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(B) At point '$A$':Electric field:There are two charges '$Q$' at distance '$a$' from '$A$' and one charge '$2Q$' at distance '$2a$' from '$A$'.The ele

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$\frac{3}{2} \frac{KQ}{a^2}, \frac{3 KQ}{a}$

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(B) At point '$A$':Electric field:There are two charges '$Q$' at distance '$a$' from '$A$' and one charge '$2Q$' at distance '$2a$' from '$A$'.The electric field due to the two charges '$Q$' at distance '$a$' makes an angle of $60^{\circ}$ with the axis of symmetry. The components perpendicular to the axis cancel out,and the components along the axis add up:$E = 2 \cdot \frac{KQ}{a^2} \cos(30^{\circ}) + \frac{K(2Q)}{(2a)^2} = 2 \cdot \frac{KQ}{a^2} \cdot \frac{\sqrt{3}}{2} + \frac{2KQ}{4a^2} = \frac{\sqrt{3}KQ}{a^2} + \frac{KQ}{2a^2} = \frac{KQ}{a^2} (\sqrt{3} + 0.5)$.Wait,re-evaluating the geometry: The two charges '$Q$' are at distance '$a$' from '$A$'. The angle between the lines connecting these charges to '$A$' is $120^{\circ}$. The resultant field of these two is $\sqrt{E_1^2 + E_1^2 + 2E_1^2 \cos(120^{\circ})} = E_1 = \frac{KQ}{a^2}$. This resultant is directed towards the center. The charge '$2Q$' is at distance '$2a$' from '$A$',creating a field $\frac{K(2Q)}{(2a)^2} = \frac{KQ}{2a^2}$ in the same direction.Total $E = \frac{KQ}{a^2} + \frac{KQ}{2a^2} = \frac{3}{2} \frac{KQ}{a^2}$.Electric potential:$V = \frac{KQ}{a} + \frac{KQ}{a} + \frac{K(2Q)}{2a} = \frac{KQ}{a} + \frac{KQ}{a} + \frac{KQ}{a} = \frac{3KQ}{a}$.

$A$ regular hexagon with side length '$a$' is shown. Find the electric field and electric potential at point '$A$'.

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