Question

The electric field at a point a is perpendicular to direction of dipole moment P of a short electric dipole the angle theta between the dipole and the point a is equal to

Answer 1

The below derivation can be used to determine the electric field at any point due to an electric dipole. Thus this is a generalized expression and can be used to determine the electric field due to dipole at equatorial and axial point too.

Consider a short electric dipole AB having dipole moment p. Let the point of interest is at a distance r from the centre O of the dipole. Let the line OP makes an angle θ with the direction of dipole moment p.

Resolve p into two components:

pcosθ along OP

psinθ perpendicular to OP

Point P is on the axial line with respect to pcosθ. So, electric field intensity at P due to short dipole is given by:

Point P is on the equatorial line with respect to psinθ. So, electric field intensity at P due to short dipole is given by:

Since, E1 and E2 are perpendicular to each other, so the resultant electric field intensity is given by:

Answer 2

Given:

Electric Field at point A due to a short dipole is E.

Point A is located at an angle of $\theta$ to the dipole.

To find:

Value of Electrostatic Field at the point A

Calculation:

Refer to the attached diagram to understand the vector components of Electrostatic Field at point A.

So net Electrostatic Field at A will be vector sum of the 2 components :

$\therefore \: E_{net} = \sqrt{ {(E_{x})}^{2} + {(E_{y})}^{2} }$

$= > E_{net} = \sqrt{ { \bigg \{ \dfrac{2p \cos( \theta) }{4\pi\epsilon_{0} {r}^{3} } \bigg \} }^{2} + { \bigg \{ \dfrac{p \sin( \theta) }{4\pi\epsilon_{0} {r}^{3} } \bigg \} }^{2}}$

$= > E_{net} = \dfrac{p}{4\pi\epsilon_{0} {r}^{3} } \sqrt{4 { \cos}^{2}( \theta) + { \sin}^{2} ( \theta) }$

$= > E_{net} = \dfrac{p}{4\pi\epsilon_{0} {r}^{3} } \sqrt{4 { \cos}^{2}( \theta) + 1 - { \cos}^{2} ( \theta) }$

$= > E_{net} = \dfrac{p}{4\pi\epsilon_{0} {r}^{3} } \sqrt{3 { \cos}^{2}( \theta) + 1 }$

So final answer is :

$\boxed{ \red{ \bold{ E = \dfrac{p}{4\pi\epsilon_{0} {r}^{3} } \sqrt{3 { \cos}^{2}( \theta) + 1 } }}}$