Question

Derive the formula for an electric field due to an electric dipole, at a point which lies at an axial line.

Answer 1

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Answer 2

To derive:

The formula for an electric field due to an electric dipole, at a point which lies at an axial line.

Derivation:

Field intensity due to +q charge at point P:

$\therefore \: E1 = \dfrac{kq}{ {(r - a)}^{2} }$

Field intensity due to -q charge at point P:

$\therefore \: E2 = \dfrac{kq}{ {(r + a)}^{2} }$

So, net field intensity:

$\therefore \: E_{net} = E1 - E2$

$= > \: E_{net} = \dfrac{kq}{ {(r - a)}^{2} } - \dfrac{kq}{ {(r + a)}^{2} }$

$= > \: E_{net} = kq \bigg \{ \dfrac{1}{ {(r - a)}^{2} } - \dfrac{1}{ {(r + a)}^{2} } \bigg \}$

$= > \: E_{net} = kq \bigg \{ \dfrac{ {(r + a)}^{2} - {(r - a)}^{2} }{ {(r - a)}^{2} {(r + a)}^{2} } \bigg \}$

$= > \: E_{net} = kq \bigg \{ \dfrac{ 4ra }{ {(r - a)}^{2} {(r + a)}^{2} } \bigg \}$

$= > \: E_{net} = kq \bigg \{ \dfrac{ 4ra }{ {( {r}^{2} - {a}^{2} )}^{2} }\bigg \}$

$= > \: E_{net} = \dfrac{ 4rakq }{ {( {r}^{2} - {a}^{2} )}^{2} }$

$= > \: E_{net} = \dfrac{ 2k(2qa)r }{ {( {r}^{2} - {a}^{2} )}^{2} }$

$= > \: E_{net} = \dfrac{ 2kMr }{ {( {r}^{2} - {a}^{2} )}^{2} }$

[M = Dipole Moment]

Putting value of Coulomb's Constant:

$= > \: E_{net} = \dfrac{ 2Mr }{4\pi\epsilon_{0} {( {r}^{2} - {a}^{2} )}^{2} }$

So, final answer is:

$\boxed{ \bold{\: E_{net} = \dfrac{ 2Mr }{4\pi\epsilon_{0} {( {r}^{2} - {a}^{2} )}^{2} }}}$