If an oil drop of weight 3.2 x 10^–13 N is balanced in an electric field of 5 x 10⁵ Vm^–1, find the charge on oil drop.

Since the weight of an oil drop is balanced in an electric field we will have

$$\begin{gathered} \mathrm{mg}=\mathrm{qE} \\ \Rightarrow 3.2\times {10}^{-13}\times 9.8=\mathrm{q}\times 5 \times {10}^5 \\ \Rightarrow \mathrm{q} =\frac{3.2\times {10}^{-13}\times 9.8}{5\times {10}^5} \\ \Rightarrow \mathrm{q}=6.272\times {10}^{-18} \mathrm{C} \end{gathered}$$
 
where, q is the charge on the oil drop.

(i)$$\begin{gathered} \mathrm{Electric} \mathrm{flux} \mathrm{passing} \mathrm{through} \mathrm{first} \mathrm{sphere}, {\varphi }_1 = \frac{2\mathrm{Q}}{{\mathrm{\epsilon }}_0} \\ \mathrm{Electric} \mathrm{flux} \mathrm{passing} \mathrm{through} \mathrm{second} \mathrm{sphere}, {\mathrm{\varphi }}_2 = \frac{2\mathrm{Q}+4\mathrm{Q}}{{\mathrm{\epsilon }}_0} = \frac{6\mathrm{Q}}{{\mathrm{\epsilon }}_0} \end{gathered}$$
                  $$\begin{gathered} \therefore \mathrm{ratio} \mathrm{of} \mathrm{flux} \mathrm{through} {\mathrm{S}}_{1 } \mathrm{and} {\mathrm{S}}_{2 } \mathrm{is} , \frac{{\mathrm{\varphi }}_1}{{\mathrm{\varphi }}_2} = \frac{2\mathrm{Q}}{{\mathrm{\epsilon }}_0}\times \frac{{\mathrm{\epsilon }}_0}{6\mathrm{Q}} \\ = \frac{1}{3} \end{gathered}$$ 

(ii) If a medium of dielectric constant k is introduced then,  
Using Gauss's theorem,

                ${\mathrm{\varphi }}_1 = \oint \overrightarrow{\mathrm{E}} · \overrightarrow{\mathrm{ds}} = \frac{2\mathrm{Q}}{{\mathrm{\epsilon }}_0}$
i.e,                $\boxed{\mathrm{\varphi }{\text{'}}_1= \frac{1}{\mathrm{k}}\oint \overrightarrow{\mathrm{E}·}\overrightarrow{\mathrm{ds}}}$

                  $= \frac{1}{\mathrm{k}}\frac{2\mathrm{Q}}{{\mathrm{\epsilon }}_0} = \frac{1}{6}\frac{2\mathrm{Q}}{{\mathrm{\epsilon }}_0} = \frac{\mathrm{Q}}{3{\mathrm{\epsilon }}_0}$

Consider the equilibrium of sphere A.
Following forces act on the sphere A.
(i) Force F of repulsion on A due to B.
(ii) Weight mg acting vertically downwards.
(iii) Tension T in the string towards the point of suspension O.

Resolving the tension T into two rectangular components:

$\mathrm{T} \cos \mathrm{\theta } \mathrm{and} \mathrm{T} \sin \mathrm{\theta }$

For equilibrium of A,
                       $\mathrm{T} \sin \mathrm{\theta } = \mathrm{F}$
                                $= \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{{\mathrm{AB}}^2}$
and
                      $\mathrm{T} \cos \mathrm{\theta } = \mathrm{mg}$

Dividing, 
                    $\tan \mathrm{\theta } = \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{{\mathrm{AB}}^2\mathrm{mg}}$

But                  $\mathrm{AB} = 2\mathrm{AC}$
                          $= 2\mathrm{l} \sin \mathrm{\theta }$

$\therefore$     $\tan \mathrm{\theta } = \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{(2\mathrm{l} \sin \mathrm{\theta }{)}^2 \mathrm{mg}}$

        $\boxed{{\mathrm{q}}^2 = (4 mg {\mathrm{l}}^2 si{n}^2 \mathrm{\theta } tan\theta ) 4\pi {\epsilon }_0.}$         

Given, 

Electric field, E = ${10}^4 {\mathrm{NC}}^{-1}$
Torque , 
$$\begin{gathered} \mathrm{\tau } = 9\times {10}^{-26} \mathrm{N} \mathrm{m} \\ \mathrm{\theta }=30° \\ \mathrm{When} \mathrm{electric} \mathrm{dipole} \mathrm{is} \mathrm{placed} \mathrm{at} \mathrm{an} \mathrm{angle} \mathrm{\theta } \mathrm{with} \mathrm{the} \mathrm{direction} \mathrm{of} \mathrm{the} \mathrm{electric} \mathrm{field}, \mathrm{torque} \mathrm{acting} \mathrm{on} \mathrm{the} \mathrm{dipole} \mathrm{is}, \\ \mathrm{\tau } = p\mathrm{E} \mathrm{sin\theta } \\ \therefore \mathrm{the} \mathrm{elctric} \mathrm{moment} \mathrm{of} \mathrm{the} \mathrm{dipole} , \\ p=\frac{\mathrm{\tau }}{\mathrm{E} \mathrm{sin\theta }}=\frac{9\times {10}^{-26}}{{10}^4\times \sin 30°}=\frac{9\times {10}^{-26}}{{10}^4\times 0.5} \\ =1.8\times {10}^{-19 }Cm \end{gathered}$$

Charge on dipole is $\pm 5\mathrm{\mu C}=\pm 5 \times {10}^{-6 }\mathrm{C}$
Distance between the charges = 1mm= ${10}^{-3 }\mathrm{m}$

Dipole moment is given by-p = q(2a)
                                        =$$\begin{gathered} 5\times {10}^{-6}\times 2\times {10}^{-3} \\ ={10}^{-8 }\mathrm{Cm} \end{gathered}$$