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(D) Let the vertices of the equilateral triangle be $A, B,$ and $O$. Let the charges be $q_A = -2q$,$q_B = +q$,and $q_O = +q$. The side length is $L$.

Vedclass · Accepted Answer

$A$ and $B$ both.

Vedclass · Answer

(D) Let the vertices of the equilateral triangle be $A, B,$ and $O$. Let the charges be $q_A = -2q$,$q_B = +q$,and $q_O = +q$. The side length is $L$.The distance from each vertex to the centroid $C$ is $r = \frac{L}{\sqrt{3}}$.The electric potential $V$ at the centroid $C$ is given by:$V = k \frac{q_A}{r} + k \frac{q_B}{r} + k \frac{q_O}{r} = k \frac{-2q + q + q}{r} = 0$.Thus,option $A$ is correct.For the dipole moment $\vec{p}$,we place the charges at coordinates: $O(0, 0)$,$B(L, 0)$,and $A(L/2, \sqrt{3}L/2)$.$\vec{p} = \sum q_i \vec{r}_i = q(0, 0) + q(L, 0) + (-2q)(L/2, \sqrt{3}L/2) = (qL - qL, 0 - \sqrt{3}qL) = (0, -\sqrt{3}qL)$.The magnitude of the dipole moment is $|\vec{p}| = \sqrt{3}qL$.Thus,option $B$ is correct.Since both $A$ and $B$ are correct,the correct option is $D$.

Three point charges are placed at the corners of an equilateral triangle of side $L$ as shown in the figure.

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