Question

The eclectic field intensity at a large distance x from an electric dipole is
proportional to​

Answer 1

Field intensity on axis of dipole:

The net field intensity will be :

$E = \dfrac{kq}{ {(x - l)}^{2} } - \dfrac{kq}{ {(x + l)}^{2} }$

$\implies E =kq \bigg \{ \dfrac{1}{ {(x - l)}^{2} } - \dfrac{1}{ {(x + l)}^{2} } \bigg \}$

$\implies E =kq \bigg \{ \dfrac{ {(x + l)}^{2} - {(x - l)}^{2} }{ {(x - l)}^{2} {(x + l)}^{2} } \bigg \}$

$\implies E =kq \bigg \{ \dfrac{ 4xl }{ {(x - l)}^{2} {(x + l)}^{2} } \bigg \}$

$\implies E =kq \bigg \{ \dfrac{ 4xl }{ {( {x}^{2} - {l}^{2} )}^{2} } \bigg \}$

• Putting 2ql to be P (dipole moment).

$\implies E =\bigg \{ \dfrac{ 2Pkx}{ {( {x}^{2} - {l}^{2} )}^{2} } \bigg \}$

• For large distance , x >>> l , we can say:

$\implies E = \dfrac{ 2Pkx}{ {x}^{4} }$

$\implies E =\dfrac{ 2Pk}{ {x}^{3} }$

$\boxed{\implies E \propto\dfrac{1}{ {x}^{3} }}$

So, Field Intensity formula is 2Pk/x³.