For a short dipole placed at origin O, the dipole moment is along x-axis, as shown in the figure. If the electric potential and electric field at A are $$V_{\circ}$$ and $$E_{\circ}$$, respectively, then the correct combination of the electric potential and electric field, respectively, at point B on the -axis is given by

dipole is along +x direction.

point A is on the axial line at distance r
point B is on the equatorial line at distance 2r (on y-axis)

at A (axial point):

for a dipole,

V ∝ p cosθ / r²

here θ = 0° ⇒ cosθ = 1

so,

V₀ = (1/4πϵ₀)(p/r²)

electric field on axial line:

E₀ = (1/4πϵ₀)(2p/r³)

at B (equatorial point):

here θ = 90° ⇒ cosθ = 0

so potential:

V_B = 0

(this is because contributions from +q and −q cancel in potential)

electric field at equatorial point:

fields from +q and −q don’t cancel completely — their components add opposite to dipole moment direction

magnitude:

E_B = (1/4πϵ₀)(p/(distance)³)

distance = 2r

so:

E_B = (1/4πϵ₀)(p/(2r)³)
= (1/4πϵ₀)(p/8r³)

now compare with E₀:

E₀ = (1/4πϵ₀)(2p/r³)

so,

E_B / E₀ = (p/8r³) / (2p/r³) = 1/16