\(0.1\mu C/m\)

\(\frac { d }{ 1+\sqrt { 2 } } ,\frac { \sqrt { 2d } }{ 1+\sqrt { 2 } } \)

\(45\times \)10^-2 J

(i) \({{10}^{-9}}C\)
(ii) r = 0.5 m

\(4.5\times {{10}^{3}}V\)

\(60\mu F\)

(a) We know that the net field inside a charged conductor is zero i.e. 
\(\overrightarrow { E } \) = 0, inside
Let us choose a gaussian surface lying wholly within the conductor and enclosing the cavity.
According to Gauss' law,
\(\oint \overrightarrow { E } .\overrightarrow { dS } =\frac { q }{ { \varepsilon }_{ 0 } } =0\)  \((\because \overrightarrow { E } =0,inside)\)
\(\therefore \) q = 0 i.e. charge inside the cavity is zero. Hence the entire charge Q on the conductor must appear on the outer surface of the conductor.
(b) The conductor B carrying a charge + q inseted in the cavity induces a charge - q on the metal surface of cavity and + q on the outside surface of the conductor A [Fig (b)]. As the outer surface of A originally had a charge Q, the total charge on it would become (Q + q).
(c) To shift a sensitive instrument from the strong electrostatic fields in its environment, enclose the instrument fully by a metallic surface.

Given \(\frac { dE }{ dz } ={ 10 }^{ 5 }{ NC }^{ -1 }\) per metre.
Electric dipole moment,
\(\overrightarrow { dp } =q\times \overrightarrow { dz } ={ 10 }^{ -7 }Cm\) along BA.
Let \(\overrightarrow { dE } \) is the increase in electric field in going from A to B.
Force on charge at A, \(\overrightarrow { { F }_{ A } } =+q\overrightarrow { E } \)
Force on charge at B, \(\overrightarrow { { F }_{ B } } =-q(\overrightarrow { E } +d\overrightarrow { E } )\)
Net force, \(\overrightarrow { F } =+q\overrightarrow { E } -(\overrightarrow { E } +\overrightarrow { dE } )\)
\(=q\overrightarrow { E } -q\overrightarrow { E } -q\overrightarrow { dE } \)
or \(\overrightarrow { F } =-q\overrightarrow { dE } \)

Magnitude of force, \(F=q\frac { dE }{ dz } .dz\)
\(=(qdz)\frac { dE }{ dz } ={ 10 }^{ -7 }\times { 10 }^{ 5 }\)
\(F={ 10 }^{ -2 }N\)
Net torque,  \(\overrightarrow { \tau } =\overrightarrow { { r }_{ 1 } } \times \overrightarrow { { F }_{ 1 } } \times \overrightarrow { { r }_{ 2 } } \times \overrightarrow { { F }_{ 2 } } \)
where \(\overrightarrow { { r }_{ 1 } } \) and \(\overrightarrow { { r }_{ 2 } } \) are the position co-ordinates of two charges.
Now \(\overrightarrow { { r }_{ 1 } } ,\overrightarrow { { r }_{ 2 } } \) are in the same direction but \(\overrightarrow { { F }_{ 1 } } \) and \(\overrightarrow { { F }_{ 2 } } \) are in the opposite directions.
\(\left| \overrightarrow { { r }_{ 1 } } \times \overrightarrow { { F }_{ 1 } } \right| ={ r }_{ 1 }{ F }_{ 1 }\) sin \({ 0 }^{ o }={ r }_{ 1 }{ F }_{ 1 }\times 0=0\)
\(\left| \overrightarrow { { r }_{ 2 } } \times \overrightarrow { { F }_{ 2 } } \right| ={ r }_{ 2 }{ F }_{ 2 }\) sin \({ 180 }^{ o }={ r }_{ 2 }{ F }_{ 2 }\times 0=0\)
Hence torque \(\tau \) is zero.

When a glass rod is rubbed with silk, the charges developed on the glass rod and the piece of silk are equal and opposite. Similar is the case in other pair of bodies. So electric charge can neither be produced nor destroyed but simply transferred from one body to another, hence is consistent with the law of conservation of charge.

(a) Quantisation of charge: It is that property of charge by virtue of which charge on a body existst in the form of discrete packet of charge e, only, where e is the charge on an electron. The charge carried by anybody would be equal to \(\pm \) ne, where n = 0, 1, 2, 3, 4, 5, etc. THe charge on a body is thus some multiple of e and cannot exist as a fraction of e.
So charge exists in the form of packets and not in continuous amunts. Thus, charge is said to have a discrete (discontinuous) nature or is said to be quantised.
(b) At macroscopic level, we deal with charges that are enormous as compared to the magnitude of minimum charge i.e. e (1.6 \(\times\) 10^-19 C).
In this case, the increase and decrease in units of e is not very different from saying that charges are continuous. So at macroscopic level we can ignore quantisation of electric charge.

