q₁ = +\(\mu C\), q₂ = +6\(\mu C\), F = 12N
\(q_{1}^{'}=+2-4=-2\mu C;\) \(q_2^{'}=+6-4=2\mu C\) 
F' = ?
\({F'\over F}={(q_1^{'})(q_2^{'})\over q_1q_2}={(-2)(2)\over(2)(6)}=-{1\over 3}\)
\(F'={-F\over 3}={-12\over 3}=-4 N\)(attractive)

On touching, charges neutralize. Therefore,
F = 0

(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
\(=\frac { 1 }{ { \varepsilon }_{ 0 } } \times \frac { Q }{ 2 } \Rightarrow \Phi =\frac { Q }{ { 2\varepsilon }_{ 0 } } \)

(i) Examples of paramagnetic material are aluminum; chromium; oxygen.
(ii) Iron and cobalt are two ferromagnetic materials. The ferromagnetic substances, but to a much larger degree. For example, relative magnetic permeability of paramagnetic substances is slightly greater than 1