 = force on charge q1 by q3.
lly, F1n = force on charge q1 by qn.
r12 = displacement vector from charge q2 to q1.
r13 = displacement vector from charge q2 to q1.

Distribution of continuous charges:

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The continuous charge distribution system is a system in which the charge is
uniformly distributed over the conductor. There are three types of the
continuous charge distribution system.
 Linear Charge Distribution
 Surface Charge Distribution
 Volume Charge Distribution

Force due to continuous charge distribution:

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 1(B) Electric Field Lines and Its Properties:
Electric Field Intensity The electric field intensity at any point due to source
charge is defined as the force experienced per unit positive test charge placed at
that point without disturbing the source charge. It is expressed as

Electric Field Intensity (EFI) due to a Point Charge

14. Electric Field due to a System of Charges

Same as the case of electrostatic force, here we will apply principle of
superposition, i.e.

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Electric Field Lines Electric field lines are a way of pictorially mapping the
electric field around a configuration of charge(s). These lines start on positive
charge and end on negative charge. The tangent on these lines at any point
gives the direction of field at that point.

Properties of electric field lines:

1. Two lines of electric field never cross each other because if they do so, it
means will be the at the intersecting point there will be two directions of
electric field which is not possible.
2. Tangent at any point in electric field represents the direction of electric field.
3. Equidistant lines represent uniform electric field and in equidistant lines
represent non uniform electric field.
4. Electric field lines due to positive and negative charge and their
combinations are shown as below:

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17. Electric Dipole Two-point charges of same magnitude and opposite nature
separated by a small distance altogether form an electric dipole.

18. Electric Dipole Moment The strength of an electric dipole is measured by

a vector quantity known as electric dipole moment (p) which is the product of
the charge (q) and separation between the charges (2a).

P = q (2a)
Direction of dipole moment is from negative to positive charge.

Electric field due to dipole on axial line:

AB is an electric dipole of two-point charges −q and +q separated by small
distance 2d. P is a point along the axial line of the dipole at a distance r from the
midpoint O of the electric dipole.
The electric field at the point P due to +q placed at B is,
q
E1 = K 2 (along BP)
(r −d )
The electric field at the point P due to −q placed at A is,
q
E2 = K 2 (along PA)
(r + d)
Therefore, the magnitude of resultant electric field (E) acts in the direction of the
vector with a greater, magnitude. The resultant electric field at P is
E=E1+(−E2)

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 1 1
E=K q ⌊ 2
− 2
⌋ along BP
(r−d) (r + d)
4 rd
E= Kq ⌊ 2 2 2 ⌋ along BP
(r −d )
If the point P is far away from the dipole, then d≪r
4d
∴E = Kq 3
r
2P
E = Kq 3 along BP
r
[∵ Electric dipole moment p=q×2d]
E acts in the direction of dipole moment.

Electric field at a point on the equatorial line of an electric dipole:

Consider an electric dipole consisting of two-point charges + q and −q separated
by a small distance AB = 2l with centre at O and dipole moment, p q = (2l) as
shown in the figure.

Resultant intensity at point Q:

On resolving EA and EB into two rectangular components , the vectors EAsinθ

and EBsinθ are equal and opposite to each other and hence cancel out each
other.
The vectors EAcosθ and EBCosθ are acting along same direction and hence add
up.
Therefore E = EAcosθ + EBcosθ = 2E cosθ

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The direction of E is along QE parallel to BA. Therefore, we can rewrite as:

Obviously, EQ is in opposite direction to P.

NOTE: Eaxial = 2 Eequatorial
Torque: Torque is the rotating effect of a force acting perpendicular.
Derivation of torque on a dipole placed in uniform electric field:
Consider a dipole with charges +q and –q forming a dipole since they are a
distance d away from each other. Let it be placed in a uniform electric field of
strength E such that the axis of the dipole forms an angle θ with the electric
field.

The force on the charges is

F+= +qE
F−= −qE
Since the force magnitudes are equal and are separated by a distance d, the
torque on the dipole is given by:
Torque (τ) = Force × Distance Separating Forces
τ= qE x 2asinθ
or ʈ = P E sinθ
or ʈ = P X E

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NOTE:
1. Minimum torque will be at θ = 0° or π i.e ʈ = 0
2. Maximum torque will be at θ = 90° i.e ʈ = PE
3. Dipole is in stable equilibrium in uniform electric field when angle between
p and E is 0° and in unstable equilibrium when angle θ= 180°.
4. Net force on electric dipole placed in a uniform electric field is zero.
5. There exists a net force and torque on electric dipole when placed in non-
uniform electric field.

