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(B) The correct option is $B$.Let the vertices of the equilateral triangle be $A$,$B$,and $C$,each having a charge $+q$. The distance from each vertex

Vedclass · Accepted Answer

Zero

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(B) The correct option is $B$.Let the vertices of the equilateral triangle be $A$,$B$,and $C$,each having a charge $+q$. The distance from each vertex to the centroid $O$ is the same,let it be $d$.The electric field intensity produced by each charge at the centroid $O$ has the same magnitude $E = \frac{kq}{d^2}$.These electric field vectors $E_A$,$E_B$,and $E_C$ are directed away from the charges along the medians of the triangle. Since the triangle is equilateral,these vectors are oriented at angles of $120^{\circ}$ with respect to each other.According to the principle of superposition,the net electric field at the centroid is the vector sum of these three fields: $\vec{E}_{net} = \vec{E}_A + \vec{E}_B + \vec{E}_C$.Since the vectors have equal magnitude and are separated by $120^{\circ}$,their vector sum is zero. Therefore,the net electric field at the centroid is zero.

Three equal charges $+q$ each are placed at the three vertices of an equilateral triangle. The electric field at the centroid of the triangle is . . . . . . . ('$r$' is the length of the side of the triangle).

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