Six point charges are placed at the vertices | vedclass

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(D) The hexagon has charges $+Q, -Q, +Q, -Q, +Q, -Q$ at its vertices. Since the charges are arranged in pairs of $(+Q, -Q)$ at opposite vertices,the t

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$0$

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(D) The hexagon has charges $+Q, -Q, +Q, -Q, +Q, -Q$ at its vertices. Since the charges are arranged in pairs of $(+Q, -Q)$ at opposite vertices,the total charge of the system is $0$. For a system with a total charge of $0$,the electric field at a large distance $x$ is dominated by the dipole moment. However,in this specific symmetric arrangement,the dipole moments of the three pairs of opposite charges cancel each other out because they are oriented at $120^{\circ}$ to each other. Alternatively,considering the potential $V$ at a point on the axis at distance $x$,$V = \sum \frac{k q_i}{r_i}$. Since the charges are equal and opposite in pairs and the distance $r_i$ from any vertex to the point on the axis is the same $(r = \sqrt{x^2 + a^2})$,the potential $V = 0$ for all $x$. Since the potential $V$ is zero everywhere on this axis,the electric field $E = -\frac{dV}{dx}$ must also be zero.

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