$$\eqalign{ & {F_{{\text{net}}}} = \int {dq} E\cos \theta \cr & = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left( {\frac{q}{{\pi R}}} \right)} Rd\theta \frac{\lambda }{{2\pi \varepsilon .R}}\cos \theta \cr & = \frac{{\lambda q}}{{2{\pi ^2}{\varepsilon _0}R}}\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\cos \theta d\theta } = \frac{{\lambda q}}{{2{\pi ^2}{\varepsilon _0}R}}\left[ {\sin \theta } \right]_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} \cr & = \frac{{\lambda q}}{{2{\pi ^2}{\varepsilon _0}R}}\left[ {1 - \left( { - 1} \right)} \right] \cr & = \frac{{\lambda q}}{{{\pi ^2}{\varepsilon _0}R}} \cr} $$

Let us consider a spherical shell of thickness $$dx$$ and radius $$x.$$ The volume of this spherical shell $$ = 4\pi {r^2}dr.$$
The charge enclosed within shell $$ = \frac{{Qr}}{{\pi {R^4}}}\left[ {4\pi {r^2}dr} \right]$$
The charge enclosed in a sphere of radius $${r_1}$$ is $$ = \frac{{4Q}}{{{R^4}}}\int\limits_0^{{r_1}} {{r^3}} dr = \frac{{4Q}}{{{R^4}}}\left[ {\frac{{{r^4}}}{4}} \right]_0^{{r_1}} = \frac{Q}{{{R^4}}}r_1^4$$
$$\therefore $$ The electric field at point $$p$$ inside the sphere at a distance $${r_1}$$ from the centre of the sphere is
$$E = \frac{1}{{4\pi { \in _0}}}\frac{{\left[ {\frac{Q}{{{R^4}}}r_1^4} \right]}}{{r_1^2}} = \frac{1}{{4\pi { \in _0}}}\frac{Q}{{{R^4}}}r_1^2$$

$$\eqalign{ & {\text{Here,}}\,\ell = 2.4\,m,r = 4.6\,mm = 4.6 \times {10^{ - 3}}\,m \cr & q = - 4.2 \times {10^{ - 7}}C \cr} $$
Linear charge density, $$\lambda = \frac{q}{\ell }$$
$$ = \frac{{ - 4.2 \times {{10}^{ - 7}}}}{{2.4}} = - 1.75 \times {10^{ - 7}}C{m^{ - 1}}$$
Electric field, $$E = \frac{\lambda }{{2\pi {\varepsilon _0}r}}$$
$$\eqalign{ & = \frac{{ - 1.75 \times {{10}^{ - 7}}}}{{2 \times 3.14 \times 8.854 \times {{10}^{ - 12}} \times 4.6 \times {{10}^{ - 3}}}} \cr & = - 6.7 \times {10^5}\,N{C^{ - 1}} \cr} $$

Electric field lines do not form closed loops. Therefore options (A), (B) and (D) are wrong.
Option (C) is correct. There is repulsion between similar charges.

$$\eqalign{ & {\text{Charges}}\,\left( q \right) = 2 \times {10^{ - 6}}C, \cr & {\text{Distance}}\,\left( d \right) = 3\,cm = 3 \times {10^{ - 2}}m\,{\text{and}}\,{\text{electric}}\,{\text{field}}\,\left( E \right) = 2 \times {10^5}\,N/C. \cr & {\text{Torque}}\,\left( \tau \right) = q.d. \cr & E = \left( {2 \times {{10}^{ - 6}}} \right) \times \left( {3 \times {{10}^{ - 2}}} \right) \times \left( {2 \times {{10}^5}} \right) \cr & = 12 \times {10^{ - 3}}N - m \cr} $$

Force of interaction
$$F = \frac{1}{{4\pi { \in _0}}}.\frac{{3{p_1}{p_2}}}{{{x^4}}}$$

Electric field due to a dipole at bisector or at a point on its broad side on position or equatorial position is given by $$E = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{p}{{{r^3}}}$$
or $$E \propto \frac{p}{{{r^3}}}\,......\left( {\text{i}} \right)$$
where, $$r$$ is the distance of that point from centre of dipole and $$p$$ is dipole moment.
So, from Eq. (i)
$$E \propto p\,\,{\text{and}}\,\,E \propto {r^{ - 3}}$$

Net electric field at the centre of thin conducting ring is zero.
$$\eqalign{ & {E_{{\text{total}}}} = 0 \cr & {E_{AKB}} + {E_{ACDB}} = 0 \cr & {E_{ACDB}} = - {E_{AKB}} \cr & {E_{ACDB}} = E\,{\text{along}}\,OK \cr} $$

The flux through the Gaussian surface is due to the charges inside the Gaussian surface. But the electric field on the Gaussian surface will be due to the charges present in side the Gaussian surface and outside it. It will be due to all the charges.

Let $$Q$$ be the total charge on the sphere. Then surface charge density is $$\frac{Q}{{4\pi {R^2}}}.$$  A hole is now cut of area $$\alpha \left( {4\pi {R^2}} \right).$$   The charge on this hole is
$$\eqalign{ & \frac{q}{{\alpha \left( {4\pi {R^2}} \right)}} = \frac{Q}{{4\pi {R^2}}} \cr & \therefore q = \alpha Q\,\,\,\,\,\,{\text{Also}}\,{V_0} = \frac{{KQ}}{R}. \cr} $$
Now we visualise this situation as a complete spherical distribution of positive charge on the surface with a negative charge of the same surface charge density on the hole. This negative charge can be treated as a point charge.

Potential
$$\eqalign{ & {V_P} = \frac{{KQ}}{R} - \frac{{K\alpha Q}}{R} = \frac{{KQ}}{R}\left( {1 - \alpha } \right) \cr & = {V_0}\left( {1 - \alpha } \right) = {V_0} - \alpha {V_0} \cr} $$
Therefore option (D) is incorrect.
$$\eqalign{ & {V_A} = \frac{{KQ}}{R} - \frac{{K\alpha Q}}{{\frac{R}{2}}} = \frac{{KQ}}{R}\left( {1 - 2\alpha } \right) \cr & \therefore \frac{{{V_P}}}{{{V_A}}} = \frac{{1 - \alpha }}{{1 - 2\alpha }} \cr} $$
Therefore option (B) is correct.
Electric field
$$\eqalign{ & {\left( {{E_B}} \right)_{initial{\text{ }}}} = \frac{{K{Q_0}}}{{{{\left( {2R} \right)}^2}}} \cr & {\left( {{E_B}} \right)_{final{\text{ }}}} = \frac{{KQ}}{{{{\left( {2R} \right)}^2}}} - \frac{{K(\alpha Q)}}{{{R^2}}} = \frac{{KQ}}{{{{\left( {2R} \right)}^2}}} - \alpha {V_0} \cr} $$
Option (A) is incorrect.
$$\eqalign{ & {\left( {{E_P}} \right)_{initial{\text{ }}}} = 0 \cr & {\left( {{E_P}} \right)_{final{\text{ }}}} = \frac{{K\left( {\alpha Q} \right)}}{{{R^2}}} = \frac{{\alpha {V_0}}}{R} \cr} $$
Option (C) is incorrect.