(3)

Here, the dipole distance is nearly equal to the distance between the point $P$ thus, we cannot apply the dipole formula directly. We have to calculate the potential. Hence, electric potential due to charge $q$ is given by,

due to individual charges

Potential due to charge $\text{'}q\text{'}$ at point $\text{'}P\text{'}$ is

         $V_q=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}=Kq\frac{1}{\left(R\right)\times {10}^{-2}}=\frac{Kq\left({10}^2\right)}{2}$                      ...(i)

Potential due to charge $\text{'}-q\text{'}$ at point $\text{'}P\text{'}$ is,

        $V_{-q}=\frac{1}{4\pi {\epsilon }_0}\frac{(-q)}{(8\times {10}^{-2})}=\frac{-Kq\left({10}^2\right)}{8}$                                 ...(ii)

From equations (i) and (ii), we get

        $V_{\text{net}}=Kq\left({10}^2\right)(\frac{1}{2}-\frac{1}{8})$

        $V_{net}=\left(\frac{3}{8}\right)qk\left({10}^2\right)\text{V}$

(2)

Since, $R\gg L,$ the given charge configuration can be treated as dipole.

Electric field due to a dipole at any arbitrary point $(R,\theta )$ is

$E=\frac{p}{4\pi {\epsilon }_0{R}^3}\sqrt{3{\cos }^2\theta +1}$

Here, $E\propto \frac{1}{R^3}.$

(1)

The polar molecules are the molecules have one end slightly positive and other end is slightly negatively charged separated by some distance.
So, they have permanent electric dipole moment.