Question

The answer to this question is '1' can someone please explain how?

Answer 1

We know,
Electric field on equatorial line of the electric dipole at distance r from the centre of of the electric dipole is given by ,

E = KP/( r² + a²)^3/2
where a is half of distance between two charge .

P = Q(2a) = 2.5 × 10^-9 × 6 = 15 × 10^-9 Cm

E = K(15 × 10^-9) /( 4² + 3²)^3/2
= 15K × 10^-9/125 -------(1)

now,
Electric field on axial line of the electric dipole at the point of distance r from the centre of electric dipole is given by ,

E = 2KPr/(r² - a²)²

put
P = 15 × 10^-9 Cm
r = 4 m
a = 3 m

E = 2K(15× 10^-9)×4 /(4² - 3²)²

= 2K(15 × 10^-9)×4/( 16 -9)²

= 120K × 10^-9/49 --------(2)

divide equation (2) ÷ (1)

Eaxial/Eequatorial = {120K × 10^-9/49 }/{15K×10^9 /125}

= 8 × 125/49

= 1000/49

hence, option (1) is correct