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Six charges are placed at the certices of a regular hexagon as shown in the figure . The electric field on the line passing through point O and perpendicular to the plane of the figure as a function of distance x from point O is $\;(x > > a)$

$(a)\;zero\qquad(b)\;\large\frac{Qa}{\pi \in_{0}x^3}\qquad(c)\;\large\frac{2Qa}{\pi\in_{0}x^3}\qquad(d)\;\large\frac{\sqrt{3}Qa}{\pi\in_{0}x^3} $

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Answer : (a) $\;\large\frac{Qa}{\pi\in_{0}x^3}$
Explanation :
Consider the charges at opposite sides of O . They form a dipole with dipole moment p=2Qa
Electrical field due to a dipole on its axis is $E=\large\frac{k\;p}{x^3}=\large\frac{Qa}{2\pi \in_{0} x^3}$
Now three fields exists as shown in figure :
Net electric field :
$=\large\frac{2Qa}{2 \pi \in_{0} x^3}=\large\frac{Qa}{\pi \in_{0}x^3}$


answered Feb 12, 2014 by yamini.v
edited Aug 14, 2014 by thagee.vedartham

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