Question

Derive an expression for eletric field at a point on the axial line and equtorial line of an eletric dipole?

Answer 1

Derivation :- 1

$\rm \implies\: \: electric \: \: field \: \: on \: \: the \: \: axis \: \: of \: \: dipole$

$\rm(at \: a \: distance \: x \: from \: centre \: of \: dipole)$

Figure:- 1 is attached

$\rm \: E _{net} = E _2 - E _1$

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

$\rm \: E _{net} = kq \bigg( \dfrac{1}{(x - l) {}^{2} } - \dfrac{1}{(x + l) {}^{2} } \bigg)$

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

$\rm \: E _{net} = kq \dfrac{( {x}^{2} + {l}^{2} + 2xl - {x}^{2} - {l}^{2} + 2xl) }{( {x}^{2} - {l}^{2} ){}^{2} }$

$\rm \: E _{net} = \dfrac{kq \times 4xl}{( {x}^{2} - {l}^{2} ) {}^{2} } = \dfrac{2kq(2l)x}{( {x}^{2} - {l}^{2}) {}^{2} }$

$\boxed{ \rm \: E _{net} = \dfrac{2kpx}{( {x}^{2} - {l}^{2} ) {}^{2} } }$

Derivation:- 2

$\rm \implies \: \: electric \: \: field \: \: on \: \: the \: \: equtorial \: \: line \: of \: \: dipole$

$\rm(at \: a \: distance \: x \: from \: centre \: of \: dipole)$

Figure:- 2 is attached

$\rm \: \to \: {r}^{2} = {x}^{2} + {l}^{2}$

$\rm \to \: \cos\theta = \dfrac{l}{r}$

So

$\rm \to \: \cos\theta = \dfrac{l}{ \sqrt{ {x}^{2} + {l}^{2} } }$

By taking component we get

$\rm \: E _{net} = 2E \cos\theta = \dfrac{2kq}{ {x}^{2} + {l}^{2} } \times \dfrac{l}{ \sqrt{ {x}^{2} + {l}^{2} } }$

$\rm \: E _{net} = \dfrac{2kql}{( {x}^{2} + {l}^{2}) {}^{ \frac{3}{2} } } = \dfrac{kq(2l)}{( {x}^{2} + {l}^{2} ) {}^{ \frac{3}{2} } }$

$\boxed{ \rm \: E _{net} = \dfrac{kp}{( {x}^{2} + {l}^{2} ) {}^{ \frac{3}{2} } } }$