Answer:
E=[tex]\frac{2K\cdot P}{r^3}[/tex]
Explanation:
We are given that a dipole consist of two charge Q and -Q and charge separated by l.Let a charge +1 C is placed at point P at distance r from the centre of dipole.
We have to find the magnitude of the electric field at point along the axis of dipole .
We know that Electric field=[tex]\frac{Force}{unit \;positive\;Charge}[/tex]
Electric filed due to positive charge Q
[tex]E_1=\frac{kQ}{(r+\frac{\rho}{2})^2}[/tex] {from A to P}
Electric field due to negative charge -Q
[tex] E_2=\frac{KQ}{(r-\frac{\rho}{2})^2}[/tex] ( Along PB)
Net electric field E=[tex]E_2-E_1[/tex]
E=[tex]\frac{kQ}{(r-\frac{\rho}{2})^2}-\frac{KQ}{(r+\frac{\rho}{2})^2}[/tex]
E=[tex]KQ\frac{r^2+\rho^2+r\rho-r^2-\rho^2+r\rho}{(r^2-\frac{\rho^2}{4})^2}[/tex]
E=[tex]\frac{2 KQ r\rho}{(r^2-\frac{\rho^2}{4})^2}[/tex]
We are given that r >> L then
E=[tex]\frac{2KQ r\rho}{r^4}=\frac{2kQ \rho}{r^3}[/tex]
E=[tex]\frac{K\cdot 2Q\rho}{r^3}[/tex]
E=[tex]\frac{2K\cdot P}{r^3}[/tex]
Where, P=Diple = Distance between two charges [tex]\times[/tex] any charge