Radius of sphere A = R
Surface charge density on sphere A =  \(\sigma \)
Radius of sphere B = 2R
Surface charge density on sphere B = \(\sigma \)
Before contact, the charge on sphere A is 
Q₁ = Surface charge density x Surface area
\(\Rightarrow\) Q₁ = \(\sigma \). \({ 4\pi R }^{ 2 }\quad \quad ...(i)\quad \quad \quad { }\)
Before contact, the charge on sphere B is 
Q₂ = Surface charge density x Surface area
Q₂ = \(\sigma 4\pi { (2R) }^{ 2 }\ =\ \sigma .{ 16\pi R }^{ 2 } ...(ii)\quad \quad \quad { }\)
Let after the contact, the charge on A be Q'₁ and the charge on B Q'₂ .
According to the conservation of charge, the charge before contact is equal to charge after contact.
Q'₁ + Q'₂ = Q₁ + Q₂
Putting the values of Q₁ and Q₂ from Eqs. (i) and (ii), we get
Q'₁ + Q'₂ = \(4\pi { R }^{ 2 }\sigma +{ 16\pi R }^{ 2 }\sigma ={ 20\pi R }^{ 2 }\sigma \quad ...(iii)\quad \quad { }\)
As they are in contact. So, they have same potential.
Potential on sphere A is V_A = \(\frac { 1 }{ { 4\pi \varepsilon }_{ 0 } } .\frac { { Q' }_{ 1 } }{ R } \)
Potential on sphere B is V_B = \(\frac { 1 }{ { 4\pi \varepsilon }_{ 0 } } .\frac { { Q' }_{ 2 } }{ 2R } \)
So,   V_A = V_B
\(\Rightarrow \ \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q_{1}^{\prime}}{R}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q_{2}^{\prime}}{2 R}\)
\(\Rightarrow \ \frac{Q_{1}^{\prime}}{R}=\frac{Q_{2}^{\prime}}{2 R}\)
\(\Rightarrow \ 2 Q_{1}^{\prime}=Q_{2}^{\prime}\)       .......(iv)
  Putting the value of Q'₂ in Eq. (iii), we get
\(Q_{1}^{\prime}+2 Q_{1}^{\prime}=20 \pi R^{2} \sigma \Rightarrow 3 Q_{1}^{\prime}=20 \pi R^{2} \sigma\)
\(\Rightarrow \ Q_{1}^{\prime}=\frac{20}{3} \pi R^{2} \sigma\)
and  \(Q_{2}^{\prime}=\frac{40}{3} \pi R^{2} \sigma\)     [from eq.(iv)]
Let the new charge densities be  \(\sigma\) ₁and \(\sigma\) ₂
\(\sigma_{1}=\frac{Q_{1}^{\prime}}{4 \pi R^{2}}=\frac{20 \pi R^{2} \sigma}{3 \times 4 \pi R^{2}}=\frac{5}{3} \sigma\)
\(\sigma_{2}=\frac{Q_{2}^{\prime}}{4 \pi(2 R)^{2}}=\frac{40 \pi R^{2} \sigma}{3 \times 4 \pi \times 4 R^{2}}=\frac{40 \sigma}{16 \times 3}\)
\(\sigma_{2}=\frac{10 \sigma}{4 \times 3}=\frac{5}{6} \sigma\)
Thus, the surface charge densities on spheres after contact are \(\frac { 5 }{ 3 } \sigma \ and \ \frac { 5 }{ 6 } \sigma .\) 

(i) (a) When a capacitor is placed in an external electric field, the free charges present inside the conductor redistribute themselves in such a manner that electric field within the conductor. This happens until a static situation is achieved,i.e. when the two fields cancel each other and the net electrostatic field in the conductor becomes zero.

(b) In contrast to conductors, dielectrics are non-conducting substance, i.e. they have no charge carriers.Thus, in a dielectric, free movement of charges in not possible.It turns out that the external field induces dipole moment by stretching molecules of the dielectric.
The collective effect of all the molecular dipole moments is the net charge on the surface of the dielectric which produces a field that opposes the external field. However, the opposing field is so induced, that does not exactly cancel the extent of the effect depends on the nature of dielectric.

