Electric field intensity at a point on axis of a dipole at a distance of 20 cm from the centre of dipole is 0.015 N/C. When the distance of the point from the centre is made double to its initial value then electric field intensity is reduced to 0.001 N/C . Calculate the length of the electric dipole .

Explanation

Electric field at axial point of dipole of length 2l and dipole moment p, at a distance r from its centre is given by :
`E=1/(4piepsilon_0).(2pr)/(r^2-l^2)^2`
Case I: `r_1`=20 cm =0.2 m , `E_1`=0.015 N/C
`E_1=1/(4piepsilon_0)(2p(0.2))/((0.2)^2 -l^2)^2=0.015`...(i)
Case II: `r_2=0.4` m ,`E_2`=0.001 N/C
`therefore E_2=1/(4piepsilon_0)(2p(0.4))/((0.4)^2-l^2)^2` =0.001 ...(ii)
Dividing (i) by (ii) ,
`0.015/0.001=((0.2)((0.4)^2-l^2)^2)/((0.4)((0.2)^2-l^2)^2)`
`rArr 15/1=1/2 (0.16-l^2)^2/(0.04-l^2)^2`
`rArr sqrt(30/1)=(0.16-l^2)/(0.04-l^2)`
`rArr l^2`=0.013 m `rArr` l=0.114 m
`therefore` Dipole length =2l = 0.228 m