Question

Two point charges + q and -2q are placed at the vertices B and C of an equilateral triangle ABC of side a as given in the figure. Obtain the expression for the direction of the resultant electric field at the vertex A due to these two charges.

• A. $\alpha ={20}^0$
• B. $\alpha ={90}^0$
• C. $\alpha ={70}^0$
• D. $\alpha ={30}^0$

Answer 1

The correct option is D $\alpha ={30}^0$

Direction of electric field at Vertex

$tan\alpha =\frac{E_{AB}\sin 120°}{E_{AC}+E_{AB}\cos 120°}=\frac{E\times \frac{\sqrt{3}}{2}}{2E+E\times (-\frac{1}{2})}tan\alpha =\frac{1}{\sqrt{3}}\alpha =ta{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)\Rightarrow \alpha =30°$

(with side AC)