Let the net electric field be zero at point P at a distance x from charge +4\(\mu\)C, then
\(\frac{1}{4 \pi \varepsilon_{0}} \frac{4 \times 10^{-6}}{x^{2}}-\frac{1 \times 10^{-6} \times 1}{4 \pi \varepsilon_{0} \times(2-x)^{2}}=0\)
\(\Rightarrow \ \frac{4}{x^{2}}=\frac{1}{(2-x)^{2}}\)
\(\Rightarrow \ \frac{2}{x}=\frac{1}{2-x}\)
\(\Rightarrow\)  x = 4-2x
\(\Rightarrow \ x=\frac{4}{3} \mathrm{~m}\)

E lectric flux over an area in an electric field represents the total number of electric field lines crossing the area. The SI unit of electric flux is Nm²C^-1.
According to Gauss' law in electrostatics, the surface integral of electrostatic field E produced by any sources over any dosed surface S enclosing a volume V in vacuum, i.e. total electric flux over the closed surface S in vacuum, is 1/ Eo times the total charge (q) contained inside S, i.e.
\({ \phi }_{ E }=\underset { S }{ \oint } E.dS={ {q }\over{ { \epsilon}_{ 0} } }\)
Gauss' law in electrostatics is true for an closed surface, no matter what its shape or size is.
So in order to justify the above statement, suppose in isolated positive charge q is situated at the centre O of a sphere of radius r. According to Coulomb's law, electric field intensity at any point P on the surface of the sphere is
E = \({ { q }\over{ 4\pi{\epsilon}_{0} } }{ { \hat{r} }\over{ {r}^{2} } }\)

where, \(\hat{r}\) is unit vector directed from O to P. Consider a small area element dS of the sphere around P. Let it be represented by the vector \(dS+\hat{r}.dS.\)
where, n is unit vector along out drawn normal to the area element.
\(\therefore\) Electric flux over the area element.
\(d{\phi}_{E}=E.dS\)
= \(\left({ {q }\over{ 4\pi{\epsilon}_{0} } }.{ {\hat{r} }\over{ {r}^{2} } } \right).(\hat{n}.dS)\)
E.dS = \({ { q }\over{ 4\pi{\epsilon}_{0} } }.{{dS}\over{{r}^{2}}}.\hat{r}.\hat{n}\)
As normal to a surface of every point is along the radius vector at that point, therefore, \(\hat{r}.\hat{n}=1\)
E.dS = \({ { q}\over{ 4\pi{\epsilon}_{0}} }.{ {dS }\over{{r}^{2} } }\)
Integrating over the closed surface area of the sphere, we get total normal electric flux over the entire sphere,
\({\phi}_{E}\underset{S}{\oint}E.dS={{q}\over{4\pi{\epsilon}_{0}{r}^{2}}}\underset{S}{\oint}dS={{q}\over{4\pi{\epsilon}_{0}{r}^{2}}}\times total\ area\)
or surface of sphere.= \({{q}\over{4\pi{\epsilon}_{0}{r}^{2}}}(4\pi{r}^{2})\)
= \({{q}\over{{\epsilon}_{0}}}\)
Hence, \(\oint_{S}E.dS={{q}\over{{\epsilon}_{0}}},\) which proves Gauss' theorem.
Electric field inside a uniformly charged spherical shell.
According to Gauss' theorem
\(\oint_{S}E.dS=\oint_{S}E\ \hat{n}.dS={{q}\over{{\epsilon}_{0}}}\)
or   \(E\oint_{s}{dS}={{q}\over{{\epsilon}_{0}}}\)
\(\therefore\)   \(E.2\pi{r}^{2}={{q}\over{{\epsilon}_{0}}}\)
\(\Rightarrow\)  E = \({{q}\over{4\pi{\epsilon}_{0}}{r}^{2}}\)                        ...(i)
In the given figure, the point P where we have to find the electric field intensity is inside the shell. The Gaussian surface is the surface of a sphere S₂ passing through P and with the centre at O. The radius of the sphere S₂ is
r < R. The electric flux through the Gaussian surface, as calculated in Eq. (i), i.e. E X \(4\pi {r}^{2}\). As, charge inside a spherical shell is zero, the Gaussian surface encloses no charge. The Gauss' theorem gives

\(E\times4\pi\ {r}^{2}={{q}\over{{\epsilon}_{0}}}=0\)
\(\therefore\) E = 0 for r < R .
Hence, the field due to a uniformly charged spherical shell is zero at all points inside the shell.

