An electric dipole is an arrangement of two equal and opposite charges placed at a distance $$2l.$$  The dipole is placed in electric field as such its dipole moment is in direction of electric field as shown in figure. Direction of dipole moment is from negative to positive charge.

Now, force on the charge $$q$$ is $${F_2} = qE$$   along the direction of $$E$$ and force on charge $$-q$$  is $${F_1} = - qE$$   in the direction opposite to $$E.$$
Since, forces on the dipole are equal and opposite, so net force on the electric dipole is zero.

Now, potential energy of the dipole.
$$U = - pE\cos \theta $$
where, $$\theta $$ is the angle between direction of electric field and direction of dipole moment.
$$\therefore \theta = {0^ \circ }$$
Hence, $$U = - pE\cos {0^ \circ } = - pE\,\,\left( {{\text{minimum}}} \right)$$

By Gauss's law $$\phi = \frac{1}{{{\varepsilon _0}}}\,\left( {{Q_{{\text{enclosed}}}}} \right)$$
$$\eqalign{ & \Rightarrow {Q_{{\text{enclosed}}}} = \phi {\varepsilon _0} = \left( { - 8 \times {{10}^3} + 4 \times {{10}^3}} \right){\varepsilon _0} \cr & = - 4 \times {10^3}\,{\varepsilon _0}\,\,{\text{coulomb}}{\text{.}} \cr} $$

After connection, $${V_1} = {V_2}$$
$$\eqalign{ & \Rightarrow K\frac{{{Q_1}}}{{{r_1}}} = K\frac{{{Q_2}}}{{{r_2}}} \cr & \Rightarrow \frac{{{Q_1}}}{{{r_1}}} = \frac{{{Q_2}}}{{{r_2}}} \cr} $$
The ratio of electric fields
$$\eqalign{ & \frac{{{E_1}}}{{{E_2}}} = \frac{{K\frac{{{Q_1}}}{{r_1^2}}}}{{K\frac{{{Q_2}}}{{r_2^2}}}} = \frac{{{Q_1}}}{{r_1^2}} \times \frac{{r_2^2}}{{{Q_2}}} \cr & \Rightarrow \frac{{{E_1}}}{{{E_2}}} = \frac{{{r_1} \times r_2^2}}{{r_1^2 \times {r_2}}} \Rightarrow \frac{{{E_1}}}{{{E_2}}} = \frac{{{r_2}}}{{{r_1}}} = \frac{2}{1} \cr} $$
Since the distance between the spheres is large as compared to their diameters, the induced effects may be ignored.

$$\eqalign{ & \phi = EA\cos {0^ \circ } = E \times \frac{{\pi {d^2}}}{4}, \cr & \therefore E = \frac{{4\phi }}{{\pi {d^2}}}. \cr} $$

$$\eqalign{ & {\text{Torque,}}\,\,\vec \tau = \vec p \times \vec E = pE\sin \theta \cr & 4 = p \times 2 \times {10^5} \times \sin {30^ \circ } \cr & {\text{or,}}\,\,p = \frac{4}{{2 \times {{10}^5} \times \sin {{30}^ \circ }}} = 4 \times {10^{ - 5}}\,Cm \cr} $$
Dipole moment, $$p = q \times l$$
$$q = \frac{p}{l} = \frac{{4 \times {{10}^{ - 5}}}}{{0.02}} = 2 \times {10^{ - 3}}C = 2\,mC$$

$$\eqalign{ & {E_1} = \frac{{nkQR}}{{{{\left( {{R^2} + {R^2}} \right)}^{\frac{3}{2}}}}},{E_2} = \frac{{nkQ\left( {2R} \right)}}{{{{\left( {{R^2} + {{\left( {2R} \right)}^2}} \right)}^{\frac{3}{2}}}}} \cr & \therefore \frac{{{E_1}}}{{{E_2}}} = \frac{{5\sqrt 5 }}{{4\sqrt 2 }} \cr} $$

As we know, the electric flux $$\left( \phi \right)$$ through any surface area is given by,
$$\phi = E \cdot ds = \left| E \right|\left| {ds} \right|\cos \theta $$
As according to question, surface area is in plane of paper and $$E$$ is also in plane of paper. So, angle between area vector and $$E$$ is $${90^ \circ }$$
So, $$\phi = \left| E \right|\left| {ds} \right|\cos {90^ \circ } = {0^ \circ }$$

Using Gauss's law, we have
$$\eqalign{ & \oint {\vec E} \cdot d\bar A = \frac{1}{{{ \in _0}}}\int {\left( {\rho dv} \right)} = \frac{1}{{{ \in _0}}}\int\limits_0^R k {r^a} \times 4\pi {r^2}dr \cr & {\text{or}}\,\,E \times 4\pi {r^2} = \left( {\frac{{4\pi k}}{{{ \in _0}}}} \right)\frac{{{R^{\left( {a + 3} \right)}}}}{{\left( {a + 3} \right)}} \cr & \therefore {E_1} = \frac{{k{R^{\left( {a + 1} \right)}}}}{{{ \in _0}\left( {a + 3} \right)}} \cr & {\text{For}}\,\,r = \frac{R}{2} \cdot {E_2} = \frac{{k{{\left( {\frac{R}{2}} \right)}^{a + 1}}}}{{{ \in _0}\left( {a + 3} \right)}} \cr & {\text{Given,}}\,\,{E_2} = \frac{{{E_1}}}{8}\,\,{\text{or}}\,\,\frac{{k{{\left( {\frac{R}{2}} \right)}^{a + 1}}}}{{{ \in _0}\left( {a + 3} \right)}} = \frac{1}{8}\frac{{k{R^{\left( {a + 1} \right)}}}}{{{ \in _0}\left( {a + 3} \right)}} \cr & \therefore \frac{1}{{{2^{a + 1}}}} = \frac{1}{8}\,\,{\text{or}}\,\,a = 2. \cr} $$

NOTE : As we move along the direction of electric field the potential decreases.
$$\therefore {V_A} > {V_B}$$

$$\alpha = {60^ \circ }.$$   Solid angle subtended by $$BCD$$  is $$\omega = 2\pi \left( {1\cos \alpha } \right) = \pi $$
Solid angle subtended by $$ABDE$$   is
$${\omega _{\left( {ABCDE} \right)}} - {\omega _{\left( {BCD} \right)}} = 2\pi - \pi = \pi $$
Hence, flux through $$ABDE$$   is
$$\phi = \frac{q}{{{\varepsilon _0}}}\frac{\pi }{{4\pi }} = \frac{q}{{4{\varepsilon _0}}}$$