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(C) The electric field due to a point charge $q$ at a distance $a$ is given by $E = \frac{1}{4\pi \varepsilon_0} \frac{q}{a^2}$.Let $k = \frac{1}{4\pi

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$\frac{\sqrt{3} q}{4\pi \varepsilon_0 a^2}$

Vedclass · Answer

(C) The electric field due to a point charge $q$ at a distance $a$ is given by $E = \frac{1}{4\pi \varepsilon_0} \frac{q}{a^2}$.Let $k = \frac{1}{4\pi \varepsilon_0}$. The magnitude of the electric field at $C$ due to charge at $A$ is $E_A = k \frac{q}{a^2}$ and due to charge at $B$ is $E_B = k \frac{q}{a^2}$.Since $ABC$ is an equilateral triangle,the angle between the vectors $\vec{E}_A$ and $\vec{E}_B$ is $60^\circ$.The magnitude of the net electric field $E_{net}$ is given by the vector sum formula:$E_{net} = \sqrt{E_A^2 + E_B^2 + 2E_A E_B \cos 60^\circ}$Substituting $E_A = E_B = E$,we get:$E_{net} = \sqrt{E^2 + E^2 + 2E^2 \cos 60^\circ} = \sqrt{2E^2 + 2E^2(0.5)} = \sqrt{3E^2} = E\sqrt{3}$Substituting $E = \frac{1}{4\pi \varepsilon_0} \frac{q}{a^2}$,we get:$E_{net} = \frac{\sqrt{3} q}{4\pi \varepsilon_0 a^2}$

Equal charges $q$ are placed at the vertices $A$ and $B$ of an equilateral triangle $ABC$ of side $a$. The magnitude of the electric field at point $C$ is:

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