Since drift velocity \({ v }_{ d }\) and current, I flowing in a conductor are related by the relation:
\({ v }_{ d }=\frac { I }{ neA } ......(1)\)
Also drift velocity in terms of average relaxation time τ is given by
\({ v }_{ d }=\frac { eE\tau }{ m } .....(2)\)
From (1) and (2), we have
\(\frac { eE\tau }{ m } =\frac { I }{ neA }\)
\( \\ or\quad \frac { E }{ I } =\frac { m }{ n{ e }^{ 2 }A\tau } \)
\(or\quad \frac { V }{ lI } =\frac { m }{ n{ e }^{ 2 }A\tau }\)
\(or\quad \frac { V }{ I } =\frac { ml }{ n{ e }^{ 2 }A\tau } .......(3)\)
The R.H.S. of Eq(3) is constant
\(\frac { V }{ I } = \ Constant\)
This is Ohm′s law
\((But \ \frac { V }{ I } =R,\ \)the resistance of the conductor)
So Equation (3) becomes
\(R=\frac { ml }{ n{ e }^{ 2 }A\tau }\)
\(But\quad R=\rho \frac { l }{ A } \)
\(\therefore \quad \rho \frac { l }{ A } =\frac { ml }{ n{ e }^{ 2 }A\tau }\)
\(\rho =\frac { m }{ n{ e }^{ 2 }A\tau } )\)

I₁ = 0
I₂ = 1/2 A
I₃ = 1/2 A

Because the electric field due to a system of discrete charges is not defined at the location of any charge.

During landing or take off, the tyres of aircrafts get charged due to the friction between tyres and ground. In case, the tyres are slightly conducting, the charge developed on the tyres will not stay on them and it finds its way to the earth.

When sphere C is brought in contact with A, then charge on sphere C,
\(q_C=\frac{q_A+0}{2}=\frac{q_A}{2}\)
and new charge on sphere A,
\(q_A^{\prime}=\frac{q_A}{2}\)
When sphere C is brought in contact with B, then charge on sphere C,
\( q_C^{\prime}=\frac{q_C+q_B}{2}=\frac{\frac{q_A}{2}+q_B}{2} \\ =\frac{q_A+2 q_B}{4} \)
\(\therefore\)  New charge on sphere B,
\(q_B^{\prime}=\frac{q_A+2 q_B}{4}\)

\(R_2=12\Omega, I_2=0.5A\)

\(R_2=25\Omega, I_2=0.025A\)
\(E=I(R+r)\)
\(E=I_1(R_1+r)\)
\(\therefore\)        \(E=0.5(12+r)\)
\(=6+.5r\)
\(E=I_2(R+r)\)
\(E=0.25(25+r)\)
\(=6.25+.25r\)
From eqns. (1) & (2)
\(6+0.5r=6.25+.25r\)
\(\Rightarrow\ \ \ \ 0.5r-0.25r=6.25-6.0r\)
\(\Rightarrow\ \ \ 0.25r=0.25\)
\(\Rightarrow \ \ \ \ r=1\Omega\)
\(E=6+0.5\times 1\)
\(E=6.5 \ V\)

Potential at the centre,
\(V=\frac{1}{4 \pi \Sigma 0}\left(\frac{q}{r}\right)\)
\(=9 \times 10^{9} \times 10^{-8} \times \frac{60}{360} \times 2 \pi r\)
\(=9 \times 10^{+9} \times 10^{-8} \times \frac{2 \times 3.14 \times 2}{6}=188.4 \mathrm{~V}\) 