(i) The equivalent capacitance of C₁ and C₂ connected in series
\(\frac { 1 }{ C' } =\frac { 1 }{ { C }_{ 1 } } +\frac { 1 }{ { C }_{ 2 } } \)
\(\Rightarrow C'=\frac { 6 }{ 2n } =3\mu F\)
∴ Charge, q' = C'V = (3μF)12 = 36μF
∴ Charge on each capadtor of C₁ and C₂ is 36 μC
∴ Charge on C₃,
q₃ = C₃V = (6μF) x 12 = 72μC
q₃ = 72μC
(ii) Equivalent capacitance of network,
\({ C }_{ eq }=\frac { { C }_{ 1 }{ C }_{ 2 } }{ { C }_{ 1 }+{ C }_{ 2 } } +{ C }_{ 3 }\)
\(=\frac { 6\times 6 }{ 6+6 } +6\)
= 3 + 6 = 9μF
C_eq = 9μF
(iii) Energy stored in the network of capacitors
U = U₁ + U₂ + U₃ = \(\frac { { q' }^{ 2 } }{ { 2C }_{ 1 } } +\frac { { q }'^{ 2 } }{ { 2C }_{ 2 } } +\frac { { q }^{ 2 } }{ { 2C }_{ 3 } } \)
∵ C₁ = C₂ = C₃ = 6μF
\(\therefore U=\frac { 1 }{ (12\mu F) } [{ q' }^{ 2 }+{ q' }^{ 2 }+{ q }^{ 2 }]\)
\(=\frac { 1 }{ (12\mu F) } [({ 36\mu C) }^{ 2 }+({ 36\mu C) }^{ 2 }+({ 72\mu C) }^{ 2 }]\)
\(U=648\mu J\)

Energy stored in 6 μF capacitor = E 

\( \therefore \ \frac{1}{2} C_{1} V_{1}^{2}=E \Rightarrow V_{1}^{2}=\frac{2 E}{C_{1}}=\frac{E}{3} \)
\(\because C_{1} \text { and } C_{2} \text { are in parallel, } \therefore V_{1}^{2}=V_{2}^{2}=\frac{E}{3} \)
Now, \(C_{12}=6+12=18 \mu \mathrm{F}\) 
\( \therefore \text { Charge on } C_{12}=q_{12}=C_{12} . V_{12}=18 \times \sqrt{\frac{E}{3}} \)
\(\therefore \text { Charge on } C_{3}=q_{12}=18 \sqrt{\frac{E}{3}} \)
(a) Now, energy stored in 12 μF capacitor
\(=\frac{1}{2} C_{2} V_{1}^{2}=\frac{1}{2} \times 12 \times \frac{E}{3}=2 E\)
(b) Energy stored in 3 μF capacitor
\(=\frac{q_{3}^{2}}{2 C_{3}}=\frac{1}{2} \times\left(\frac{18}{3}\right)^{2} E=18 E\)
\((c) \because C_{e q}=\left(\frac{18 \times 3}{18+3}\right) \mu \mathrm{F}=\frac{18}{7} \mu \mathrm{F}, Q=18 \sqrt{\frac{E}{3}}\)
Energy drawn from the battery,
\(U=\frac{Q^{2}}{2 C_{e q}}=\frac{18 \times 18 \times E \times 7}{3 \times 18}=42 E\)

Concentration & involvement in the work by the doctor, reply by the doctor answering the querry without forgetting, observation of the medical attendant.
Yes, charge.  
Electrostatic force.

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 \(\theta\)
  \(\overrightarrow\tau = \overrightarrow{p}\times\overrightarrow{E}\) 

Yes, it is true. Every conductor/insulator is electrically neutral, as it contains equal amounts of positive charge and negative charge.
This statement is true. Charging lies in charge imbalance. When a body loses some electrons, it becomes positively charged because it has excess of protons over electrons. The reverse is also true.
Nature/God has created the universe. In original, all bodies are neutral with no forces of attraction/repulsion. When interests of any two persons clash (i.e. two bodies are rubbed against each other), they become charged. From the charging, the forces of attraction/repulsion arises, i.e. pulls and pressures of life.
Nature/God wants us to live in peace without stress and tensions in life. We get charged over petty things in life and invite all sorts of pulls, pressures and tensions.