Let us consider a differential element $$dl.$$ charge on this element.

$$\eqalign{ & dq = \left( {\frac{q}{{\pi r}}} \right)dl \cr & = \frac{q}{{\pi r}}\left( {rd\theta } \right)\left( {\because dl = rd\theta } \right) \cr & = \left( {\frac{q}{\pi }} \right)d\theta \cr} $$
Electric field at $$O$$ due to $$dq$$  is
$$dE = \frac{1}{{4\pi { \in _0}}} \cdot \frac{{dq}}{{{r^2}}} = \frac{1}{{4\pi { \in _0}}} \cdot \frac{q}{{\pi {r^2}}}d\theta $$
The component $$dE\cos \theta $$   will be counter balanced by another element on left portion. Hence resultant field at $$O$$ is the resultant of the component $$dE\sin \theta $$  only.
$$\eqalign{ & \therefore E = \int d E\sin \theta = \int\limits_0^\pi {\frac{q}{{4{\pi ^2}{r^2}{ \in _0}}}} \sin \theta d\theta \cr & = \frac{q}{{4{\pi ^2}{r^2}{ \in _0}}}\left[ { - \cos \theta } \right]_0^\pi = \frac{q}{{4{\pi ^2}{r^2}{ \in _0}}}\left( { + 1 + 1} \right) \cr & = \frac{q}{{2{\pi ^2}{r^2}{ \in _0}}} \cr} $$
The direction of $$E$$ is towards negative $$y$$-axis.
$$\therefore \overrightarrow E = - \frac{q}{{2{\pi ^2}{r^2}{ \in _0}}}\hat j$$

As shown in the figure, the resultant electric fields before and after interchanging the charges will have the same magnitude, but opposite directions.
Also, the potential will be same in both cases as it is a scalar quantity.

The given point is at axis of $$\frac{{\vec p}}{2}$$ dipole and at equatorial line of $${\vec p}$$ dipole so that field at given point is $${{\vec E}_1} + {{\vec E}_2}$$
$$\eqalign{ & {{\vec E}_1} = \frac{{2k\left( {\frac{p}{2}} \right)}}{{{2^3}}} = \frac{{Kp}}{8}\left( { + \hat k} \right) \cr & {{\vec E}_2} = \frac{{Kp}}{1}\left( { - \hat k} \right) \cr & {{\vec E}_1} + {{\vec E}_2} = - \frac{7}{8}Kp\left( { - \hat k} \right) = - \frac{{7p}}{{32\pi { \in _0}}}\hat k \cr} $$

The electric field inside a thin spherical shell of radius $$R$$ has charge $$Q$$ spread uniformly over its surface is zero.

Outside the shell the electric field is $$E = k\frac{Q}{{{r^2}}}.$$
These characteristics are represented by graph (A).

$$\eqalign{ & {E_{in}} \propto r \cr & {E_{out}} \propto \frac{1}{{{r^2}}} \cr} $$

$$\eqalign{ & T\sin \theta = \frac{\sigma }{{{\varepsilon _0}K}}.q\,......\left( {\text{i}} \right) \cr & T\cos \theta = mg\,......\left( {{\text{ii}}} \right) \cr} $$
Dividing (i) by (ii),
$$\eqalign{ & \tan \theta = \frac{{\sigma q}}{{{\varepsilon _0}K.mg}}\,\,\,\, \cr & \therefore \sigma \, \propto \tan \theta \cr} $$

The field at $$O$$ due to small element $$dx$$ is
$$dE = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{\lambda dx}}{{{x^2}}}$$
Hence, due to one wire,
$$\eqalign{ & {E_1} = \int\limits_a^\infty {\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{\lambda dx}}{{{x^2}}}} \cr & {E_1} = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{\lambda }{a}\,{\text{towards}}\,{\text{left}}{\text{.}} \cr} $$
Electric field at $$O$$ due to other wire,
$${E_2} = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{\lambda }{a}\,{\text{towards}}\,{\text{left}}$$
$$\therefore $$ Net field at $$O$$ is $$E = 2 \times \frac{1}{{4\pi {\varepsilon _0}}}\frac{\lambda }{a} = \frac{\lambda }{{2\pi {\varepsilon _0}a}}$$

Charge resides on the outer surface of a conducting hollow or solid sphere of radius $$R$$ (say). We consider a spherical surface of radius $$r < R.$$

By Gauss’ theorem, $$\sum {E \cdot ds = \frac{{{q_{{\text{inside}}}}}}{{{\varepsilon _0}}}} $$
or $$E \times 4\pi {r^2} = \frac{1}{{{\varepsilon _0}}} \times {q_{{\text{inside}}}}$$
and we know that $${q_{{\text{inside}}}} = 0$$
So, $$E = 0$$
i.e. electric field inside a hollow sphere is zero.

$$\eqalign{ & \int {\overrightarrow E .\overrightarrow {dA} = \phi } \cr & \Rightarrow \phi = {E_0}\left( {a \times a\sqrt 2 } \right)\cos {45^ \circ } = {E_0}{a^2} \cr} $$

According to Gauss's law, the electric flux through a closed surface is equal to $$\frac{1}{{{\varepsilon _0}}}$$  times the net charge enclosed by the surface.
As the charge enclosed $$ = \frac{q}{8}$$
So, electric flux $$ = \frac{{{q_{{\text{enclosed}}}}}}{{{\varepsilon _0}}} \Rightarrow \phi = \frac{q}{{8{\varepsilon _0}}}$$