Potential energy of dipole in uniform electric field.

τ=p×E
work done in rotating dipole by small angle
dw = τdθ
= −pεsinθdθ
The change in potential energy
dU=−dW = pEsinθdθ
If θ→90° to 0°
0°

U(0)−U(90°) = ∫ PEsinθ
90 °

=−P.E
So
U(θ)=−P.E
Work done in rotating the electric dipole from θ 1 to θ2 is W = pE (cos θ1 – cos θ2)
Potential energy of electric dipole when it rotates from θ1 = 90° to θ2 =0
U = pE (cos 90° – cosθ) = -pE cos θ = – p. E
Work done in rotating the dipole from the position of stable equilibrium to
unsable equilibrium, i.e.
when θ1 = 0° and θ2 = π.
W = 2 pE
Work done in rotating the dipole from the position of stable equilibrium to the
position in which dipole experiences maximum torque, i.e.
when θ1 = 0° and θ2 = 90°.
W = pE

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 1(C): Gauss law and its application
Electric Flux
 Electric flux is the measure of flow of the electric field through a given area.
Electric flux is proportional to the number of electric field lines going through
a normally perpendicular surface. ΦE=E⋅S=EScosθ
The vector associated with every area element of a closed surface is taken to
be in the direction of the outward normal.
Area element vector ΔS = ΔSn̂, ΔS is magnitude of area element and n̂ is
unit vectorin the direction of outward normal.
 Electrical flux has SI units of volt metres (V m).

Gauss law:
According to the Gauss law, the total flux linked with a closed surface is
1/ε0 times the charge enclosed by the closed surface.
q
ΦE=∫ E . ds = ε
0

For example, a point charge q is placed inside a cube of edge ‘a’. Now as
per Gauss law, the flux through each face of the cube is q/6ε0.
Proof of gauss law:
Let us consider a sphere of radius r having center o. let ds is the area vector on
which we want to calculate flux. Let E is the electric field at that surface due to
charge q placed at the center.

ΦE=∫ E . ds
q
= 4 π ε ∫ E .ds cosθ
0

Θ = 0° and we know that cos 0 =1

Therefore:
q
ΦE = 4 π ε x 4πr2
0
q
Or, ΦE = ε
0

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Application of gauss law:
1. Electric field due to infinite uniform straight wire at any point:
Consider an infinitely long thin straight wave with uniform linear charge q
density λ. Let P be a point at ⊥r distance r from the wire.

To calculate the E at P, imagine a cylindrical Gaussian surface.

∴ The surface area of the curved part S = 2πrl
Total charge enclosed by the Gaussian surface q = λl
Electric flix through the end Surfaces of the cylinder is Φ = 0
Electric flux through the curved Surfaces of the cylinder is Φ2 = Ecosθ.s
Φ2 = E x 1 x 2πrl
The total electric flux Φ = Φ1 + Φ2
Φ = 0 + E2πrl
Φ2 = 2πrlE ………(1)
q λl
A/C to Gauss law, ΦE = ε = ε ……..(2)
0 0

from (1) and (2)

λl
2πrlE = ε
0
λ
E = 2 πrε
0

2. Electric field due to thin sheet:

Let us consider a thin sheet of uniform charge density. To calculate electric field
at any point we will consider a Gaussian surface.

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3. Electric field due to two thin sheets having same uniform charge:

In region 1:
E = -E1 - E2
σ1 σ1
E = -2ε -2ε
0 0
−σ 1
E= ε
0

In region 11:
E = E1 - E 2
σ1 σ1
E = 2ε -2ε
0 0

E=0

In region 111:
E = E1 + E 2
σ1 σ1
E = 2ε +2ε
0 0
σ1
E= ε
0

Similarly, we can calculate uniform charge density due to two opposite uniform
charged sheets:

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 1(d): Electric potential
Electrostatic Potential or absolute electric potential:
The electrostatic potential at any point in an electric field is equal to the
amount of work done per unit positive test charge or in bringing the unit
positive test charge from infinite to that point, against the electrostatic force
without acceleration.
work done (w)
Electrostatic potential V= charge(q)
Its S.I. unit is volt (v) = 1J/1C and its dimensions is [ML 2 T-3 A-1].
It is a scalar quantity.

NOTE: Electrostatic potential is a state dependent function as electrostatic

forces are conservative forces.

