Question

The ratio of the electric field intensitydue to an electric dipole at an axialpoint to that at an equatorial point atsame distance from the centre is(i) 1:1(ii)2:1(iii)1:2(iv)1:4​

Answer 1

Answer:

2 : 1

ii. option is correct.

Explanation:

Electric field due to dipole any point.

$\sf E_{net}= K \ P \ \dfrac{\sqrt{1+3\cos^2 \theta}}{x^3}$

Now :

We know electric field due to dipole at axial :

Ф = 0°

$\sf E_{net}= K \ P \ \dfrac{\sqrt{1+3\cos^2 0}}{x^3}$

$\sf E_{net}= \sqrt{4} \ \dfrac{K \ P }{x^3}$

$\sf E_{axial}= 2 \ \dfrac{K \ P }{x^3}$

We also know electric field due to dipole at equatorial point :

Ф = 90°

$\sf E_{net}= K \ P \ \dfrac{\sqrt{1+3\cos^2 90}}{x^3}$

$\sf E_{net}= K \ P \ \dfrac{\sqrt{1}}{x^3}$

$\sf E_{equi.}=\dfrac{K \ P }{x^3}$

Now their ratio :

$\sf \dfrac{E_{axial}}{E_{equi.}} =\left(\dfrac{2 \ \dfrac{K \ P }{x^3}}{\dfrac{K \ P }{x^3}} \right)$

$\sf \dfrac{E_{axial}}{E_{equi.}} =\dfrac{2}{1}$

Therefore we get required answer.