Question

. Show that ⃗E(axial) = −2×⃗E(Equa)
.

Answer 1

Given:

A short dipole has been provided.

To prove:

Electrostatic field intensity at axial point is twice the electrostatic field intensity at an equatorial point.

E_(ax) = 2 × E_(eq)

Proof:

We shall consider a short dipole such that the inter charge distance will be insignificant as compared to the distance at which the field intensity is measured .

$\boxed{ \sf{2a < < x}}$

Now , field intensity at axial point :

$\therefore \: E_{ax} = \dfrac{2kp}{ {r}^{3} }$

Similarly , field intensity at equatorial point :

$\therefore \: E_{eq} = \dfrac{kp}{ {r}^{3} }$

Dividing the two equations :

$\therefore \: \dfrac{E_{ax}}{E_{eq}} = \dfrac{ (\dfrac{2kp}{ {r}^{3} }) }{ (\dfrac{kp}{ {r}^{3} } )}$

$= > \: \dfrac{E_{ax}}{E_{eq}} = 2$

$= > \: E_{ax} = 2 \times E_{eq}$

Considering the direction of the field vectors in both the cases :

$= > \: E_{ax} = -2 \times E_{eq}$

[Hence proved]