Electrostatic Potential Difference The electrostatic potential difference

between two points in an electric field is defined as the amount of work done in
moving a unit positive test charge from one point to the other point against of
electrostatic force without any acceleration (i.e., the difference of electrostatic
potentials of the two points in the electric field).
W AB
VB – VA = q
0

where, is work done in taking charge q0 from A to B against of electrostatic

force.
Also, the line integral of electric field from initial position A to final position B
along any path is termed as potential difference between two points in an
electric field, i.e.
B

VB – VA = -∫ E . dl
A

NOTE: As, work done on a test charge by the electrostatic field due to any given
charge configuration is independent of the path, hence potential difference is
also same for any path.
For the diagram given as below, potential difference between points A and B will
be same for any path.

Relationship between electric field and potential gradient

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where, negative sign indicates that the direction of electric field is from higher
potential to lower potential, i.e. in the direction of decreasing potential.
NOTE:
(i) Electric field is in the direction of which the potential decreases steepest.
(ii) Its magnitude is given by the change in the magnitude of potential per unit
displacement normal to the equipotential surface at the point.

Electrostatic potential due to a point charge q :

Let us consider we want to calculate potential at at any point P lying at a
distance r from charge q:
r

V = -∫ E . dl ……(1)
∞

Electric field due to positive point charge q is:

q
E = 4 πε r 2
0
r
1 q
V=- ∫
4 πε 0 ∞ r 2
.dr
q
V =- 4 πε r
0

The potential at a point due to a positive charge is positive while due to

negative charge, it is negative.
When a positive charge is placed in an electric field, it experiences a force
which drives it from points of higher potential to the points of lower potential.
On the other hand, a negative charge experiences a force driving it from lower
potential to higher.
Electrostatic potential due to an electric dipole:
To find electric potential due to a dipole consider charge -q is placed at point P
and charge +q is placed at point Q as shown below in the figure.

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 Since electric potential obeys superposition principle so potential due to
electric dipole as a whole would be sum of potential due to both the charges +q
and -q. Thus

Where r1 and r2 respectively are distance of charge +q and -q from point R.

 Now draw line PC perpendicular to RO and line QD perpendicular to RO as
shown in figure. From triangle POC
cosθ=OC/OP = OC/a
therefore OC=acosθ similarly OD=acosθ
Now ,
r1 = QR≅RD = OR-OD = r-acosθ
r2 = PR≅RC = OR+OC = r+acosθ

Since magnitude of dipole is

|p| = 2qa

 If we consider the case where r>>a then

Again, since pcosθ= p·rˆ where, rˆ is the unit vector along the vector OR then
electric potential of dipole is

for r>>a
 From above equation we can see that potential due to electric dipole is
inversely proportional to r2 not 1/r which is the case for potential due to
single charge.
 Potential due to electric dipole does not only depends on r but also depends
on angle between position vector r and dipole moment p.

NOTE: The electrostatic potential on the perpendicular bisector (equator) due

to an electric dipole is zero.

Electrostatic potential due to a thin charged spherical shell carrying charge q

and radius R respectively, at any point P lying
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10. Graphical representation of variation of electric potential due to a charged
shell at a distance r from centre of shell is given as below:

Equipotential Surface:
A surface which has same electrostatic potential at every point on it, is known
as equipotential surface.
The shape of equipotential surface due to
(i) line charge is cylindrical.
(ii) point charge is spherical as shown alongside:
(a) Equipotential surfaces do not intersect each other as it gives two directions
of electric field E at intersecting point which is not possible.
(b) Equipotential surfaces are closely spaced in the region of strong electric field
and vice-versa.
(c) Electric field is always normal to equipotential surface at every point of it
and directed from one equipotential surface at higher potential to the
equipotential surface at lower potential.
(d) Work done in moving a test charge from one point of equipotential surface to
other is zero.

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Electrostatic Potential Energy The work done against electrostatic force gets
stored as potential energy. This is called electrostatic potential energy.
∆U = UB-UA =WAB

Electrostatic potential energy of a system of two-point charges is given by

The work done in moving a unit positive test charge over a closed path in an
electric field is zero. Thus, electrostatic forces are conservative in nature.

Potential Energy in an External Field

(i) Potential Energy of a single charge in external field Potential energy of a
single charge q at a point with position vector r, in an external field is qV(r),
where V(r) is the potential at the point due to external electric field E.
(ii) Potential Energy of a system of two charges in an external field

Potential energy of a dipole in a uniform electric field E is given by

Potential energy = -p.E

Electrostatic Shielding The process which involves the making of a region free
from any electric field is known as electrostatic shielding.

It happens due to the fact that no electric field exist inside a charged hollow
conductor. Potential inside a shell is constant. In this way we can also conclude
that the field inside the shell (hollow conductor) will be zero.

