A point dipole is located at the origin in some orientation. The electric field at the point (10 cm, 10 cm) on the x-y plane is measured to have a magnitude 1.0 × 10^–3 V/ m. What will be the magnitude of the electric field at the point(20 cm, 20 cm)
(a) 5.0 × 10^–4 V/m               
(b) 2.5 × 10^–4 V/m             
(c) It will depend on the orientation of the dipole
(d) 1.25 × 10^–4 V/m

Dear Student,

           As given in the question, the electric field E = 1 x 10^-3 V/m at a point (10cm,10cm) so the distance o this point from origin,   $r =\sqrt{(10{)}^2+(10{)}^2}= 10\sqrt{2} cm = 0.1\sqrt{2}m$
                       So the electric field due to a dipole of dipole moment 'p' at a distance 'r' is given as.
,
                                                          $$\begin{gathered} E=\frac{1}{4\mathrm{\pi }{\in }_{\mathrm{o}}}\frac{p}{r^3} \\ \Rightarrow p = E. 4\mathrm{\pi }{\in }_{\mathrm{o}}{\mathrm{r}}^3 ..........\left(1\right) \end{gathered}$$

Now value of electric field at point(20cm, 20cm)i.e. at a distance, $r\text{'}=\sqrt{(20{)}^2+(20{)}^2} = 20\sqrt{2} cm =0.2\sqrt{2} m$
                                                                  ​$$\begin{gathered} E\text{'}=\frac{1}{4\mathrm{\pi }{\in }_{\mathrm{o}}}\frac{p}{r{\text{'}}^3} \\ \Rightarrow E\text{'} =\frac{1}{4\mathrm{\pi }{\in }_{\mathrm{o}}}\frac{E.4\mathrm{\pi }{\in }_{\mathrm{o}}{\mathrm{r}}^3}{(0.2\sqrt{2}{)}^3} [ u\sin g (1\left)\right] \\ = \frac{1 \times {10}^{-3}\times (0.1{)}^3\times (\sqrt{2}{)}^3}{(0.2{)}^3\times (\sqrt{2}{)}^3} \\ = 0.125 \times {10}^{-3} \\ = \mathbf{1}\mathbf{.}\mathbf{25}\mathbf{ }\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{ }\mathit{V}\mathbf{/}\mathit{m}\mathbf{.} \end{gathered}$$

                              Hence correct option is(d).

Regards