Question

In the derivation of electric field due to an electric dipole on an equatorial line, why do we put cos $\theta =\frac{a}{\sqrt{2}+r^2}$

Answer 1

Step 1: Given data and diagram

The electric field at A due to the charges q and −q is shown in the above figure. We resolve E into horizontal and vertical components. The vertical components (Esinθ) cancel out each other and only the horizontal components survive to give a net electric field at A as 2Ecosθ. The electric field at points A, E

A =2Ecosθ

Step 2: Calculation of Electric field at point A

Electric field at point A $E_A=2Ecos\theta$

Where

$E=2Ecos\theta$

we get $E_A=\frac{2q}{4\pi {ϵ}_{o(r^2+a^2)}}cos\theta$

$cos\theta =\frac{a}{\sqrt{a^2+r^2}}$

on putting the value

$E_A=\frac{2qa}{4\pi {ϵ}_o{(r^2+a^2)}^{\frac{3}{2}}}$

we know the dipole moment p=2qa

hence $E_A=\frac{p}{4\pi {ϵ}_o{(r^2+a^2)}^{\frac{3}{2}}}$

For a<<r, we can neglect $a^2$ compared to $r^2$

Hence final answer is $E_A=\frac{p}{4\pi {ϵ}_o{x}^3}$