Question

Show that E.F intensity at any point due to an electric dipole at distance r from the dipole is given by E=1/4pi efsilon P root[3 cos^2 thita =1]/r^3,where thita is the angle between the radius vector r and the direction of dipole moment p?

Answer 1

To show:

Electric field intensity at any point due to a short electric dipole is

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

Proof:

Let the point be located at a distance of "r" from the short dipole and at an angle of $\theta$.

Axial component at that point:

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

Equatorial component at that point:

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

Net Electrostatic Field:

$\therefore \: E = \sqrt{ {(E_{ax}) }^{2} + {(E_{eq}) }^{2} }$

$= > \: E = \sqrt{ {( \dfrac{2kp \cos( \theta) }{ {r}^{3} } ) }^{2} + {( \dfrac{kp \sin( \theta) }{ {r}^{3} } ) }^{2} }$

$= > \: E = \sqrt{( { \dfrac{kp}{ {r}^{3}})}^{2} \{4 { \cos}^{2}( \theta) + { \sin}^{2}( \theta) \} }$

$= > \: E = \dfrac{kp}{ {r}^{3} } \sqrt{ \{4 { \cos}^{2}( \theta) + { \sin}^{2}( \theta) \} }$

$= > \: E = \dfrac{kp}{ {r}^{3} } \sqrt{ 4 { \cos}^{2}( \theta) + 1 - { \cos}^{2}( \theta) }$

$= > \: E = \dfrac{kp}{ {r}^{3} } \sqrt{ 3 { \cos}^{2}( \theta) + 1 }$

[Hence proved]