Question

Show that ⃗E(axial) = −2×⃗E(Equa)
.

Answer 1

Given:

A short dipole has been provided.

To show:

Electrostatic field at axial position is related to the the electrostatic field at equatorial position as follows :

$\boxed{E_{ax} = - 2 \times E_{eq}}$

Proof:

Since a short dipole has been provided we can assume the following :

$\boxed{2a < < x}$

This means that the inter charge distance will be negligible as compared to the distance at which the electrostatic field intensity has to be found.

At axial position:

$\vec E_{ax} = \dfrac{2kp}{ {x}^{3} } \: \hat{i}$

So , direction of electrostatic field intensity at the axial point is along the direction of the dipole.

At equatorial position:

$\vec E_{eq} = \dfrac{kp}{ {x}^{3} } \: ( - \hat{ i})$

So , direction of electrostatic field intensity at the equatorial point is opposite to the direction of the dipole.

Hence , we can say that :

$\boxed{ \bold{ \vec E_{ax} = - 2 \times \vec E_{eq}}}$