This Slide explains basic theories in electrostatics, i.e. Coulomb's law, Electric field, electric potential, electric dipole, electric field due to electric dipole, etc. Visit: https://phystudypoint.blogspot.com

1. Electrostatic - I
Dr. Chhagan D. Mungmode
Associate Professor
M.G. College, Armori

2. Coulomb’s Law:
It states that the electrical force (attractive or repulsive) between two charged objects is directly
proportional to the product of the quantity of charge on the objects and inversely proportional to
the square of the separation distance between the two objects.
where,
Electric Field:
The region around an electric charge or a group of electric charges in which other charge
experience an electrostatic force (either attractive or repulsive) is called electric
f
ield of that
source.
Electric Field Intensity:
The electric
f
ield intensity or strength of an electric
f
ield at a point is de
f
ined as the electrostatic
force experienced by a unit positive charge placed at that point.
E = F/q2
F = k
q1q2
r2
=
1
4πε0
q1q2
r2
1
4πε0
= 9 × 109
Nm2
/C2
E = k
q1
r2
=
1
4πε0
q1
r2

3. Electric Potential:
• The electric potential at a point in electric
f
ield is de
f
ined as the work done in moving a
unit positive charge from in
f
inity to that point against electrostatic force.
• It is a scalar quantity.
Electric Dipole:
• A pair of equal and opposite point charges, +q and -q,
separated by small distance is called as electric dipole.
• Total charge of the dipole is zero but electric
f
ield of the dipole is not zero as charges q and
-q are separated by some distance and electric
f
ield due to them when added is not zero.
• Examples of electric dipole:- Dipoles are common in nature. Molecules like
are electric dipoles and have permanent dipole moments. They
have permanent dipole moments because the centre of their positive charges does not fall
exactly over the center of their negative charges.
Vb =
W∞→B
q0
H2O, HCl, CH3COOH

4. • Electric dipole moment de
f
inition :-
• The dipole moment of an electric
f
ield is a vector whose magnitude is charge times the
separation between two opposite charges.
• Direction of dipole moment is along the dipole axis from negative charge to positive charge.
• Consider the
f
igure given below which shows an electric dipole consisting of charges +q
and -q separated by a small distance 2a.
• Its electric dipole moment is given by,
• The SI unit of dipole moment is Coulomb-meter.
⃗
p = q × 2 ⃗
a = 2q ⃗
a

5. Electric
f
ield intensity due to electric dipole:
• Electric
f
ield due to dipole on axial line:
Electric
f
ield due to +q charge along BP:
Electric
f
ield due to -q charge along PA:
Total
f
ield at P on axial line
= along BP
E1 =
1
4πε0
q
(r − d)
2
E2 =
1
4πε0
q
(r + d)
2
E = E1 + (−E2) = E1 − E2
E =
q
4πε0
1
(r − d)
2
−
1
(r + d)
2
q
4πε0
4rd
(r2 − d2
)
2

6. Dipole moment is p = 2qd
When r is very large than d, then
………… (1)
Along the equatorial line:
Vertical components cancels each other, only horizontal components contribute. Total electric
f
ield at P
E =
1
4πε0
2pr
(r2 − d2
)
2
E =
1
4πε0
2p
r3
E+ =
1
4πε0
q
r2
+
=
1
4πε0
q
(r2 + d2
)
E− =
1
4πε0
q
r2
−
=
1
4πε0
q
(r2 + d2
)
E = E+cosθ + E−cosθ

7. But
at r greater and greater than d, neglect d.
………….. (2)
From equation (1) and (2), electric
f
ield intensity at axil line is double that on equatorial line at the same
distance.
cosθ =
d
(r2 + d2
)
1/2
E =
1
4πε0
2q
(r2 + d2
)
d
(r2 + d2
)
1
2
=
1
4πε0
2qd
(r2 + d2
)
3
2
E =
1
4πε0
p
r3

8. • Electric Field as negative gradient of Potential:
The electric potential at point P in an electric
f
ield is given by,
………………… (1)
V is the function of x, y, z I.e. V(x, y, z). represents rate of change of V along x, y, and z axis
respectively.
…………. (2)
But
………………. (3)
From equation (1) and (3)
………. (4)
V = −
∫
r
∞
E ⋅ dr dV = − E ⋅ dr
δV
δx
,
δV
δy
and
δV
δz
∴ dV =
δV
δx
dx +
δV
δy
dy +
δV
δz
dz
∴ dV = (
δV
δx
̂
i +
δV
δy
̂
j +
δV
δz
̂
k ) ⋅ (dx ̂
i + dy ̂
j + dz ̂
k)
(dx ̂
i + dy ̂
j + dz ̂
k) = dr and (
δV
δx
̂
i +
δV
δy
̂
j +
δV
δz
̂
k ) = (
δ
δx
̂
i +
δ
δy
̂
j +
δ
δz
̂
k )V = ∇V
∴ dV = ∇V ⋅ dr
−E ⋅ dr = ∇V ⋅ dr E = − ∇V

