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

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

Charge on the dipole (q) = 500 PC = 500 × 10–6 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 (E)

$$\begin{gathered} =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{2(\mathrm{q}.2\mathrm{l})\mathrm{r}}{{\left({\mathrm{r}}^2-{\mathrm{l}}^2\right)}^2} \\ =9\times {10}^9\times \frac{2\left(500\times {10}^{-6}\times 0.1\right)\times 0.25}{{\left[(0.25{)}^2-(0.05{)}^2\right]}^2} \\ =\frac{225\times {10}^3}{3.6\times {10}^{-3}}=6.25\times {10}^7 \mathrm{N}/\mathrm{C}(\mathrm{k}=1 \mathrm{for} \mathrm{air}) \end{gathered}$$