Let a charge $$2q$$  be placed at $$P,$$ at a distance $$I$$ from $$A$$ where charge $$q$$ is placed, as shown in figure.
The charge $$2q$$  will not experience any force, when force, when force of repulsion on it due to $$q$$ is balanced by force of attraction on it due to $$-3q$$  at $$B$$ where $$AB = d$$
$$\eqalign{ & {\text{or}}\,\,\frac{{\left( {2q} \right)\left( q \right)}}{{4\pi {\varepsilon _0}{\ell ^2}}} = \frac{{\left( {2q} \right)\left( { - 3q} \right)}}{{4\pi {\varepsilon _0}{{\left( {\ell + d} \right)}^2}}} \cr & {\left( {\ell + d} \right)^2} = 3{\ell ^2}\,\,{\text{or}}\,\,2{\ell ^2} - 2\ell d - {d^2} = 0 \cr & \therefore \ell = \frac{{2d \pm \sqrt {4{d^2} + 2{d^2}} }}{4} = \frac{d}{2} \pm \frac{{\sqrt 3 d}}{2} \cr} $$

Let charges on an electron and hydrogen nucleus be $${q_1}$$ and $${q_2}.$$ Then, Coulomb's force between them at a distance $$r$$ is, $$F = - \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}{q_2}}}{{{r^2}}}\hat r$$
Putting, $$\frac{1}{{4\pi {\varepsilon _0}}} = k\,\,\left( {{\text{given}}} \right)$$
$$F = - k\frac{{{q_1}{q_2}}}{{{r^2}}}\hat r$$
Since, the nucleus of hydrogen atom has one proton, so charge on nucleus is $$e$$ i.e. $${q_2} = + e$$
also $${q_1} = e$$  for electron
So, $$F = - k\frac{{e.e}}{{{r^2}}}\hat r = - k\frac{{{e^2}}}{{{r^2}}}\hat r$$
but $$\hat r = \frac{r}{{\left| r \right|}} = \frac{r}{r}.$$
Hence, $$F = - k\frac{{{e^2}}}{{{r^2}}}.\frac{r}{r} = - k\frac{{{e^2}}}{{{r^3}}}.r$$

$$\left| {\overrightarrow E } \right| = \frac{{Kq}}{{{r^2}}}$$
Electric field due to $$P$$ on $$O$$ is cancelled by electric field due to $$S$$ on $$O.$$ Similarly Electric field due to $$Q$$ on $$O$$ is cancelled by electric field due to $$T$$ and $$O.$$ The electric field due to $$R$$ on $$O$$ is in the same direction as that of $$U$$ on $$O.$$ Therefore the net electric field is $$2\overrightarrow E .$$

Potential outside the shell, $${V_{{\text{outside}}}} = \frac{{KQ}}{r}$$
where $$r$$ is distance of point from the centre of shell
Potential inside the shell, $${V_{{\text{inside}}}} = \frac{{KQ}}{R}$$
where $$'R'$$ is radius of the shell

$$\eqalign{ & {V_B} = \frac{{K{q_A}}}{{{r_b}}} + \frac{{K{q_B}}}{{{r_b}}} + \frac{{K{q_C}}}{{{r_c}}} \cr & {V_B} = \frac{1}{{4\pi { \in _0}}}\left[ {\frac{{\sigma 4\pi {a^2}}}{b} - \frac{{\sigma 4\pi {b^2}}}{b} + \frac{{\sigma 4\pi {c^2}}}{c}} \right] \cr & {V_B} = \frac{\sigma }{{{ \in _0}}}\left[ {\frac{{{a^2} - {b^2}}}{b} + c} \right] \cr} $$

Electric potential is given by,
$$V = \int\limits_L^{2L} {\frac{{kdq}}{x} = } \int\limits_L^{2L} {\frac{1}{{4\pi {\varepsilon _0}}}\frac{{\left( {\frac{q}{L}} \right)dx}}{x} = \frac{q}{{4\pi {\varepsilon _0}L}}\ln \left( 2 \right)} $$

Force due to charge $$+Q,$$
$${F_1} = \frac{{KQQ}}{{{d^2}}}$$
Force due to charge $$q,$$
$${F_2} = \frac{{KQq}}{{{{\left( {\frac{d}{2}} \right)}^2}}}$$
For equilibrium,
$$\frac{{kQQ}}{{{d^2}}} + \frac{{kQq}}{{{{\left( {\frac{d}{2}} \right)}^2}}} = 0\quad \therefore q = - \frac{Q}{4}$$

Force on charge $${{q_1}}$$ due to $${{q_2}}$$ is
$${F_{12}} = k\frac{{{q_1}{q_2}}}{{{b^2}}}$$
Force on charge $${{q_1}}$$ due to $${{q_3}}$$ is
$${F_{13}} = k\frac{{{q_1}{q_3}}}{{{a^2}}}$$
The $$X$$- component of the force $$\left( {{F_x}} \right)$$  on $${q_1}$$ is $${F_{12}} + {F_{13}}\sin \theta $$
$$\eqalign{ & \therefore {F_x} = k\frac{{{q_1}{q_2}}}{{{b^2}}} + k\frac{{{q_1}{q_2}}}{{{a^2}}}\sin \theta \cr & \therefore {F_x} \propto \frac{{{q_2}}}{{{b^2}}} + \frac{{{q_3}}}{{{a^2}}}\sin \theta \cr} $$

Here, $$D = 2r = 4.4{\text{ }}m,\,\,{\text{or}}\,\, = r = 2.2\,m$$
$$\sigma = 60\,\mu C\,{m^{ - 2}}$$
Charge on the sphere, $$q = \sigma \times 4\pi {r^2}$$
$$ = 60 \times {10^{ - 6}} \times 4 \times \frac{{22}}{7} \times {\left( {2.2} \right)^2} = 3.7 \times {10^{ - 3}}C$$

When a positive point charge is placed outside a conducting sphere, a rearrangement of charge takes place on the surface. But the total charge on the sphere is zero as no charge has left or entered the sphere.

The distance of point $$A\left( {\sqrt 2 ,\sqrt 2 } \right)$$   from the origin,
$$OA = \left| {{{\vec r}_1}} \right| = \sqrt {{{\left( {\sqrt 2 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} = \sqrt 4 = 2\,units.$$
The distance of point $$B\left( {2,0} \right)$$  from the origin,
$$OB = \left| {{{\vec r}_2}} \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( 0 \right)}^2}} = 2\,units.$$
Now, potential at $$A,{V_A} = \frac{1}{{4\pi { \in _0}}} \cdot \frac{Q}{{\left( {OA} \right)}}$$
Potential at $$B,{V_B} = \frac{1}{{4\pi { \in _0}}} \cdot \frac{Q}{{\left( {OB} \right)}}$$
$$\therefore $$ Potential difference between the points $$A$$ and $$B$$ is zero.