The electric intensity due to a dipole of length 10 cm and having a charge of 500 µC, at a point on the axis at a distance 20 cm from one of the charges in air, is 

Given: Length of the dipole (2/) =10 cm = 0.1m or l= 0.05 m 

Charge on the dipole ( q) = 500 µC $=500\times {10}^{-6}\mathrm{C}$ and distance of the point on the axis from the mid-point of the dipole (r) = 20 + 5 = 25 cm= 0.25 m. We know that the electric field intensity due to dipole on the given point$\begin{array}{l}\left(\mathrm{E}\right)=\frac{1}{4\pi {\epsilon }_0}\times \frac{2\left(\mathrm{q}.2\mathrm{l}\right)\mathrm{r}}{{\left({\mathrm{r}}^2-{\mathrm{l}}^2\right)}^2}\\ =9\times {10}^9\times \frac{2\left(500\times {10}^{-6}\times 0.1\right)\times 0.25}{{\left[(0.25{)}^2-(0.05{)}^2\right]}^2}\\ =\frac{225\times {10}^3}{36\times {10}^{-3}}=6.25\times {10}^7\mathrm{N}/\mathrm{C}\end{array}$