Question

Equal charge $q$ are placed at the vertices A and B of an equilateral triangle ABC of side $a$. The magnitude of electric field at point C is
$(K=\frac{1}{4\pi {ϵ}_0})$

• A. $\frac{kq}{a^2}$
• B. $\frac{\sqrt{3}kq}{a^2}$
• C. $\frac{\sqrt{2}kq}{a^2}$
• D. $\frac{2kq}{a^2}$

Answer 1

The correct option is B $\frac{\sqrt{3}kq}{a^2}$

$E=\frac{Kq}{a^2}$

Field at C due to charge at A and B are inclined at ${60}^o$ to each other.

$E_{net}=\sqrt{E^2+E^2+2E.E\cos {60}^0}$

$⟹E_{net}=\sqrt{3}E$

$⟹E_{net}=\frac{\sqrt{3}Kq}{a^2}$