(2)

According to Coulomb's law, $F=\frac{k{q}^2}{R^2}$

By touching an uncharged sphere C with charged sphere A, charge on each sphere A and C becomes $\frac{q}{2}$.

[IMAGE 130]------------------------------------------------------

Now, when charged sphere C is kept in contact with sphere B,
then total charge is, $\frac{q}{2}+\frac{q}{1}=\frac{3q}{2}$

Then, 

Charge on B = $\frac{3q}{4}$

Charge on C = $\frac{3q}{4}$

Now, charge on A is $\frac{q}{2}$ and charge on B is $\frac{3q}{4}$. 

So, $F\text{'}=\frac{k}{R^2}\left(\frac{{\displaystyle q}}{{\displaystyle 2}}\right)\left(\frac{{\displaystyle 3q}}{{\displaystyle 4}}\right)=k\frac{3}{8}\frac{q^2}{R^2} \text{or} F\text{'}=\frac{3}{8}F$

(3)

[IMAGE 131]---------------------------------------

In Case I :

$F=-\frac{1}{4\pi {\epsilon }_0}\frac{Q^2}{r^2}$                                               ...(i)

In Case II : $Q_A=Q-\frac{Q}{4}, Q_B=-Q+\frac{Q}{4}$

$\therefore$    $F\text{'}=\frac{1}{4\pi {\epsilon }_0}\frac{(Q-\frac{Q}{4})(-Q+\frac{Q}{4})}{r^2}$

$=\frac{1}{4\pi {\epsilon }_0}\frac{{\displaystyle \left(\frac{{\displaystyle 3}}{{\displaystyle 4}}Q\right)\left(\frac{{\displaystyle -3}}{{\displaystyle 4}}Q\right)}}{r^2}=-\frac{1}{4\pi {\epsilon }_0}\frac{9}{16}\frac{Q^2}{r^2}$                    ...(ii)

From equations (i) and (ii), $F\text{'}=\frac{9}{16}F$

(2)

A hydrogen atom consists of an electron and a proton.

$\therefore$  $\text{Charge on one hydrogen atom }=q_e+q_p=-e+(e+\Delta e)=\Delta e$

Since a hydrogen atom carries a net charge $\Delta e$

$\therefore$  $\text{Electrostatic force, }{F}_e=\frac{1}{4\pi {\epsilon }_0}\frac{{\left(\Delta e\right)}^2}{d^2}$                          ...(i)

will act between two hydrogen atoms.

[IMAGE 132]------------------------------------

The gravitational force between two hydrogen atoms is given as $F_g=\frac{G{m}_h{m}_h}{d^2}$                      ...(ii)

Since the net force on the system is zero, $F_e=F_g$
Using eqns. (i) and (ii), we get

$\frac{{\left(\Delta e\right)}^2}{4\pi {\epsilon }_0{d}^2}=\frac{G{m}_h^2}{d^2} \text{or,} {\left(\Delta e\right)}^2=4\pi {\epsilon }_{0}G{m}_h^2$

           $=6.67\times {10}^{-11}\times {(1.67\times {10}^{-27})}^2\left[\frac{{\displaystyle 1}}{{\displaystyle (9\times {10}^9)}}\right]$

$\Delta e\approx {10}^{-37}$ C

(1)

[IMAGE 133]-----------------------------------

From figure, $T\cos \theta =mg$                                  ...(i)

                      $T\sin \theta =\frac{k{q}^2}{x^2}$                             ...(ii)

From eqns. (i) and (ii), $\tan \theta =\frac{k{q}^2}{x^{2}mg}$

Since $\theta$ is small,      $\therefore$   $\tan \theta \approx \sin \theta =\frac{x}{2l}$

$\therefore$    $\frac{x}{2l}=\frac{k{q}^2}{x^{2}mg} \Rightarrow q^2=x^3\frac{mg}{2lk} \text{or} q\propto x^{3/2}$

$\Rightarrow \frac{dq}{dt}\propto \frac{3}{2}\sqrt{x}\frac{dx}{dt}=\frac{3}{2}\sqrt{x} v$

Since, $\frac{dq}{dt}=\text{constant}$ ;     $\therefore$   $v\propto \frac{1}{\sqrt{x}}$