Given $$\vec E = {E_0}\hat x$$
This shows that the electric field acts along $$ + x$$ direction and is a constant. The area vector makes an angle of 45° with the electric field. Therefore the electric flux through the shaded portion whose area is
$$a \times \sqrt 2 a = \sqrt 2 {a^2}\,{\text{is}}\,\phi = \vec E.\vec A = EA\cos \theta = {E_0}\left( {\sqrt 2 {a^2}} \right)\cos {45^ \circ } = {E_0}\left( {\sqrt 2 {a^2}} \right) \times \frac{1}{{\sqrt 2 }} = {E_0}{a^2}$$

For a thin uniformly positive charged spherical shell
(i) Inside the shell at any point
$$E = 0\,{\text{and}}\,V = \frac{1}{{4\pi { \in _0}}}\frac{q}{R} = constt.$$
where $$q =$$  charge on sphere
$$R =$$  Radius of sphere
(ii) Outside the shell at any point at any distance $$r$$ from the centre $$E \propto \frac{1}{{{r^2}}}\,{\text{and}}\,V \propto \frac{1}{r}$$

Given, $${E_{\max }} = 3 \times {10^6}\,V/m$$     and $$R = 3\,m$$
We know that,
$$\eqalign{ & E = \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{Q}{{{R^2}}} \cr & \Rightarrow {Q_{\max }} = 4\pi {\varepsilon _0}{R^2}{E_{\max }} \cr & = \frac{{3 \times 3 \times 3 \times {{10}^6}}}{{9 \times {{10}^9}}} \cr & = 3 \times {10^{ - 3}}C \cr} $$

Total flux through any enclosed surface is given by, $${\phi _{{\text{net}}}} = \frac{{{\text{net charge enclosed}}}}{{{\varepsilon _0}}}$$
Hence, we can say that the electric flux depends only on net enclosed charge by surface.

Field intensity on axial line of electric dipole is given by
$$E = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{2p}}{{{r^3}}}\,......\left( {\text{i}} \right)$$
and electric field at equatorial position is given by
$$E = \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{p}{{{r^3}}}\,......\left( {{\text{ii}}} \right)$$
where, $$p$$ is electric dipole moment.
From Eqs. (i) and (ii), we get
$$E \propto \frac{1}{{{r^3}}}$$

Net field at $$A$$ should be zero
$$\eqalign{ & \sqrt 2 {E_1} + {E_2} = {E_3} \cr & \therefore \frac{{kQ \times \sqrt 2 }}{{{a^2}}} + \frac{{kQ}}{{{{\left( {\sqrt 2 a} \right)}^2}}} = \frac{{kq}}{{{{\left( {\frac{a}{{\sqrt 2 }}} \right)}^2}}} \cr} $$

$$ \Rightarrow \frac{{Q\sqrt 2 }}{1} + \frac{Q}{2} = 2q \Rightarrow q = \frac{Q}{4}\left( {2\sqrt 2 + 1} \right)$$

From the figure it is clear that the charge enclosed in the cubical surface is $$3C + 2C - 7C = - 2C.$$     Therefore the electric flux through the cube is
$$\phi = \frac{{{q_{{\text{in}}}}}}{{{\varepsilon _0}}} = \frac{{ - 2C}}{{{\varepsilon _0}}}$$

Electric field on the axis of a ring of radius $$R$$ at a distance $$h$$ from the centre,
$$E = \frac{{kQh}}{{{{\left( {{h^2} + {R^2}} \right)}^{\frac{3}{2}}}}}$$
Condition for maximum electric field
We have $$\frac{{dE}}{{dh}} = 0$$
$$ \Rightarrow \frac{d}{{dh}}\left[ {\frac{{kQh}}{{{{\left( {{R^2} + {h^2}} \right)}^{\frac{3}{2}}}}}} \right] = 0$$
On solving we get, $$h = \frac{R}{{\sqrt 2 }}$$

Unit positive charge at $$O$$ will be repelled equally by three charges at the three corners of triangle. By symmetry, resultant $${\vec E}$$ at $$O$$ would be zero.

$$\eqalign{ & {\phi _{{\text{plain}}}} + {\phi _{{\text{curve}}}} = 0\,\,{\text{or}}\,\,{\phi _{{\text{plain}}}} = - {\phi _{{\text{curve}}}} \cr & {{\vec A}_1} = - \frac{{\pi {R^2}}}{2}\hat i,{{\vec A}_2} = - \frac{{\pi {R^2}}}{2}\hat j \cr & \vec E = E\cos {45^ \circ }\hat i + E\sin {45^ \circ }\hat j \cr & = \frac{E}{{\sqrt 2 }}\hat i + \frac{E}{{\sqrt 2 }}\hat j\,\,{\text{and}} \cr & \phi = \vec E \cdot \left( {{{\vec A}_1} + {{\vec A}_2}} \right) \cr & = \frac{{ - E}}{{\sqrt 2 }}\frac{{\pi {R^2}}}{2} - \frac{E}{{\sqrt 2 }}\frac{{\pi {R^2}}}{2} \cr & = \frac{{ - \pi {R^2}E}}{{\sqrt 2 }} \cr} $$

This is the flux entering. So flux is $$\frac{{\pi {R^2}E}}{{\sqrt 2 }}$$