\(1.8\times 10^{-4}J\)

Given: \(q_{1}=4 \mu \mathrm{C}=4 \times 10^{-6} \mathrm{C}\)
\(q_{2}=-6 \mu \mathrm{C}=-6 \times 10^{-6} \mathrm{C}, r=20 \mathrm{~cm}\)
Let the electric potential be zero at a point P, a distance x (in cm) from q₁ Then
\(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1}}{x}+\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{2}}{(r-x)}=0\)
\( \therefore \frac{4}{x} =-\frac{(-6)}{(20-x)} \)
\(\Rightarrow 4(20-x=6 x \)
\(x =8 \mathrm{~cm}\)
i.e. 8 cm from 4 μC charge.

\( \text { (i) Force on }+q, \ \overrightarrow{F_{B}}=+q \vec{E} \)
\( { Force on }-q, \ \overrightarrow{F_{A}}=-q \vec{E} \)
As forces are equal in magnitude and opposite in direction, therefore, net force = 0.
Thus, no translatory motion. 

\(\text { (ii) } \tau=F \times\) perpendicular distance between \(\overrightarrow{F_{A}} \text { and }\overrightarrow{F_{B}}\)
\(\tau=F(2 l \sin \theta)=q E 2 l \sin \theta\)
\(\vec{\tau}=\vec{p} \times \vec{E}\) \((\because \vec{p}=q \overrightarrow{2 l})\)

Given, p = 4 \(\times\)10^-9 C-m, E = 5 \(\times\) 10⁴,
\(\theta\) = 30°
\(\therefore\) \(\tau\) = pE sin \(\theta\)
= 4 \(\times\)10^-9 \(\times\)5 \(\times\)10⁴ \(\times\)sin 30°
\(=4\times 10^{-9}\times 5\times 10^{4}\times \frac{1}{2}\)    [\(\because\)sin 30° =\(\frac{1}{2}\)]
= 10 \(\times\)10^-5 = 10^-4 N-m

We find V₁ = 600 Volt and V₂ = 400 Volt. As outer plate of first capacitor is at 1000 volt, its inner plate must be at 1000 - 600 = 400 Volt
Further, as outer plate of second capacitor is earthed, i.e., at zero potential, therefore inner plate of second capacitor must be at
400 - 0 = 400 Volt
Charge on inner plate of each condenser q= \(C_1V_1=(2\times 10^{-6})\times 600=1.2\times 10^{-3}C\) 

When current flows through the battery from its negative to positive terminal, (i.e., current is drawn from battery) then
V = \(\epsilon - Ir \ \ or \ \ 9.0 = \epsilon - 3.5 r\)  .....(i)
When current flows through the battery from its positive to negative terminal (i.e., battery is charged). then
\(V = \epsilon + Ir \ \ or \ \ 12 = \epsilon - 2r\)        ........(ii)
On solving (i) and  (ii), we get;
r = \(2 \ \Omega\)  and \(\epsilon\)  = 16 V

v_d  = \(\frac{I}{nAe}\)
 =  \(\frac{2.7}{(9\times10^{28})\times(2.5\times10^{-7})\times(1.6\times10^{-19})}\)
= 0.75 x 10^-3 ms^-1
= 0.75 mms^-1

Number density of free electrons in a copper conductor, n = 8.5 x 10²⁸ m^-3 Length of the copper wire, l = 3.0 m
Area of cross-section of the wire, A = 2.0 x 10^-6 m²
Current carried by the wire, I = 3.0 A, which is given by the relation,
I = nAeV_d
Where,
e = Electric charge = 1.6 x 10⁻¹⁹ C
V_d = Drift velocity = \(\frac{\text { Length of the wire (l) }}{\text { Time taken to cover l(t) }}\)
\(I=n A e \frac{l}{t}\)
\(t=n A e \frac{l}{I}\)
\(=\frac{3 \times 8.5 \times 10^{28} \times 2 \times 10^{-6} \times 1.6 \times 10^{-19}}{3.0}\)
\(=2.7 \times 10^{4} s\)
Therefore, the time taken by an electron to drift from one end of the wire to the other is 2.7 x 10⁴ s.

