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The electric field on the axis of a dipole of dipole moment $\;\overrightarrow{P}\;$ at a distance x from the line joining the charges and $\;(x > > d)\;$

$(a)\;\large\frac{k\overrightarrow{P}}{x^3}\qquad(b)\;-\large\frac{k\overrightarrow{P}}{x^3}\qquad(c)\;\large\frac{2 k\overrightarrow{P}}{x^3}\qquad(d)\;-\large\frac{2 k\overrightarrow{P}}{x^3}$

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Answer : (b) $\;-\large\frac{k\overrightarrow{P}}{x^3}$
Explanation :
$E_{D}=\large\frac{2 k\;q }{((\large\frac{d}{2})^{2}+x^2)^{\large\frac{3}{2}}}\times \large\frac{d}{2}$
This is because only horizontal components will add vertical cancel each other being equal and opposite
$E_{D} \approx \large\frac{2 \;k \;q\; d}{2x^3}$
$E_{D} \approx \large\frac{k \; P}{x^3}$
From the diagram it is clear the $\;E_{D} \;$ & P are in opposite direction
Therefore
$\overrightarrow{E}_{D}=-\large\frac{k\;\overrightarrow{P}}{x^3}\;.$

 

answered Feb 7, 2014 by yamini.v
edited Aug 14, 2014 by thagee.vedartham
 

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