Four point charges are placed at the vertices | vedclass

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(C) Let the side length of the regular hexagon be $a$. The distance from any vertex to the centre is $a$.$1$. Consider the two $+Q$ charges placed at

Vedclass · Accepted Answer

$\frac{kQ\sqrt{3}}{a^2}$

Vedclass · Answer

(C) Let the side length of the regular hexagon be $a$. The distance from any vertex to the centre is $a$.$1$. Consider the two $+Q$ charges placed at opposite vertices. The electric field produced by each at the centre is equal in magnitude $(E = \frac{kQ}{a^2})$ but opposite in direction. Thus,they cancel each other out.$2$. Now consider the two $-Q$ charges. Each produces an electric field of magnitude $E = \frac{kQ}{a^2}$ at the centre,directed towards the respective charge. The angle between these two field vectors is $60^{\circ}$.$3$. The resultant electric field $E_R$ is given by the vector sum:$E_R = \sqrt{E^2 + E^2 + 2E^2 \cos 60^{\circ}}$$E_R = \sqrt{2E^2 + 2E^2(0.5)} = \sqrt{3E^2} = E\sqrt{3}$$4$. Substituting $E = \frac{kQ}{a^2}$,we get:$E_R = \frac{kQ\sqrt{3}}{a^2}$

Four point charges are placed at the vertices of a regular hexagon as shown in the figure below. What will be the net electric field at the centre?

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