Question

An electric dipole of dipole moment p consists of point charges +q and -q separated by a distance 2a apart. Deduce the expression for the electric field E due to the dipole at a distance x from the centre of the dipole on its axial line in terms of the dipole moment p. Hence show that in the limit
x >>a, E= 2p /(4πεοχ3)​

Answer 1

Field intensity on axis of dipole :

$\sf E_{net} = E_{ (+ q)} - E_{( - q)}$

$\sf \implies E_{net} = \dfrac{kq}{ {(x - a)}^{2} } - \dfrac{kq}{ {(x + a)}^{2} }$

$\sf \implies E_{net} = kq \bigg \{ \dfrac{1 }{ {(x - a)}^{2} } - \dfrac{1}{(x + a)} \bigg \}$

$\sf \implies E_{net} = kq \bigg \{ \dfrac{{(x + a)}^{2} - {(x - a)}^{2} }{ {({x}^{2} - {a}^{2} )}^{2} } \bigg \}$

$\sf \implies E_{net} = kq \bigg \{ \dfrac{4xa}{ {({x}^{2} - {a}^{2})}^{2} } \bigg \}$

$\sf \implies E_{net} = \dfrac{2k(q \times 2a)x}{ {({x}^{2} - {a}^{2})}^{2} }$

$\boxed{ \sf \implies E_{net} = \dfrac{2kPx}{ {({x}^{2} - {a}^{2})}^{2} } }$

Now, for x >>> a , we can say:

$\sf \implies E_{net} = \dfrac{2kPx}{ {x}^{4} }$

$\sf \implies E_{net} = \dfrac{2kP}{ {x}^{3} }$

$\sf \implies E_{net} = \dfrac{2P}{ 4\pi \epsilon_{0} {x}^{3} }$

$\boxed{ \sf \implies E_{net} = \dfrac{P}{ 2\pi \epsilon_{0} {x}^{3} } }$

Hope It Helps.