Since \({ F }_{ e }=-\frac { { ke }^{ 2 } }{ { r }^{ 2 } } \)
So     \({ ke }^{ 2 }={ -F }_{ e }{ r }^{ 2 }\)
And    \({ F }_{ G }=-G\frac { { m }_{ p }{ m }_{ e } }{ { r }^{ 2 } } \)
So \({ Gm }_{ p }{ m }_{ e }={ -F }_{ G }{ r }^{ 2 }\)
So dimensions of \(\frac { { ke }^{ 2 } }{ { Gm }_{ e }{ m }_{ p } } =\frac { { F }_{ e }{ r }^{ 2 } }{ { F }_{ G }{ r }^{ 2 } } =\frac { { F }_{ e } }{ { F }_{ G } } \)
 \(=\frac { { [MLT }^{ -2 }] }{ { [MLT }^{ -2 }] } =[{ M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 }]\)
 = No dimensions
The value of \(\frac { { ke }^{ 2 } }{ { Gm }_{ e }{ m }_{ p } } \)
  \(=\frac { 9\times { 10 }^{ 9 }\times (1.6\times { 10 }^{ -19 })^{ 2 } }{ 6.67\times { 10 }^{ -11 }(1.67\times { 10 }^{ -27 })(9.1\times { 10 }^{ -31 }) } \)
  \(=2.9\times { 10 }^{ 39 }\)
From Eq. (1) and (2), we have
  \(\left| \frac { { F }_{ e } }{ { F }_{ G } } \right| =\frac { { ke }^{ 2 } }{ { Gm }_{ p }{ m }_{ e } } =2.9\times { 10 }^{ 39 }\)
The ratio of the two forces shows that electrical forces are enormously stronger than the gravitational forces.

(i) (b)
(ii) (a): \(R=\frac{V}{I}=\frac{2}{10^{-6}}=2 \times 10^{6} \Omega\)
(iii) (d): Specific resistance depends upon the nature of material and is independent of mass and dimensions of the material
(iv) (a)
(v) (d): l = 1.0 m; D = 0.4 mm = 4 x 10^-4m
\(R=2 \Omega\)
\(A=\frac{\pi D^{2}}{4}=\frac{\pi \times\left(4 \times 10^{-4}\right)^{2}}{4}=4 \pi \times 10^{-8} \mathrm{~m}^{2}\)
Now, \(\rho=\frac{R A}{l}=\frac{2 \times 4 \pi \times 10^{-8}}{1}=2.55 \times 10^{-7} \Omega \mathrm{m}\)

(i) (b): The proportionality constant k depends on the nature of the medium between the two charges.
(ii) (c): \({ As, }\left[\varepsilon_{0}\right]=\frac{1}{4 \pi F} \cdot \frac{q_{1} q_{2}}{r^{2}} =\frac{[\mathrm{AT}]^{2}}{\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}\right]\left[\mathrm{L}^{2}\right]} \)
\(=\left\lfloor\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]\)
(iii) (b)
(iv) (d): \(F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{d^{2}}\)
\(\therefore\left(10 \times 10^{-3}\right) \times 10=\frac{\left(9 \times 10^{9}\right) \times q^{2}}{(0.6)^{2}}\)
\(\text { or } \ q^{2}=\frac{10^{-1} \times 0.36}{9 \times 10^{9}}=4 \times 10^{-12}\)
\(\text { or } \ q=2 \times 10^{-6} \mathrm{C}=2 \mu \mathrm{C}\)
(v) (b)