Five point charges +q each are placed at the five vertices of a regular hexagon of side a. The electric field at the centre of the hexagon is

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$\frac{1}{4 π ε_{0}} \frac{5 q}{a^{2}}$

$\frac{1}{4 π ε_{0}} \frac{3 q}{a^{2}}$

$\frac{1}{4 π ε_{0}} \frac{q}{a^{2}}$

zero

answer is C.

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Detailed Solution

If another identical charge is located at the sixth vertices of the regular hexagon. The field at the centre would have been zero. That means the net field due to fire charge must be equal in magnitude and opposite in direction to the field due to sixth one

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