When we rub two bodies, due to friction, some electrons are transferred from one body to another. The body which gains electrons becomes negatively charged and which loses electrons becomes positively charged by equal amount.

\(60\mu F\)

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

(a) Electric flux over an area in an electric field represents the total number of electric field lines crossing this area and is given by theproduct of surface area and the component of electric field intensity normal to the area.
The S.1.unit of flux is Nm² C^-1
(b) Let point P₁ is at a distance R from the centre O. S₁ is the Gaussian surface, then according to Gauss's theorem
\(\oint_s \overrightarrow{E}.\overrightarrow{ds}= \oint_s\overrightarrow{E}.\hat{n}ds=\frac{q}{\epsilon_0}\)
or \( {E}\oint_s {ds}= \frac{q}{\epsilon_0}\)
or  \( {E} = \frac{q}{\epsilon_0\times \oint_s {ds}}\)
= \( \frac{q}{4\pi\epsilon_0R^2}\) [As\(\oint{s} ds = 4\pi r^2\)]
Inside the shell the net charge is zero, so the field is also zero.

(c) The direction of electric field is shown in figure.

(a) (i) Ability to take prompt decision.
(ii) Knowledge of subject matter.
(iii) Proper responsibility.
(b) If there is no current in the circuit, the capacitor may have charge. So, by handling a charged capacitor a person may get a severe shock. Thus, the circuit containing a capacitor must be handled carefully.

The magnitudes of the electric fields due to the two charges + q and - q are given by
\(E_{+q} = \frac{q}{4\pi\epsilon_0}\frac{1}{r^2+2a^2}\)
\(E_{-q} = \frac{q}{4\pi\epsilon_0}\frac{1}{r^2+2a^2}\)
and both are equal The directions of E_+q and E_-q are as shown in fig.  Clearly, the components normal to the dipole axis cancel away. The components along the dipole axis add up. The total electric field is opposite to \(\hat{p}\) . We have
E = - (E_+q+ E_-q) cos \(\theta\) \(\hat{p}\)
   \(= \frac{-2qa}{4\pi\epsilon_0(r^2+a^2)^\frac{3}{2}} \hat{p}\)
At large diatances (r > > a), this reduces to
\(= \frac{-2qa}{4\pi\epsilon_0(r^3)} \hat{p}\) (r > > a)
Thus, the graph takes the form as shown below:

Electric dipole of charges +q and -q separated by distance 2a is shown in figure. It is placed in uniform electric field at an angle \(\theta\) with it.

Torque on dipole= force x perpendicular distance
 = qE x 2a sin\(\theta\)
 = 2 qa E sin \(\theta\)
 = pEsin \(\theta\)
  \(\overrightarrow\tau = \overrightarrow{p}\times\overrightarrow{E}\) 

(i) \(\frac { \left( 2\sqrt { 2 } +1 \right) q }{ 8\Pi \varepsilon { \alpha }^{ 2 } } \)
(ii) 19.4 N , 21.8⁰ above X-axis

\(We \ know \ that,{ E }_{ axial }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } .\frac { 2\rho r }{ { ({ r }^{ 2 }-{ r }^{ 2 }) }^{ 2 } } \)
\(CaseI,\)
\( When \ r=10cm=0.1m\)
\(\\ { E }_{ axial }=3.75\times { 10 }^{ 5 }N/C\)
\(\\ 3.75\times { 10 }^{ 5 }=9\times { 10 }^{ 5 }\times \frac { 2\rho \times 0.1 }{ { { [(0.1) }^{ 2 }-{ a }^{ 2 }] }^{ 2 } } \ .....(i)\)
\(Case \ II,\)
\( When \ r=20cm=0.2m\)
\(\\ { E }_{ axial }=3\times { 10 }^{ 4 }\times \frac { 2\rho \times 0.2 }{ { { [(0.2) }^{ 2 }-{ a }^{ 2 }] }^{ 2 } } \ .......(ii)\)
Solving the Eqs. (i) and (ii), we get 
a = 0.05 m
Therefore, lemgth of the dipole is 2z.
So, 2a = 2 x 0.05
or  2a = 0.1 m

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