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

Options

A. `6.25xx10^7` N/C
B. `9.28xx10^7` N/C
C. `13.1xx10^11` N/C
D. `20.5xx10^7` N/C

Answer

A

Explanation

Electric field intensity at a point on the axis of the dipole of length 2a is given by the following relation:
`E=1/(4piepsilon_0)[Q/(r-a)^2 -Q/(r+a)^2]`
here r is the distance of the point from the centre of dipole
Here, r-a=20 cm
r+a=30 cm
`Q=5xx10^(-4)` C
Hence , E can be calculated as follows :
`E=9xx10^9 [(5xx10^(4))/(0.2)^2 -(5xx10^(-4))/(0.3)^2]`
`=6.25xx10^7` N/C