The dipole is an electric field will be in stable equilibrium if the following conditions are satisfied:
(i) The resultant force in dipole is zero, i,e, there is no translatory motion of dipole.
(ii) the torque on dipole is zero, ie,e there is no rotary motion of dipole.
(iii) the potential energy of dipole is minimum.
It will be so when dipole is aligned along the direction of electric field.
The dipole will be in unstable equilibrium if
(i) the resultant force on dipole is zero.
(ii) The torque on the dipole is zero. 
(iii) the potential energy of dipole is maximum. it will be so when dipole is aligned opposite to the direction of electric filed.

The electric field due to point charge has spherical symmetry with point charge at the center. Is is so because, at equal distances from the point charge, filed intensity is equal.
The electric field due to dipole has a cylindrical symmetry. The axis of the cylinder passes through the dipole axis. Is is so because the electric field due to dipole will be same at every point on the surface of a right circular cylinder with electric dipole as the axis.

For a Point on the axial line of dipole :  The direction of electric field intensity \(\overrightarrow { E } \) is along a line parallel to the axis of dipole directed along the direction of dipole moment \(\overrightarrow { p } \) 
For a point on the equatorial line of dipole : The direction of electric field \(\overrightarrow { E } \) is along a line parallel to the axis of dipole directed opposite to the direction of dipole moment \(\overrightarrow { p } \)

As in vector form \(\hat { r_{ 12 } } =-\hat { r_{ 21 } } \)  \(\overrightarrow { F_{ 12 } } \) and \(\overrightarrow { F_{ 21 } } \)
It shows two charges exert equal and opposite forces on each other. So Newton's third law is obeyed.
As  Coulombian force acts along \(\overrightarrow { F_{ 12 } } \) and \(\overrightarrow { F_{ 21 } } \)
i.e., along the line joining the centres of two charges, so they are central fores.

(i) Like charges repel and unlike charges attract each other.
(ii) Charge is quantized
(iii) Charge is conserved 
(iv) Charge on a body is not affected by its motion.

Additivity of charge means the total charge on a system is the algebraic sum (with proper signs) of all individual charges in the system.

Charge on any body or particle can be integral multiple of charge on an electron (-e), i.e.,
q = \(\pm ne\), where n = 1, 2, 3,....

The ebonite rod acquires a negative charge and fur/ wool acquires an equal positive charge.

Its mass decreases slightly as it losses some electrons.

The cause of charging is actual transfer of electrons from one body to the other.

Electric field of 1st spherical conductor on its surface
\({ E }_{ 1 }\propto \frac { 1 }{ a } \)
Electric field of 2nd spherical conductor on its surface
\({ E }_{ 2 }\propto \frac { 1 }{ b }\)
\(\frac { { E }_{ 1 } }{ { E }_{ 2 } } =\frac { b }{ a } \) 
Since a flat portion may be considered as a spherical surface of large radius (i.e. lower charge density, as charge density \(=\left( \frac { charge }{ area } \right) \) and a pointed portion as of small radius (i.e. highest charge density).

(a) When a charge +q is held at the centre of the shell, it will induce a charge -q on the inner surface of the shell of radius r and the charge +q on the outer surface of the outer shell.
Surface charge density on the inner surface of shell, 
As total charge on the outer surface of the outer shell = Q + q

There are three capacitors cach of capacitance 9 pF.
\(\therefore\) C₁ = C₂ = C₃ = 9 pF
and voltage, V = 120 V
(i) The total capcitance in series combination,
\(\frac{1}{C_{s}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}\)
\(\Rightarrow \frac{1}{C_{s}}=\frac{3}{9} \Rightarrow C_{s}=3 pF\)
(ii) Let the charge across the system be q and potentials across C₁, C₂ and C₃ be V₁, V₂ and V₃, respectively.
Charge, q = C_s. V = 3 \(\times\)120 = 360 pC
Potential difference across C₁,
\(V_{1}=\frac{q}{C_{1}}=\frac{360}{9}=40 V\)
Potential difference across C₂,
\(V_{2}=\frac{q}{C_{2}}=\frac{360}{9}=40 V\)
Potential difference across C₃,
\(V_{3}=\frac{q}{C_{3}}=\frac{360}{9}=40 V\)
Thus, the potential difference across each capacitor is 40 V.

\(4\mu C\)

r_m = r/4

\(5.8\times {{10}^{-9}} s\)

Given, capscitance, C = 2 F
and separation between plates,
d = 0.5 cm = 0.5 \(\times\) 10^-2 m
Capscitance of a parallel plate capacitor, C=\(\frac{\varepsilon_{0}A}{d}\)
or A=\(\frac{Cd}{\varepsilon_{0}}=\frac{2\times 0.5\times 10^{-2}}{8.854\times 10^{-12}}\)
= 1.13 \(\times\)10⁹ m² = 1130 km²
This area is very large, so it is not possible that the capacitance of a capacitor is too large as 2F. So, the capacitance of any capacitor should be the range of 2 \(mu\)F.

(1) Given,
Radius of spherical conductor, r = 12cm = 0.12m
Charge is distributed uniformly over the surface, q = 1.6 x 10^-7 C.
The electric field inside a spherical conductor is zero.
(2) Electric field E, just outside the conductor is given by the relation
\(\mathrm{E}=\frac{1}{4 \pi \epsilon_{o}} \cdot \frac{q}{r^{2}}\)
Here, permittivity of free space and \(\frac{1}{4 \pi \epsilon_{o}}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}\)
Therefore,
\(\mathrm{E}=\frac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{(0.12)^{2}}=10^{5} \mathrm{NC}^{-1}\)
Therefore, just outside the sphere the electric field is 4.4 x 10⁴ NC^-1.
(3) From the centre of the sphere the electric field at a point 18m = E₁.
From the centre of the sphere, the distance of point d = 18 cm = 0.18m
\(\mathrm{E}_{1}=\frac{1}{4 \pi \epsilon_{o}} \cdot \frac{q}{d^{2}}=\frac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{\left(1.8 \times 10^{-2}\right)^{2}}=4.4 \times 10^{4} \mathrm{NC}^{-1}\)
So, from the centre of sphere the electric field at a point 18 cm away is 4.4 x 10⁴ NC^-1.