The electric field intensity at a point on the axis of an electric dipole in air is 4 N`C^(-1)` .Then the electric field intensity at a point on the equatorial line which is at a distance equal to twice the distance on the axial line and if the dipole is in a medium of dielectric constant 4 is

Options

A. 1 N`C^(-1)`
B. `1/8 NC^(-1)`
C. 16 N`C^(-1)`
D. `1/16 NC^(-1)`

Answer

D

Explanation

As, electric field intensity on the axis ,
`E_("axis") = (2kP)/r^(3) rArr 4 = (2kP)/r^(3)" for " r gt gt a`
`(because" given " E_("axis")= 4)` `rArr (kP)/r^(3) = 2 " "…(i)`
Electric field intensity on equatorial line,
`E_(eq) = (k.P)/r_(1)^(3)`
where, `r_(1) = 2r and k. = k/4`
So, `E_(eq) =(kP)/(4xx 8 r^(3)) `
Now, from the Eq. (i) we get
`E_(eq) = 2/(4xx 8 ) = 1/16 NC^(-1)`
Hence, the correct option is (d).