9. • Conservative Electric Field:
A vector
f
ield for which the work done by it depends only on end points but is independent of the actual path
is called as conservative
f
ield.
I.e. A vector
f
ield A is said to be conservative if
∮
A ⋅ dl = 0

10. • Torque acting on a Dipole in uniform Electric Field:
• Consider a dipole having charges +q and -q separated by distance 2l
placed in a uniform electric
fi
eld E making an angle θ with the direction
of electric
fi
eld.
• A torque acts on the dipole which will try to align dipole in the
direction of electric
fi
eld.
• Force on the charge +q will be
• Force on the charge - q will be
• The components of force perpendicular to the dipole are:
and
• Since the force magnitudes are equal and are separated by a distance
2l, the torque on the dipole is given by:
Torque (
𝜏
) = Force × Distance Separating Forces
𝜏
= 2l
𝑞
𝐸
sin
𝜃
• Since dipole moment is given by: p = q2l
F+ = + qE
F− = − qE
F⊥
+ = + qE sin θ F⊥
− = − qE sin θ

11. • And the direction of the dipole moment is from the positive to the negative charge, it can see from the above
equation that the torque is the cross product of the dipole moment and electric
fi
eld. Notice that the torque is
in the clockwise direction (hence negative) in the above
fi
gure. If the direction of Electric Field is positive.
𝜏
= −
𝑝
𝐸
sin
𝜃
Or
• If the dipole is placed perpendicular to the electric
fi
eld i.e. θ = 900 ,
𝜏
=
𝑝
𝐸
it means maximum torque acts on
the dipole.
• If the dipole become parallel to the electric
fi
eld then θ = 00 ,
𝜏
= 0 thus no torque acting on the dipole and
dipole is called in stable equilibrium.
τ = p × E

12. • Potential Energy of an Electric Dipole:
• Potential energy of an electric dipole in an electric
fi
eld is equal to the work done in bringing the dipole
from the in
fi
nity to that position within the electric
fi
eld and placing it in desired orientation.
• Consider a dipole placed in a uniform electric
fi
eld and it is in equilibrium position. If we rotate this dipole
from its equilibrium position , work has to be done.
• Suppose electric dipole of moment p is rotated in uniform electric
fi
eld E through an angle θ from its
equilibrium position. Due to this rotation couple acting on dipole changes.
• If at any instant dipole makes an angle φ with uniform electric
fi
eld then torque acting on dipole is
𝜏
= pE sinφ
• again work done (change in potential energy) in rotating this dipole through an in
fi
nitesimally small angle
dφ is
dW = dU = torque x angular displacement
= pE sinφ dφ

13. • Total work done in rotating the dipole through an angle θ from its equilibrium position is
• This is the required formula for work done in rotating an electric dipole placed in uniform electric
fi
eld
through an angle θ from its equilibrium position.
• We have choosen the value of φ going from π/2 to θ because at π/2 we can take potential energy to be zero
(axis of dipole is perpendicular to the
fi
eld). Thus U(90) = 0 and above equation becomes
•

14. • Electrostatic Field Energy:
• Electric
f
ield do work on the charge placed in electric
f
ield.
• To do work energy is required. Hence, energy is stored in the electric
f
ield.
• Consider two charges q1 and q2 . The charge q2 is bringing from in
f
inity to point B in the
f
ield of charge q1 .
The work done in this process is given by,
• This work done is stored in the form of electrostatic
f
ield energy.
• Total electrostatic
f
ield energy of the system of n charges is given by,
U = W =
∫
r12
∞
− F ⋅ dr
=
∫
r12
∞
−
q1q2
4πϵ0r2
dr =
q1q2
4πϵ0r12
=
n
∑
i=1
n
∑
j = i+1
qiqj
4πϵ0rij

15. • Flux of Electric Field:
• The electric
f
ield is represented by electric lines of force. Tangent drawn at any point to the electric lines
of force gives the direction of electric
f
ield at that point.
• Intensity of electric
f
ield is shown by number of electric lines of force.
• The total number of electric line of force passing normally through a surface imagined in an electric field is
called the electric flux linked with that surface.
• Electric
f
lux linked with small element will be the product of the component of
electric
f
ield and area of the element.
The total
f
lux linked with whole surface will be,
• If surface is normal to the electric
f
ield,
f
lux linked with surface will be
maximum.
• If surface is parallel to the electric
f
ield,
f
lux linked with surface will be zero.
dϕ = E . dA = E dA cos θ
ϕ =
∬
E ⋅ dA

19. • Electric Field Intensity due to Quadrupole on Equatorial Line: