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(A) Let the three charges $q$ be placed at vertices $A, B,$ and $C$ of an equilateral triangle of side $L$. Let a charge $Q$ be placed at the centroid

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Zero

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(A) Let the three charges $q$ be placed at vertices $A, B,$ and $C$ of an equilateral triangle of side $L$. Let a charge $Q$ be placed at the centroid $O$ of the triangle.The distance from each vertex to the centroid $O$ is $r = \frac{L}{\sqrt{3}}$.The electrostatic force exerted by each charge $q$ on the charge $Q$ at the center is given by Coulomb's Law: $F = \frac{1}{4\pi \varepsilon _0} \frac{qQ}{r^2} = \frac{1}{4\pi \varepsilon _0} \frac{qQ}{(L/\sqrt{3})^2} = \frac{3}{4\pi \varepsilon _0} \frac{qQ}{L^2}$.Since the charges are identical and placed symmetrically at the vertices of an equilateral triangle,the three force vectors $\overrightarrow{F_A}, \overrightarrow{F_B},$ and $\overrightarrow{F_C}$ acting on $Q$ have equal magnitudes and are directed away from (or towards) the vertices at angles of $120^\circ$ to each other.The vector sum of three equal forces acting at $120^\circ$ to each other is zero. Therefore,the net electrostatic force on the charge $Q$ at the center is zero.

Three charges each of magnitude $q$ are placed at the corners of an equilateral triangle. The electrostatic force on the charge $Q$ placed at the center is (each side of the triangle is $L$):

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