Three charges $- q$ , $+ q$ and $- q$ are placed at the corners of an equilateral triangle of side ‘a’. The resultant electric force on a charge $+ q$ placed at the centroid O of the triangle is

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$\frac{q^{2}}{4 π ϵ_{o} a^{2}}$

$\frac{3 q^{2}}{2 π ϵ_{o} a^{2}}$

$\frac{3 q^{2}}{4 π \in_{o} a^{2}}$

$\frac{q^{2}}{2 π \in_{o} a^{2}}$

answer is D.

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Detailed Solution

$F_{n e t} = 2 \times F_{- q} \cos 60^{o} + F_{+ q} = \frac{2 k q^{2}}{d^{2}}$

$∵ d = \frac{a}{\sqrt{3}}$

$\Rightarrow F = 2 \times \frac{1}{4 π \in_{o}} \frac{q^{2}}{a^{2}} \times 3 = \frac{3 q^{2}}{2 π \in_{0} a^{2}}$

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