Five point charges, $+ q$ each, are placed at the five vertices of a regular hexagon. The distance of the centre of the hexagon from any of the vertices is a. The electric field at the centre of the hexagon is

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$\frac{q}{8 π ε_{0} a^{2}}$

$\frac{q}{4 π ε_{0} a^{2}}$

$\frac{q}{16 π ε_{0} a^{2}}$

zero

answer is A.

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Detailed Solution

$E = \frac{1}{4 π ε_{0}} \frac{q}{a^{2}}$

Suppose the charge is present at the sixth vertex also, then electric field at center would be zero. Now, if charge is not present at this vertex, the electric field at center would be because of other five charges, which should be equal and opposite to the field produced due to single charge at the sixth vertex.