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 1(E): Capacitors
Dielectrics:
Dielectrics are non-conducting substances. In contrast to conductors, they have
no (or negligible number of) free charge carriers.
Types of dielectrics:
Dielectrics are of two types:
(i) Non-Polar Dielectrics: When the centre of positive charge coincides with the
centre of negative charge in a molecule, e.g., Nitrogen, Oxygen, CO 2 etc.
(ii) Polar Dielectrics: When the centre of positive and negative charges do not
coincide because of the asymmetric shape of the molecules, e.g., NH 3, HCl etc.

Dielectric Polarization:
In a dielectric, this free movement of charges is not possible. It turns out that
the external field induces dipole moment by stretching or re-orienting molecules
of the dielectric. The collective effect of all the molecular dipole moments is net
charges on the surface of the dielectric which produce a field that opposes the
external field. Unlike in a conductor, however, the opposing field so induced
does not exactly cancel the external field. It only reduces it. The extent of the
effect depends on the nature of the dielectric.

Capacitor:
It is an arrangement of primarily two conductors for storing large amount of
electric charge.
Capacitance of a Capacitor:
Capacitance (C) of a capacitor is defined as ratio of charge (Q) given to the
potential difference (V) applied across the conductors, i.e., C = Q/V.
The SI unit of capacitance is farad (F).
Capacitance of an Isolated Conducting Sphere:
Consider a Spherical conductor with center as o and radius r. Give some
positive charge say q to charge the conductor. When we will give the charge to
the conductor it will start spreading equally over its outer surface. Charge
spreading does not depend upon the type of conductor used.

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In the above equation Capacitance C is in farad.
We know earth is spherical. So putting the values according to the earth in the
above equation
Let’s take the radius r=6400 Km.
So, it becomes 6.4 X 10 6 m.
After substituting the values:
C= Electrostatic figure 2.55 X r
= 6.4 X 10 6/9 X 10 9
=0.711 X 10 -3 farad.
So, the value after calculation will be 711 uF.

Parallel Plate Capacitor and Its Capacitance:

A parallel plate capacitor consists of two large plane parallel conducting plates
separated by a small distance. The two plates have charges Q and –Q.

Electric field between the plates of capacitor:

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 σ Q
E = ε = Aε
0 0

Q. d
V = E.d = Aε
0

Q Aε
C= V = 0
d

NOTE:
When the two charged spherical conductors are connected by a conducting
wire they acquire the same potential: V1 = V2

Effect of dielectric and Conductors in Capacitor:

Let A be the plate area and d be the distance between the plates,
1) A dielectric slab of thickness t is inserted,

Due to polarization inside the dielectric electric field will reduce,

So outside of dielectric, field = E₀

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inside of dielectric, field = E = E₀/k
where k is the dielectric constant,
So the net potential will be,
V = Et + E₀(d-t)
=> V = E₀t/k + E₀(d-t)
=> V = E₀(d - t + t/k)
we know that, E₀ = q/ε₀A
=> V = q(d - t + t/k)/ε₀A
=> C = q/V
=> C = ε₀A/(d - t + t/k)
which is the required expression for capacitance,

2) A metallic plate of thickness t is inserted,

when a conducting metallic plate is inserted then electric field inside the plate
will be zero
=> E = E₀/k = 0
=> k = ∞
hence putting the value of k in the above equation we get,
C = ε₀A/(d - t)
which is the required expression for capacitance with metallic plate

Combination of Capacitors
We can combine several capacitors of capacitance C 1, C2,…, Cn to obtain a
system with some effective capacitance C. The effective capacitance depends on
the way the individual capacitors are combined. Two simple possible ways are:

Capacitors in Series:
In series arrangement the magnitude Q of charge on each plate is same

Capacitors in Parallel
In parallel arrangement the potential difference across each capacitor will
remain same.

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 CEquivalent = C1 + C2 + C3 . . . + Cn

Energy Stored in Capacitor

Energy stored (U) in capacitor (C) charged to a potential difference V is given
as,

Total Energy of the combination of capacitor:

Series combination:

Parallel combination:

Energy Density or Energy per Unit Volume:

The energy density or energy stored per unit volume of a charged capacitor is
given by

Common Potential of charged capacitors

If C1and C2 are capacitors charged to potentials V 1 and V2 respectively. These

capacitors are connected by a conducting wire, charges flow from higher
potential to lower potential. This flow of charge will continue till their potentials
become equal. There will no charge is lost in sharing and their common
potential is given by,

Energy dissipated when two charged capacitors are connected:

If U1 is the total energy before sharing of charges and U2 is total energy after
sharing of charges then,

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