ABC is an equilateral triangle. Charges +q ar | vedclass

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(C) The electric field intensity due to a point charge $q$ at a distance $r$ is given by $E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}$.Since the tri

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(C) The electric field intensity due to a point charge $q$ at a distance $r$ is given by $E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}$.Since the triangle $ABC$ is equilateral and the charges at each corner are identical $(+q)$,the distance from each corner to the centroid $O$ is the same $(r)$.Therefore,the magnitude of the electric field at $O$ due to each charge is $E_A = E_B = E_C = E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}$.These three electric field vectors are directed away from the charges (outward from the corners) and are oriented at an angle of $120^\circ$ to each other.According to the principle of superposition,the net electric field at $O$ is the vector sum: $\vec{E}_{net} = \vec{E}_A + \vec{E}_B + \vec{E}_C$.Since the vectors are equal in magnitude and symmetrically distributed at $120^\circ$,their resultant is zero.Thus,the electric intensity at $O$ is zero.

$ABC$ is an equilateral triangle. Charges $+q$ are placed at each corner. The electric intensity at $O$ will be

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