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(D) The system consists of three electric dipoles,each with a dipole moment $p = Q(2a)$,where $2a$ is the distance between the charges $+Q$ and $-Q$ a

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(D) The system consists of three electric dipoles,each with a dipole moment $p = Q(2a)$,where $2a$ is the distance between the charges $+Q$ and $-Q$ along the diagonals of the hexagon.For a point on the axis perpendicular to the plane of the hexagon at a distance $x$ from the center $O$,the electric field due to a single dipole is given by $E = \frac{1}{4\pi\varepsilon_0} \frac{p}{x^3} = \frac{1}{4\pi\varepsilon_0} \frac{2Qa}{x^3}$.Since the three dipoles are oriented at $120^{\circ}$ to each other in the plane of the hexagon,their components along the perpendicular axis are all equal in magnitude and direction.However,the electric field vector of a dipole at a point on its equatorial plane is directed opposite to the dipole moment. For a point on the axis perpendicular to the plane,the field is directed parallel to the dipole moment vector.Summing the components,the net electric field is $E_{net} = 3 	imes E \cos(90^{\circ}) = 0$ is incorrect because the field is along the axis. Actually,the field due to each dipole at a point on the axis perpendicular to the plane is $E = \frac{1}{4\pi\varepsilon_0} \frac{p}{x^3}$. Since the dipoles are in the plane,the field at a point on the perpendicular axis is directed opposite to the dipole moment. The sum of three such vectors at $120^{\circ}$ is zero. Thus,the net electric field is zero.

Six charges are placed at the vertices of a regular hexagon as shown in the figure. The electric field on the line passing through point $O$ and perpendicular to the plane of the figure at a distance of $x (>> a)$ from $O$ is

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