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Five point charges $\;(q\;each)\;$ are placed at five vertices of a regular hexagon of side 2a . What is the magnitude of net electric field at the centre of hexagon .

$(a)\;\large\frac{kq}{a^2}\qquad(b)\;\large\frac{kq}{4 a^2}\qquad(c)\;\large\frac{kq}{2 a^2}\qquad(d)\;\large\frac{kq}{16 a^2}$

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Answer : (b) $\;\large\frac{kq}{4 a^2}$
Explanation :
Electric field on centre will be only due to $\;q_{3}\;$ and field due to $\;q_{1}\;and\;q_{4}\;and\;q_{2}\;and\;q_{5}\;$ being equal and opposite cancel each other .
$Therefore$
$E = \large\frac{k q}{4 a^2}\;.$
answered Jan 30, 2014 by yamini.v
edited Aug 22, 2014 by meena.p
 

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