Both polar and non-polar dielectrics develop net dipole moment in the presence of an external field. The dipole moment per unit volume is called polarisation and is denoted by P for linear isotropic dielectrics.
P = X^E
Where, X is constant of proportionality and is called electric susceptibility of the electric slab.
(ii) (a) At point C, inside the shell. Electric field inside a spherical shell is zero.
Thus, the force experienced by charge at centre C will also be zero.
\(\because \)  F_c = qE (E_inside  the shell = 0)
\(\therefore \) F_c = 0
At point A, | F_A | = 2Q \(\left[ \frac { 1 }{ { 4\pi \varepsilon }_{ 0 } } .\frac { 3Q/2 }{ { x }^{ 2 } } \right] \\.\)
\( F \ = \ \frac { { 3Q }^{ 2 } }{ { { 4\pi \varepsilon }_{ 0 } }{ x }^{ 2 } } ,\) away from shell
Electric flux through the shell,
\(\Phi =\frac { 1 }{ { \varepsilon }_{ 0 } } \) x magnitude of charge enclosed by shell
\(=\frac { 1 }{ { \varepsilon }_{ 0 } } \times \frac { Q }{ 2 } \Rightarrow \Phi =\frac { Q }{ { 2\varepsilon }_{ 0 } } \)

Given, C₁ = C₂ = 600 pF
= 600 \(\times\)10^-12F
= 6 \(\times\)10^-10 F
V₁ = 200 V, V₂ = 0
\(\begin{aligned} \therefore \text { Energy lost } & =\frac{C_1 C_2\left(V_1-V_2\right)^2}{2\left(C_1+C_2\right)} \\ \end{aligned}\)
\(\begin{aligned} =\frac{\left(6 \times 10^{-10}\right)^2(200-0)^2}{2 \times 12 \times 10^{-10}} \end{aligned}\)
= 6 \(\times\)10^-6 J

Given,
Capacitance, C = 8pF.
In the first case, the parallel plates are at a distance ‘d’  and is filled with air.
Air has dielectric constant, k = 1
Capacitance, C\(=\frac{k \times \epsilon_{o} \times A}{d}=\frac{\epsilon_{o} \times A}{d}\)          .......(i)
Here,
A = area of each plate
ϵ_o​ = permittivity of free space.
Now, if the distance between the parallel plates is reduced to half, then d₁ = d/2
Given, dielectric constant of the substance, k₁ = 6
Hence, the capacitance of the capacitor,
\(\mathrm{C}_{1}=\frac{k_{1} \times \epsilon_{o} \times A}{d_{1}}=\frac{6 \epsilon_{0} \times A}{d / 2}=\frac{12 \epsilon_{o} A}{d}\)         .......(ii)
Taking ratios of eqns. (1) and (2), we get,
C₁ = 2 x 6 C = 12 C = 12 x  8 pF = 96pF.
Hence, the capacitance between the plates is 96pF.

This is because net electric field intensity at this point is zero, the field intensities due to two charges being equal and opposite

The principle of superposition states that total force on a given charge is the vector sum of the individual forces exerted on it by all other charges, the force between two charges being exerted in such a manner as if all other charges were absent
\(\overrightarrow { F } =\overrightarrow { F_{ 12 } } +\overrightarrow { F_{ 13 } } +....+\overrightarrow { F_{ 1N } } \)

(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.

Yes, charge conservation is a global phenomenon.

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,....

Given,
The area of plate of the capacitor, A = 6 x 10^-3 m²
Distances between the plates, d = 3mm = 3 x 10^-3 m
Voltage supplied, V = 100V
Capacitance of a parallel plate capacitor is given by, \(C=\frac{\epsilon \times A}{d}\) 
Here,
ε = permittivity of free space = 8.854 x10^-12 N^-1 m ^-2 C^-2
\(C=\frac{8.854 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}}=17.81 \times 10^{-12} \mathrm{~F}=17.71 \mathrm{pF}\)
Therefore, each plate of the capacitor is having a charge of
q = VC = 100 x 17.81 x 10^-12 C = 1.771 x 10^-9 C